a collection of algebraic identities

A Collection of Algebraic Identities is a 300+ page book I wrote that was inspired, in part, by the unusual Ramanujan 6-10-8 Identity, 64[(a+b+c)6+(b+c+d)6+(a-d)6-(c+d+a)6-(d+a+b)6-(b-c)6] [(a+b+c)10+(b+c+d)10+(a-d)10-(c+d+a)10-(d+a+b)10-(b-c)10] =45[(a+b+c)8+(b+c+d)8+(a-d)8-(c+d+a)8-(d+a+b)8-(b-c)8]2 which is true if ad = bc. Its form and use of certain high exponents is certainly intriguing (a quality that is characteristic of most of Ramanujan’s work). Later we will see that this can be generalized. Trying to look for other examples of unusual algebraic identities in the Internet, or at least collections of which, I couldn’t find anything substantial. So I decided to upload my work. It covers an eclectic range and includes the basics, like the Bramagupta-Fibonacci Two-Square Identity (BTS), (a2+b2)(c2+d2) = (ac+bd)2 + (ad-bc)2

generalizations like Vladimir Arnold's perfect forms, an example being, (2a2+ab+3b2)(2c2+cd+3d2) = 2x2+xy+3y2, where {x,y} = {a(-c+d)+b(c+2d), a(c+d)+b(c-d)} hence the product of two {2,1,3} quadratic forms is of like form (analogous to the BTS), to more unusual and exotic examples, including one of the most complete collection of identities for Equal Sums of Like Powers for 4th and higher powers. The ancient Greeks only knew the Pythagorean theorem, 2nd powers (a2-b2)2 + (2ab)2 = (a2+b2)2 but, more than 2000 years later, we know much more. Who knows what we'll discover in the next thousand years. A few examples are:

3rd powers i) Vieta (a4-2ab3)3 + (a3b+b4)3 = (b4-2a3b)3 + (ab3+a4)3 ii) Ramanujan (3a2+5ab-5b2)3 + (4a2-4ab+6b2)3 + (5a2-5ab-3b2)3 = (6a2-4ab+4b2)3 (which gives rise to 33+43+53 = 63) 4th powers i) Fauquembergue (a4-4b4)4 + 2(2a3b)4 + 2(4ab3)4 = (a4+4b4)4 ii) Gerardin (a+3a2-2a3+a5+a7)4 + (1+a2-2a4-3a5+a6)4 =(a-3a2-2a3+a5+a7)4 + (1+a2-2a4+3a5+a6)4 5th powers i) Choudhry (a-a3-2a5+a9)5 + (1+a2-2a6+2a7+a8)5 + (2a3+2a4-2a7)5 =(a+3a3-2a5+a9)5 + (1+a2-2a6-2a7+a8)5 + (-2a3+2a4+2a7)5 (Differences between LHS and RHS are highlighted in blue.)

ii) Chernick If 11a2+4b2 = 9c2, and -4a2+b2 = 9d2, then for k = 1,2,3,5, (b-d)k + (-b-d)k + (2d)k = (a+2d)k + (-a+2d)k + (-c-2d)k + (c-2d)k 6th powers i) Brudno, Delorme: For k = 2,6, (-n4-n3-5n2+8n+8)k + ((n3+7n-2)(n+2))k + (9n2+6n+12)k = ((n2-n+3)(n+2)2)k + (-4n3-5n2-8n+8)k + (-n4+n2+14n+4)k ii) Chernick If u2+uv+v2 = 7, then for k = 1,2,4,6, (-u+7)k + (u-2v+1)k + (-3u-1)k + (3u+2v+1)k = (u+7)k + (-u+2v+1)k + (3u-1)k + (-3u-2v+1)k

7th powers (Piezas): 1) If x2-10y2 = 9, then for k = 1,3,5,7, 1 + 5k + (3+2y)k + (3-2y)k + (-3+3y)k + (-3-3y)k = (-2+x)k + (-2-x)k + (5-y)k + (5+y)k

2) If x2+6y2 = 1, then for k = 1,2,3,5,7, (3x+2y)k + (-3x+7y)k + (3x-2)k + (3+2x)k + (-5x-3y)k = (-3x+2y)k + (3x+7y)k + (-3x-2)k + (3-2x)k + (5x-3y)k

8th powers i) Letac, Sinha If a2+12b2 = c2, and 12a2+b2 = d2, then for k = 1,2,4,6,8, (a+c)k + (a-c)k + (3b+d)k + (3b-d)k + (4a)k = (3a+c)k + (3a-c)k + (b+d)k + (b-d)k + (4b)k

ii) Wroblewski If 17a2-33b2 = 8c2, and 3a2-3b2 = 8d2, then for k = 1,2,4,6,8, (a+b)k + (-a+b)k + (a+c)k + (a-c)k + (b+3d)k + (-b+3d)k = (b+c)k + (b-c)k + (b+d)k + (-b+d)k + (2a)k + (4d)k

9th powers i) Piezas Let 283a2-475b2 = 2072c2, and 5491a2-47347b2 = 2072d2. If {p,q} = 17c ± d, {r,s} = 7c ± 5d, {t,u} = 39c ± 5d, then for k = 1,2,3,9, (aq+br)k + (bs-ap)k + (bu+ap)k + (-aq+bt)k + (2ad-78bc)k + (-2ad-46bc)k =(-aq+br)k + (bs+ap)k + (bu-ap)k + (aq+bt)k + (-2ad-78bc)k + (2ad-46bc)k

10th powers

i) Choudhry, Wroblewski If 45a2-11b2 = c2, and -11a2+45b2 = d2, then for k = 1,2,4,6,8,10, (-a-3b-c)k + (-a-3b+c)k + (3a-b+d)k + (3a-b-d)k + (2a+8b)k + (-8a+2b)k =(a-3b-c)k + (a-3b+c)k + (3a+b+d)k + (3a+b-d)k + (-2a+8b)k + (-8a-2b)k

ii) Wroblewski, Piezas If a2+b2 = c2, and a2+52b2 = d2, (a pair of conditions which is a concordant form). Then for k = 1,2,4,6,8,10, (8b)k + (5a-4b)k + (-a-2d)k + (a-2d)k + (-5a-4b)k + (-12b+4c)k + (12b+4c)k =(4a+8b)k + (3a-2d)k + (-3a-2d)k + (-4a+8b)k + (-16b)k + (a+4c)k + (-a+4c)k

As the equations are homogenous, it does not matter if the variables are to be solved either in the integers or the rationals, though trivial values should be avoided. And since two quadratics define an elliptic curve, all these examples have an infinite number of rational solutions. These are just a few given in the book, but not much is really known for higher powers. For example, no identity is yet known for k = 1,3,5,7,9 with a minimal number of terms. It is hoped that this collection will spur others to find more; some identities have a simplicity, clarity, and beauty all their own. II. Why? “Politics is for the moment, but an equation is forever.” – Albert

Einstein The argument can be given that identities are simply tautologies A = A and hence do not impart anything new. In fact, that is not the case. One can algebraically deform each side of the equation, separate it into components, give it structure, to such an extent that it is no longer immediately apparent the two sides are equal to each other. (One need only look at the 6-10-8 Identity.) As another easy example, the structure of the Two-Square Identity given above implies that the product of the sums of two squares is itself the sum of two squares, and this in fact belongs to a finite family of identities important to division algebras. Still another is Hirschhorn’s Odd-Even Identity which proves that the sum of four distinct odd squares is the sum of four distinct even ones. And so on. The great thing about mathematical truths, other than their unreasonable effectiveness in the sciences as discussed in a famous essay by the physicist E. Wigner is that, in contrast to physical truths (such as the statement, “There are still polar bears.”), they are eternally and universally true. Thus, if you discover a mathematical truth like a new algebraic identity (okay, so I’m a Platonist), then you are looking at an object which is true forever. For me, that is reason enough to look for them. Admittedly though, there can be more prosaic reasons. Another nice aspect to algebraic identities is that they usually need only elementary mathematics and, as such, is accessible to a broad audience. (However, one may sometimes come across concepts like elliptic curves, resultants, Pell equations, quadratic forms, etc.) With the advent of computer algebra systems (CAS) like Mathematica, Maple, and others, it’s now more convenient to find – and verify – algebraic identities. Of course, it would take some ingenuity to solve the more complicated cases, especially those involving powers higher than the fourth. However, for anyone interested in finding new identities, I highly suggest taking a look at what CAS have to offer. (Mathematica's Home Edition is offered at

just $295, or one can go to the free but limited site www.quickmath.com.)

III. Submissions If you have:

a) a new algebraic identity b) a new solution to an old formc) a new form that needs a solutiond) questionse) correctionsf) comments

feel free to email at [email protected] and help enlarge and improve this database of algebraic identities. When submitting something new, kindly include, if possible, the general principle by which it was found since it would be nice to know if it can be generalized. Quite a lot of the material in this book is independent work by this author. If a particular result turns out to be a rediscovery, pls email me so I can give proper credit to that person (kindly also include a citation to the relevant book or paper). This is just a first draft – there is still a lot of work to be done – but one has to start somewhere.

Pi Formulas I. Class number h(-d) = 1II. Class number h(-d) = 2, where d = 4mIII. Class number h(-d) = 2, where d = 3m I. Class number h(-d) = 1. It is quite well-known that, eπ√7 ≈ 153 + 697

eπ√11 ≈ 323 + 738eπ√19 ≈ 963 + 743eπ√43 ≈ 9603 + 743.999…eπ√67 ≈ 52803 + 743.99999…eπ√163 ≈ 6403203 + 743.99999999999… with the last also known as the Ramanujan constant. The discriminants d under the square root are the six highest of the nine Heegner numbers. As I pointed out in a 2008 sci.math.research post, it is not so commonly known there is another internal structure, eπ√7 ≈ 33(32-22)3 + 697eπ√11 ≈ 43(32-1)3 + 738eπ√19 ≈ 123(32-1)3 + 743eπ√43 ≈ 123(92-1)3 + 744eπ√67 ≈ 123(212-1)3 + 744eπ√163 ≈ 123(2312-1)3 + 744 The reason for the squares of {3, 9, 21, 231} is due to a certain Eisenstein series, and they will appear again below. Before going to the pi formulas, it can be pointed out that these transcendental numbers, in addition to being closely approximated by integers (which are simply algebraic numbers of degree 1), can also be closely approximated by algebraic numbers of degree 3, eπ√11 ≈ x24 – 24; (x3-2x2+2x-2 = 0)eπ√19 ≈ x24 – 24; (x3-2x-2 = 0)eπ√43 ≈ x24 – 24; (x3-2x2-2 = 0)eπ√67 ≈ x24 – 24; (x3-2x2-2x-2 = 0)eπ√163 ≈ x24 – 24; (x3-6x2+4x-2 = 0) The corresponding cubic equations are extremely simple and have a

very similar form. In fact, x can be exactly expressed in terms of the Dedekind eta function η(τ), a modular function which involves a 24th root, and explains the 24 in the approximations. (Note: As Rich Schroeppel points out, John Brillhart, in the early days of computers, looked at continued fractions of cubic irrationalities and found a few had some interesting properties, one of which was x3-6x2+4x-2 = 0. If you look at the first 200 terms, it is curiously pepppered with large terms. Of course, the fact that it is related to eπ√163 gives an inkling of the reason why.) These transcendental numbers can also be closely approximated by algebraic numbers of degree 4, eπ√19 ≈ 35(3-√(2v1))-2 - 12.00006…

eπ√43 ≈ 35(9-√(2v2))-2 - 12.000000061…

eπ√67 ≈ 35(21-√(2v3))-2 - 12.00000000036…

eπ√163 ≈ 35(231-√(2v4))-2 - 12.00000000000000021… (notice how the numbers {3, 9, 21, 231} appear again) and where, v1 = -3 + 1√(3*19)v2 = -39 + 7√(3*43)v3 = -219 + 31√(3*67)v4 = -26679 + 2413√(3*163) and that, 3*26(-32 + 3*19*12) = 962

3*26(-392 + 3*43*72) = 9602

3*26(-2192 + 3*67*312) = 52802

3*26(-266792 + 3*163*24132) = 6403202

which, with the appropriate fractional power, are precisely the j-invariants. As well as for algebraic numbers of degree 6, eπ√19 ≈ (5x)3 - 6.000010… eπ√43 ≈ (5x)3 - 6.000000010… eπ√67 ≈ (5x)3 - 6.000000000061… eπ√163 ≈ (5x)3 - 6.000000000000000034… where the x’s are given respectively by the appropriate root of the sextics, 5x6-96x5-10x3+1 = 05x6-960x5-10x3+1 = 05x6-5280x5-10x3+1 = 05x6-640320x5-10x3+1 = 0 with the j-invariants appearing again. Another interesting feature of these sextics is that they are not only algebraic, but are also solvable in radicals. These factor into two cubics, with the exception of the first which has a quadratic factor, over the extension Q(φ) where φ = (1+√5)/2, or the golden ratio. The last three, respectively, have the cubic factors, 5x3 - 5(53+86φ)x2 + 5(8+13φ)x - (18+29φ) = 05x3 - 20(73+118φ)x2 - 20(21+34φ)x - (47+76φ) = 05x3 - 20(8849+14318φ)x2 + 20(377+610φ)x - (843+1364φ) = 0 Amazingly, pairs of consecutive Fibonacci numbers F(n) = {0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610,…} appear in the x term, as well as Lucas numbers L(n) = {2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364,…} in the constant term. Why this is so I have no idea as it came as a surprise while I was looking at their

factorization over Q(φ). If you have Mathematica, the factorization is easily verified by the Resultant[] function, Resultant[5x3 - 5(53+86φ)x2 + 5(8+13φ)x - (18+29φ), φ2-φ-1, φ] which eliminates the variable φ between the two equations and recovers the original sextic. (Likewise for the other two.) (Update, 5/23/11) Qiaochu Yuan from MathStackExchange gave the identity for Fibonacci numbers, Fn-1 + Fn φ = φn

A related one for Lucas numbers is, Ln-1 + Ln φ = φn√5 Hence, the cubic factors simplify as, 5x3 - 5 φ6(4+3√5) x2 + 5φ7 x - φ7√5 = 05x3 - 20 φ9(-1+2√5) x2 - 20φ9 x - φ9√5 = 05x3 - 20 φ15(19+2√5) x2 + 20φ15 x - φ15√5 = 0 or the coefficients neatly contain powers of φ. (End update) These algebraic approximants of degree 3,4,6 can be exactly expressed as Dedekind eta η(τ) quotients. For example, let τ = (1+√-163)/2, then, x1 = eπi/24 η(τ)/η(2τ)

x2 = eπi/12 η(τ)/η(3τ)

x3 = eπi/6 η(τ)/η(5τ) leading to,

eπ√163 ≈ x124 - 24.00000000000000105…

eπ√163 ≈ x212 - 12.00000000000000021…

eπ√163 ≈ x36 - 6.000000000000000034…

where the eta quotients xi are the algebraic numbers given above. There are also nice Diophantine relationships for the approximants of degree 1, 7 (123 + 153) = 32 *632

11 (123 + 323) = 42 *1542

19 (123 + 963) = 122 *3422

43 (123 + 9603) = 122 *162542

67 (123 + 52803) = 122 *2617022

163 (123 + 6403203) = 122 *5451401342

So the sums are perfect squares. And where else do the blue and red numbers appear? Of all places, in formulas for π. Let c = (-1)n(6n)! / ((n!)3(3n)!), then, 1/π = 3 Σ c (63n+8)/(153)n+1/2

1/π = 4 Σ c (154n+15)/(323)n+1/2

1/π = 12 Σ c (342n+25)/(963)n+1/2

1/π = 12 Σ c (16254n+789)/(9603)n+1/2

1/π = 12 Σ c (261702n+10177)/(52803)n+1/2

1/π = 12 Σ c (545140134n+13591409)/(6403203)n+1/2

where the sum Σ goes from n = 0 to ∞. Beautiful, aren’t they? The modular function responsible for this is the j-function which has the q-series expansion, j(q) = 1/q + 744 + 196884q + 21493760q2 + 864299970q3 + ...

which explains why, especially for these discriminants, the number eπ√d has an excess close to the integer "744". This is sequence A000521 of the OEIS and, other than the constant term, is the McKay-Thompson series class 1A of the Monster simple group. These pi formulas were discovered by the Chudnovsky brothers inspired by Ramunujan’s formulas given in the next section (which uses d with class number 2). For more details, see “Pi formulas, Ramanujan, and the Baby Monster Group”, by this author (File 07). II. Class number h(-d) = 2, where d = 4m. There are exactly 18 negative fundamental discriminants d with class number h(-d) = 2, 7 of which are even, namely d = 4m, for m = {5, 13, 37} and m = {6, 10, 22, 58}. As m increases, the expression eπ√m gets intriguingly close to an integer with a fixed excess "104", eπ√5 ≈ 322 + 100eπ√13 ≈ 2882 + 103.9…eπ√37 ≈ 141122 + 103.9999… eπ√6 ≈ 482 - 106eπ√10 ≈ 1442 - 104.2…eπ√22 ≈ 15842 - 104.001…eπ√58 ≈ 1568162 - 104.0000001… Just like for d with class number h(-d) = 1, there are also nice Diophantine relationships between these numbers, 5 (44 + 322) = 42 *202

13 (44 + 2882) = 42 *2602

37 (44 + 141122) = 42 *214602

6 (-44 + 482) = 82 *3*82

10 (-44 + 1442) = 82 *2*402

22 (-44 + 15842) = 82 *11*2802

58 (-44 + 1568162) = 82 *2*1055602

So the sums for the first set are perfect squares, while the second set are almost-squares. The blue and red numbers again appear in pi formulas courtesy of, who else, but Ramanujan. Let r = (4n)! / (n!4), then, 1/π = 4 Σ r (-1)n (20n+3)/(322)n+1/2

1/π = 4 Σ r (-1)n (260n+23)/(2882)n+1/2

1/π = 4 Σ r (-1)n (21460n+1123)/(141122)n+1/2

1/π = 8√3 Σ r (8n+1)/(482)n+1/2

1/π = 8√2 Σ r (40n+4)/(1442)n+1/2

1/π = 8√11 Σ r (280n+19)/(15842)n+1/2

1/π = 8√2 Σ r (105560n+4412)/(1568162)n+1/2

where n = 0 to ∞. The modular function, call it r(q), that is responsible for this has the q-series expansion, r(q) = 1/q + 104 + 4372q + 96256q2 + 1240002q3 + … and explains why the number eπ√m has an excess close to the integer "104". This is sequence A007267 and, excepting the first term, is the McKay-Thompson series of Class 2A for the Monster group. III. Class number h(-d) = 2, where d = 3m. There are exactly 5 discriminants of this form, namely d = 3m, for m = {5, 8, 17, 41, 89}. This time, the expression eπ√(m/3) gets close to an integer with a fixed excess "42", though I will focus only on the three highest m,

eπ√(17/3) ≈ 123 + 41.6…eπ√(41/3) ≈ 483 + 41.99…eπ√(89/3) ≈ 3003 + 41.9999… and three non-fundamental odd m = {9, 25, 49}, eπ√(9/3) ≈ 3*43 + 38.8…eπ√(25/3) ≈ 5*123 + 41.91…eπ√(49/3) ≈ 7*363 + 41.998… Since these non-fundamental m are the squares of the first 3 odd primes, the approximations look nice, being the product of that prime and a cube. As usual, there are nice Diophantine relationships, 3*17 (2233 + 123) = 62 *512

3*41 (2233 + 483) = 62 *6152

3*89 (2233 + 3003) = 62 *141512

3 (2233 + 3*43) = 62 *52

3 (2233 + 5*123) = 62 *272

3 (2233 + 7*363) = 62 *1652

so the sums are perfect squares. Define r2 = (2233)n (1/2)n(1/3)n(2/3)

n / (n!3) where (a)n is the rising factorial, or Pochhammer symbol, such that (a)n = (a)(a+1)(a+2)…(a+n-1). Equivalently, let r2 = (2n)!

(3n)!/ (n!5), then, 1/π = 2 Σ r2 (-1)n (51n+7)/(123)n+1/2

1/π = 2 Σ r2 (-1)n (615n+53)/(483)n+1/2

1/π = 2 Σ r2 (-1)n (14151n+827)/(3003)n+1/2

1/π = 2*3 Σ r2 (-1)n (5n+1)/(3*43)n+1/2

1/π = 2*5 Σ r2 (-1)n (27n+3)/(5*123)n+1/2

1/π = 2*7 Σ r2 (-1)n (165n+13)/(7*363)n+1/2

where n = 0 to ∞. These have been derived by this author (On Ramanujan's Other Pi Formulas, file 08), though Chan and Liaw gave a different form for the first one in their paper, "Cubic Modular Equations and New Ramanujan-Type Series for 1/π" (2000). The modular function responsible, call it h(q), has q-series expansion, h(q) = 1/q + 42 + 783q + 8672q2 + 65367q3 + … This is sequence A030197 and, perhaps not surprisingly, is the McKay-Thompson series of class 3A for the Monster group. To summarize, there is this deep relationship between transcendental numbers of the form eπ√d, the modular functions j(q), r(q), h(q), pi formulas, and the Monster group. (End update.)

Abstract: The McKay-Thompson series for the Monster group, along with certain Dedekind eta quotients, are used in a q-expansion to explain why transcendental constants of form eπ√d with class number h(-d) ≠ 1 are also close to integers.

I. IntroductionII. The j-function, j(τ)III. The Dedekind eta function, η(τ)IV. The r-functions rp(τ)V. The eta quotients fpVI. Examples and Mathematica ComputationsVII. Close ApproximationsVIII. Conclusion: Pi formulas

I. Introduction The approximations, eπ√43 ≈ 123(92-1)3 + 743.9997…eπ√67 ≈ 123(212-1)3 + 743.999998…eπ√163 ≈ 123(2312-1)3 + 743.9999999999992… are well known, the result of a beautiful theory involving complex multiplication and the modular function called the j-function. These are the three largest Heegner numbers and have class number h(-d) = 1. The pattern continues for some discriminants d with h(-d) = 2, with the highest being, eπ√427 ≈ 123(x2-1)3 + 743.99999999999999999999998…, where x = 3(2405+308√61) (note that d = 427 = 7*61), and so on for d with higher class numbers h(-d) = n involving algebraic integers of degree n. (As long as d ≠ 3m or d ≠ 12n-1.) However, how do we explain that there are others that are also close to integers, eπ√37 ≈ 141122 + 103.99997…eπ√10 ≈ 124 -104.2…eπ√58 ≈ 3964 -104.0000001… where the “excess” hovers not around 744, but 104? Furthermore, given the fundamental unit Ud, U37 = 6+√37U5 = (1+√5)/2

U29 = (5+√29)/2 which is involved in Pell equations, x2-dy2 = {±1, ±4}, then, 26(U37

3- U37(-3))2 = 141122

26(U56+ U5

(-6))2 = 124

26(U296+ U29

(-6))2 = 3944

The answer is an intriguing elliptic function called the Dedekind eta function. II. The j-function, j(τ) Before going into the Dedekind eta function, first, the basics of the j-function. This is a modular form of weight zero (equivalently, a modular function) defined by, j(τ) = 1/q + 744 + 196884q + 21493760q2 + 864299970q3 + … (A000521) where q = e2πiτ = exp(2πiτ), and τ is the half-period ratio. This can be calculated in Mathematica as, j(τ) = N[123KleinInvariantJ[τ], n] for arbitrary n decimal places. We can distinguish two forms of τ. Let d be some positive integer, Case 1. τ = (1+√-d)/2 (Associated with odd discriminant d) q = e2πi τ = e2πi (1+√-d)/2 = eπi (1+√-d) = eπi eπi√-d = (-1) e-π√d = -1/

(eπ√d) Case 2. τ = √-m (Associated with even discriminant d = 4m) q = e2πi τ = e2πi√-m = e-2π√m = 1/(e2π√m) Thus as d increases, q becomes a very small real number, negative for the first case, and positive for the second. For the first, we have q = -1/(eπ√d) so, j(τ) = -eπ√d + 744 + 196884q + 21493760q2 + 864299970q3 + … But since q is very small, the contribution of subsequent terms rapidly falls off. To illustrate for d = 163, the sum of the first n terms are, 2 terms = -6403203 + 7 x 10-13

3 terms = -6403203 - 3 x 10-28

4 terms = -6403203 + 4 x 10-44

One can see the contribution of the other terms is just getting smaller and smaller. Thus, we can effectively truncate the j-function as, j(τ) ≈ -eπ√d + 744, or equivalently, eπ√d ≈ -j(τ) + 744 It is well-established that, Theorem: “Let d be a positive integer. Then j(τ) is an algebraic integer of degree n = h(-d), where h(-d) is the class number of d.”

If d has class number h(d) = 1, then j(τ) is an integer and explains, especially for the larger d = {43, 67, 163}, why eπ√d is so close to an integer. Similarly, for the second case, j(τ) ≈ e2π√m + 744 so the fact that j(√-7) = 2553 explains, eπ√28 ≈ 2553 - 744.01… though, while d = 28 has class number 1, it is not a fundamental discriminant. Both can be calculated in Mathematica for, say, n = 1000 decimal places as, j(τ) = N[1728KleinInvariantJ[τ] /. τ → (1+√-163)/2, 1000] = -6403203

j(τ) = N[1728KleinInvariantJ[τ] /. τ → √-7, 1000] = 2553

Taking note that a functional equation of the j-function is, j(τ) = j(τ+1) then for τ = (1+√-d)/2 or τ = √-m, with d or m as positive integers, there are exactly 13 points such that j(τ) is an integer: 9 fundamental discriminants and 4 non-fundamental ones: j(√-1) = 123

j(√-2) = 203

j((1+√-3)/2) = 0j((1+√-7)/2) = -153

j((1+√-11)/2) = -323

j((1+√-19)/2) = -963

j((1+√-43)/2) = -9603

j((1+√-67)/2) = -52803

j((1+√-163)/2) = -6403203

with the first two corresponding to even discriminants d = {4, 8}. And, j(√-3) = 2*303

j(√-4) = 663

j(√-7) = 2553

j((1+√-27)/2) = -3*1603

with the first three corresponding to even discriminants d = {12, 16, 28}. III. The Dedekind eta function, η(τ) Just like the j-function, the Dedekind eta function is also a modular form, but it is of weight ½. (Note: My thanks to Michael Somos for pointing out an error in a previous draft). It was introduced in 1877 by Wilhelm Dedekind (1831-1916) and is usually defined in terms of the infinite product,

where q = e2πiτ = exp(2πiτ). As one can see, it involves a 24th root (which “explains” why the integer 24 will play such a significant role later). This can be calculated in Mathematica as, η(τ) = N[DedekindEta[τ], n] for arbitrary n decimal places. A remarkable feature of the η(τ) is that

while for complex τ it generally yields a transcendental number, certain eta quotients turn out to be algebraic numbers. Furthermore, one can express the j-function j(τ) discussed in the previous section using the simple formula, j(τ) = (f2

24 + 16)3 / f224

where f2 = η[τ/2]/η[τ]. The eta quotient f2 is one of the Weber functions. (Unfortunately, Weber labeled it as f1 though why it is labeled f2 in this paper will become clearer later.) There are in fact several other formulas using various eta quotients, but that will be for another article. What we’ll focus on is how some eta quotients can be used to define other interesting modular forms similar to j(τ). IV. The r-functions, rp(τ) Ramanujan did preliminary work on these functions which can be defined in terms of eta quotients. A lot of his results were on the orders p = 2,3,5,7, but there are several more. As was pointed out, the Dedekind eta function involved a 24th root. Amazingly, the orders p of the r-functions rp(τ) depend on simple arithmetic properties of the integer 24, namely, they are the positive integers p such that, k = 24/(p-1) is also an integer. It is easily seen there are eight, p = {2, 3, 4, 5, 7, 9, 13, 25}, and can be divided into 2 kinds: a) p ≠ 1 mod 4: p = {2, 3, 4, 7}b) p = 1 mod 4: p = {5, 9, 13, 25}

To recall, the j-function can be given the formula, j(τ) = (f2

24 + 16)3 / f224 = 1/q + 744 + 196884q + 21493760q2 +

864299970q3 + … (A000521, A007240) Excluding the constant term 744, this sequence of coefficients (given by the second OEIS link) is the McKay-Thompson series of class 1A for the Monster group. The r-functions, rp(τ) of order p, have the similar form, I. p ≠ 1 mod 4; p = {2, 3, 4, 7} r2(τ) = (f2

24 + 64)2 / f224 = 1/q + 104 + 4372q + 96256q2 +

1240002q3 + … (A007267) r3(τ) = (f3

12 + 27)2 / f312 = 1/q + 42 + 783q + 8672q2 + 65367q3 +

… (A030197) r4(τ) = (f4

8 + 16)2 / f48 = 1/q + 24 + 276q + 2048q2 + 11202q3 + …

(A097340, A107080) r7(τ) = (f7

4 + 7)2 / f74 = 1/q + 10 + 51q + 204q2 + 681q3 + …

(A030183) And no wonder, as the sequences are, for p = {2, 3, 4, 7}, the McKay-Thompson series of class pA for the Monster group! These functions rp(τ), as eta quotients fp, have a common form, rp(τ) = (fp

k + pk/4)2 / fpk

where k = 24/(p-1), hence k = {24, 12, 8, 4}. Interestingly, the constant term c = {104, 42, 24, 10} also has a common form, c = 2pk/4-k but I haven’t been able to figure out if the subsequent terms can be expressed in {p, k}. The other set of p are, II. p = 1 mod 4; p = {5, 9, 13, 25} r5(τ) = (f5

12 + 125) / f56 = 1/q - 6 + 134q + 760q2 + 3345q3 + …

(A007251) r9(τ) = (f9

6 + 27) / f93 = 1/q - 3 + 27q + 86q2 + 243q3 + …

(A07266) r13(τ) = (f13

4 + 13) / f132 = 1/q - 2 + 12q + 28q2 + 66q3 + …

(A034318) r25(τ) = (f25

2 + 5) / f25 = 1/q - 1 + 4q + 5q2 + 10q3 + … (A058594) The sequences are the McKay-Thompson series of class pA for the Monster group for p = {5, 9, 13, 25}. Let k = 24/(p-1), and we get k = {6, 3, 2, 1}. These functions also have a common form, rp(τ) = (fp

2k + pk/2) / fpk

where fp as usual are eta quotients. Notice the subtle differences from the first set, like the fact that the numerator is no longer squared. The constant term c = {-6, -3, -2, -1} are just -k.

V. The Eta Quotients, fp For the purposes of this paper, let the q-expansion in the previous section involve q = exp(2πiτ/p). Then the eta quotients fp are, A. Even order p f2 = η[τ/2] / η[τ], where τ = √-m or τ = 1+√-mf4 = η[τ/4] / η[τ], where τ = √-m, or τ = 6+√-m B. Odd order p = 2b+1 Case 1: If τ = √-m, then fp = η[τ/p] / η[τ]. Case 2: If τ = (1+√-d)/2, then fp = ωk η[(τ+b)/p] / η[τ]. where ωk = (-1)1/k = exp(πi/k), and k = 24/(p-1). It is easy to evaluate these eta quotients using Mathematica. An interesting feature of fp is that its kth power where k = 24/(p-1) also has a q-expansion with integer coefficients. Let fp = fp(τ), then, f2(τ)24 = 1/q - 24 + 276q - 2048q2 + 11202q3 - 49152q4 + … (A007191) f3(τ)12 = 1/q - 12 + 54q - 76q2 - 243q3 + 1188q4 + … (A030182) f4(τ)8 = 1/q - 8 + 20q - 62q3 + 216q5 - 641q7 + … (A007248) f5(τ)6 = 1/q - 6 + 9q + 10q2 - 30q3 + 6q4 + … (A007252)

f7(τ)4 = 1/q - 4 + 2q + 8q2 - 5q3 - 4q4 - 10q5 + …(A052240) f9(τ)3 = 1/q - 3 + 5q2 - 7q5 + 3q8 + 15q11 + … (A058091) f13(τ)2 = 1/q - 2 + q + 2q2 + q3 + 2q4 - 2q5 + … (A058496) f25(τ) = 1/q - 1 - q + q4 + q6 - q11 - q14 + … (A096563) And what are these sequences when normalized without the constant term? These are mostly the McKay-Thompson series of class pB for Monster! (With the exception of f4 which is 4C and f25 which is 25a.) VI. Examples and Mathematica Computations A few examples will be given to highlight the differences when using forms τ = √-m, τ = a+√-m (a an integer), or τ = (1+√-d)/2. A. Even order p r2(τ) = N[(f2

24 + 64)2 / f224 /. f2 → DedekindEta[τ/2]/DedekindEta

[τ] /. τ → √-58, 1000] = 3964

r2(τ) = N[(f224 + 64)2 / f2

24 /. f2 → DedekindEta[τ/2]/DedekindEta

[τ] /. τ → 1+√-37, 1000] = -141122

B. Odd prime order p r3(τ) = N[(f3

12 + 27)2 / f312 /. f3 → (-1)1/12DedekindEta[(τ+1)/3]/

DedekindEta[τ] /. τ → (1+√-267)/2, 1000] = -3003

r5(τ) = N[(f512 + 125) / f5

6 /. f5 → (-1)1/6DedekindEta[(τ+2)/5]/

DedekindEta[τ] /. τ → (1+√-235)/2, 1000] = -53(122)r7(τ) = N[(f7

4 + 7)2 / f74 /. f7 → (-1)1/4DedekindEta[(τ+3)/7]/

DedekindEta[τ] /. τ → (1+√-427)/2, 1000] = -7(392)r13(τ) = N[(f13

4 + 13) / f132 /. f13 → (-1)1/2DedekindEta[(τ+6)/13]/

DedekindEta[τ] /. τ → (1+√-403)/2, 1000] = -13(10) These d have class number h(-d) = 2 and there are exactly 18 fundamental discriminants that have it, d = {15, 20, 24, 35, 40, 51, 52, 88, 91, 115, 123, 148, 187, 232, 235, 267, 403, 427} With the exception of d = 187 = 11(17), all are divisible by any of the primes p = {2, 3, 5, 7, 13}. Using appropriate p, these yield integer rp(τ), r2(√-6) = 482

r2(√-10) = 124

r2(√-22) = 15842

r2(√-58) = 3964

r2(1+√-5) = -322

r2(1+√-13) = -2882

r2 (1+√-37) = -141122

r3((1+√-15)/2) = -33

r3((1+√-51)/2) = -123

r3((1+√-123)/2) = -483

r3((1+√-267)/2) = -3003

r5((1+√-35)/2) = -52(2)

r5((1+√-115)/2) = -52(34)

r5((1+√-235)/2) = -53(122) r7((1+√-91)/2) = -7(32)

r7((1+√-427)/2) = -7(392) r13((1+√-403)/2) = -13(10) Testing it with fundamental discriminants with class number h(-d) = 4, the appropriate rp(τ) were algebraic integers of degree 2. For h(-d) = 6, they were of degree 3. There seems to be a pattern, Conjecture: “Given primes p = {2, 3, 5, 7, 13} and a composite fundamental discriminant d = pv with class number h(-d) = 2n. Then, for appropriately chosen τ, rp(τ) is an algebraic integer of degree n.” 1. For odd d: let τ = (1+√-d)/2 2. For even d = 4m: a. odd m, let τ = (1+√-m)/2; (But τ = 1+√-m for p = 2.) b. even m, let τ = √-m VII. Close Approximations The largest of the r-functions rp(τ) in the previous section give rise to

close integer approximations to transcendental numbers of form e(π/p)

√d , namely, e(π/2)√148 = eπ√37 ≈ 141122 + 103.99997…e(π/2)√232 = eπ√58 ≈ 3964 - 104.0000001… e(π/3)√123 = eπ√(41/3) ≈ 483 + 41.992…e(π/3)√267 = eπ√(89/3) ≈ 3003 + 41.99997… e(π/5)√235 = eπ√(47/5) ≈ 53(122) - 6.008… e(π/7)√427 = eπ√(61/7) ≈ 7(392) + 9.995… The reason why is analogous to the one for the j-function. To recall, the q-expansion of rp(τ) for p = 2 is, r2(τ) = 1/q + 104 + 4372q + 96256q2 + 1240002q3 + … For τ = √-m, then, q = e2πi τ/p = eπi√-m = e-π√m = 1/(eπ√m) As before, as m increases, q is a very small real number and the contribution of subsequent terms of the q-expansion becomes minuscule. To illustrate, for τ = √-58, we saw that the formula using the eta quotients yields an integer value as r2(√-58) = 3964. But using the q-expansion, the sum of the first n terms are, 2 terms = 3964 - 1.7 x 10-7

3 terms = 3964 - 1.6 x 10-16

4 terms = 3964 - 8.3 x 10-26

where the difference is getting smaller and smaller. Hence, we can just truncate r2(τ) as, r2(τ) ≈ 1/q + 104 ≈ eπ√m + 104 explaining why, eπ√58 ≈ 3964 - 104.0000001… and, by analogy, eπ√(89/3) ≈ 3003 + 41.99997… as well as the other examples, are so close to an integer. VIII. Conclusion: Pi Formulas Before concluding this article, there are many other surprising aspects to the functions j(τ) and rp(τ), which I plan to write about in a series of articles. The next one involves their role in pi formulas. For example, define the factorial quotients, h2 = (4n)! / (n!4)

h3 = (2n)!(3n)! / (n!5) Then, for n = 0 to ∞, 1/π = 32√2 Σ h2 (26390n+1103)/(3964)n+1/2

1/π = 2 Σ h3 (-1)n (14151n+827)/(3003)n+1/2

where the blue numbers are the exact values of r2(√-58) and r3((1+√-267)/2), respectively. It turns out that r2(τ) for certain τ may involve familiar constants like the golden ratio, tribonacci constant, and the plastic constant, all of which can be used in pi formulas! But that’s another story.

Abstract: Ramanujan-type pi formulas using the silver ratio, golden ratio, tribonacci constant, and plastic constant will be given. The constants’ commonality as being all: 1) Pisot numbers, 2) limiting ratios of special integer sequences, and 3) appearing in the metric properties of certain geometric solids will also be high-lighted. Contents:I. Silver and Golden RatioII. Tribonacci ConstantIII. Plastic Constant I. Silver and Golden Ratio As a basic introduction, recall that the silver ratio S = 1+√2 = 2.41421… is the positive root of the quadratic, S2-2S-1 = (S-1)2-2 = 0 (eq.1) and is the limiting ratio of the Pell numbers y = {1, 2, 5, 12, 29, 70…}. These are the denominators of the best rational approximations to √2, being the “y” values of the Pell equation, x2-2y2 = ±1. This sequence has the recurrence relation S(n) = 2S(n-1) + S(n-2). The golden ratio φ = (1+√5)/2 = 1.61803… is the positive root of, φ2-φ-1 = 0 (eq.2)

and is best known in the context of the Fibonacci numbers {1, 1, 2, 3, 5, 8, 13,….}. Each term is the sum of the previous two, and φ is its limiting ratio, as well as for the Lucas numbers, a fact which will be relevant later. However, the constants S and φ also appears in other contexts. Recall the well-known approximations, eπ√43 ≈ 123(92-1)3 + 743.9997…eπ√67 ≈ 123(212-1)3 + 743.999998…eπ√163 ≈ 123(2312-1)3 + 743.9999999999992… with the last being the so-called Ramanujan constant. Not so well-known are the ff ones. Given the silver ratio S = 1+√2, and the golden ratio φ = (1+√5)/2, then, 1) eπ√6 ≈ 26S4 + 23.8…2) eπ√8 ≈ 29S3 + 23.9…3) eπ√16 ≈ 2(21/2)S6 + 23.999…4) eπ√22 ≈ 26S12 + 23.9999… 1) eπ√5 ≈ 26φ6 - 24.25…2) eπ√10 ≈ 26φ12 + 23.98…3) eπ√15 ≈ 212φ8 - 24.001…4) eπ√25 ≈ 26φ24 - 24.00004… The fact that the excess hovers around the integer “24” is not coincidence, as the blue numbers are the exact values of certain Dedekind eta quotients (to be given later), a modular form which involves a 24th root. All these can be used for pi formulas, though for brevity we will give only one example using S. Define h4 = (2n)!3 / (n!6), and n = 0 to ∞. Then,

4) 1/π = 1/(2S3) Σ (-1)n h4 (6*22n + (50-17S)) / (26S12)n

as well as, 1) 1/π = 1/(φ√φ) Σ h4 ((2√5)n + (φ-1)) / (26φ6)n

2) 1/π = (1/2) Σ (-1)n h4 ((-48φ+84)n + (-20φ+33)) / (26φ12)n

3) 1/π = (1/16) Σ h4 ((42φ-6)n + (5φ-3)) / (212φ8)n

4) 1/π = (10/51/4) Σ h4 (6(18φ-29)n + (47φ-76)) / (26φ24)n

Amazingly, for the last, the coefficients involve the four consecutive Lucas numbers (in red) as {2, 1, 3, 4, 7, 11, 18, 29, 47, 76,…}! Why that is so, I have no idea. For the formulas using the golden ratio, (1) is by this author, (3) is by Ramanujan, while (2) & (4) are by J. Guillera, though I have tweaked the numerators so that they are polynomials in φ. (Update, 5/23/11) Qiaochu Yuan from MathStackExchange gave the identity for Fibonacci numbers, Fn-1 + Fn φ = φn

A related one for Lucas numbers is, Ln φ - Ln+1 = (1/φn-1) √5 Hence, Guillera's pi formula [4] simplifies as, 1/π = (10/51/4) Σ h4 (6√5/φ6n + √5/φ8) / (26φ24)n

or just simply, 1/π = 10(51/4)/φ6 Σ h4 (6n + 1/φ2) / (26φ24)n

One mystery solved. (End update) Of the five Platonic solids, the silver ratio appears in three: the tetrahedron (4-face), the cube (6-face), and the octahedron (8-face). The golden ratio figures prominently in the last two: the dodecahedron (12-face) and icosahedron (20-face). However, the next two constants also have their own roles in other geometric forms. The associated pi formulas are by this author. II. Tribonacci constant The tribonacci constant T = 1.83929… is the real root of the cubic eqn, T3-T2-T-1 = 0 (eq.3) Analogous to the Fibonacci numbers, the tribonacci numbers {1, 1, 2, 4, 7, 13, 24,….} is a sequence where each term is the sum of the previous three, and the tribonacci constant is its limiting ratio. Eq. 3 can be solved as,

with discriminant d = 11. This has a beautiful infinite nested radical expression,

Compare this to the one for the golden ratio given in the next section. Furthermore, define the constant’s reciprocal as v = 1/T. Then, analogous to the golden ratio, we have the approximation, eπ√11 ≈ (v+1)24 - 24.008…

where the power and excess now both involve the integer 24. The expression v+1 = 1.543689… is the exact value of a special eta quotient, hence we can also use it in a pi formula. Define, A = (1/2)(1+2v)(1+4v)B = (1/4)(-1+2v+4v2)C = (v+1)24

h4 = (2n)!3 / (n!6) where v = 1/T, then, 1/π = Σ h4 (An+B) / Cn

from n = 0 to ∞. There is an interesting and orderly q-continued fraction for T. Let q = -1/(eπ√11), then,

which gives the precise relationship between eπ√11 and its approximation (1/T + 1)24. Also, as I pointed out (in 2006) to E. Weisstein of MathWorld, the tribonacci constant appears, not in the Platonic solids, but in the Archimedean solid, the snub cube,

Explicitly, the Cartesian coordinates for the vertices of a snub cube are all the even and odd permutations of, {± 1, ± 1/T, ± T} with an even and odd number of plus signs, respectively. Furthermore, T can also be used to express the hard hexagon entropy constant. There is certainly more to this constant than meets the eye. Note: (9/27/10) I just learned that my observation about the tribonacci constant and the snub cube was preceded by John Sharp in his 1998 article in the Mathematical Gazette, "Have You Seen This Number?" III. Plastic constant The plastic constant P = 1.32472.… is the real root of the cubic eqn, P3-P-1 = 0 (eq.4) Recall that the Fibonacci numbers {1, 1, 2, 3, 5, 8,...} have the recurrence relation F(n) = F(n-1) + F(n-2). This forms a square spiral,

On the other hand, the Padovan sequence {1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16,…} has P(n) = P(n-2) + P(n-3), and forms a triangular spiral,

The role of the plastic constant is, just like for the previous three constants, it is a limiting ratio, namely, for the Padovan and Perrin sequences. Furthermore, the golden ratio φ and the plastic constant P have the similar beautiful infinite nested radical representation,

The radical soln of eq. 4 is,

with discriminant d = 23. From the previous two sections, I guess one can figure out that it appears in a similar approximation, eπ√23 ≈ 212P24 - 24.00007… as well as the pi formula, 1/π = Σ h4 (An+B) / Cn

A = (1/8)(5P-4)(5P+8)B = (1/16)(9P2+4P-16)C = 212P24

h4 = (2n)!3 / (n!6),

from n = 0 to ∞. Just like for the tribonacci constant, there is a q-continued fraction for the plastic constant P. Let q = -1/(eπ√23), then,

which gives the relationship between eπ√23 and the approximation (√2 P)24. While the tribonacci constant is implicit in the snub cube, I found that the same can be said between the plastic constant and the snub icosidodeca-dodecahedron (2006),

(See also here.) How these and other pi formulas were derived will be discussed in the next article.

Addendum: There are other affinities between these four constants: the silver ratio S, the golden ratio φ, tribonacci constant T, and plastic constant P. 1) All four are Pisot numbers. P is the smallest Pisot number, and φ is the smallest accumulation point. 2) All can be expressed in terms of the Dedekind eta function, η(τ), a modular form which, as was pointed out, involves a 24th root. Given the root of negative unity eπi/24 = Exp(πi/24), define the eta quotient f2(τ) = η(τ/2)/ η(τ). Then, φ1/1 = (1/21/4) Exp(πi/24) f2(1+√-25) = 1.61803…

φ1/2 = (1/21/4) f2(√-10)

φ1/3 = (1/21/2) Exp(πi/24) f2(1+√-15)

φ1/4 = (1/21/4) Exp(πi/24) f2(1+√-5) S1/2 = (1/21/4) f2(√-22)

S1/4 = (1/27/16) f2(√-16)

S1/6 = (1/21/4) f2(√-6)

S1/8 = (1/23/8) f2(√-8) 1/T + 1 = Exp(πi/24) f2(1+√-11) = 1.543689…

P = (1/21/2) Exp(πi/24) f2(1+√-23) = 1.324717… Raising these to the 24th power gets rid of the root of unity and explains the denominators of the pi formulas given in the article. 3). As infinite nested radicals.

The expressions for the golden ratio and the Plastic constant have been given previously. For the silver ratio S, not surprisingly it involves the number 2,

but surprisingly so does for the tribonacci constant T as,

4) In basic algebraic relations: a) φ and T: φ + 1/φ 2 = 2T + 1/T 3 = 2 b) φ and P: 1/φ + 1 = φ1/P + 1 = P 2 c) T and P: (1-u)3 = (2u)2 , where u = (T-1)/(T+1)(1-v)3 = (v)2 , where v = 1/(P+1)

Abstract: The McKay-Thompson series of class 1A, 2A, 3A, and 4A

for the Monster group, along with certain expressions involving Dedekind eta quotients, are used as a unifying framework for the four kinds of pi formulas found by Ramanujan, the Chudnovskys, and the Borweins. Formulas for four kinds of singular moduli using the same eta quotients will also be given. I. IntroductionII. The Dedekind eta functionIII. The variable C: The j-function j(τ) and r-functions rp(τ).IV. The variables A and BV. Ramanujan and class number h(-d) = 1VI. Formulas for Singular ModuliVII. Going beyond I. Introduction Ramanujan gave many beautiful and unusual formulas for pi and, inspired by these, the Chudnovsky and Borwein brothers came up with similar ones. First though, define the factorial quotients, h1 = (6n)! / ((3n)! n!3)

h2 = (4n)! / (n!4)

h3 = (2n)!(3n)! / (n!5)

h4 = (2n)!3 / (n!6) or, equivalently, as Pochhammer symbol products, h1 = 1728n (1/2)n(1/6)n(5/6)n / (n!3)

h2 = 256n (1/2)n(1/4)n(3/4)n / (n!3)

h3 = 108n (1/2)n(1/3)n(2/3)n / (n!3)

h4 = 64n (1/2)n(1/2)n(1/2)n / (n!3) where (a)n = (a)(a+1)(a+2)…(a+n-1). Starting with n = 0, these generate the following integer sequences (with limiting ratio Lp between successive terms as n → ∞), h1[n] = {1, 120, 83160, 81681600,…} A001421 (L1 = 1728)h2[n] = {1, 24, 2520, 369600,…} A008977 (L2 = 256)h3[n] = {1, 12, 540, 33600,…} (none) (L3 = 108)h4[n] = {1, 8, 216, 8000,…} A002897 (L4 = 64) As J. Guillera points out in his paper, "A class of conjectured series representations for 1/π", these sequences satisfy a first-order recurrence relation whose coefficients are 3rd degree polynomials in n, n3 h1[n]-24(2n-1)(6n-1)(6n-5) h1[n-1] = 0

n3 h2[n]- 8(2n-1)(4n-1)(4n-3) h2[n-1] = 0

n3 h3[n]- 6(2n-1)(3n-1)(3n-2) h3[n-1] = 0

n3 h4[n]- 8(2n-1)(2n-1)(2n-1) h4[n-1] = 0 For example, let n = 2 for h1[n], and the A001421 sequence yields, h1[2] = 83160; h1[1] = 120, Thus, for n = 2, n3 (83160) - 24(2n-1)(6n-1)(6n-5) (120) = 0 Nice, isn't it? One can also see the affinity between the hp (as

Pochhammer symbols) and the recurrence relations, as well as the appearance of the limiting ratio Lp. Below is the complete list of 36 pi formulas using the four kinds of hp and three constants {A,B,C} where C is an integer, and {A,B} are either integers or have a square root. The formulas beginning in red are 16 of Ramanujan’s 17 pi formulas found in his 1914 paper, Modular Equations and Approximations to π (one had a radical C as it involved the golden ratio) . The formula beginning in blue was also given by Ramanujan in his Notebooks. With sum Σ going from n = 0 to ∞, then, 1. j(τ) Let h1 = (6n)! / ((3n)! n!3) = 1728n (1/2)n(1/6)n(5/6)n / (n!3) (none) eπ√3 ≈ 03 + …1/π = 3 Σ h1 (-1)n (63n+8)/(153)n+1/2 eπ√7 ≈ 153 + 6961/π = 4 Σ h1 (-1)n (154n+15)/(323)n+1/2 eπ√11 ≈

323 + 7381/π = 12 Σ h1 (-1)n (342n+25)/(963)n+1/2 eπ√19 ≈

963 + 7431/π = 36 Σ h1 (-1)n (506n+31)/(3*1603)n+1/2 eπ√27 ≈

3*1603 + 743.9…1/π = 12 Σ h1 (-1)n (16254n+789)/(9603)n+1/2 eπ√43 ≈

9603 + 743.99…1/π = 12 Σ h1 (-1)n (261702n+10177)/(52803)n+1/2 eπ√67 ≈

52803 + 743.99999…1/π = 12 Σ h1 (-1)n (545140134n+13591409)/(6403203)n

+1/2 eπ√163 ≈ 6403203 + 743.999999999… (none) eπ√4 ≈ 123 - …1/π = 8 Σ h1 (28n+3)/(203)n+1/2, eπ√8 ≈ 203 - 771

1/π = 72 Σ h1 (11n+1)/(2*303)n+1/2 eπ√12 ≈ 2*303 - 7471/π = 24√2 Σ h1 (63n+5)/(663)n+1/2 eπ√16 ≈ 663 - 7441/π = 162 Σ h1 (133n+8)/(2553)n+1/2 eπ√28 ≈ 2553 -744.01… 2. r2(τ) Let h2 = (4n)! / (n!4) = 256n (1/2)n(1/4)n(3/4)n / (n!3) 1/π = 4 Σ h2 (-1)n (20n+3)/(322)n+1/2 eπ√5 ≈ 322 + 1001/π = 4 Σ h2 (-1)n (260n+23)/(2882)n+1/2 eπ√13 ≈ 2882 + 103.9…1/π = 4 Σ h2 (-1)n (21460n+1123)/(141122)n+1/2 eπ√37 ≈ 141122 + 103.9999… 1/π = √7 Σ h2 (-1)n (65n+8)/(632)n+1/2 eπ√7 ≈ 632 + 102.9...1/π = 4*3 Σ h2 (-1)n (28n+3)/(3*642)n+1/2 eπ√9 ≈ 3*642 + 103...

1/π = 4*5 Σ h2 (-1)n (644n+41)/(5*11522)n+1/2 eπ√25 ≈

5*11522 + 103.999.. 1/π = 4√2 Σ h2 (7n+1)/(2*182)n+1/2 eπ√4 ≈ 2*182 - 1121/π = 8√3 Σ h2 (8n+1)/(482)n+1/2 eπ√6 ≈ 482 - 1061/π = 32√2 Σ h2 (10n+1)/(124)n+1/2 eπ√10 ≈ 124 - 104.2…1/π = 48√3 Σ h2 (40n+3)/(7842)n+1/2 eπ√18 ≈ 7842 -104.00…1/π = 8√11 Σ h2 (280n+19)/(15842)n+1/2 eπ√22 ≈ 15842 - 104.001…1/π = 32√2 Σ h2 (26390n+1103)/(3964)n+1/2 eπ√58 ≈ 3964 - 104.00000… 3. r3(τ) Let h3 = (2n)!(3n)! / (n!5) = 108n (1/2)n(1/3)n(2/3)n / (n!3) (none) eπ√(5/3) ≈ 33 +1/π = 2 Σ h3 (-1)n (51n+7)/(123)n+1/2 eπ√(17/3) ≈ 123 + 41.6…1/π = 2 Σ h3 (-1)n (615n+53)/(483)n+1/2 eπ√(41/3) ≈ 483 + 41.99…1/π = 2 Σ h3 (-1)n (14151n+827)/(3003)n+1/2 eπ√(89/3) ≈ 3003 + 41.9999… 1/π = 2*3 Σ h3 (-1)n (5n+1)/(3*43)n+1/2 eπ√(9/3) ≈ 3*43

+ 38.8…1/π = 2*5 Σ h3 (-1)n (27n+3)/(5*123)n+1/2 eπ√(25/3) ≈

5*123 + 41.91…1/π = 2*7 Σ h3 (-1)n (165n+13)/(7*363)n+1/2 eπ√(49/3) ≈

7*363 + 41.998… (none) eπ√(4/3) ≈ 22

(33) - …1/π = 2√2 Σ h3 (6n+1)/(63)n+1/2, eπ√(8/3) ≈ 63 - 471/π = 4√2 Σ h3 (15n+2)/(2*93)n+1/2 eπ√(16/3) ≈

2*93- 42.5…1/π = 2√5 Σ h3 (33n+4)/(153)n+1/2 eπ√(20/3) ≈ 153 - 42.2… 4. r4(τ) Let h4 = (2n)!3 / (n!6) = 64n (1/2)n(1/2)n(1/2)n / (n!3) (none) eπ√1 ≈ 23 + 15.1…1/π = 4 Σ h4 (-1)n (4n+1)/(26)n+1/2 eπ√2 ≈ 26 + 21.0…1/π = 8 Σ h4 (-1)n (6n+1)/(29)n+1/2 eπ√4 ≈ 29 + 23.5…1/π = 4 Σ h4 (6n+1)/(28)n+1/2 eπ√3 ≈ 28 - 25.2…1/π = 4 Σ h4 (42n+5)/(212)n+1/2 eπ√7 ≈ 212 -

24… The functions j(τ) and rp(τ) will be discussed in a short while. (Surprisingly, Ramanujan missed two of the r2(τ) which were later found by Berndt, Guillera, et al.) The associated power of Gelfond’s constant eπ is given at the right side for comparison. The “excess” {744, 104, 42, 24} is also significant, as this can be a clue to what modular form is being used. In his 1914 paper, Ramanujan gave little explanation on how he came up with the formulas other than saying that there were “corresponding theories”. He apparently wasn’t aware of it, but he was in fact exploring certain natural consequences of the Monster group (predicted sixty years later in 1973 by Fischer and Griess, and constructed by Griess in 1980). The general form of these pi formulas F is, 1/π = Σ hp (An+B) / Cn

where {A,B,C} are algebraic numbers, and n = 0 to ∞ (as in the rest of this paper). There is an infinite number of F, a special few use just integers while some involve involve familiar irrational constants. For example, one found by Ramanujan involved the golden ratio φ = (1+√5)/2, 1/π = (1/16) Σ h4 ((-6+42φ)n + (-3+5φ)) / (212φ8)n

Another one by J. Guillera, simplified from its original form in Mathworld's Pi Formulas [117] is, 1/π = (10/51/4) Σ h4 (6(18φ-29)n + (47φ-76)) / (26φ24)n

For this article, the author tweaked the numerators so that they are polynomials in φ and found that the second involve four consecutive Lucas numbers (in blue) as {2, 1, 3, 4, 7, 11, 18, 29, 47, 76,…}! Why

that is so, I have no idea. (Update, 5/23/11) Qiaochu Yuan from MathStackExchange gave the identity for Fibonacci numbers, Fn-1 + Fn φ = φn

A related one for Lucas numbers is, Ln φ - Ln+1 = (1/φn-1) √5 Hence, Guillera's pi formula simplifies as, 1/π = (10/51/4) Σ h4 (6√5/φ6n + √5/φ8) / (26φ24)n

or just simply, 1/π = 10(51/4)/φ6 Σ h4 (6n + 1/φ2) / (26φ24)n One mystery solved. (End update) Surprisingly, the tribonacci constant T which is the limiting ratio of the tribonacci numbers (in the same manner as the golden ratio is for the Fibonacci and Lucas numbers) can also be used in a pi formula. The constant T is the real root of x3-x2-x-1 = 0. Define v = 1/T, A = (1/2)(1+2v)(1+4v)B = (1/4)(-1+2v+4v2)C = (v+1)24

h4 = (2n)!3 / (n!6) then 1/π = Σ h4 (An+B)/Cn. See Article 2, “A Tale of Four Constants”, with the other two being the silver ratio and the plastic constant. So how do we find these

formulas? An easy approach is, given the factorial quotient hp, if any two of {A,B,C} is known, then one can solve for the third using enough terms of the summation, easily done by computer algebra systems like Mathematica. But first of all, what are the {A,B,C}’s? II. The Dedekind eta function, η(τ) The Dedekind eta function is a modular form usually defined in terms of the infinite product, η(τ) = q1/24 Π (1-qm) where m = 1 to infinity, q = e2πiτ = exp(2πiτ), and τ is the half-period ratio. To point out again, it involves a 24th root (which “explains” why the integer 24 plays such a significant role in certain contexts). This can be calculated in Mathematica as, η(τ) = N[DedekindEta[τ], n] for arbitrary n decimal places. III. The variable C: The j-function j(τ) and r-functions rp(τ). First define, qp = e2πiτ = exp(2πiτ) It is easily seen that for imaginary values τ = (1+√-d)/2, or τ = √-m, (and similar ones) where d or m is a positive real, then qp is a real number.

Proof: Case 1. τ = (1+√-d)/2 (Associated with odd discriminant d) qp = e2πi τ = e2πi (1+√-d)/2 = eπi (1+√-d) = eπi eπi√-d = (-1) e-π√d = -1/(eπ√d) Case 2. τ = √-m (Associated with even discriminant d = 4m) qp = e2πi τ = e2πi√-m = e-2π√m = 1/(e2π√m) and other cases. Thus as d increases, qp becomes a very small real number, negative for the first case, and positive for the second. To find q, it is a simple matter of taking nth roots of a real number which, by the Fundamental Theorem of Algebra (FTA), is guaranteed to have n roots. See also Mathworld’s j-function for a similar derivation. (End proof.) Then using qp = e2πiτ = exp(2πiτ), define the ff functions j(τ) and rp(τ), 1. j(τ) = (f2

24 + 16)3 / f224 = 1/q + 744 + 196884q + 21493760q2 +

864299970q3 + … (A000521, A007240) 2. r2(τ) = (f2

24 + 64)2 / f224 = 1/q + 104 + 4372q + 96256q2 +

1240002q3 + … (A007267) 3. r3(τ) = (f3

12 + 27)2 / f312 = 1/q + 42 + 783q + 8672q2 + 65367q3 +

… (A030197) 4. r4(τ) = (f4

8 + 16)2 / f48 = 1/q + 24 + 276q + 2048q2 + 11202q3 +

… (A097340, A107080)

and perhaps also relevant, 5. r7(τ) = (f7

4 + 7)2 / f74 = 1/q + 10 + 51q + 204q2 + 681q3 + …

(A030183) The rp(τ) have a common form. Let k = 24/(p-1), and we get k = {24, 12, 8, 4}. Then, rp(τ) = (fp

k + pk/4)2 / fpk

Furthermore, the constant term c = {104, 42, 24, 10} of their q-series expansion also has a common form in terms of {p, k} as, c = 2pk/4-k Using appropriate Dedekind eta quotient fp, then j(τ) and rp(τ) are real algebraic numbers. The first is the well-known j-function and its q-expansion is the McKay-Thompson series of class 1A for the Monster group, while the r-functions rp(τ) involve class pA for p = {2, 3, 4, 7}. Examples: Bearing in mind that qp = e2πiτ = exp(2πiτ): 1. For j(τ), let p = 1, τ = (1+√-163)/2, hence q = -e-π√163. Substituting this q into the j-function's q-series above, we get, j((1+√-163)/2) = -6403203

2. Let p = 2, and τ = 1+√-37, hence q2 = e-2π√37. By FTA, this has 2 roots, namely q = ±e-π√37. Using the negative root q = -e-π√37,

r2(1+√-37) = -141122 = -(84√2)4

Note: Fundamental discriminant d = -4·37 has class number h(d) = 2. 3. Let p = 3, and τ = (1+√-267)/2, hence q3 = -e-π√267. By FTA, this has 3 roots. Using the real root, we have q = (-e-π√267)1/3 = -e-π√

(89/3), hence, r3((1+√-267)/2) = -3003

Note: When verifying these identities using computer algebra software (CAS) like Mathematica or Maple, one should be careful since Mathematica gives the cube root of a negative real number r, or N[(-r)1/3], as a complex value. Of course, by FTA this is still valid but becomes a problem when the real root, like q above, is desired. Of course, a small tweak can easily give the real root. 4. Let p = 4, and τ = 2+√-16, hence q4 = e-2π√16. This has 4 roots, namely q = ±(e-2π√16)1/4 and q = ±i(e-2π√16)1/4, Using the real negative root, q = -e-2π, r4(2+√-16) = -29

5. Let p = 7, and τ = (1+√-427)/2, hence q7 = -e-π√427. Again using the real root, we have q = (-e-π√427)1/7 = -e-π√(61/7), hence, r7((1+√-427)/2) = -7(392) Alternatively, instead of using the q-series expansion, one can employ the Dedekind eta quotients fp. (Note: In Mathematica, the syntax [] is

simply used to introduce a function’s argument.) Recalling that η[τ] is the Dedekind eta function, then, 1. j(τ) = (f2

24 + 16)3 / f224, where f2 = η[τ/2] / η[τ], for τ = √-m, or τ

= (1+√-d)/2. 2. r2(τ) = (f2

24 + 64)2 / f224, where f2 = η[τ/2] / η[τ], for τ = √-m, or τ

= 1+√-m. 3. r3(τ) = (f3

12 + 27)2 / f312,

a) f3 = η[τ/3] / η[τ], for τ = √-mb) f3 = (-1)1/12 η[(τ+1)/3] / η[τ], for τ = (1+√-d)/2 4. r4(τ) = (f48 + 16)2 / f48, where f4 = η[τ/4] / η[τ], for τ = √-m, or τ = 2+√-m 5. r7(τ) = (f74 + 7)2 / f74, a) f7 = η[τ/7] / η[τ], for τ = √-mb) f7 = (-1)1/4 η[(τ+3)/7] / η[τ], for τ = (1+√-d)/2 (Notice the qualitative difference of fp for odd and even p.) Mathematica examples: 1. j(τ) j(τ) = N[(f2

24 + 16)3 / f224 /. f2 → DedekindEta[τ/2]/DedekindEta[τ] /.

τ → √-4, 1000] = 663

j(τ) = N[(f224 + 16)3 / f2

24 /. f2 → DedekindEta[τ/2]/DedekindEta[τ] /.

τ → (1+√-7)/2, 1000] = -153

2. r2(τ) r2(τ) = N[(f2

[τ] /. τ → √-58, 1000] = 3964

r2(τ) = N[(f224 + 64)2 / f2

24 /. f2 → DedekindEta[τ/2]/DedekindEta

[τ] /. τ → 1+√-37, 1000] = -141122 = -(84√2)4 3. r3(τ) r3(τ) = N[(f3

[τ] /. τ → √-6, 1000] = 63

r3(τ) = N[(f312 + 27)2 / f3

12 /. f3 → (-1)1/12DedekindEta[(τ+1)/3]/

DedekindEta[τ] /. τ → (1+√-267)/2, 1000] = -3003

4. r4(τ) r4(τ) = N(f4

8 + 16)2 / f48 /. f4 → DedekindEta[τ/4]/DedekindEta[τ] /. τ

→ √-28, 1000] = 212

r4(τ) = N(f48 + 16)2 / f4

8 /. f4 → DedekindEta[τ/4]/DedekindEta[τ] /. τ

→ 2+√-16, 1000] = -29

5. r7(τ) r7(τ) = N[(f7

4 + 7)2 / f74 /. f7 → DedekindEta[τ/7]/DedekindEta[τ] /. τ

→ √-28, 1000] = 14(32)r7(τ) = N[(f7

4 + 7)2 / f74 /. f7 → (-1)1/4DedekindEta[(τ+3)/7]/

DedekindEta[τ] /. τ → (1+√-427)/2, 1000] = -7(392)

So what is the variable C in the pi formula? It turns out that Cp = j(τ) or rp(τ)! For more about the r-functions rp(τ), see also Article 1, “The j-function and Its Cousins”. IV. The variables A and B Given the eqn, 1/π = Σ hp (An + B) / Cn, Conveniently, A can be expressed in terms of C as, 1. for h1 and j-function j(τ): A1 = √a1, a1 = d(1 – 123/C1); and C1 = j(τ).2. for hp and r-functions rp(τ): Ap = (1/p)√ap, ap = d(1 – 4p6/(p-1)/Cp); and Cp = rp(τ). hence, A1 = (1/1)√(d(1 – 1728/C1))A2 = (1/2)√(d(1 – 256/C2))A3 = (1/3)√(d(1 – 108/C3))A4 = (1/4)√(d(1 – 64/C4)) The discriminant d depends on the form of τ: Case 1: If τ = a+√-m, for a = {0,1,2}, then d = 4m.Case 2: If τ = (1+√-d)/2, then d = d.

Examples. Using the τ’s from the previous section: Let τ = {√-4, √-58, √-6, √-28}, C1 = j(τ) = 663, a1 = 4*4(1 – 1728/C1), so A1 = (1/1)√a1 = (84/121)√33C2 = r2(τ) = 3964, a2 = 4*58(1 – 256/C2), so A2 = (1/2)√a2 = (52780/9801)√2C3 = r3(τ) = 63, a3 = 4*6(1 – 108/C3), so A3 = (1/3)√a3 = (2/3)√2C4 = r4(τ) = 212, a4 = 4*28(1 – 64/C4), so A4 = (1/4)√a4 = 21/8 Let τ = {(1+√-7)/2, 1+√-37, (1+√-267)/2, 2+√-16}, C1 = j(τ) = -153, a1 = 7(1 – 1728/C1), so A1 = (1/1)√a1 = (21/25)√15C2 = r2(τ) = -141122, a2 = 4*37(1 – 256/C2), so A2 = (1/2)√a2 = 5365/882C3 = r3(τ) = -3003, a3 = 267(1 – 108/C3), so A3 = (1/3)√a3 = (4717/1500)√3C4 = r4(τ) = -29, a4 = 4*16(1 – 64/C4), so A4 = (1/4)√a4 = (3/2)√2 We now have A and C. Unfortunately, B is a complicated function in terms of C, and beyond the scope of this paper. To find B, one can just numerically solve for it using enough terms of the summation. If C is an algebraic number of degree n, then B should be of degree either n or 2n, and an integer relations algorithm can then find its minimal polynomial. For example, using the last entry with C4 = -29,

x = NSolve[Sum[h4(An+B)/Cn /. A →(3/2)√2 /. C→ -29 /. h4 →

(2n)!3 / n!6, {n, 0, 50}] = 1/Pi, B, 30] = 0.353553390593… The Mathematica add-on Recognize[x, k, v] tries to recognize the real number x as a root of an kth degree equation in some variable v. Applying the command on x for k = 2 yields the simple equation 1-8B2 = 0, so B = (1/4)√2, hence, 1/π = Σ h4 ((3/2)√2n + (1/4)√2) / (-29)n

simplified as, 1/π = (1/4)√2 Σ h4 (6n+1) / (-29)n

Equivalently, 1/π = 8 Σ h4 (-1)n (6n+1) / (29)n+1/2

The fact that A applies a square root on an expression involving C explains the “n+1/2” exponent in the denominator though, if C is negative, then (-1)n should be factored out as in the above example. V. Ramanujan and Class number h(-d) = 1 Why didn’t Ramanujan find the Chudnovsky family using the j-function j(τ) and class number h(-d) = 1? He almost did. To see this, first define v1 = (1/2)n(1/6)n(5/6)n / (n!3) where (a)n is the Pochhammer symbol. The original form of two of his 17 formulas were, 1/π = (2√3)/(5√5) Σ v1 (11n+1)(4/125)n

1/π = (18√3)/(85√85) Σ v1 (133n+8)/(4/85)3n

However, this can be transformed to, 1/π = 72 Σ h1 (11n+1)/(2*303)n+1/2, eπ√12 ≈ 2*303 - 747…1/π = 162 Σ h1 (133n+8)/(2553)n+1/2, eπ√28 ≈ 2553 -744.01… where, as before, h1 = (6n)! / ((3n)! n!3). Thus, Ramanujan also found formulas using non-fundamental discriminants d = {12, 28} with class number h(-d) = 1! In fact, in his Lost Notebook, he wrote down some of his results on d = 11, 19, 43, 67, 163, fundamental d which have h(-d) = 1. I’m sure that had he only lived long enough (he died at age 32), he would have found back in the 1920s the formulas discovered by the Chudnovsky brothers only in 1987. VI. Formulas for Singular Moduli Other than pi formulas, the same eta quotients fp are also useful for calculating singular moduli. To recall, the hp can also be defined as Pochhammer symbol products, h1 = 1728n (1/2)n(1/6)n(5/6)n / (n!3)

h2 = 256n (1/2)n(1/4)n(3/4)n / (n!3)

h3 = 108n (1/2)n(1/3)n(2/3)n / (n!3)

h4 = 64n (1/2)n(1/2)n(1/2)n / (n!3) Expressed in this way, the connection to four kinds of singular moduli

become more transparent. First, given the hypergeometric function 2F1,

2F1(a,b;c;z):= Σ g zn/n! where, g = (a)n(b)n / (c)n with |z| < 1, and n = 0 to ∞. In Mathematica, this is Hypergeometric2F1[a,b,c,z]. Define, m1 = 2F1(a, b; 1; 1-αN)m2 = 2F1(a, b; 1; αN) Then, for positive numbers {a,b} such that a+b = 1, the singular modulus αN is the unique positive real number 0 < αN < 1 satisfying, m1/m2 = √N To find αN, it can be observed that, 1. Let {a,b} = {1/6, 5/6}. Then p = 1, αN = (1-√(1-u1))/2, and u1 = 1728 / j(τ)2. Let {a,b} = {1/4, 3/4}. Then p = 2, αN = 1/(u2+1), and u2 = (1/26)

3. Let {a,b} = {1/3, 2/3}. Then p = 3, αN = 1/(u3+1), and u3 = (1/33)

4. Let {a,b} = {1/2, 1/2}. Then p = 4, αN = 1/(u4+1), and u4 = (1/42)

where fp are the Dedekind eta quotients, f2 = η[τ/2] / η[τ]f3 = η[τ/3] / η[τ]f4 = η[τ/4] / η[τ] and τ = √(-pn). Example. In his second letter to Hardy, Ramanujan gave the amazing radical, α210 = (8-3√7)2 (2-√3)2 (6-√35)2 (√15-√14)2 (1-√2)4 (3-√10)4 (4-√15)4 (√7-√6)4 = 2.7066… x 10(-19)

This applies to the case {a,b} = {1/2, 1/2}, and N = 210. To confirm this, set p = 4, so, α210 = 1/(u4+1), where u4 = (1/42) f4

8 = (1/42)(η[τ/4]/η[τ])8 and τ = √(-210p) = √(-210*4), and we find that indeed, α210 = 2.7066… x 10(-19)

such that, m1/m2 = √210 To express αN as a product of simple radicals (in fact, the factors are units in some quadratic field) just like what Ramanujan ingeniously did is another matter entirely, and requires much mathematical dexterity.

VII. Going Beyond The natural generalization of Ramanujan’s pi formulas is, 1/πk = Σ h P(n) / Cn (eq.1) where h is some factorial quotient and the numerator P(n) is a polynomial in n of degree k. Ramanujan’s was just the case k = 1. A semi-trivial solution to eq.1 is taking kth powers of known formulas for k = 1, with the resulting P(n) as a complicated mess. However, without squaring, Jesús Guillera has found many simple and elegant examples for k = 2 using various h, and Boris Gourevitch has found a single instance of k = 3. To give examples, first define, v2 = (2n)!/(n!2) then, again for n = 0 to ∞, 1. 1/π = 22 Σ v2

3 (6n+1) / (28)n+1/2,

2. 1/π = 22 Σ v23 (42n+5) / (212)n+1/2

3. 1/π2 = 23 Σ v25 (-1)n (20n2+8n+1) / (212)n+1/2

4. 1/π2 = 23 Σ v25 (-1)n (820n2+180n+13) / (220)n+1/2

5. 1/π3 = 25 Σ v27 (168n3+76n2+14n+1) / (220)n+1/2

6. 1/π4 = 27 Σ v29 P(n)(?) / (C)n+1/2

The first two were found by Ramanujan, the next two by Guillera (correcting two typos in Mathworld), and the fifth is by Gourevitch. Does the pattern go on for all 1/πk using polynomials of degree k? If so, is there a function that generates the variable C just like for (1) and

(2)? This phenomenon is simply begging for an explanation. Anybody knows how to find (6), where P(n) is a 4th deg polynomial and C is a power of 2?

Abstract: “We consider the asymptotic behavior of the infinite nested radical defining the golden ratio φ and plastic constant P,

and give a conjecture on the general case.” I. Introduction While trying to find more commonalities between the golden ratio φ and plastic constant (see article, A Tale of Four Constants), this author came across the Paris constant. In 1987, R. Paris proved that the nested radical expression for φ given above approaches φ at a constant rate. For example, defining φn as using n = {5, 6, 7} terms respectively, then, (1/2)(φ-φ5)(2φ)5 = 1.0977…

(1/2)(φ-φ6)(2φ)6 = 1.0983…

(1/2)(φ-φ7)(2φ)7 = 1.0985… which is approaching the Paris constant R = 1.09864196… It seems

this result can be generalized to radicals of kth power, hence includes the plastic constant and other values. Conjecture 1: “Given the infinite nested radical,

for integer k > 1, non-zero a, and the equation (eq.1),

Let x be the root of eq.1 such that x = xn as n → ∞. Define,

for some constant C(a,k).” II. Particular cases For quadratic radicals k = 2, since y = a/x + 1 = x, then this is simply, (x-xn)(2x)n → C(a,2) which apparently is relatively easy to prove. The special case a = 1,

yields C(1,2) = 2R = 2.19728392…, with the Paris constant R = 1.09864196… (R. Paris, An Asymptotic Approximation Connected with the Golden Number, Amer. Math. Monthly, 1987). For a = 2,

as Denis Feldmann points out, the constant C(a,k) has a nice closed-form expression as, C(2,2) = (π/2)2 = 2.46740110… For cubic radicals, say, {a,k} = {1,3}, we get the plastic constant ,

where P = 1.32472…, the real root of x3 = x+1. Since y = 1/P+1 = 1.75488…, then for small n, (P-P5)(3y)5 = 1.81673…

(P-P6)(3y)6 = 1.81685…

(P-P7)(3y)7 = 1.81687…

As n increases, one can see this is approaching the constant, C(1,3) = 1.8168834242447… mentioned in the title and accurate to 12 decimal places. And so on for other values {a,k}. Note, however, that for the special case a = 1, C(1,2) = 2.19728…C(1,3) = 1.81688…C(1,4) = 1.68029…...C(1,152) = 1.39246…C(1,153) = 1.39242…C(1,154) = 1.39238… Conjecture 2: “The sequence of constants C(a,k) for a = 1 as k → ∞ is also approaching a constant.” Question 1: Anybody knows how to prove/disprove Conjectures 1 and 2? If indeed Conjecture 2 is true, what is this constant, or, at least, what are its first dozen or so decimal digits? (Update, 10/4/10): Courtesy of Adam Strzebonski and Daniel Lichtblau of Wolfram Research, I managed to find C(1,k) for very high k. The convergence is VERY slow, taking powers of 10 to barely move one decimal digit: C(1,10

14) = 1.3862943611198999…

C(1,1015

) = 1.3862943611198915…

C(1,1016

) = 1.3862943611198907… Using Plouffe’s Inverse Symbolic Calculator, one finds that, 2 ln(2) = 1.3862943611198906… so it is a reasonable conjecture that the sequence of constants C(1,k) as k → ∞ has its limit 2 ln(2). Note that the expression 2 ln(2) also appears in Lévy’s constant γ,

another constant involved in asymptotic behavior, this time for the denominators of the convergents of continued fractions. (End update)

Question 2: Is there a closed-form expression for C(a,k) for other values {a,k} ?

-- End -- P.S. The next article will be on Ramanujan's beautiful continued fractions and its interesting connection to Platonic solids.

Abstract: We give a short overview on how some j-function formulas using three kinds of Ramanujan’s continued fractions are connected to the Platonic solids. I. IntroductionII. Platonic Solids, Dedekind eta function, and the j-functionIII. Ramanujan’s octic continued fraction (p = 2) and the octahedral

equationIV. Ramanujan’s cubic continued fraction (p = 3) and the tetrahedral equationV. Rogers-Ramanujan continued fraction (p = 5) and the icosahedral equation I. Introduction It is well-known that the continued fraction representation of quadratic irrationals is periodic. The simplest is for the golden ratio φ,

However, Ramanujan came up with beautiful analogues to this using the base of the natural logarithms e,

In 1913, the British mathematician G.H. Hardy, after reading the letter (which included examples of these extraordinary continued fractions) that Ramanujan sent to him, remarked, “…the [theorems] defeated me completely; I had never seen anything in the least like them before.” Hardy would have been even more amazed had he known that these

continued fractions were connected, among all things, to geometry. (See also Article 7, “Ramanujan's Continued Fractions, Apery's Constant, and More”. II. Platonic Solids, Dedekind eta function, and the j-function On one side of the coin are the Platonic solids, polyhedra with equivalent faces composed of regular polygons. There are only five such solids: the tetrahedron, cube, octahedron, dodecahedron, and icosahedron, with 4, 6, 8, 12, 20 faces, respectively. However, for every polyhedron, there is a dual, that is, another polyhedron in which faces and vertices have complementary positions. The dual of a Platonic solid is a Platonic solid. For 1) the tetrahedron, it is self-dual; for 2) the octahedron, it is the cube; and for 3) the icosahedron, it is the dodecahedron. Thus, for the symmetry groups of the Platonic solids, there are just three polyhedral groups: the tetrahedral group of order 12, the octahedral group of order 24, and the icosahedral group of order 60. On the other side of the coin are Ramanujan’s three q-continued fractions (and perhaps other similar ones) which can be expressed as special forms of the Dedekind eta function η(τ),

where q = e2πiτ = exp(2πiτ), in the form of the eta quotient,

Not surprisingly, the number 24 appears again. Of course, there are only three p such that 24/(p2-1) is an integer, namely p = 2, 3, 5.

Fact: We have three polyhedral groups, and three eta quotients with the above form. So here’s the interesting part: each of the three polyhedral groups can be associated with an eta quotient of order p, hence, to one (or other similar) form of q-continued fractions. It was Ramanujan’s genius that enabled him to find an example for all three orders, though presumably he was not thinking about polyhedra. What can make this relationship clearer is the well-known j-function, j(τ). This is the modular function defined by, j(τ) = 1/q + 744 + 196884q + 21493760q2 + 864299970q3 + … (A000521) where (as in the rest of this paper), q = e2πiτ = exp(2πiτ), and is responsible for remarkable approximations such as, eπ√163 ≈ 123(2312-1)3 + 743.9999999999992… While there are formulas for j(τ) using the Jacobi theta functions and others, it turns out these q-continued fractions can also be used in j-function formulas where the numerator and denominator involve polynomial invariants of polyhedral groups, as will be seen in the next section. III. Ramanujan’s octic continued fraction (p = 2) and the octahedral equation The general form is given by,

(The example given in the Introduction, as well as the two others, was just the case τ = √-1.) For convenience, let u(τ) = u. For p = 2, since 24/(p2-1) = 24/3 = 8, then a formula for u in terms of eta quotients is,

or equivalently,

or just,

This q-continued fraction can be used in the j-function formula,

Or let u = √v, and we get,

Its connection to the 8-faced octahedron,

can be made apparent when one looks at the octahedral equation, a set of related equations derived from its projective geometry. Assume an octahedron centered (0,0,0) on the intersection of three planes (x,y,z). Using a stereographic projection, by projecting the vertices with unit inradius, circumradius, and midradius, respectively, and expressing these vertex locations as roots of an algebraic equation, one gets the following, Inradius: z8 + 14z4 + 1 = 0Circumradius: z5 – z = 0Midradius: z12 – 33z8 – 33z4 + 1 = 0 For a clearer presentation, please refer to the link given above. Do the first two polynomials look familiar? These are the numerator and denominator of the j-function formula using Ramanujan’s octic continued fraction in the variable v. And the third polynomial can be derived using the formula, –Numerator + 123Denominator, hence,

the same formula which will appear in the tetrahedral and icosahedral equations as well. Also, considering the fact that the tetrahedron is composed of 8 triangles, perhaps it is not surprising that the polynomial x8+14x4y4+y8 can be expressed in terms of the Pythagorean triple {x2-y2, 2xy, x2+y2} as the sum of three 8th powers,

Lastly, the j-function formula in v has 24 roots, and the octahedral group is of order 24. IV. Ramanujan’s cubic continued fraction (p = 3) and the tetrahedral equation The general form is,

As usual, let c(τ) = c for convenience. For p = 3, since 24/(p2-1) = 24/8 = 3, then,

This q-continued fraction can also used in a j-function formula. Define,

Its connection to the tetrahedron,

can be seen in the tetrahedral equation (which is just analogous to the one in the previous section), Circumradius: z4 – 2√2z = 0Inradius: 2√2z3 + 1 = 0Midradius: z6 + 5√2z3 – 1 = 0 which again involves the numerator and denominator of the j-function formula and where the polynomial for the midradius can be derived as –Numerator + 123Denominator, hence,

The associated j-function formula has 12 roots, and the tetrahedral group is of order 12.

V. Rogers-Ramanujan continued fraction (p = 5) and the icosahedral equation The general form is,

Let r(τ) = r. For p = 5, since 24/(p2-1) = 24/24 = 1, then,

with j-function formula,

As expected, its connection to the icosahedron,

is given by the icosahedral equation, Inradius: z20 – 228z15 + 494z10 + 228z5 + 1 = 0Circumradius: z(z10 + 11z5 – 1) = 0Midradius: z30 + 522z25 – 10005z20 – 10005z10 – 522z5 + 1 = 0 where –Numerator + 123Denominator yields,

The j-function formula has 60 roots, and the icosahedral group is of order 60. Thus, the relationship between the j-function j(τ) and the continued fractions u(τ), c(τ), r(τ) imply that, for arguments τ = √-d or τ = (1+√-d)/2 for some positive integer d, since it is well known that j(τ) is an algebraic number, then so are u(τ), c(τ), r(τ). Also, since the three continued fractions involve the expressions q1/8, q1/3, q1/5, respectively, then these in fact are multi-valued functions, having n = {8, 3, 5} values, some of which are complex numbers. This is particularly relevant in the case τ = (1+√-d)/2 since the eta quotient is a complex number. For a more technical discussion of some of the points discussed in this paper, pls refer to Continued Fractions and Modular Equations by W.

Duke (Bulletin of the AMS, Vol. 42, 2005)

Abstract: A certain function related to Ramanujan’s pi formulas is explored at arguments k = {-½, 0, ½} and a conjecture will be given. I. IntroductionII. Fundamental d with class number h(-d) = 1,2III. Non-fundamental d with class number h(-d) = 1,2IV. Higher class numbers h(-d) = 4,6V. Conclusion: Conjecture I. Introduction In his 1914 paper, “Modular Equations and Approximations to π”, Ramanujan gave 17 formulas for pi, one of which was,

with the Pochhammer symbol (a)n = (a)(a+1)(a+2)…(a+n-1). He didn’t provide a proof, just stating that there were “corresponding theories”. It was not until the 1980s work by the Chudnovsky and Borwein brothers that these formulas were placed on a rigorous footing. In 2003, J. Guillera observed in his paper “Series closely related to Ramanujan formulas for pi” that, for k = ½,

It is easily seen that Ramanujan’s is just the case k = 0. Guillera found that most of Ramanujan’s formulas (as well as those by the Chudnovskys, Borweins, and others) if modified in a similar manner, then evaluated to mln(x), where m and x are rationals. Aptly enough, there is still no proof that the relationships are true, though it holds for arbitrary thousands of decimal digits. What the “corresponding theories” look like remains to be seen.

II. Fundamental d with class number h(-d) = 1,2 In Guillera’s paper, the evaluation mln(x) had a fluctuating m. To have a more well-behaved function, we will use the form,

for some constant k, discriminant d, Pochhammer products hp, and algebraic numbers {A,B,C}. Define the hp as,

For k = 0, the formulas below neatly sum to,

However, for k = ½, then, p = 1:

p = 2:

p = 3:

For p = 1, the denominators of the formulas are the exact values of the negated j-function j(τ), while for p = 2,3, these are for the related modular functions r2(τ), r3(τ). (See article, Pi formulas and the Monster group). Note the prime factorizations,

as compared to the function’s respective ln(x), for example,

and one can see an extremely interesting “coincidence”. There is also a similar correspondence between the prime factorization of the denominators of the formulas for p = 2,3, and ln(x). III. Non-fundamental d with class number h(-d) = 1,2 Let k = ½, then, p = 1:

p = 2:

Evaluated as mln(x), these have a slightly different m from those with fundamental d. If k = 0, then F(0)d = 1/π. IV. Higher class numbers h(-d) = 4,6 A. Quadratic irrationals Guillera pointed out that {A,B,C} may be algebraic and gave a few examples. For fundamental d, we’ll use the smallest with class number h(-d) = 2, and the smallest d = {4u, 3v} with h(-d) = 4, namely d = {15, 68, 39}, respectively. Let {φ, α, β} be the fundamental units,

All evaluate as mln(x) where x is an algebraic number of degree 2 but, as it turns out, the second one does not have the expected m = (1/2)4, but has m = (1/2)5, which will be relevant for the conjecture in the conclusion. B. Cubic irrationals

We’ll use the smallest d with class number h(-d) = 3, and the smallest d = {4u, 3v} with h(-d) = 6, namely d = {23, 116, 87}. Let {x,y,z} be the real roots of,

respectively, then,

As expected, the general form is mln(x’) where x’ now is an algebraic number of degree 3. And so on (apparently) for other d with higher class numbers. As usual, all six formulas have F(0)d = 1/π. V. Conclusion: Conjecture Conjecture: “Given the function,

with the Pochhammer products hp as defined previously, {A,B,C} as algebraic numbers of degree N, and fundamental discriminant d. If,

where {x, y} are algebraic numbers of degree N.” Some points: 1. Since ln(ab) = bln(a), the exponent 6 may be reduced to either {5, 4, 3} if x is a square, fourth power, or eighth power, respectively, over a field of the same degree N, as was seen in the examples. 2. If indeed the “equalities” are true, there should be a closed-form expression for x as a function of d, presumably complicated, similar to how the Chudnovskys and Borweins derived {A,B,C} from d. 3. Furthermore, as Guillera pointed out, there might be interesting evaluations over other constant k, one of which he tried was k = ¼. I tried negative values and found that, for k = -½, “y” apparently is an algebraic number that has the same degree as {A,B,C}. Thus, we have, F(-½)d = y = {-70.5, -248, -1752, -88056, -144088, -3107743896} for d = {7, 11, 19, 43, 67, 163}, respectively. I do not know what these integers are, as their factorizations are not divisible by d. For

d with higher class number, F(-½)d = y = {-15(89+45√5)/4, -4(143+34√17), -(31+13√13)/4}

for d = {3(5), 4(17), 3(13)} using the formulas given above. If {A,B,C} are algebraic numbers of degree > 1, it seems the minimal polynomial of y and {A,B,C} share the same factor of the discriminant d. For the background behind this article, see J. Guillera’s 2003 paper, “Series closely related to Ramanujan formulas for pi” in his homepage.

I. IntroductionII. Euler’s Continued FractionIII. Ramanujan and Apery’s ConstantIV. ConstantsV. Between Two Continued FractionsVI. Infinite SeriesVII. Gamma FunctionVIII. Generalized Hypergeometric FunctionIX. Integrals I. Introduction In the article “Ramanujan’s Continued Fractions and the Platonic Solids” we discussed three kinds of beautiful continued fractions (for brevity, “cfrac”) which involve the argument q = e2πiτ = exp(2πiτ) and their unexpected connections to geometry. Ramanujan, however, had many other kinds of cfracs up his sleeve. In “Chapter 12 of

Ramanujan’s Second Notebook” [1], authors Berndt et al gave more than 30 others, many with several free variables. There are others in various places in his Notebooks but, for the moment, this author will include mostly those in [1]. The primary motivation for this article is to present Ramanujan’s cfracs in the original, visually-pleasing form. While not the most space-saving, in this modern era where information is measured in terabytes and we have convenient math formats like LaTex, it seems there is no reason not to do so. So without further ado… II. Euler’s Continued Fraction Formula Two very general results are Euler’s continued fraction (1748) and Gauss’ continued fraction (1813). (In the quote at the start of this article, Hardy was careful in specifying Ramanujan’s role as he was preceded by Euler and Gauss. It would have been interesting had they all met in the same century.) Gauss’ results will be appropriate for the section on hypergeometric functions. Euler’s cfrac can be given as,

from which one can derive other forms, such as,

This automatically gives a continued fraction representation for the Riemann zeta function ζ(s), or the more general Hurwitz zeta function, though Ramanujan gave his own version in the next section. III. Ramanujan and Apery’s constant Entry 32iii: For any x > 0. Define v = 2(x2+x), then, Form 1:

We can also express x in terms of v,

then for v ≥ 0,

with the case v = 1 specifically given by Ramanujan. Form 2:

still with v = 2(x2+x), and the sequence P(m) = {1, 3, 7, 13,…} generated by P(m) = m2+m+1. This form, for x = v = 0, reduces to Euler’s cfrac for the Riemann zeta function ζ(s) at s = 3. This particular case naturally has, (2m+1)(m2+m+1) = m3 + (m+1)3

or the sum of two consecutive cubes. (The sum of two consecutive squares will appear in the next section.) Apery would later give a similar formula,

where P(n) = 17n2+17n+5 yields the sequence {5, 39, 107, 209,…}, though it works only if x = y = 0. It is interesting that in Ramanujan’s version, there is this polynomial relationship between x and v, given by v = 2x2+2x, since there doesn’t seem to be anything analogous using Euler’s cfrac for the zeta function ζ(s) with odd exponent s > 3 of simple form, v = P(x) where v is a polynomial in x, though I didn't check the more complicated version, P(v) = P(x) where each side is a polynomial with rational coefficients. The rest of this article will give a selection from Chap 12 with little commentary, starting with the simpler ones. I’ll add the more complicated results from other parts of his Notebooks in another article. The interested reader is encouraged to find generalizations, special cases, or simple proofs (if possible) for these beautiful mathematical objects. IV. Constants

Entry 7: For x > 0,

While simple in form, this holds a hidden surprise. For x = 1, the convergents are {2/1, 4/5, 20/19, 100/101, 620/619,…} where the numerators N(n) are “weighted Fibonacci numbers” (sequence A153229) and denominators D(n) are alternating factorials with the formula for both as,

Entry 13: If a < b,

Entry 12: If x ≠ 0,

Entry 18: Let x be outside the real interval [-1,1], then,

V. Between Two Continued Fractions Ramanujan gave several examples of two continued fractions converging to the same value (such as the one given for the Hurwitz zeta function in Part III). A few others are, Entry 27: Let x,y > 0, then,

Notice that x and y just exchange places. A similar one is, Entry 28: Let x > 0. Define,

One can just admire the artistry by which Ramanujan conjures up his continued fractions. The next (slightly modified by this author) is a transformation formula not in Chap 12, but in the Lost Notebook (p. 46). Let k ≥ 0, and, u = (1+√(1+4k))/2v = (1-√(1+4k))/2

For any |q| < 1, then,

One feature of this identity is the RHS accelerates the convergence of the LHS. Note that for k = 0, then both sides reduce to the Rogers-Ramanujan continued fraction without the factor q1/5. Also, it is easily noticed that u and v are the roots of the quadratic equation x2-x-k = 0. A conjecture by this author is then, Conjecture: “Given x2-ax-b = 0, a2+4b ≥ 0, and roots {x1, x2} where x1 ≥ x2. For any |q| < 1, then,”

For various {a,b} that satisfy the constraint a2+4b ≥ 0, one can easily check with Mathematica that the two cfracs apparently are converging on the same value. If indeed generally true, it should be interesting if this has a cubic version for x3-ax2+bx-c = 0 and its roots {x1, x2, x3}.

VI. Infinite Series (Note: Ramanujan’s infinite series Σ start with either k = 0 or 1, so care should be exercised.) Entry 8:

Corollary: For x = 0, we get,

A similar form, but where the denominators of the infinite series involve only odd numbers is, Entry 43: For x > 0, then,

For x = 1, this is faintly reminiscent of the previous entry. (Regarding this particular example, Kevin Brown of the Mathpages remarked, “Is there any other mathematician whose work is so instantly recognizable?”) Entry 16: If {m,n} are not negative integers,

Corollary: For m = n = 0, this alternating series reduces to the Dirichlet eta function η(s) for s = 2, hence,

Entry 29: Let x > 0 and n2 < 1.

Corollary: For n = 0, this reduces to a version of the Dirichlet eta function η(s) for s = 1,

This cfrac has the nice property that, depending on whether x is odd or even, then it evaluates to an expression involving either log(2) or π. For general x though, Conjecture 1: For any x ≥ 1,

which, for integer x, then odd x involve log(2), while even x involve π. Specifically, f(1) = log(2), f(2) = (4-π)/2, f(3) = 1-log(2), and so on. Entry 31: Let x > 0 and n2 < 1. (Or |x| > 1 and n2 is real.)

Corollary: As n → 0, then,

For integer x, just like for Entry 29, this involves two kinds of constants: 1) Catalan's constant C for even x, and 2) π for odd x > 1. Specifically, f(2) = 2-2C, f(3) = (12-π2)/24, and so on. Entry 30: Let x > 0 and n2 is real.

Corollary: As n → 0, then,

For positive integer x, this involves only one constant, namely f(1) = π2/12, f(2) = (-8+π2)/4, etc. The next one involves the square of the denominators of the previous entry. Entry 38: Let |x| > 1 and n2 is real. Define u = x2-n2, then,

where the integer sequence P(m) = {1, 5, 13, 25,…} is the same as in Entry 40, or the sum of two consecutive squares, P(m) = m2 + (m+1)2. Entry 32i & 32ii: If x > 0,

VII. Gamma Functions

To recall, the gamma function Γ(n) is an extension of the factorial n! to complex and real number arguments n. Entry 25: If |x| > 1 and n2 is real. (Or x > 0 and n2 < 1.)

Corollary: If n = 0, then,

and for x an odd number, f(x) is a multiple of π, with the first x = 1 giving the well-known continued fraction,

Amazingly, there is also an expression for the square of entry 25,

Entry 26: If |x| > 1 and n2 is real.

The next three entries are one of Ramanujan’s more fascinating results on gamma functions. Entry 33: Given the (22 = 4) sign changes of,

Let p be the product of the gamma functions where the argument has an even number of minus signs, and q for an odd number of minus signs, namely, p = Γ((x+a+b+1)/2) Γ((x-a-b+1)/2)q = Γ((x-a+b+1)/2) Γ((x+a-b+1)/2) then,

Entry 35: Given the (23 = 8) sign changes of,

in contrast to Entry 33, let p be the product of the gamma functions where the argument has an odd number of minus signs, and q for an even number of minus signs. Suppose either a,b,c is a positive integer, then,

where, u1 = x2-a2-b2-c2+1

u2 = x2-a2-b2-c2+5

u3 = x2-a2-b2-c2+13

u4 = x2-a2-b2-c2+25 and so on, with the sequence {1, 5, 13, 25,…) as P(m) = m2 + (m+1)2. Entry 40: Given the (24 = 16) sign changes of,

Like the previous entry, let p be the product of the gamma functions where the argument has an odd number of minus signs, and q for an even number of minus signs. Suppose at least one of {a,b,c,d} is a positive integer. Then,

where, u1 = 2(a4+b4+c4+d4+x4+1) - (a2+b2+c2+d2+x2-1)2 - 22

u2 = 2(a4+b4+c4+d4+x4+1) - (a2+b2+c2+d2+x2-5)2 - 62

u3 = 2(a4+b4+c4+d4+x4+1) - (a2+b2+c2+d2+x2-13)2 - 142

u4 = 2(a4+b4+c4+d4+x4+1) - (a2+b2+c2+d2+x2-25)2 - 262

with the same sequence {1, 5, 13, 25,…} as P(m) = m2 + (m+1)2. Incredible, isn’t it? As Berndt points out, Entry 40 is one of the crowning achievements of Ramanujan’s work on cfracs. There is only one proof for (by G. N. Watson, 1935) using certain assumptions. How Ramanujan knew in the first place that such an unusual continued fraction existed, or if it can be generalized further using 2n variables, is unknown. VIII. Generalized Hypergeometric Functions Define the ff functions,

where (p)k is a Pochhammer symbol. One can see that the first two are just degenerate cases of the hypergeometric function 2F1 (which, historically, was the first one to be studied). For certain {a,b,c}, Gauss’ continued fraction involve the ratios of these functions. With sometimes different {a,b,c}, Ramanujan also gave cfracs with a simpler form, Entry 19:

Entry 21 (Corollary 1):

Entry 22 (Limited case): If |x| < 1,

These are just particular cases of the more general entries which involve ratios. In this limited form, one can more easily see the “family resemblance” of the three. For Entry 21, the case n = x simplifies the cfrac, and an alternative form is,

Entry 42: If x > 0,

For x a positive integer, the function evaluates to an expression involving the xth power of e and a rational number,

In fact, for this case, the summation above and the cfrac are equivalent, as a positive integer x truncates the latter. IX. Integrals Ramanujan gave several cfracs that evaluate to an integral (one of which was given in Section III). Another is, Entry 44: For x > 0,

Compare to the similar,

which is derivable from Euler's cfrac, and perhaps may not be surprising since the log function satisfies the integral,

For a more technical discussion, refer to Berndt et al’s paper, “Chapter 12 of Ramanujan’s Second Notebook” (open access). Prologue (2009): Back in 2008, I dedicated an algebraic identity I

found to the astronomer and science writer Carl Sagan (1934-1996). I read his "Broca's Brain" and "The Dragons of Eden" in my late teens, and read and eventually saw "Contact". Since I already dedicated one article I wrote on Degen's Eight-Square Identity to the novelist Katherine Neville, author of the amazing book "The Eight", I thought it was only fitting I name one of the identities I found after Sagan. After all, it has billions and billions of solutions. (You need to be a Sagan fan to understand the previous remark.) This article was originally posted in one of my Geocities pages. Since Yahoo unfortunately closed that domain (last October 2009), I'm transfering it, with some edits, to my Google Sites pages since today, November 7, 2009, is the first Carl Sagan Day. I. IntroductionII. Ramanujan’s 6-10-8 IdentityIII. Hirshhorn’s 3-7-5 IdentityIV. Sagan’s IdentityV. Conclusion: Going Beyond I. Introduction The remarkable Ramanujan 6-10-8 Identity is given by: 64[(a+b+c)6+(b+c+d)6+(a-d)6-(c+d+a)6-(d+a+b)6-(b-c)6][(a+b+c)10+(b+c+d)10+(a-d)10-(c+d+a)10-(d+a+b)10-(b-c)10] = 45[(a+b+c)8+(b+c+d)8+(a-d)8-(c+d+a)8-(d+a+b)8-(b-c)8]2 where ad = bc. Its form and use of high exponents is certainly unusual. This can be more concisely expressed as, define, Fk: = (a+b+c)k + (b+c+d)k + (a-d)k - (c+d+a)k - (d+a+b)k - (b-c)k

If ad = bc, then 64F6F10 = 45F82. Note also that the terms can be

related, (b+c+d) + (a-d) = (a+b+c),(c+d+a) + (b-c) = (d+a+b), as well as F2 = F4 = 0. These details will be important later. On the other hand, Sagan’s Identity, named by this author after the astronomer and science writer Carl Sagan (1934-1996), is given by the multi-grade equation, 1 + 5k + (3+2y)k + (3-2y)k + (-3+3y)k + (-3-3y)k = (-2+x)k + (-2-x)k + (5-y)k + (5+y)k

for k = 1,3,5,7 where x2-10y2 = 9. Such equations appear in the context of the Prouhet-Tarry-Escott Problem. While Ramanujan’s 6-10-8 Identity involves even exponents and Sagan’s has odd ones, and seem to be of very different forms, it can be shown that both fundamentally have their basis in the same simple system of equations, call this S4, as ak+bk+ck = dk+ek+fk, for k = 2,4. II. Ramanujan’s 6-10-8 Identity Though employing high exponents, it has its roots on the much simpler S4. First, some preliminaries. Theorem 1: If ak+bk+ck = dk+ek+fk for k = 2,4. Define {x,y} = {a2+b2+c2, a4+b4+c4}. Then,

3(a8+b8+c8-d8-e8-f8) = 4(a6+b6+c6-d6-e6-f6)(x) (eq.1) 6(a10+b10+c10-d10-e10-f10) = 5(a6+b6+c6-d6-e6-f6)(x2+y) (eq.2) One can combine eq.1 and eq.2. Theorem 2: If ak+bk+ck = dk+ek+fk for k = 2,4. Then, 32(a6+b6+c6-d6-e6-f6)(a10+b10+c10-d10-e10-f10) = 15(n+1)(a8+b8+c8-d8-e8-f8)2 where n is defined by a4+b4+c4 = n(a2+b2+c2)2. This is the more general case of Ramanujan’s 6-10-8 Identity. Proof: Define Ek := ak+bk+ck – (dk+ek+fk). So eq.1 and eq.2 become, 3E8 = 4E6(x) (eq.1)

6E10 = 5E6(x2+y) (eq.2) Square eq.1, multiply eq.2 by E6, and then eliminate E6

2 between eq.1 and eq.2 to get: 32E6E10 = 15E8

2 + 15E82y/x2

Or, 32E6E10 = 15E8

2 (1+y/x2)

32E6E10 = 15E82 (1+n)

(End of proof) Corollary: If ak+bk+ck = dk+ek+fk for k = 2,4 where a+b = c. Then 64(a6+b6+c6-d6-e6-f6)(a10+b10+c10-d10-e10-f10) = 45(a8+b8+c8-d8-e8-f8)2 This is Ramanujan’s 6-10-8 Identity phrased differently. Proof: It is easily shown that, a4+b4+c4 = n(a2+b2+c2)2 when a+b = c, then n = 1/2. Substituting n = 1/2 into Theorem 2, one gets this corollary. (End of proof) Thus, as claimed, this remarkable identity has its roots on the properties of S4 with the variables subject to certain constraints. III. Hirshhorn’s 3-7-5 Identity Inspired by the 6-10-8, there's also Hirshhorn's 3-7-5 Identity: 25[(a+b+c)3-(b+c+d)3-(a-d)3+(c+d+a)3-(d+a+b)3+(b-c)3][(a+b+c)7-(b+c+d)7-(a-d)7+ (c+d+a)7-(d+a+b)7+(b-c)7] = 21[(a+b+c)5-(b+c+d)5-(a-d)5+(c+d+a)5-(d+a+b)5+(b-c)5]2 again with ad = bc. In fact, the two can be combined: Theorem: If ak+bk+ck = dk+ek+fk, k = 2,4 where a+b+c = d+e+f = 0, then,

64(a6+b6+c6-d6-e6-f6)(a10+b10+c10-d10-e10-f10) = 45(a8+b8+c8-d8-e8-f8)2 25(a3+b3+c3-d3-e3-f3)(a7+b7+c7-d7-e7-f7) = 21(a5+b5+c5-d5-e5-f5)2 Proof: By doing the substitution {a,b,c,d,e,f} = {p, q, -p-q, r, s, -r-s}, all three equations have the common factor p2+pq+q2-(r2+rs+s2) = 0, so the problem is reduced to finding expressions {p,q,r,s} that satisfy this. IV. Sagan’s Identity Compared to the ones above, this has the very different form: 1 + 5k + (3+2y)k + (3-2y)k + (-3+3y)k + (-3-3y)k = (-2+x)k + (-2-x)k + (5-y)k + (5+y)k, for k = 1,3,5,7 where x2-10y2 = 9. This has billions and billions (Sagan's favorite catchphrase) of solutions. In fact, there is an infinite number of them, whether rational, or integral if the condition is to be solved as a Pell-like equation. Again, this identity can be shown to depend on the properties of S4, but with its variables subject to different constraints. As before, some preliminaries, Sinha’s Theorem: If, (a+3c)k + (b+3c)k + (-a-b+2c)k = (c+d)k + (c+e)k + (2c-d-e)k, for k = 2,4, then,

ak + (a+2c)k + bk + (b+2c)k + (-c+d+e)k = (a+b-c)k + (a+b+c)k + dk + ek + (3c)k, for k = 1,3,5,7 excluding the trivial case c = 0. (Sinha stated his theorem differently, but in this article it was re-phrased it to make it simpler.) Note that the variables of S4, if expressed as {xi, yi}, have the constraint x1+x2+x3 = 2(y1+y2+y3) = 8c, excluding c = 0, hence all solutions applicable to the 6-10-8 and 3-7-5 Identities must be avoided. Sinha gave only one solution in terms of a binary quadratic form (with discriminant D = 70), but this author found others with a different square-free discriminant (D = 10), one of which is, (y+6)k + (3y+4)k + (4y-2)k = (x+y-1)k + (-x+y-1)k + (2y+6)k, for k = 2,4 where x2-10y2 = 9. This obeys the constraints of Sinha's theorem. By solving for {a,b,c,d} and constructing the terms of the k = 1,3,5,7 system, one can recover Sagan's Identity after some minor algebraic tweaking. (This author knows of only four solutions as binary quadratic forms that satisfy Sinha’s theorem, all with discriminant D that, square-free and unsigned, is either D = {10, 70}. Are there others with different D?) V. Conclusion: Going Beyond The next step after S4 would be S5 and S6, or ak+bk+ck+dk = ek+fk+gk

+hk, for k = 1,3,5, or 2,4,6. There are solutions for these, one of which for the latter is the beautifully simple result given by Chernick and Escott:

If, 1k + 2k + 3k = uk + vk + (u+v)k for k = 2,4, then, (-u+7)k + (u-2v+1)k + (-3u-1)k + (3u+2v+1)k = (u+7)k + (-u+2v+1)k + (3u-1)k + (-3u-2v+1)k, for k = 1,2,4,6. This system depends on solving u2+uv+v2 = 7 in the rationals, which is easily done. Analogous general identities similar to the 6-10-8 can be found using even higher powers, but none are known that have the simple form found by Ramanujan. If we define S7 and S8 as ak+bk+ck+dk+ek = fk+gk+hk+ik+jk, for k = 1,3,5,7, or 2,4,6,8, there are several parametric solutions known for the former, but only one for the latter, the beautifully simple, (a+c)k + (a-c)k + (3b+d)k + (3b-d)k + (4a)k = (3a+c)k + (3a-c)k + (b+d)k + (b-d)k + (4b)k

for k = 1,2,4,6,8, where, a2+12b2 = c2 12a2+b2 = d2

This was found by Letac and Sinha and can be reduced to an elliptic curve, hence has an infinite number of solutions. Is there a solution to S8 in terms of binary quadratic forms? No one knows. Then there’s S9, S10, etc…. P.S. Here is an interesting Newscientist article on naming mathematical theorems.

I. Introduction This is a compilation of recent developments by various authors for Ramunujan-type pi formulas of form,

for m > 1 where,

1) sv is a finite product of Pochhammer symbols (x)n for some rational x < 12) P(n) is a polynomial in n.3) C is a rational constant

For m = 1, an infinite number of formulas are known (where C is algebraic) since there is a well-developed theory behind it. However, for m > 1, ONLY 12 are known so far (updated, Dec. 19, 2010), found by: 9 – Jésus Guillera (m = 2)1 – Jésus Guillera, Gert Almkvist (m = 2)1 – Boris Gourevitch (m = 3)1 – Jim Cullen (m = 4) Without a general guiding principle, the few examples have been found mostly by clever use of integer relations algorithms, with only four rigorously proven (by Guillera). By compiling these in one place and arranging them in a reasonable order, perhaps it can give – in one glance – an overview of what is known, and motivate others to find

more of a similar kind. (In fact, four of the formulas have been discovered only in 2010. I’ll be updating this website each time there is a new one discovered.) II. Type 1 Define the Pochhammer symbol products av as,

These are the basic combinations used by Ramanujan and a famous example is,

For a complete listing of the 36 formulas which use the av, rational C,

and its associated eπ√d, refer to the article, “Pi Formulas and the Monster Group”. III. Type 2 To generalize Ramanujan’s results, Guillera considered five Pochhammer symbols. There may be as much as eight different combinations that can be used in a pi formula. The first four are, define,

which are simply the av multiplied by ((1/2)n/n!)2, then,

Guillera’s early formulas (pre-2010) had a constant C that was a unit fraction (with a numerator N = 1). However, the third formula with C = (3/4)3 (from his “A New Ramanujan-Like Series for 1/π2 ”, 2010) proves that it need not be so, hence opens possibilities for b4 and others. (Note: Guillera and Almkvist have searched hard for formulas using b4, and subsequently by J. Cullen who extended the search, but they haven't found any yet.) <Update, 12/18/10>: They have also considered a variant of the general form which uses a ratio of polynomials A(n)/B(n). One by Almkvist is,

Compare it with the coeffiicients of the third formula. That is surely

more than a coincidence. (Guillera has noticed a similar phenomenon between some of his formulas and Ramanujan's.) <End> The next four Pochhammer products are, define,

The third and fourth (both also found in 2010) also contains a C that is not a unit fraction. The latter, when translated into factorials, has the especially nice form,

Though not yet rigorously proven, this implies, as was discussed in “Ramanujan-Like Series for 1/π2 and String Theory” (J. Guillera and Gert Almkvist, 2010), that any decimal digit of the number 1/π2 can be computed without computing all the preceding decimals, similar to the Bailey-Borwein-Plouffe formula. (Note: This observation is courtesy

of Wilfried Pigulla, since (1/3)(6n)!/n!6 is an integer for all n > 0.) The last one, b8, involves Pochhammer symbols not used by Ramanujan and is the only one known so far. IV. Type 3 This goes beyond five Pochhammer symbols and hence involves 1/pim for m > 2. Define,

The first is by Boris Gourevitch (2002), while the second is by Jim Cullen (Dec, 2010). The third is a logical extension of b8 where P(n)

as a hypothetical 4th degree polynomial. (Though Almkvist has pointed out that the probability for c3 is probably zero.) (Update, 12/19/10): Consistent with Guillera’s observations (see Ramanujan’s Pi Formulas with a Twist), I noticed that we can transform Cullen's formula by letting,

then, for k = ½,

Almkvist simplied this by using binomial sums to find the Cullen-Almkvist formula,

Note: To find more examples of a usable cv, it might be productive to consider multiplying the Pochhammer products av or bv with powers

of ((1/2)n/n!)k, since the two known cv have that form. Any other combinations (whether for positive or alternating series)?

I. IntroductionII. Three SquaresIII. Two TrianglesIV. One CuboidV. Some Formulas I. Introduction In 1644, at the tender age of 18, the Italian mathematician Pietro Mengoli (1626-1686) proposed what would later be known as the Basel Problem. He asked for the exact sum of, Σ 1/n2 = 1/12 + 1/22 + 1/32 + 1/42 + … as n goes to infinity. Nearly a century later, this was solved by Euler (at age 28) who found that, Σ 1/n2 = π2/6 = 1.644934… This would not be the only instance that two lives, separated by a century, would intersect. Mengoli also asked for three positive integers {x,y,z} such that the sum and difference of any two are all squares, {x+y, y+z, x+z} = {a1

2, b12, c1

{x-y, y-z, x-z} = {a22, b2

2, c22}

This is known as Mengoli’s Six-Square Problem, or MSP for short.

He was able to find a rather large solution, but Euler would find a smaller one which, in fact, turns out to be the smallest. The three smallest primitive solns {x,y,z} with all terms less than 106 are, {434657, 420968, 150568} (Euler’s){496625, 474993, 428175}{733025, 488000, 418304} with the first one yielding, x+y = 9252

x-y = 1172

y+z = 7562

y-z = 5202

x+z = 7652

x-z = 5332

Note also that, (925*117)2 + (756*520)2 = (765*533)2

More generally, given the tautology, (x2-y2) + (y2-z2) = (x2-z2) then MSP implies the special Pythagorean triple, (a1a2)2 + (b1b2)2 = (c1c2)2

hence {a1a2, b1b2, c1c2} are the three sides of a right triangle as the numerical example above clearly shows. Courtesy of Randall Rathbun, Jarek Wroblewski, and Raymond Manzoni who independently did an exhaustive search on face cuboids, we now know

Euler’s solution is the smallest. Before going into the proof, it should be pointed out that MSP can be transformed into three related problems:

1. Three squares {u2, v2, w2} such that the difference of any two is a square2. Two right triangles such that the products of their legs, or hypotenuses and legs, is a square, or abcd(a2±b2)(c2±d2) = t2.3. One face cuboid with sides {a,b,c} wherein {a2+b2, a2+c2, a2+b2+c2} are all squares.

II. Three Squares T. Leybourn showed that MSP can be reduced to finding three squares {u2, v2, w2} such that the difference of any two is a square. Proof: Set the first half of MSP as, {x+y, x+z, y+z} = {u2, v2, w2} Solve for {x,y,z} to get, {2x, 2y, 2z} = {u2+v2-w2, u2-v2+w2, -u2+v2+w2} Substitute these into the other half of MSP, or{x-y, y-z, x-z}, and they transform into the expressions which must be made squares, specifically, u2-v2 = s1

2 (eq.1)

u2-w2 = s22 (eq.2)

v2-w2 = s32 (eq.3)

(End proof.) The system eq.1,2,3 can be partially reduced to the problem, abcd(a2±b2)(c2±d2) = t2 (eq.0) also to be discussed (with solutions) in the next section. Let, {u,v,w} = {(p2+q2)(r2-s2), (p2-q2)(r2+s2), (p2-q2)(r2-s2)}. This makes eq.2 and 3 as squares while eq.1 becomes, 4(p2r2-q2s2)(q2r2-p2s2) = (pr+qs)(pr-qs)(qr+ps)(qr-ps) = s1

2 (eq.1b) Eq.1b can be transformed into the form of eq.0 in two ways: 1. Let {p,q,r,s} = {ac-bd, ad+bc, ac+bd, ad-bc}, to get 16 abcd(a2-b2)(c2-d2) (a2+b2)2(c2+d2)2 = t2 2. Let {p,q,r,s} = {ac+bd, ad+bc, ac-bd, ad-bc}, to get 16 abcd(a2+b2)(c2+d2) (a2-b2)2(c2-d2)2 = t2 A direct soln to eq.1b was given by W. Lenhart using (after scaling variables) what is essentially a Pythagorean triple. Let x2+9y2 = z2. Define {p,q,r,s} = {4y, z, 4xy, x2+5y2}. Then, 4(p2r2-q2s2)(q2r2-p2s2) = 16y2(x2-3y2)2(x2+25y2)2 III. Two triangles Part 1. The equation discussed previously, abcd(a2±b2)(c2±d2) = t2 (eq.0)

can also be interpreted in terms of a right triangle. Since the sides is the Pythagorean triple {x1

2-y12, 2x1y1, x1

2+y12}, then eq.0 is the

problem of finding two right triangles such that either, 1) the product of their legs is a square (– case)2) or the product of two legs and the two hypotenuses is a square (+ case) Note thay solving eq.0 for either sign generates a third Pythagorean triple since, 4abcd(a2-b2)(c2-d2) = (ac-bd)2(ad+bc)2 - (ac+bd)2(ad-bc)2 = t1

2 (eq.0a)

4abcd(a2+b2)(c2+d2) = (ac+bd)2(ad+bc)2 - (ac-bd)2(ad-bc)2 = t22

(eq.0b) Two small solns to eq.0a are, {a,b,c,d} = {(u2+4v2)/2, u2-2v2, 6v2, u2-2v2}{a,b,c,d} = {u+v, u-v, (u2+v2)2, 4uv(u2-v2)} for arbitrary {u,v}. Part 2. MSP has two broad solutions. Recall the problem is to find {x,y,z} such that, {x+y, y+z, x+z} = {a1

2, b12, c1

{x-y, y-z, x-z} = {a22, b2

2, c22}

1. Mengoli: Let pqrs and (p2+q2)(r2+s2) both be squares. Then, {2x,

2y, 2z} = {a2+b2, c2+d2, a2-b2}, where {a,b,c,d} = {pr+qs, ps+qr, pr-qs, ps-qr}.2. Euler: Let pqrs and (p2-q2)(r2-s2) both be squares. Then, {2x, 2y, 2z} = {a2+b2, a2-b2, c2-d2}, where {a,b,c,d} = {pr+qs, ps-qr, pr-qs, ps+qr}. Notice the subtle sign differences between the two and how the variables are to be defined. The two are related such that a soln to [1] automatically leads to one for [2], and vice-versa, as will be shown in Note 3. Note 1: Mengoli gave two particular solns for [1] but a smaller one is {p,q,r,s} = {12, 35, 21, 20} which, after scaling, yields {x,y,z} = {3713858, 891458, 88642}. This soln has the form, {p,q,r,s} = {mu, nv, mv, nu} which takes care of the first condition since it becomes pqrs = (mnuv)2 and reduces the second to finding {m,n,u,v} such that, (m2u2+n2v2)(n2u2+m2v2) = y2

This can be treated as an elliptic curve in either {u,v}, hence there are an infinite number of solns. Note 2: Euler gave {p,q,r,s} = {4, 9, 49, 81} for [2] which yields the smallest solution to MSP as {x,y,z} = {434657, 420968, 150568}. This obviously has the form, {p,q,r,s} = {t2, u2, v2, w2} which takes care of the first condition and modifies the second to, (t4-u4)(v4-w4) = y2

Likewise, this can also be treated as an elliptic curve. In fact, there are many small solns to this where GCD(t,u) = GCD(v,w) = 1 other than Euler’s with a few others being {t,u,v,w} = {1, 17, 3, 7}, or {2, 3, 2, 11}, and so on. Note 3: One can use the formula for Pythagorean triples {x1

2-y12,

2x1y1, x12+y1

2} to prove that a soln to [1] leads to one for [2], and vice versa. Let, {p,q,r,s} = {e2-f2, 2ef, g2-h2, 2gh} This satisfies the 2nd condition of [1], and transforms the first into finding, 4efgh(e2-f2)(g2-h2) = y1

which is fulfilled by [2]. This is also eq.0a and two simple solns were given in the previous section. For a particular example, one can use Euler’s {e,f,g,h} = {4, 9, 49, 81}. Similarly, let, {p,q,r,s} = {e2+f2, 2ef, g2+h2, 2gh} This satisfies the 2nd condition of [2], and transforms the first into, 4efgh(e2+f2)(g2+h2) = y2

which is now fulfilled by [1]. For example, with {e,f,g,h} = {12, 35, 21, 20}. IV. One Face Cuboid

Following Randall Rathbun, there are 4 kinds of cuboids with sides {a,b,c} depending on which dimension is irrational. These are: 1) Euler brick, 2) edge cuboid, and 3) face cuboid wherein the space diagonal, an edge, or a face diagonal is irrational, respectively. The last is 4) perfect cuboid where no dimension is irrational. It is yet unknown if it does or doesn’t exist.

Expressed in terms of 3-dimensional Cartesian coordinates,

then, equivalently, a perfect cuboid is a point {x,y,z} such that, x2+y2 = t1

x2+z2 = t22

y2+z2 = t32

x2+y2+z2 = t42

defined by seven positive integers {t1, t2, t3, t4, x, y, z}. It can be proven that the Mengoli's Six-Square Problem (MSP) is equivalent to finding a rational “face cuboid” {a,b,c} such that, a2+b2 = s1

2 (eq.1)

a2+c2 = s22 (eq.2)

a2+b2+c2 = s32 (eq.3)

call this system as M. Proof: Define MSP’s {x,y,z} as, {2x, 2y, 2z} = {2a2+b2+c2, b2+c2, -b2+c2} (Def. 1) One can easily solve for {a,b,c} as, {a2, b2, c2} = {x-y, y-z, y+z} Substituting these values into a face cuboid’s definition {a2+b2, a2+c2, a2+b2+c2} will yield {x-z, x+z, x+y}. Thus, if the six expressions {x-y, y-z, y+z, x-z, x+z, x+y} are all squares (which is precisely the Six-Square Problem), then one gets integer {a,b,c} which gives a rational face cuboid. Conversely, given a rational face cuboid {a,b,c}, one can

easily find {x,y,z} using Def.1 given above. (End proof.) Thus, the problem of determining if Euler’s soln is the smallest is reduced to a finite calculation of much smaller magnitude, namely, to find all primitive face cuboids with sides {a,b,c} < 1000. Rathbun, Wroblewski, and Manzoni independently did an exhaustive search and found, {104, 153, 672}{117, 520, 756} (Euler’s){448, 264, 975}{495, 264, 952}{840, 448, 495} While there is a smaller face cuboid, its {b,c} do not have the same parity and hence the sum ±b2+c2 is not evenly divisible by 2. The second has {b,c} with the same parity and, after dividing by 2, yields the smallest {x,y,z} for Mengoli’s Six-Square Problem. To recall, a face cuboid {a,b,c} has {a2+b2, a2+c2, a2+b2+c2} as all squares. A remarkable triplet found by Wroblewski share the common side “a”, {1680, 1925, 2052}{1680, 819, 3740}{1680, 3404, 4653} hence is the common leg of six different Pythagorean triples and three Pythagorean quadruples. This is the only one with a < 105, though using an identity (in the formulas section below), and the eqn p4+q4 = r4+s4, it can be proven that the value “2pqrs” always appears in a triplet of face cuboids, though not necessarily the side “a”. (Update, 2/9/11): R. Rathbun has provided the complete list of 24 primitive face cuboid n-tuplets sharing the common side “a” less than

ten billion (1010), with the next four as, {683760, 132349, 3581820} {683760, 614180, 1893771} {683760, 1522180, 4036851} {683760, 726869, 7505820} -------------------------------------{4084080, 1487772, 5110525} {4084080, 2024207, 5975424} {4084080, 4776541, 10668060} {4084080, 5619537, 6145984} --------------------------------------{5525520, 2042975, 10043712}{5525520, 4055360, 6844761}{5525520, 6616640, 20282031}{5525520, 9746825, 14511168} ---------------------------------------{6126120, 3150048, 9088625} {6126120, 13316839, 14048160} {6126120, 7274625, 24095008} and a quintuplet (#15 of the 24), {232792560, 55306628, 117515475}{232792560, 71608131, 135412420}{232792560, 135423925, 447886692}{232792560, 153939420, 414785371}{232792560, 180609955, 1219785588} which is the only one known so far. With more data, an interesting trend can be observed, 1680 = 24·3·5·7683760 = 24·3·5·7·11·374084080 = 24·3·5·7·11·13·17

5525520 = 24·3·5·7·11·13·236126120 = 23·32·5·7·11·13·17232792560 = 24·32·5·7·11·13·17·19 Why are they so highly factorable? This can be seen for all the 24 face cuboid n-tuplets given by Rathbun. (End update.)

V. Formulas There are interesting formulas by various authors that use any of the simple eqns, 1) x2+y2 = z2,2) x2+3y2 = z2,3) p4+q4 = r4+s4

to find a face cuboid {a,b,c}. However, the formulas are not in their original form as this author used Mathematica to find a more aesthetic presentation that reduced the degree of the polynomials, and their connections to the simple conditional equations (1), (2), (3). I. x2+y2 = z2

1. M. Rolle, Lowry: Define {a,b,c} = {y(3x4-4y4), x(4y2z2-x4), 4xyz(y2+z2)}. Then, a2+b2 = (x4+4y4)2z2

a2+c2 = (5x4+8x2y2+4y4)2y2

a2+b2+c2 = (x4+8x2y2+4y4)2z2

For ex, let {x,y,z} = {3,4,5}, then {a,b,c} = {3124, 4557, 9840}.

2. W. Lenhart: Define the scaled Pythagorean triple x2+(3y)2 = z2. Then, {a,b,c} = {8xy(x2-7y2)(x2+5y2), 8y2(x2-3y2)(x2+25y2), (x2-7y2)(x4-6x2y2+25y4)} Let {x,y,z} = {4,1,5}, then {a,b,c} = {6048, 4264, 1665}. II. x2+3y2 = z2

To recall, the smallest primitive face cuboids {a,b,c} < 1000 as calculated by Rathbun et al are, {104, 153, 672}{117, 520, 756}{495, 264, 952}{448, 264, 975}{840, 448, 495} It can be noticed the last three is a triplet that share common terms. It turns out this can be explained by three identities that, interestingly enough, depend on the simple equation, x2+3y2 = z2 (eq.1) 1. Euler. Define {a,b,c} = {(x2-y2)(y2-z2), 2xy(y2-z2), 2yz(x2+y2)}. Then, a2+b2 = ((x2+y2)(y2-z2))2

a2+c2 = (4y4+x2z2)2

a2+b2+c2 = ((x2+y2)(y2+z2))2

2. Piezas. Define {a,b,c} = {(4xy2z, 2xy(y2-z2), 4y4-x2z2}. Then, a2+b2 = (2xy(y2+z2))2

a2+c2 = (4y4+x2z2)2

a2+b2+c2 = ((x2+y2)(y2+z2))2 3. Piezas. Define {a,b,c} = {(2yz(x2-y2), 4xy2z, (x2-y2)(y2-z2)}. Then, a2+b2 = (2yz(x2+y2))2

a2+c2 = (4y4-x2z2)2

a2+b2+c2 = (4y4+x2z2)2 Thus there is a triplet of face cuboids that, two at a time, share a common addend, with all three having a common sum (in red). Using the smallest soln of eq.1 as {x,y,z} = {1,4,7}, this yields the triplet (according to color). Since eq.1 can be given the parametrization {x,y,z} = {u2-3v2, 2uv, u2+3v2}, then there is an infinite number of such triplet face cuboids. (Note: Ultimately, however, a bivariate quadratic solution can be expressed in terms of Pythagorean triples {a,b,c}. For example, it is the case that (2a-c)2 + 3(b)2 = (a-2c)2, though the resulting expressions in {a,b,c} are messier compared to the ones in {x,y,z}.) III. p4+q4 = r4+s4

1. Let {a,b,c} = {p2q2-r2s2, p4-r4, 2pqrs}. Then, a2+b2 = (p2s2-q2r2)2

a2+c2 = (p2q2+r2s2)2

a2+b2+c2 = (p2s2+q2r2)2

2. Let {a,b,c} = {p2q2-r2s2, q4-r4, 2pqrs}. Then, a2+b2 = (q2s2-p2r2)2

a2+c2 = (p2q2+r2s2)2

a2+b2+c2 = (q2s2+p2r2)2 3. Let {a,b,c} = {2pqrs, p2s2-q2r2, p2r2-q2s2}. Then, a2+b2 = (p2s2+q2r2)2

a2+c2 = (p2r2+q2s2)2

a2+b2+c2 = (p4+q4)(r4+s4) The first two are by Piezas, while the third is by R. Beauregard and E. Suryanarayan. This is a triplet of face cuboids that share a common term, 2pqrs, while the first two share an additional term. (For that pair, p and q just swap places since it does not affect the conditional equation.) An example is given by {p,q,r,s} = {59, 158, 133, 134}. Since there are an infinite number of identities for the conditional eqn, then so is there for face cuboids. Any other simple formula for a) three squares, b) two triangles, c) one face cuboid and, by extension, to Mengoli’s Six-Square Problem?

I. Euler BrickII. Generalized Euler BrickIII. Euler QuadruplesIV. Formulas for QuadruplesV. A General Method

I. Euler Brick In 1740, a noted blind mathematician by the name of Nicholas Saunderson found a simple parametrization to what is now called a Euler brick. To recall, this is defined by a triple of positive integers {a,b,c} such that, {a2+b2, a2+c2, b2+c2} are all squares, the smallest being {a,b,c} = {44, 117, 240}. Saunderson’s solution (rediscovered by Euler and others) was, given x2+y2 = z2, define, {a,b,c} = {x(x2-3y2), y(3x2-y2), 4xyz} [1] then, a2+b2 = (z3)2

a2+c2 = x2(x2+5y2)2

b2+c2 = y2(5x2+y2)2 with the first sum as a 6th power. Since {x,y,z} is the Pythagorean triple {p2-q2, 2pq, p2+q2}, then {a,b,c} are 6th degree polynomials in {p,q}. For a particular example, let {x,y,z} = {3, 4, 5} and this yields the smallest numerical example given above. (Author’s note: Saunderson, the 4th Lucasian professor of mathematics – Newton being the 2nd – was blind from infancy.) If {a,b,c} is a Euler brick, then so is {ab, ac, bc}, call this Rule 1. Saunderson’s solution, then yields after removing common factors, {a’,b’,c’} = {4xz(x2-3y2), 4yz(3x2-y2), (x2-3y2)(3x2-y2)} [2]

which are 8th deg polynomials in {p,q}. No other 6th deg (or smaller) solutions are known, but there are others of 8th deg. First, recall that the four smallest Euler bricks are, {44, 117, 240}{85, 132, 720}{88, 234, 480}{132, 351, 720} Curiously, the second and fourth bricks share two terms. There are an infinite number of such Euler brick pairs. Let u2+v2 = 5w2, then, W. Lenhart: {a,b,c} = {(u2-w2)(v2-w2), 4uvw2, 2uw(v2-w2)} [3]Piezas: {a,b,c} = {(u2-w2)(v2-w2), 4uvw2, 2vw(u2-w2)} [4] (Notice the variables u and v just swap places.) A parametric solution is {u,v,w} = {p2+4pq-q2, 2(p2-pq-q2), p2+q2}. For this pair, one member can be derived from the other using Rule 1 and removing common factors. This phenomenon of shared terms can also be found in face cuboids (explainable by three related identities discussed in the previous article), Euler quadruples (by a change of sign in a single identity), and in generalized Euler bricks though for the last case, no identities have yet been found that explains the shared terms. However, it can be proven that there are an infinite number of identities of the kind [1]-[4]. A. Euler’s method Let {a,b,c} = {(p2-1)/(2p), (q2-1)/(2q), 1}. This makes a2+c2, b2+c2 as squares. To make a2+b2 a square as

well, find {p,q} such that, p2(q2-1)2 + q2(p2-1)2 = t2 This can be treated as an elliptic curve hence, from an initial rational point, one can compute an infinite number of other rational points. Define the conditional equation Q as x2+y2 = z2, then, {p,q} = { 2y/z, z/(2x) }{p,q} = { 2y/z, y(2x+z)/((x+z)(2x-z)) } and so on. After simplifying, these two yield identities [1] and [2], respectively. B. Lenhart’s method Let {a,b,c} = {(p2-1)/(2p), 2q/(q2-1), 1}. Notice this is a very similar transformation to Euler. Likewise, two of the diagonals become squares, but to make a2+b2 a square as well, solve, (p2-1)2(q2-1)2 + 16p2q2 = t2 Again, this can be treated as an elliptic curve. Define a different conditional equation Q as u2+v2 = 5z2, then, {p,q} = { uv/(2w2), (v+w)/(v-w)}{p,q} = { uv/(2w2), (v2-4w2)(v2+w2)/((v2-w2)(v2-6w2)) } and so on. The first point yields identity [3]. It seems to be unknown if there are other conditional quadratics Q that can be used in Euler’s or Lenhart’s method. (Note: Ultimately,

however, a quadratic solution to any Q can be expressed in terms of Pythagorean triples {x,y,z}. For example, it is the case that (x+2y)2 + (2x-y)2 = 5z2, though the resulting expressions for Identity 3 in {x,y,z} are messier compared to the ones in {u,v,w}.) In Andrew Bremner’s 1988 paper, The rational cuboid and a quartic surface, he gives a general method for finding Euler brick parametrizations and points out evidence suggest these may be found for every even deg ≥ 6 (though are not necessarily expressible by some Q) . C. Leudesdorf showed that a Euler brick {a,b,c} is equivalent to finding three positive rationals {u,v,w} such that, 2(u2+v2-w2) = a2 (eq.1)2(u2-v2+w2) = b2 (eq.2)2(-u2+v2+w2) = c2 (eq.3) Proof: Solving for {u,v,w}, we get, a2+b2 = (2u)2

a2+c2 = (2v)2

b2+c2 = (2w)2

which defines a Euler brick. The smallest example would yield {u,v,w} = { 125/2, 122, 267/2}. (End proof). Adding eqns 1,2,3 together, a perfect Euler brick must then have a space diagonal d whose square is both the sum of three squares and double the sum of another three squares, a2+b2+c2 = 2(u2+v2+w2) = d2 (eq.4) By itself, eq.4 is easily solved, the smallest in distinct and positive integers is,

{a,b,c,d} = {4, 6, 12, 14}{u,v,w} = {3, 5, 8} though solving eq.1-4 simultaneously is another matter. II. Generalized Euler Brick Similar to Leudesdorf’s, Euler also considered the system, u2+v2-w2 = a2

u2-v2+w2 = b2

-u2+v2+w2 = c2

Solving for {u,v,w}, the problem is equivalent to finding generalized Euler bricks {a,b,c} for the case n = 2 of, a2+b2 = nu2 (eq.1)a2+c2 = nv2 (eq.2)b2+c2 = nw2 (eq.3) call this system Sn. Jarek Wroblewski pointed out that Sn has non-trivial solutions in the integers only for n = {1, 2}. Excluding the trivial a = b = c, the smallest for S2 is {a,b,c} = {1,1,7}. Solutions with a = b can be parameterized as, {a,b,c} = {p2-2q2, p2-2q2, p2+4pq+2q2} for arbitrary {p,q} with the smallest being the case {p,q} = {1,1}. For distinct and primitive {a,b,c}, Wroblewski (2010) found for bound B < 6000, only seven, namely, {329, 191, 89}

{527, 289, 23}{833, 553, 97}{1081, 833, 119}{1127, 697, 17}{4991, 2263, 287}{5609, 4991, 1871} The smallest (in red) was known to Euler as an instance of a parametric family that depended, perhaps not surprisingly, on the simple conditional equation, x2+y2 = 2z2 as, {a,b,c} = {x(y2+4yz-4z2), x(y2-4yz-4z2), y(3y2-4z2)} (This has been slightly modified by this author. One can also swap x and y since this does not affect the conditional eqn.) Let {x,y,z} = {1, 7, 5} and it yields the smallest with distinct {a,b,c}. Note also how some triples share a common term, a phenomenon also present with face cuboids and was explained by three identities that, two at a time, share a common term. However, I haven't yet found corresponding identities for these pairs of “Euler bricks” over √2. And whether, a2+b2+c2 = nt2 (eq.4) for n = 2 is solvable along with eq.1,2,3 is also unknown. III. Euler Quadruple A Euler quadruple, on the other hand, are four positive integers {a,b,c,d} such that, {a2+b2+c2, a2+c2+d2, a2+b2+d2, b2+c2+d2} are all squares. This can also be given a geometric interpretation as the

space diagonals (of four cuboids) given by the blue line,

If the four cuboids are positioned so that one end of each diagonal all meet at an apex, then this defines an irregular pyramid with four triangular faces and a slanted quadrilateral base. (Interestingly, Euler would solve a quadruple in terms of one or two triangles.) The smallest, found by Wroblewski (2010), is {a,b,c,d} = {49, 72, 72, 84} so, a2+b2+c2 = 1132

a2+c2+d2 = 114

a2+b2+d2 = 114

b2+c2+d2 = 1322

He also gave all {a,b,c,d} < 1000 with gcd = 1 as, {49, 72, 72, 84}{21, 28, 120, 120}{60, 105, 168, 280}{313, 336, 336, 492}{237, 336, 336, 952}

Apparently, Euler missed the smallest while the second and third were known to him as instances of two parametric families which uses Pythagorean triples. It is not known if the smallest belongs to a family. There are also an infinite number of quadruple pairs that share a common side as can be proven by an identity in the next section. IV. Formulas for Quadruples Just like face cuboids (discussed in the previous article, “Mengoli’s Six Square Problem and Face Cuboids”), there are interesting formulas for Euler quadruples by various authors that use any of the simple eqns, 1) x2+y2 = z2,2) x2+3y2 = z2,3) p4+q4 = r4+s4

though the formulas have been modified for a more aesthetic presentation. 1. x2+y2 = z2

Euler: i) Define {a,b,c,d} = {2xyz, x(x2-y2), y(x2-y2), 2xyz}. Then, a2+b2+c2 = z6

a2+c2+d2 = x2(x2+3y2)2

a2+b2+d2 = y2(3x2+y2)2

b2+c2+d2 = z6

Just like for Euler bricks, a sum is a 6th power. Another formula, derivable from a general method given by Euler in the next section is,

ii) Define {a,b,c,d} = {2pxyz, px(x2-y2), qy(x2-y2), 2qxyz}, where {p,q} = {x4-6x2y2-3y4, 3x4+6x2y2-y4}. Notice the affinity between the two formulas, with the first simply as the case p = q = 1. I am not aware of any other polynomial for {p,q}, though there may be. For {x,y,z}= {3, 4, 5}, this yields {a,b,c,d} = {120, 21, 28, 120} which is the second smallest solution, and {a,b,c,d} = {186120, 32571, 23828, 102120}, respectively. 2. x2+3y2 = z2

Euler, S. Tebay: Define {a,b,c,d} = {y(x2-y2), z(x2-y2), 2yz(x-y), 2yz(x+y)}. Then, a2+b2+c2 = (xz2-4y3)2

a2+b2+d2 = (xz2+4y3)2

a2+c2+d2 = (3yz2-4y3)2

b2+c2+d2 = z6

The last sum again is a sixth power. Let {x,y,z} = {1, 4, 7} and this gives {a,b,c,d} = {60, 105, 168, 280}, which is the third smallest Euler quadruple. 3. p4+q4 = r4+s4

Piezas: Define {a,b,c,d} = {2pruv, 2qsuv, uw, 2pqrsw}, where {u,v,w} = {p4-r4, p2r2-q2s2, pr(q4+s4) ±qs(p4+r4)}. This yields a pair of quadruples, {a,b,c1,d1} and {a,b,c2,d2}, with two common sides: a and b. (One can change the sign of q without

affecting the conditional eqn.) The sums are too tedious to be explicitly write down, but one can test it with any parametric solution, or particular ones like {p,q,r,s} = {59, ±158, 133, 134}, though {a,b,c,d} typically will be large values. V. A General Method Theorem (Euler): Assume, a2+b2+c2 = (bx3+dx2)2/x1

a2+c2+d2 = (bx2+dx3)2/x12

a2+b2+d2 = (ax3-cx1)2/x22

b2+c2+d2 = (ax1-cx3)2/x22

Given two Pythagorean triples {x1, x2, x3} and {y1, y2, y3}. If the

product of the legs is a square x1x2y1y2 = m2, and set n = x2y1, then a quadruple is, {a,b,c,d} = {1, m/n, d(mx2)/(nx1), (mx3y1+nx1y3)/(nx1y2-nx2y1)} Proof: One can substitute n = x2y1, {x1, x2, x3} = {e2-f2, 2ef, e2+f2},

and {y1, y2, y3} = {g2-h2, 2gh, g2+h2}, and the four eqns are true if, x1x2y1y2 = m2

or, equivalently, 4efgh(e2-f2)(g2-h2) = m2 (eq.1)

(End proof) Example: Eq.1 is also discussed in “Mengoli’s Six Square Problem and Face Cuboids”. However, there are many small solutions, one of which is {e,f,g,h} = {5, 2, 6, 1} yielding, {x1, x2, x3} = {21, 20, 29}{y1, y2, y3} = {35, 12, 37} {m,n} = {420, 735} After scaling, this gives {a,b,c,d} = {105, 60, 168, 280} which is the third smallest Euler quadruple. For a parametrization, let x2+y2 = z2: i) If {x1,x2,x3} = {y,x,z}; {y1,y2,y3} = {x,y,z}; hence x1x2y1y2 =

(xy)2, then, {a,b,c,d} = {2xyz, x(x2-y2), y(x2-y2), 2xyz}. ii) If {x1,x2,x3} = {y,x,z}; {y1,y2,y3} = {z4-4x2y2, 4xyz2,

z4+4x2y2}; hence x1x2y1y2 = (2xyz(x2-y2))2, then, {a,b,c,d} = {2pxyz, px(x2-y2), qy(x2-y2), 2qxyz}, where {p,q} = {x4-6x2y2-3y4, 3x4+6x2y2-y4}. Both were given in the previous section. Any other simple formulas for generalized Euler bricks and Euler quadruples?

(Note: Instead of using superscripts as I normally do, in this short

article I’ve used the hat symbol ^ so the equations can easily be copied and pasted onto Mathematica or Maple for those who wish to see the 16th-deg resolvent themselves.) Here's a “natural” solvable 17th-deg eqn with small coefficients: x^{17}-6 x^{16}-24 x^{15}-42 x^{14}-31 x^{13}-23 x^{12}-7 x^{11}-x^{10}-4 x^9-11 x^8-7 x^7-13 x^6-x^5+x^3+x^2+x-1 = 0 (eq.1) Its unique real root is exactly given by (in Mathematica) as x = zeta_48 DedekindEta[tau]/(Sqrt[2]DedekindEta[2tau]) = 9.1630942..., with the root of unity zeta_48 = exp(2Pi I/48), tau = (1+Sqrt[-d])/2, and d = 383. This d has class number h(-d) = 17. To solve this, depress eq.1 (get rid of its xn-1 term), by letting x = (y+6)/17 to get, y^{17}-11832 y^{15}-1124346 y^{14}-55393735 y^{13}-1784741617 y^{12}-41171464807 y^{11}-711423456455 y^{10}-9455898295636 y^9-99724287747103 y^8-887992943070295 y^7-7665207188897171 y^6-70479807472769473 y^5-592167373130143650 y^4-3496187093606980919 y^3-8695712981307573757 y^2+68265051092799270505 y-427806967360317821039 = 0 (eq.2) Its 16th-deg resolvent, a polynomial with INTEGER coefficients, call this R_16, has roots, z_k = ((y1 + w^{k}y2 + w^{2k}y8 + w^{3k}y7 + w^{4k}y16 + w^{5k}y4 + w^{6k}y12 + w^{7k}y15 + w^{8k}y11 + w^{9k}y10 + w^{10k}y14 + w^{11k}y13 + w^{12k}y5 + w^{13k}y17 + w^{14k}y6 + w^{15k}y9 + w^{16k}y3)/17)^17 for k = {1 to 16}, where w is any complex 17th root of unity. This

particular R_16 has all z_k as real numbers. Note the specific arrangement of the y_n. There are 16! ≈ 2 x 10^13 possible permutations of the y_n, and out of that huge number, there are only 16 such that R_16 has integer coefficients, and we have given one of them. Of course, I used a short cut to find it, because even if your computer can check a million permutations a second, it would still take about 8 months to go through them all. The short cut took less than two hours to find R_16. The y_n follows the root object Root[poly, n] ordering in Mathematica. Approximately, these are, y1 = 149.7726 {y2, y3} = -27.62 -/+ 18.49 i {y4, y5} = -21.61 -/+ 7.52 i {y6, y7} = -16.58 -/+ 6.34 i {y8, y9} = -10.57 -/+ 15.32 i {y10, y11} = -5.02 -/+ 13.71 i {y12, y13} = -2.34 -/+ 13.15 i {y14, y15} = 2.57 -/+ 2.60 i {y16, y17} = 6.31 -/+ 7.04 i R_16 has extremely large integer coefficients, with the largest being the 248-digit constant term 429534618434587^17 which, naturally enough, is a 17th power. (Note: R_16 can easily be formed using 500-digit precision or more on the y_n, and multiplying the 16 factors together to form the polynomial. However, since it involves extremely large integers of 200+ digits, it is understandable if I do not explicitly and tediously type them down here. But if you have Mathematica or

Maple, you can easily re-construct R_16 using the details described in this article.) The polynomial R_16 can then be factored into two octics over the extension Sqrt[17]. This, in turn, can be factored into 2 quartics over Sqrt[2(17+Sqrt[17])]. This can be factored further into 2 quadratics using an expression involved in the 17th root of unity. Apparently, in general, to solve the 16-deg resolvent of an irreducible but solvable 17-deg equation, only square roots of square roots of square roots, etc, are needed. The real root of eq.2 in radicals is then, y1 = z1^(1/17) + z2^(1/17) + z3^(1/17) + … + z16^(1/17) = 149.7726… The next step, of course, is degree p = 19. Unfortunately, to find the 18th-deg resolvent R_18 of an irreducible but solvable 19th-deg equation with rational coefficients, there is a small matter of 18! ≈ 6.4 x 10^15 possible permutations of the y_n, only 18 of which will be such that R_18 is a polynomial with rational coefficients…

Keywords: moonshine functions, Monster group, prime-generating polynomials, McKay-Thompson series, pi formulas, 163. Abstract: An expository work on 1) Conway, Norton, and Atkin’s intriguing discovery that the moonshine functions span a linear space of dimension 163; and 2) the functions’ connection to Ramanujan-type pi formulas. I. IntroductionII. The 172 McKay-Thompson Series and Linear Dependencies

III. More Pi Formulas?IV. A Continued Fraction Using eπ√163

I. Introduction Consider the following prime-generating polynomials of form P(n) = an2-an+c, P(n) = n2-n+41P(n) = 2n2-2n+19P(n) = 3n2-3n+23P(n) = 4n2-4n+5 distinctly prime for n = {1 to c-1}, with the first (studied by Euler in 1772 in the form n2+n+41) for a remarkable 40 consecutive n. And the following related functions, j(τ) = 1/q + 744 + 196884q + 21493760q2 + 864299970q3 + … (A007240) r2A(τ) = 1/q + 104 + 4372q + 96256q2 + 1240002q3 + … (A101558) r3A(τ) = 1/q + 42 + 783q + 8672q2 + 65367q3 + … (A030197) r4A(τ) = 1/q + 24 + 276q + 2048q2 + 11202q3 + … (A097340) where (throughout this article),

q = e2πiτ = exp(2πiτ) The first is the j-function j(τ), a modular form of weight zero, but without the constant term, all these are simply the McKay-Thompson series T1A , T2A, T3A, and T4A of the Monster group. Solving the polynomials at the point P(n) = 0, then, n = τ = (1+√-163)/2n = τ = (2+√-148)/4 = (1+√-37)/2n = τ = (3+√-267)/6n = τ = (4+√-64)/8 = (1+2√-1)/2 where, consistent with the requirements for modular forms, we have chosen the root τ that is on the upper half of the complex plane. The first has fundamental discriminant -d = 163, the largest (in absolute value) with class number h(d) = 1, while the next two -d = {148, 267} = {4·37, 3·89} have class number h(d) = 2. Plugging τ into q = exp(2πiτ), we get the very small real numbers, q = -exp(-π√163)q = -exp(-π/2√148) = -exp(-π√37)q = -exp(-π/3√267)q = -exp(-π/4√64) = -exp(-2π) which, when substituted into the appropriate series, yields the integers, j((1+√-163)/2) = -6403203

r2A((1+√-37)/2) = -141122

r3A((3+√-267)/6) = -3003

r4A((1+√-2)/2) = -29

The first integer appears in the Chudnovsky brothers’ pi formula,

where one can see the 163, as well as in the famous, eπ√163 ≈ 6403203 + 743.9999999999992… The second integer, -141122, appears in a pi formula found by Ramanujan in 1912 (before the Monster group was even discovered),

where 4(37) also appears, as well as in the almost-integer, eπ/2√148 = eπ√37 ≈ 141122 + 103.99997… This, and similar formulas, inspired the one found by Chudnovsky brothers. The third, -3003, appears in another kind of pi formula,

where, one can see |d| = 267, and in the almost-integer,

eπ/3√267 = eπ√(89/3) ≈ 3003 + 41.99997… Finally, -29 appears in a fourth kind,

Using his own approach, Ramanujan found examples for all four types. (See Pi Formulas and the Monster Group for the complete list of 36 known formulas where the denominator C is an integer.) We instead used four functions based on T1A, T2A, T3A, T4A, but there are in fact many more of them. And here’s an amazing fact: these moonshine functions span a linear space of dimension exactly 163! Interesting “coincidence” that it had to be that number, isn’t it? II. The 172 McKay-Thompson Series and Linear Dependencies To recall, the Monster group has 194 conjugacy classes so its character table has 194 rows and 194 columns. Each column represents one conjugacy class and linear combinations of the entries gives its McKay-Thompson series which yields, to quote John McKay or Mark Ronan, a “moonshine function” (hence the title of this article). Some are the same so is reduced to 172 series, with the two distinct conjugacy classes A & B of order 27 still included as an exceptional case. (Author note: My thanks to Michael Somos for clarifying certain points.)

Table 1. The 172 McKay-Thompson series of Monstrous Moonshine

The first dozen coefficients of each of the 172 series are given in Table 4, Values of Head Characters, (p. 334) of John Conway and Simon Norton’s 1979 paper, Monstrous Moonshine. Conveniently, an index of the McKay-Thompson series is also in the Online Encyclopedia of Integer Sequences. (The moonshine functions, in accordance with Atlas notation, have the letter corresponding to the conjugacy class in upper-case, while the non-monstrous ones have it in lower-case. By accident, T24B is missing in the index, but is still in the OEIS.) Also, David Madore has calculated the first 3500 coefficients of all the 172 moonshine McKay-Thompson series, while Michael Somos has an extensive database about the relationships between the coefficients and more. But there are linear dependencies between the 172 series,

# Linear dependency 1

T27A = T27B

T6A + 2T6E = T6B + T6C + T6D

T10A + 2T10E = T10B + T10C + T10D

T12A + 2T12I = T12B + T12E + T12H

T18B + 2T18D = T18A + T18C + T18E

T30B + 2T30G = T30A + T30C + T30F

2T8E = T4C + T4D

2T16B = T8D + T8E

T12C + T12E - T12F - T12G = 2(T12I + T24B - T24C - T24I)

# Linear dependency 1

T27A = T27B

T6A + 2T6E = T6B + T6C + T6D

T10A + 2T10E = T10B + T10C + T10D

T12A + 2T12I = T12B + T12E + T12H

T18B + 2T18D = T18A + T18C + T18E

T30B + 2T30G = T30A + T30C + T30F

2T8E = T4C + T4D

2T16B = T8D + T8E

T12C + T12E - T12F - T12G = 2(T12I + T24B - T24C - T24I)

Table 2. The 9 Linear Dependencies

Taking into account these 9 linear dependencies, then the dimension can be reduced to 172 – 9 = 163. To bring this insight to life, in his blog, Lieven le Bruyn quotes p. 227 of Mark Ronan’s book Symmetry and the Monster, “Conway recalls that, ‘As we went down into the 160s, I said ‘...let’s guess what number we will reach.’ They guessed it would be 163 – which has a very special property in number theory – and it was! There is no explanation for this. We don’t know whether it is merely a coincidence, or something more....” The first six were found by Conway and Norton, while the last three

were discovered by Oliver Atkin who also showed that there are no more. These nine are found in Conway and Norton’s paper, but in rather technical form. For example, dependency # 7 is given as, T4- + T4|2 = 2T8- so I had to translate it into English, so to speak, i.e. convert it into the notation used in the OEIS (as well as in Madore’s and Somos’ databases). We then find from the OEIS, T4C(q) = 1/q + 20q - 62q3 + 216q5 - 641q7 + … (A007248) T4D(q) = 1/q - 12q + 66q3 - 232q5 + 639q7 + … (A007249) T8E(q) = 1/q + 4q + 2q3 - 8q5 - q7 + … (A029841) and we can easily see from the first few coefficients that indeed, T4C + T4D = 2T8E though it is proven by Atkin that the equality holds for all terms. Thus, just as one can consider the series T27B as redundant being equal to T27A, and since there are nine linear dependencies, then only 163 series as “moonshine functions” are really necessary to linearly express all possible values of the 172 series. Or, as was mentioned, these functions span a linear space of dimension 163. In an email, McKay calls this as a “delicious coincidence”, while Norton opines this may be merely a coincidence. However, in Conway and Norton’s paper, in the last paragraph of Section 9 which deals with moonshine for other groups, it was asked if there is “…a period three automorphism for the case Γ0(67)+”. (The discriminant

d = -67 is, of course, the baby brother of d = -163 in the family of Heegner numbers.)

III. More Pi Formulas? In addition to the very interesting fact that the moonshine functions span a linear space of dimension 163, one question we can ask is, since there are so many of them, it is possible to use other McKay-Thompson series TpA in Ramanujan-type pi formulas? Ramanujan’s four kinds of pi formulas are of the form,

where hp is one of various products of three Pochhammer symbols (their factorial equivalents are given in this article) and C is either j(τ), r2A(τ), r3A(τ), r4A(τ) or equivalently, the McKay-Thompson series T1A, T2A, T3A, T4A with an appropriate constant term. It is tempting to speculate that, perhaps using different Pochhammer symbols and an appropriate constant term, one can use, say, T7A as well. In fact, J. Guillera has extended Ramanujan’s results and found pi formulas of form,

where gp is one of various products of five Pochhammer symbols, but

apparently the denominators are not values of any McKay-Thompson series. See “A Compilation of Ramanujan-Type Formulas for 1/πm” for m > 1 for a list. IV. A Continued Fraction Using eπ√163

First, given the curious prime-generating polynomial, P(n) = 4n2+163 which is prime for n = {0 to 19}. Solve for P(n) = 0, hence n = (1/2)√-163, plug this into r4A(n) and we find that it is an algebraic number, r4A((1/2)√-163) = x24

where x ≈ 5.31863… is the real root of the simple cubic equation x3-6x2+4x-2 = 0. Second, given the Ramanujan constant R, R = eπ√163 ≈ 6403203 + 743.9999999999992… and we see that R and x have the beautiful relationships,

where q = -1/R = -1/eπ√163. The root x has the exact value, let τ = (1+√-163)/2,

where η(τ) is the Dedekind eta function which partly explains the two relationships between R and x. While this continued fraction is not unique to d = 163 (the general form is by E. Heine, 1846), I thought of including it in this article since R is a well-known constant. The significance of the integer 24 will be further elaborated in other articles. Furthermore, it is also the case that, eπ√43 ≈ 123(92-1)3 + 743.999…eπ√67 ≈ 123(212-1)3 + 743.99999…eπ√163 ≈ 123(2312-1)3 + 743.999999999999… The reason for the squares within the cubes is a certain Eisenstein series. But that’s another story ... Part 2: Prime-Generating Polynomials and 43 Moonshine Functions Keywords: moonshine functions, Monster group, prime generating polynomial, McKay-Thompson series, class numbers, 43, 163.

Abstract: An expository work on how 1) prime-generating polynomials are connected to integer values of some moonshine functions rp(τ); and 2) for p > 1 satisfying certain constraints, it is conjectured exactly 43 of the moonshine functions obey certain rules regarding class numbers. I. IntroductionII. Class number h(d) = 1III. Class number h(d) = 2nIV. Class number h(d) = 4nV. Class number h(d) = 8nVI. ConjecturesVII. Rogers-Ramanujan Continued Fraction I. Introduction In “Part 1: The 163 Dimensions of the Moonshine Functions”, it was discussed how Conway, Norton, and Atkin showed that the moonshine functions curiously span a linear space of dimension 163. (For details, it is recommended one reads Part 1 first.) These functions assume the form of McKay-Thompson series, Tp(q) = 1/q + a0 + a1q + a2q2 + a3q2 + … where (throughout this article), q = e2πiτ = exp(2πiτ) the powers of q are either consecutive or some other arithmetic progression, and a0 is normalized as zero or another integer. But other than having a dimension of 163, another interesting aspect to them is

the values they assume for τ a complex root of a quadratic, especially if this quadratic happens to be a maximal prime-generating polynomial. For example, define r2A(τ) as T2A with a0 = 104, hence, r2A(τ) = 1/q + 104 + 4372q + 96256q2 + 1240002q3 + … (A101558) Then consider the prime-generating polynomial, P(n) = 2n2+29 which is distinctly prime for its maximum possible range of n = {0 to 28}. Solving for P(n) = 0, hence n = τ = (1/2)√-58 and plugging this into the series, one finds that, r2A((1/2)√-58) = 3964

This particular integer can be expressed in terms of the fundamental unit U29 = (5+√29)/2 as, 26(U29

6 + 1/U296)2 = 3964

and also appears in a pi formula found by Ramanujan a century ago in 1912,

where one can see the 58 as well. And, of course, in the almost integer,

eπ√58 = 3964 - 104.0000001… This article, part 2 of a series, will look at an interesting section of the 163 dimensions of the moonshine functions and discuss its relationship to quadratic prime-generating polynomials. Since the class number h(d) of our example d = -4·58 = -232 is h(d) = 2, we will start with h(d) = 1. II. Class number h(d) = 1 Using T1A with a0 = 744, we have the classical j-function j(τ), j(τ) = 1/q + 744 + 196884q + 21493760q2 + 864299970q3 + … (A007240) There are exactly 9 negative fundamental discriminants d such that h(d) = 1, namely, -d = {4, 8, 3, 7, 11, 19, 43, 67, 163} (The list of d with class number up to h(d) = 25 can be found in Mathworld.) The table below shows the associated prime-generating polynomial P(n) = an2+bn+c, its discriminant d, and j(τ) using a root τ of P(n) = 0 which, of course, is given by the quadratic formula,

For negative d, the root on the upper half of the complex plane is chosen per the requirements of modular forms. Hence,

P(n) = an2+bn

d = b2-4ac

n2+1 -4 123

n2+2 -8 203

n2-n+1 -3 0

n2-n+2 -7 -153 = -13

(42-1)3

n2-n+3 -11 -323 = -43

(32-1)3

n2-n+5 -19 -963 = -123

(32-1)3

n2-n+11 -43 -9603 = -123

(92-1)3

n2-n+17 -67 -52803 = -123(212-1)3

n2-n+41 -163 -6403203 = -123(2312-1)

Table 1

Some remarks: 1. They have a beautifully consistent “internal” structure, don’t they? The reason for the squares within the cubes is due to certain Eisenstein series (to be discussed in some other article). This pattern also carries over to higher class numbers for some d. For example, the largest d (in absolute value) with h(d) = 2 is d = -427 = -7·61, hence letting τ = (1+√-427)/2, then,

j(τ) = -123(w2-1)3

where w = 7215+924√61, an algebraic number of deg 2 as expected, since the j-function detects the class number of negative fundamental d. This also implies, eπ√427 ≈ 123(w2-1)3 + 743.9999999999999999999…. which goes on for 22 nines, more than the 12 nines of d = -163. Later it will be seen that the moonshine functions, in a special way, can also detect h(d). 2. The P(n) = an2+bn+c in Table 1 for c > 1 all have the maximum possible range for distinct primes as n = {1 to c-1}. 3. It is not so well-known that, starting with some negative n, then every third value of these P(n) is prime for the same length of consecutive n. And for higher d, the primes within this span are different.

n P(m) Prime for

→9m2-43(3m)+473

{1 to 10}

→9m2-67(3m)+1139

{1 to 16}

→9m2-163(3m)+6683

{1 to 40}

n P(m) Prime for

→9m2-43(3m)+473

{1 to 10}

→9m2-67(3m)+1139

{1 to 16}

→9m2-163(3m)+6683

{1 to 40}

For example, the sequence of 40 primes for Euler’s polynomial P(n) is {41, 43, …,1601}, while the 40 primes for P(m) will start and end differently as {6203, 5741, … ,1523}. Also, note that 2 ·{21, 33, 81} + 1 = {43, 67, 163}. Furthermore, if we solve P(m) = 0 and plug the correct root τ into the McKay-Thompson series T9A with constant term a0 = -3, r9A(τ) = 1/q - 3 + 27q + 86q2 + 243q3 + 594q4 + … A007266 we get complex deg 4 algebraic numbers. Given a cube root of unity, u = e2πi/3 = (-1+√-3)/2, then,

with the integers, {j43, j67, j163} = {960/12, 5280/12, 640320/12} Recall that, eπ√43 - 744 ≈ 9603 = 123(92-1)3

eπ√67 - 744 ≈ 52803 = 123(212-1)3

eπ√163 - 744 ≈ 6403203 = 123(2312-1)3

Interesting that the blue numbers should appear using r9A(τ) and prime-

generating polynomials of form P(n) = 9n2+bn+c. III. Class number h(d) = 2n The Monster group has the extremely large order, O = 24· 34· 54· 74· 114·134· 17· 19· 23· 29· 31· 41· 47· 59· 71 ≈ 8· 1053

The 15 primes that divide its order are called the supersingular primes, and there is a McKay-Thompson series TpA for each one. Most of these primes also divide the fundamental discriminants d with class number h(d) = 2. Define the following functions, r2A(τ) = 1/q + 104 + 4372q + 96256q2 + 1240002q3 + … (A101558) r3A(τ) = 1/q + 42 + 783q + 8672q2 + 65367q3 + … (A030197) r5A(τ) = 1/q - 6 + 134q + 760q2 + 3345q3 + … (A007251) r7A(τ) = 1/q + 10 + 51q + 204q2 + 681q3 + … (A030183) r11A(τ) = 1/q + 0 + 17q + 46q2 + 116q3 + … (A030183) r13A(τ) = 1/q - 2 + 12q + 28q2 + 66q3 + 132q4 + …

(A034318) and so on. (For brevity, we will focus only on the smaller p.) Without the constant term, these are simply the series T2A, T3A, T5A, T7A, T11A, and T13A, and the first two have already been given in Part 1. There are exactly 18 negative fundamental discriminants d such that h(d) = 2, -d = {15, 20, 24, 35, 40, 51, 52, 88, 91, 115, 123, 148, 187, 232, 235, 267, 403, 427} The table below shows their associated prime-generating polynomials P(n), and the value of rp(τ) using the root τ of P(n) = 0:

p P(n) = an2+bn

d = b2-4ac

rpA(τ)

2 2n2-2n

+3-20 = -4·5

-26·42

2 2n2-2n+7

-52 = -4·13

-26·362

2 2n2-2n+19

-148 = -4·37

-26·17642

2 2n2+3 -24 = -4·6

26·62

2 2n2+5 -40 = -4·10

26·182

2 2n2+11 -88 = -4·22

26·1982

2 2n2+29 -232 = -4·58

26·196022 = 3964

3 3n2-3n

+2-15 = -3·5

-33·16

3 3n2-3n+5

-51 = -3·17

-33·26

3 3n2-3n+11

-123 = -3·41

-33·46

3 3n2-3n+23

-267 = -3·89

-33·106 = -3003

3 3n2+2 -24 = -3·8

33·23

5 5n2-5n

+2-15 = -5·3

5 5n2-5n+3

-35 = -5·7

5 5n2-5n+7

-115 = -5·23

5 5n2-5n+13

-235 = -5·47

-15250

5 5n2+1 -20 = -5·4

10√5

5 5n2+2 -40 = -5·8

7 7n2-7n

+3-35 = -7·5

-7·12

7 7n2-7n+5

-91 = -7·13

-7·32

7 7n2-7n+17

-427 = -7·61

-7·392

11n2-11n+7

-187 = -11·17

13n2-13n+5

-91 = -13·7

13n2-13n+11

-403 = -13·31

13n2+1 -52 = -13·4

2√13

p P(n) = an2+bn

d = b2-4ac

rpA(τ)

2 2n2-2n

+3-20 = -4·5

-26·42

2 2n2-2n+7

-52 = -4·13

-26·362

2 2n2-2n+19

-148 = -4·37

-26·17642

2 2n2+3 -24 = -4·6

26·62

2 2n2+5 -40 = -4·10

26·182

2 2n2+11 -88 = -4·22

26·1982

2 2n2+29 -232 = -4·58

26·196022 = 3964

3 3n2-3n

+2-15 = -3·5

-33·16

3 3n2-3n+5

-51 = -3·17

-33·26

3 3n2-3n+11

-123 = -3·41

-33·46

3 3n2-3n+23

-267 = -3·89

-33·106 = -3003

3 3n2+2 -24 = -3·8

33·23

5 5n2-5n

+2-15 = -5·3

5 5n2-5n+3

-35 = -5·7

5 5n2-5n+7

-115 = -5·23

5 5n2-5n+13

-235 = -5·47

-15250

5 5n2+1 -20 = -5·4

10√5

5 5n2+2 -40 = -5·8

7 7n2-7n

+3-35 = -7·5

-7·12

7 7n2-7n+5

-91 = -7·13

-7·32

7 7n2-7n+17

-427 = -7·61

-7·392

11n2-11n+7

-187 = -11·17

13n2-13n+5

-91 = -13·7

13n2-13n+11

-403 = -13·31

13n2+1 -52 = -13·4

2√13

p P(n) = an2+bn

d = b2-4ac

rpA(τ)

2 2n2-2n

+3-20 = -4·5

-26·42

2 2n2-2n+7

-52 = -4·13

-26·362

2 2n2-2n+19

-148 = -4·37

-26·17642

2 2n2+3 -24 = -4·6

26·62

2 2n2+5 -40 = -4·10

26·182

2 2n2+11 -88 = -4·22

26·1982

2 2n2+29 -232 = -4·58

26·196022 = 3964

3 3n2-3n

+2-15 = -3·5

-33·16

3 3n2-3n+5

-51 = -3·17

-33·26

3 3n2-3n+11

-123 = -3·41

-33·46

3 3n2-3n+23

-267 = -3·89

-33·106 = -3003

3 3n2+2 -24 = -3·8

33·23

5 5n2-5n

+2-15 = -5·3

5 5n2-5n+3

-35 = -5·7

5 5n2-5n+7

-115 = -5·23

5 5n2-5n+13

-235 = -5·47

-15250

5 5n2+1 -20 = -5·4

10√5

5 5n2+2 -40 = -5·8

7 7n2-7n

+3-35 = -7·5

-7·12

7 7n2-7n+5

-91 = -7·13

-7·32

7 7n2-7n+17

-427 = -7·61

-7·392

11n2-11n+7

-187 = -11·17

13n2-13n+5

-91 = -13·7

13n2-13n+11

-403 = -13·31

13n2+1 -52 = -13·4

2√13

Table 3

Some remarks: 1. The values are also beautifully consistent in form. While the table used class number h(d) = 2, one can also use h(d) = 4, 6, etc. For ex, let d = -4·82 with h(d) = 4, and τ = (1/2)√-82, then one gets an algebraic number just of deg 2, r2A(τ) = 124(51+8√41)4

2. Using large d, one gets almost-integers (or in general, almost-algebraics), eπ/2√232 = eπ√58 ≈ 3964 - 104.0000001…. eπ/3√267 = eπ√(89/3) ≈ 3003 + 41.99997… eπ/5√235 = eπ√(47/5) ≈ 15250 - 6.008…

eπ/7√427 = eπ√(61/7) ≈ 7·392 + 9.995… eπ/13√403 = eπ√(31/13) ≈ 130 - 2.09… and, eπ/2√328 = eπ√82 ≈ 124(51+8√41)4 - 104.000000001… IV. Class number h(d) = 4n In contract to the ones in the previous sections, a prime-generating polynomial of form P(n) = 6n2-6n+c cannot be prime for n = c-1 since it factors at that value. For P(n) = 10n2-10n+c, it is n = c-2, and so on. The most impressive for the first kind is, P(n) = 6n2-6n+31 which is distinctly prime for the maximum n = {1 to 29}. This has discriminant d = -708 = -6·118, and we find this d also appears in an almost-integer, eπ/6√708 = eπ√(118/6) ≈ 10602 + 9.99992… But this d has class number h(d) = 4. If 10602 is the value of a moonshine function, then there must be some yielding, for appropriate d, algebraic numbers that are one-fourth the class number. It turns out these apparently are the McKay-Thompson series of order p where p is the product of two distinct supersingular primes. There are 23 such orders, 2 ·{3, 5, 7, 11, 13, 17, 19, 23, 31, 47} = {6, 10, 14, 22, 26, 34, 38, 46, 62, 94}3 ·{5, 7, 11, 13, 17, 19, 23, 29, 31} = {15, 21, 33, 39, 51, 57, 69, 87,

93}5 ·{7, 11, 19} = {35, 55, 95}7 ·{17} = 119 though p = 57 and 93 are to be excluded. (For the relevant McKay-Thompson series in this family, these two are the only ones where the powers of q are not consecutive but are in the progression 3m+2.) So what remains are 21 series. For brevity, we will focus only in the case p = 2m for small m. Define the following functions, r6A(τ) = 1/q + 10 + 79q + 352q2 + 1431q3 + … (A007254) r10A(τ) = 1/q + 4 + 22q + 56q2 + 177q3 + … (A058097) r14A(τ) = 1/q + 2 + 11q + 20q2 + 57q3 + … (A058497) r22A(τ) = 1/q - 5 + 5q + 6q2 + 16q3 + 20q4 … (A058567) r26A(τ) = 1/q + 0 + 4q + 4q2 + 10q3 + 12q4 … (A058596) and so on, which, needless to say, without the constant terms are the McKay-Thompson series T6A, T10A, T14A, T22A, and T26A. There are exactly 54 negative fundamental discriminants d with h(d) = 4. However, only some are divisible by 6, 10, 14, 22, or 26. These, and their associated P(n), and rpA(τ) are listed in the following table:

p P(n) = an2+bn

d = b2-4ac

rpA(τ)

6 6n2+5 -120 = -12·10

25·10

6 6n2+7 -168 = -12·14

25·28

6 6n2+13 -312 = -12·26

25·325

6 6n2+17 -408 = -12·34

25·352

6 6n2-6n+5

-84 = -12·7

-24·7

6 6n2-6n+7

-132 = -12·11

-24·52

6 6n2-6n+11

-228 = -12·19

-24·132

6 6n2-6n+17

-372 = -12·31

-24·31

6 6n2-6n+31

-708 = -12·59

-24·2652

10n2+3 -120 = -20·6

10n2+7 -280 = -20·14

10n2+13

-520 = -20·26

10n2+19

-760 = -20·38

10n2-10n+11

-340 = -20·17

14n2+3 -168 = -28·6

14n2+5 -280 = -28·10

14n2-14+5

-168 = -28·3

14n2-14+13

-280 = -28·19

-7·52

22n2-22+7

-132 = -44·3

22n2-22+17

-1012 = -44·23

-11·32

26n2+3 -312 = -52·6

26n2+5 -520 = -52·10

p P(n) = an2+bn

d = b2-4ac

rpA(τ)

6 6n2+5 -120 = -12·10

25·10

6 6n2+7 -168 = -12·14

25·28

6 6n2+13 -312 = -12·26

25·325

6 6n2+17 -408 = -12·34

25·352

6 6n2-6n+5

-84 = -12·7

-24·7

6 6n2-6n+7

-132 = -12·11

-24·52

6 6n2-6n+11

-228 = -12·19

-24·132

6 6n2-6n+17

-372 = -12·31

-24·31

6 6n2-6n+31

-708 = -12·59

-24·2652

10n2+3 -120 = -20·6

10n2+7 -280 = -20·14

10n2+13

-520 = -20·26

10n2+19

-760 = -20·38

10n2-10n+11

-340 = -20·17

14n2+3 -168 = -28·6

14n2+5 -280 = -28·10

14n2-14+5

-168 = -28·3

14n2-14+13

-280 = -28·19

-7·52

22n2-22+7

-132 = -44·3

22n2-22+17

-1012 = -44·23

-11·32

26n2+3 -312 = -52·6

26n2+5 -520 = -52·10

Table 4

An example using h(d) = 8 is d = -1380, hence τ = (6+√-1380)/12, giving the deg 2 algebraic, r6A(τ) = -42(2093+540√15)2

and implying, eπ/6√1380 = eπ√(230/6) ≈ 42(2093+540√15)2 + 9.9999997…

V. Class number h(d) = 8n On a suggestion by Simon Norton who asked about p a product of three distinct supersingular primes, it seems there are indeed moonshine functions that could detect h(d) = 8n. There are 7 such p, namely p = {30, 42, 66, 70, 78, 105, 110}, r30B(τ) = 1/q + 0 + 4q + 2q2 + 6q3 + … (A058613) r42A(τ) = 1/q + 0 + 2q + 2q2 + 3q3 + … (A058671) r66A(τ) = 1/q + 0 + 2q + 0q2 + q3 + … (A058739) r70A(τ) = 1/q + 0 + q + 0q2 + 2q3 + … (A058744) r78A(τ) = 1/q + 0 + q + q2 + q3 + … (A058754) r105A(τ) = 1/q + 0 + q + q2 + 0q3 + … (A058773) r110A(τ) = 1/q + 0 + 0q + q2 + q3 + … (A058774) which, since the constant term a0 = 0, are simply McKay-Thompson series. (Note that the first uses the conjugacy class 30B.) There are exactly 131 negative fundamental d with h(d) = 8, but only some are divisible by the seven p. For brevity, we will give at most only two examples per function,

p P(n) = an2+bn

d = b2-4ac

rp(τ)

30n2+7 -840 = -60·14

30n2+11 -1320 = -60·22

42n2+5 -840 = -84·10

42n2+11 -1848 = -84·22

66n2+5 -1320 = -132·10

66n2+7 -1848 = -132·14

70n2+3 -840 = -140·6

78n2-78n+23

-1092 = -156·7

105n2-105n+31

-1995 = -105·19

110n2+3 -1320 = -220·6

Table 5

Remarks: 1. Just like in the previous three tables, all the given prime-generating P(n) in Table 4 are distinctly prime for its maximum range n. For example, the form P(n) = 105n2-105n+c can only be distinctly prime for n = {1 to c-24} since P(c-23) will factor. 2. For class number h(d) = 16, an example is d = -5208 = -84·62. Let τ = (1/84)√-5208, and we get an algebraic number just of deg 2, r42A(τ) = (1/4)(15+√217)2

VI. Conjectures Given composite negative fundamental discriminant d = pm, but excluding d such that pd = a2, or pd = a2v where v divides p: Conjecture 1: If d = pm has class number h(d) = 2n, then for p a supersingular prime, the moonshine function rpA(τ) is an algebraic number of degree one-half the h(d). There are 15 such functions. Conjecture 2: If d = pm has class number h(d) = 4n, then for p a product of 2 distinct supersingular primes (excepting products p = 57 and 93), the rp(τ) of the appropriate conjugacy class is an algebraic number of degree one-fourth the h(d). There are 21 such functions. Conjecture 3: If d = pm has class number h(d) = 8n, then for p a product of 3 distinct supersingular primes, the rp(τ) of the appropriate conjugacy class is an algebraic number of degree one-eighth the h(d). There are 7 of these.

Conjecture 4: Thus, there is a total of 15 + 21 + 7 = 43 such functions. Note 1: All the functions are from the series TpA, excepting three: T30B, T33B, and T46C. Note 2: To recall, there are 172 McKay-Thompson series for the Monster, all distinct except the special case of T27A = T27B. And 172 = 4·43. In fact, the Monster has at least 43 conjugacy classes of maximal subgroups. If the numbers 43, 67, 163 turn out to be significant for moonshine functions (none is yet known for the second), then that would be an interesting triple “coincidence”. VII. Rogers-Ramanujan Continued Fraction In one of his letters to Hardy, Ramanujan gave the beautiful continued fraction,

We saw how supersingular primes play an important role in the moonshine functions. It turns out that simple arithmetic properties of these primes allow it to be also connected to the Rogers-Ramanujan continued fraction. But again, that is another story…

This page is still under construction. (But the next page is already finished.)

Abstract: Infinite series using the reciprocals of the j-function and other moonshine functions can be used to express the three Watson’s triple integrals. I. IntroductionII. Moonshine FunctionsIII. McKay-Thompson series 1AIV. McKay-Thompson series 2AV. McKay-Thompson series 3AVI. McKay-Thompson series 4AVII. ConjectureVIII. Watson and Ramanujan I. Introduction In 1939, G.N. Watson considered the following beautiful triple integrals,

Interestingly, these have a simple closed-form expression in terms of the gamma function Γ(n),

But it turns out these integrals can also be expressed in terms of values that appear in Ramanujan-type pi formulas of form 1/pi. To illustrate, first define the factorial quotients, h1 = (6n)! / ((3n)! n!3)h2 = (4n)! / (n!4)h3 = (2n)!(3n)! / (n!5)h4 = (2n)!3 / (n!6) or, equivalently, the Pochhammer symbol products, h1 = 1728n (1/2)n (1/6)n (5/6)n / (n!3)h2 = 256n (1/2)n (1/4)n (3/4)n / (n!3)h3 = 108n (1/2)n (1/3)n (2/3)n / (n!3)h4 = 64n (1/2)n (1/2)n (1/2)n / (n!3) where (a)n = (a)(a+1)(a+2)…(a+n-1). Then,

Compare to the Ramanujan-type pi formulas,

where one can see the summation for both sets have common denominators and hi. These are not isolated results, and apparently there are an infinite number of such formulas for the Watson triple

integrals. The common values are given by a few moonshine functions, namely the McKay-Thompson series of class 1A, 2A, 3A, 4A with an appropriately chosen constant term, and will be discussed in the next section. II. Moonshine Functions In “The 163 Dimensions of the Moonshine Functions”, Conway, Norton, and Atkin showed that these functions, rather curiously, span a linear space of dimension 163. However, for the purposes of this article, we need only four, namely, r1A(τ) = j(τ) = 1/q + 744 + 196884q + 21493760q2 + 864299970q3 + … (A007240) r2A(τ) = (f2 + 64)2 / f2 = 1/q + 104 + 4372q + 96256q2 + 1240002q3 + … (A101558) r3A(τ) = (f3 + 27)2 / f3 = 1/q + 42 + 783q + 8672q2 + 65367q3 + … (A030197) r4A(τ) = (f4 + 16)2 / f4 = 1/q + 24 + 276q + 2048q2 + 11202q3 + … (A097340) with, q = e2πiτ = exp(2πiτ) While the first is essentially the j-function, given in Mathematica as j(τ) = N[123KleinInvariantJ[τ], n], for arbitrary precision n, the other three can also be conveniently defined in terms of quotients fp of the Dedekind eta function, η(τ), where, f2 = (η(τ)/η(2τ))24

f3 = (η(τ)/η(3τ))12 f4 = (η(τ)/η(4τ))8 also easily calculated in Mathematica as η(τ) = N[DedekindEta[τ], n]. This then gives a convenient way to calculate the rpA(τ). Ramanujan’s pi formulas are simply of the form,

where C = rpA(τ) for p = {1,2,3,4} and, for appropriate τ, then A,B,C are algebraic numbers. We conjecture and give heuristic evidence that Watson’s triple integrals are,

which, for carefully chosen τ, then w is also an algebraic number. Part III. McKay-Thompson Series 1A In the Introduction, it was shown that the j-function j(√-4) = 663 appears in the Watson integral I1. However, for carefully chosen τ, it turns out one can use j(τ) for all three triple integrals. (A few similar formulas were found by C.H.Brown, but he didn’t connect them to the Ii.) Recall that, r1A(τ) = j(τ) = 1/q + 744 + 196884q + 21493760q2 + 864299970q3 + …

and, h1 = (6n)! / ((3n)! n!3) = 1728n (1/2)n (1/6)n (5/6)n / (n!3) then,

These uses the four arguments τ = {√-4, √-3, (1+3√-3)/2, √-6}, respectively. (However, there are many more formulas using j(τ) and I merely chose the simplest ones.) Part IV. McKay-Thompson Series 2A Likewise, the moonshine function r2A(τ), at certain τ, can be used for all three Watson integrals. Recall that, r2A(τ) = (f2 + 64)2 / f2 = 1/q + 104 + 4372q + 96256q2 + 1240002q3 + …

where, f2 = (η(τ)/η(2τ))24 For I1: τ = {i, (1+3i)/2, (1+5i)/2}For I2: τ = {√-3}For I3: τ = {(1/2)√-6} Since h2 = (4n)! / (n!4) = 256n (1/2)n (1/4)n (3/4)n / (n!3), then,

Part V. McKay-Thompson Series 3A Define,

r3A(τ) = (f3 + 27)2 / f3 = 1/q + 42 + 783q + 8672q2 + 65367q3 + … where, f3 = (η(τ)/η(3τ))12 For I1: τ = iFor I2: τ = {(1/6)√-3, (3+3√-3)/6, (3+5√-3)/6, (3+7√-3)/6}For I3: τ = (1/3)√-6 Since h3 = (2n)!(3n)! / (n!5) = 108n (1/2)n (1/3)n (2/3)n / (n!3), then,

Part VI. McKay-Thompson Series 4A Lastly, r4A(τ) = (f4 + 16)2 / f4 = 1/q + 24 + 276q + 2048q2 + 11202q3 + … where, f4 = (η(τ)/η(4τ))8 For I1: τ = (1/2)(1+√-4)For I2: τ = (1/2)√-3For I3: τ = (1/2)√-6 Since h4 = (2n)!3 / (n!6) = 64n (1/2)n (1/2)n (1/2)n / (n!3), then,

VII. Conjecture It should be pointed out that not just any τ will do in expressing the

Watson triple integrals Ii similar to the forms above. For example, we have, r4A((1+√-2)/2) = -26

and using this and h4 in an infinite series, it was recognized by Mathematica that,

or, the neat equivalent form found by Ramanujan using double factorials,

However, the ratio of x with any of the three Ii needs more than algebraic numbers. Thus,

Hence, Conjecture: “For p = {1,2,3,4}, let the moonshine function rpA(τ) and Pochhammer products hp be as defined in the previous sections. Let C = rpA(τ). If τ is chosen as,

for some integer mk, and if the infinite series below converges to a Watson triple integral,

then w is an algebraic number.” VIII. Watson and Ramanujan As an afterword, it can be mentioned that Watson was very familiar with Ramanujan’s work. G. N. Watson (1886-1965) co-wrote with Whittaker the classic, A Course of Modern Analysis (1915) which influenced a generation of Cambridge mathematicians, including Littlewood and Hardy. (Hardy, of course, was the one who discovered Ramanujan.) After Ramanujan’s early death, his notebooks, including the so-called Lost Notebook, found its way to Watson who attempted to organize them. He would subsequently spend many years on Ramanujan’s formulas, especially on mock theta functions and others.

Abstract: Infinite series using the reciprocals of the j-function and other moonshine functions can be used to find values of the complete elliptic integral of the first kind. I. IntroductionII. Some Formulas for the Complete elliptic integral of the first kind K(k)III. Gamma functionsIV. Elliptic ModulusV. Moonshine FunctionsVI. ConjecturesVII. Some Remarks I. Introduction In Part 4: Watson’s Triple Integrals and Ramanujan-Type Pi Formulas, it was discussed how the j-function and other moonshine functions can be used to express the three beautiful Watson triple integrals Ii. For example, given I2,

it’s long known that it evaluates to,

where K(kd) is the complete elliptic integral of the first kind. However, it can also be expressed, among others, by the j-function j(√-3) = 2·303 as,

One can easily calculate these values in Mathematica or http://www.wolframalpha.com/ using the commands, K(k3) := EllipticK[ModularLambda[√-3]] j(√-3) := 1728KleinInvariantJ[√-3] The fact that the Watson triple integrals are just special values (which used d = 1,3,6) of a family was the clue that the formulas using the j-function and others in Part 4 were not so much about them, but in fact were more about the K(kd). Thus. there should be similar formulas for other d. In fact there are, and some interesting ones are,

where φ is the golden ratio, T is the tribonacci constant, and P is the plastic constant. II. Some Formulas for the Complete Elliptic Integral of the First Kind K(k) A. Class number h(-d) = 1 Recall that,

where (a)n is the Pochhammer symbol. Using the j-function j(τ), then,

where u(τ) is the Dedekind eta quotient,

(Note: It is understood that all appearances of "tau", τ, in this article refer to this form.) For positive integer d, u(τ) is a real algebraic number. For h(-d) = 1, with the exception of d = 7, u(τ) is the real root of simple cubics with small coefficients, namely,

d u(τ) is a root of:7 x2-2 = 011 x3-2x2+2x-2 = 019 x3-2x-2 = 043 x3-2x2-2 = 067 x3-2x2-2x-2 = 0163 x3-6x2+4x-2 = 0

For the special case d = 11, then x = (T+1)/T where T is the tribonacci constant. (See also, A Tale of Four Constants by this author.) These cubic roots x have the beautiful continued fraction,

where q is the negative real number q = -exp(-π√d) = -1/eπ√d. In general, the cfrac is valid for u(τ) and τ as defined above. Furthermore, for d with h(-d) = 1, especially the three largest d, then the 24th power of x is a near-integer. In particular,

One can also see the j(τ) in the denominators of the K(kd) formulas, i.e., j((1+√-163)/2) = 1728KleinInvariantJ[(1+√-163)/2] = -6403203

and easily discern a common form which will be given in the Conjectures section. B. Class number h(-4d) = 2 Let,

The three formulas below, however, involve another moonshine function.

which now uses the cube of u(τ),

The values of the u(τ) are easily calculated as,

d y = u(τ)5 (1+√5)1/4

13 (3+√13)1/4

37 21/4(6+√37)1/4

Consistent with the results above, then it is also the case that,

III. Gamma Functions

It’s long been proven that K(k), if k has a singular value, then K(k) can be expressed in terms of the gamma function. Focusing only on the infinite series part given above, then for odd fundamental d > 3 with class number h(-d) = 1, this is simply,

where the exponent of the gamma function Γ(n) is the Kronecker symbol. This can be implemented in Mathematica as KroneckerSymbol[-d,m]. Since it yields only the values {-1,0,1}, then it determines whether Γ(n) is in the numerator, denominator, or if it vanishes. For example, for d = 7, then,

where the simplification results from the fact that,

and similarly for d = {11, 19, 43, 67, 163}. Of course, after suitable algebraic manipulation, one can easily replace the infinite series with the K(kd). Explicitly, for d = -163, this is,

where x is the real root of, x3-6x2+4x-2 = 0 and the 81 numerators N are, N163 = {1, 4, 6, 9, 10, 14, 15, 16, 21, 22, 24, 25, 26, 33, 34, 35, 36, 38, 39, 40, 41, 43, 46, 47, 49, 51, 53, 54, 55, 56, 57, 58, 60, 61, 62, 64, 65, 69, 71, 74, 77, 81, 83, 84, 85, 87, 88, 90, 91, 93, 95, 96, 97, 100, 104, 111, 113, 115, 118, 119, 121, 126, 131, 132, 133, 134, 135, 136, 140, 143, 144, 145, 146, 150, 151, 152, 155, 156, 158, 160, 161} IV. The Elliptic Modulus The complete elliptic integral of the first kind K(k) is defined for 0 < k < 1 by,

or equivalently by the infinite series,

where k is the elliptic modulus, and 2F1(a,b;c;x) is the hypergeometric function. The complementary modulus is k’ = √(1-k2), where 0 < k’ < 1. Let d be a positive integer, then the equation,

defines a unique real number kd called the singular modulus. In Mathematica, when k has a singular value, then k2 can be given by the elliptic lambda function, λ(n), k2 = λ(n) = ModularLambda[n] for n = √-d. Exact values of kd for small values of d are given in the link above and labeled as λ*(d). For example, the first four kd are, k1 = λ*(1) = 1/√2k2 = λ*(2) = -1+√2k3 = λ*(3) = ¼ (-1+√3)√2k4 = λ*(4) = 3-2√2 K(k) can then be calculated in terms of the parameter m = k2 as: K(k) = EllipticK[m] = EllipticK[ModularLambda[n]]. Example: Let n = √-4, and we find that, k2 = λ(n) = ModularLambda[√-4] = (3-2√2)2

hence k4 = 3-2√2 and, K(k4) = K(3-2√2) = EllipticK[(3-2√2)2] = (1+√2) Γ2[1/4] / (27/2√π) = 1.58255…

Exact values of other K(kd) are given in the page, Elliptic integral singular value. V. Moonshine Functions Let q = e2πiτ = exp(2πiτ) and, as before, define four of the moonshine functions as, r1A(τ) = j(τ) = 1/q + 744 + 196884q + 21493760q2 + 864299970q3 + … (A007240) r2A(τ) = (f2 + 64)2 / f2 = 1/q + 104 + 4372q + 96256q2 + 1240002q3 + … (A101558) r3A(τ) = (f3 + 27)2 / f3 = 1/q + 42 + 783q + 8672q2 + 65367q3 + … (A030197) r4A(τ) = (f4 + 16)2 / f4 = 1/q + 24 + 276q + 2048q2 + 11202q3 + … (A097340) with, f2 = (η(τ)/η(2τ))24 f3 = (η(τ)/η(3τ))12 f4 = (η(τ)/η(4τ))8 Also define, h1 = (6n)! / ((3n)! n!3)h2 = (4n)! / (n!4)

h3 = (2n)!(3n)! / (n!5)h4 = (2n)!3 / (n!6) or, equivalently, h1 = 1728n (1/2)n (1/6)n (5/6)n / (n!3)h2 = 256n (1/2)n (1/4)n (3/4)n / (n!3)h3 = 108n (1/2)n (1/3)n (2/3)n / (n!3)h4 = 64n (1/2)n (1/2)n (1/2)n / (n!3) VI. Conjectures Given the moonshine functions defined above and still let,

with d a positive integer above a bound. Then, we conjecture that,

where,

For j(τ), r2A(τ), r4A(τ), then d ≥ {7, 4, 2} respectively, otherwise the series diverges. Unfortunately, I have not yet been able to find the correct formulation using r3A. On the other hand, there is a much simpler form using only the 24th power of u(τ) and h4. Given a 48th root of unity ζ48 = exp(2π i/48) = exp(π i/24), then,

for d ≥ 1, nice examples of which for d = {5, 11, 23} were given in the Introduction. It’s been known since Watson (1908) that, given the complementary modulus, k’, then,

where 0 < k ≤ 1/√2. But this can also be expressed as,

By comparing the two formulas, it must be the case that,

where, as in the rest of this article,

Note: As pointed out by Michael Somos, the sequence A002897 generated by [(2n)! / n!2]3 = [C(2n,n)]3 = {1, 8, 216, 8000, 343000,...}, or the cube of the central binomial coefficients, can also be interpreted as the expansion of [K(k) / (π/2)]2 in powers of x = (k k'/4)2, or, [K(k) / (π/2)]2 = 1+ 8x + 216x2 + 8000x3 + 343000x4… VII. Some Remarks Of the four moonshine functions mentioned, only r3A(τ) remains unformulated in relation to K(k). It seems in the formula,

that x, like for the other three functions, is always an algebraic number

for integer d above a bound, in this case, d ≥ 3. For example, for d = 3, then,

and similarly for other d, though a closed-form expression for general x has not yet been found. However, it turns out that r3A(τ) may yet have a role in another function, one similar to K(k). But that’s another story.

1. Various Forms I. Maillet’s IdentityIIa. Lucas-Liouville Polynomial IdentityIIb. Boutin’s IdentityIIIa. Lagrange’s Polynomial IdentityIIIb. Lamé-Type IdentitiesIV. Waring-like Problems I. Maillet's Identity (E. Maillet) (a+b)3+(a+c)3+(a+d)3+(a-b)3+(a-c)3+(a-d)3 = 6a(a2+b2+c2+d2) IIa. Lucas-Liouville Polynomial Identity (a+b)k+(a+c)k+(a+d)k+(b+c)k+(b+d)k+(c+d)k+(a-b)k+(a-c)k+(a-d)k+(b-c)k+(b-d)k+(c-d)k = 6(a2+b2+c2+d2)k/2, for k = 2,4

Note: It seems E. Lucas knew this identity before Liouville. Anything else for k > 4? Update 9/26/09: William Ellison pointed out that David Hilbert has proven (x1

2+x22+ ... +xn

2)k can be written as a rational combination of (2k)th powers of linear forms in the xi. (You can read his paper on Waring's Problem and how Hilbert solved it in the Attachments section at the bottom of this page.) For example, the Lucas-Liouville Polynomial Identity above can be concisely written as, 6 (x1

2+x22+x3

2+x42)2 = ∑ (xi ± xj)

To determine the number of terms in the summation, if we are to choose 2 objects from 4, this gives 6 ways. (This can be calculated in Mathematica as Binomial[4, 2] = 6.) Since there are 2 sign changes, this gives a total of 6 x 2 = 12 terms on one side of the equation, which can be seen in the explicit identity. This in fact belongs to an infinite family where the addends involve 2 variables (xi ± xj) taken at a time and the xi taken singly: 6 (x1

2+x22)2 = ∑ (xi ± xj)

4 + 4 ∑ xi4

6 (x12+x2

2+x32)2 = ∑ (xi ± xj)

4 + 2 ∑ xi4

6 (x12+x2

2+x32+x4

2)2 = ∑ (xi ± xj)4 + 0 ∑ xi

6 (x12+x2

2+ ... + x52)2 = ∑ (xi ± xj)

4 - 2 ∑ xi4

and so on. For 3 variables taken at a time, 12 (x1

2+x22+x3

2)2 = ∑ (xi ± xj ± xk)4 + 8 ∑ xi4

24 (x12+x2

2+x32+x4

2)2 = ∑ (xi ± xj ± xk)4 + 12 ∑ xi4

36 (x12+x2

2+ ... + x52)2 = ∑ (xi ± xj ± xk)4 + 12 ∑ xi

48 (x12+x2

2+ ... + x62)2 = ∑ (xi ± xj ± xk)4 + 8 ∑ xi

60 (x12+x2

2+ ... + x72)2 = ∑ (xi ± xj ± xk)4 + 0 ∑ xi

72 (x12+x2

2+ ... + x82)2 = ∑ (xi ± xj ± xk)4 - 12 ∑ xi

Note how for seven squares, the second summation has a zero coefficient like for the Lucas-Liouville identity. For 4 variables taken at a time, 24 (x1

2+x22+x3

2+x42)2 = ∑ (xi ± xj ± xk ± xm)4 + 24 ∑ xi

72 (x12+x2

2+ ... + x52)2 = ∑ (xi ± xj ± xk ± xm)4 + 40 ∑ xi

144 (x12+x2

2+ ... + x62)2 = ∑ (xi ± xj ± xk ± xm)4 + 64 ∑ xi

240 (x12+x2

2+ ... + x72)2 = ∑ (xi ± xj ± xk ± xm)4 + 80 ∑ xi

360 (x12+x2

2+ ... + x82)2 = ∑ (xi ± xj ± xk ± xm)4 + 80 ∑ xi

I was not yet able to compute when the second summation has a zero coefficient but, as it seems to be reversing just like for 3 variables, I'm guessing it will be for 12 squares. (Anyone can check if this conjecture is true, as well as provide the identities for the next level of 5 variables taken at a time?) See the complete soln by Gerry Martens at the 9/30 update below. Update 9/29/09: I guessed wrong, it should be 10 squares, and the infinite family is given by: 6 (x1

2+x22+x3

2+x42)2 = ∑ (xa ± xb)4

60 (x12+x2

2+ ... + x72)2 = ∑ (xa ± xb ± xc)4

672 (x12+x2

2+ ... + x102)2 = ∑ (xa ± xb ± xc ± xd)4

7920 (x12+x2

2+ ... + x132)2 = ∑ (xa ± xb ± xc ± xd ± xe)4

96096 (x12+x2

2+ ... + x162)2 = ∑ (xa ± xb ± xc ± xd ± xe ± xf)

and so on. My thanks to Daniel Lichtblau of Wolfram Research (makers of Mathematica), and Renzo Benedetti, Martin Rubey, and Gerry Martens from sci.math.symbolic who answered my questions. One can see that the number of squares {4, 7, 10, 13, ...} is just in arithmetic progression and Benedetti observed that the numerical factor is generated by 2m-1 Binomial[3(m-1), m-1] where m is the number of objects taken at a time per term in the RHS. Including the leading coefficient of the trivial identity 1(x1

2)2 = xa4, this sequence {1, 6, 60,

672, 7920,...} is, as of this date, not yet in the Online Encyclopedia of Integer Sequences, but is there without the powers of 2 as sequence {1, 3, 15, 84, 495,...}. (End update, 9/29) For a combination of 3 and 2 variables, these may be derived by combining the identities given previously and we get: 72 (x1

2+x22+ ...+ x5

2)2 = ∑ (xi ± xj ± xk)4 + 6 ∑ (xi ± xj)4

60 (x12+x2

2+ ...+ x62)2 = ∑ (xi ± xj ± xk)4 + 2 ∑ (xi ± xj)

For seven squares, the second summation would have a zero coefficient. A general identity when the first summation involves choosing n objects from n squares is given by, 3*2n-1 (x1

2+x22+ ... + xn

2)2 = ∑ (xa ± xb ± ... ± xn)4 + 2n ∑ xn4

(Id.1) Going higher, for 3rd powers, 60 (x1

2+x22+x3

2+x42)3 = ∑ (xi ± xj ± xk)6 + 2 ∑ (xi ± xj)

6 + 36 ∑ xi6

120 (x12+x2

2+x32+x4

2)3 = ∑ (xi ± xj ± xk ± xm)6 + 6 ∑ (xi ± xj)6 +

∑ (2xi)6

60 (x12+x2

2+x32+x4

2+x52)3 = ∑ (xi ± xj ± xk)6 + 36 ∑ xi

For 4th powers, 5040 (x1

2+x22+x3

2+x42)4 = 6 ∑ (xi ± xj ± xk ± xm)8 + ∑ (2xi ± xj ±

xk)8 + 60 ∑ (xi ± xj)8 + 6 ∑ (2xi)

and so on for other k powers. (End update, 9/26.) Update 9/30/09: Gerry Martens gave the complete solution to, p (x1

2+x22+ ... + xn

2)2 = ∑ (xa ± xb ± ... ± xm)4 + q ∑ xn4

where we choose m out of n objects. First, define r1 = Pochhammer[2-n, m-2] / (m-2)!, then, p = (3/2)(-2)m r1

q = (1/2)(-2)m r1 (-n+3m-2)/(m-1) Mathematica's Pochhammer[a, n] gives the rising factorial (a)n = a(a+1)...(a+n-1). Note that by choosing some constant m, we can set q = 0 by letting n = 3m-2, which then yields the infinite family starting with the Lucas-Liouville identity at m = 2. If m = n, this is the same as (Id.1). For 6th powers, Renzo Benedetti solved, p (x1

2+x22+ ... + xn

2)3 = ∑ (xa ± xb ± ... ± xm)6 + q ∑ xn6

Define r2 = Binomial[3(m-2), m-2] and n = 3m-4, then, p = (5/2) 2m r2q = 2m r2 m/(m-1)

Note that, unlike for 4th powers, n depends on m and one also can't set q = 0 except for the trivial case m = 0. Starting with m = 2 to prevent division by zero, we get, 10 (x1

2+x22)3 = (x1+x2)6 + (x1-x2)6 + 8(x1

6 +x26)

then the identity involving 5 squares given above, and so on. (End update, 9/30.) IIb. Boutin’s Identity S ± (x1 ± x2 ±…± xk)k = k! 2k-1x1x2…xk where the exterior sign is the product of the interior signs. (In other words, the term is negative if there is an odd number of negative interior signs; positive if even.) For the first few k: (a+b)2 - (a-b)2 = 4ab (a+b+c)3 - (a-b+c)3 - (a+b-c)3 + (a-b-c)3 = 24abc (a+b+c+d)4 - (a-b+c+d)4 - (a+b-c+d)4 - (a+b+c-d)4 + (a-b-c+d)4 + (a-b+c-d)4 + (a+b-c-d)4 - (a-b-c-d)4 = 192abcd and so on for other kth powers. This then generalizes the difference of two squares to a sum and difference of 2k-1 kth powers. (Update, 11/18/09): I realized that Boutin's Identity is behind some unusual but elegant identities as the ratio between exponents k and k+2 is simply a rational multiple of (x1

2+x22+...+xk

2). For the first few k, 2[a2+b2][(a+b)2 - (a-b)2] = (a+b)4 - (a-b)4

10[a2+b2+c2][(a+b+c)3 - (a-b+c)3 - (a+b-c)3 + (a-b-c)3] = 3[(a+b+c)5 - (a-b+c)5 - (a+b-c)5 + (a-b-c)5] 5[a2+b2+c2+d2][(a+b+c-d)4 + (a+b-c+d)4 + (a-b+c+d)4 + (-a+b+c+d)4 - (2a)4 - (2b)4 - (2c)4 - (2d)4] =(a+b+c-d)6 + (a+b-c+d)6 + (a-b+c+d)6 + (-a+b+c+d)6 - (2a)6 - (2b)6 - (2c)6 - (2d)6

and so on, with the last example tweaked a bit and also discussed in Section 8.4. (End update.) IIIa. Lagrange’s Polynomial Identity ( ∑ ak

2) ( ∑ bk2) = ( ∑ akbk)2 + ∑ (akbj-ajbk)2

where the first three summations are from k = 1 to n, while the fourth is 1≤k<j≤n. For n = 2,3,4 these are, (a1

2+a22)(b1

2+b22) = (a1b1+a2b2)2 + (a1b2-a2b1)2

2+a22+a3

2)(b12+b2

2+b32) = (a1b1+a2b2+a3b3)2 + (a1b2-a2b1)2 +

(a1b3-a3b1)2 + (a2b3-a3b2)2

2+a22+a3

2+a42)(b1

2+b22+b3

2+b42) = (a1b1+a2b2+a3b3+a4b4)2 +

(a1b2-a2b1)2 + (a1b3-a3b1)2 + (a1b4-a4b1)2 + (a2b3-a3b2)2 + (a2b4-

a4b2)2 +(a3b4-a4b3)2

and so on. Note that for n = 4, while it is the sum of seven squares by Lagrange's Identity, it is just the sum of four squares by Euler's Four-

Square Identity. The case n = 3 shows that the product of two sums of three squares is likewise the sum of four squares. (Update, 12/21/09): IIIb. Lamé-Type Identities (x+y+z)3 - (x3+y3+z3) = 3(x+y)(x+z)(y+z)(x+y+z)5 - (x5+y5+z5) = 5(x+y)(x+z)(y+z)(x2+y2+z2+xy+xz+yz)(x+y+z)7 - (x7+y7+z7) = 7(x+y)(x+z)(y+z)((x2+y2+z2+xy+xz+yz)2+xyz(x+y+z)) The last was used by Gabriel Lamé in his proof for Fermat’s Last Theorem for k = 7. (I do not know a simple form for k = 11 and higher primes.) The form is simpler when z = 0. Define {a,b,c} = {xy, x+y, x2+xy+y2}. Then, (x+y)3 - x3 - y3 = 3ab(x+y)5 - x5 - y5 = 5abc(x+y)7 - x7 - y7 = 7abc2

(x+y)11 - x11 - y11 = 11abc(a2b2+c3)(x+y)13 - x13 - y13 = 13abc2(2a2b2+c3)(x+y)17 - x17 - y17 = 17abc(a4b4+5a2b2c3+c6)(x+y)19 - x19 - y19 = 19abc2(3a2b2+7a2b2c3+c6) and so on. Note the slight difference between primes of form 6n-1 and 6n+1. (End update.) (Update, 12/23/09): A.Verghote, Piezas Define Fk = xk+yk+zk. Then, F1F7 - F3F5 = (x+y)(x+z)(y+z) (2F5-F1F4+xyz(5F2-F1

F1F9 - F3F7 = (x+y)(x+z)(y+z) (2F7-F1F6+xyz(5F4-2F1F3+F22)/2)

F1F11 - F3F9 = (x+y)(x+z)(y+z) (2F9-F1F8+xyz(5F6-2F1F5+2F2F4-

F32)/2)

F1F13 - F3F11 = (x+y)(x+z)(y+z) (2F11-F1F10+xyz

(5F8-2F1F7+2F2F6-2F3F5+F42)/2)

and so on. When z = 0, the form is much simpler as the long expression on the RHS vanishes. Verghote provided the first identity, and I used Mathematica to find the higher ones. After finding the next two, the pattern was easily seen, though I have no proof it goes on for all F1Fn+2 - F3Fn with odd n > 3, though it will be odd if it stops. (End update) IV. Waring-like Problems 1. Summary The following identities are good for Waring-like problems since they involve rational, instead of integral, terms. Known results are, "Any non-zero rational N, in an infinite number of non-trivial ways, is the sum/difference: 1) of 2 rational 2nd powers (easy)2) of 3 rational 3rd powers (Ryley)3) of 4 rational 4th powers (Norrie)4) of 6 rational 5th powers (Choudhry)5) of 8 rational 6th powers (Piezas)6) of 8 rational 7th powers (Choudhry)7) of 12 rational 8th powers (Choudhry)." It would be nice if it would be proven in a general way that, "...any

rational N is non-trivially the sum/difference of k rational kth powers", but these are the best results so far. If you know of others, pls submit them. (An update on sums of increasing kth powers is given at the end of this section.)

2. Form: a2+b2+c2+d2 = N By Langrange's Four-Square Theorem, any positive integer N is the sum of four integral squares. If we extend this to rational numbers, then, Theorem: "Any positive rational N is the sum of four rational squares in an infinite number of ways." Proof: x1

2+x22+x3

2+x42 = (y1

2+y22+y3

2+y42)(z1

2+z22+z3

2+z42)2

where {x1, x2, x3, x4} = {uy1-vz1, uy2-vz2, uy3-vz3, uy4-

vz4} and {u,v} = {z12+z2

2+z32+z4

2, 2(y1z1+y2z2+y3z3+y4z4)} with initial solution yi to y1

2+y22+y3

2+y42 = N and four arbitrary

zi. Since there is always yi for any positive rational N, then by

dividing the eqn with the square factor of the RHS, one can find an infinite number of xi. This is just a particular case of a general identity involving n sums of squares discussed more here. Note: It is easily proven that the rational number p/q is the sum of four rational squares. Multiplying this by q/q, the denominator becomes a square and the problem is reduced to expressing pq as the sum of four squares. Also, four squares can't generally be reduced to three since one can't solve a2+b2+c2 = Ny2, where N =

8m+7. Thus, Lagrange's Four-Square Theorem applies whether the variables are integral or rational. 3a. Form: x3+y3+z3 = N Ryley's Theorem: “Any non-zero rational number N is the sum of three rational cubes in an infinite number of non-trivial ways.” (S. Ryley, 1825) Proof: (p3+qr)3 + (-p3+pr)3 + (-qr)3 = N (6Nvp2)3, where {p,q,r}= {N2+3v3, N2-3v3, 36N2v3}, for arbitrary v. Corollary: "Any positive rational N is the sum of three positive rational cubes in an infinite number of ways." Since v is an arbitrary variable, one can choose it such that all the terms are positive. For small integral N, the ranges roughly are: N = 1: v = {1.32 - 1.42}N = 2: v = {1.20 - 1.40}N = 3: v = {2.8 - 3.0}N = 4: v = {3.4 - 3.7} and so on. Q: Does anyone know how to calculate v and express it in terms of N such that all terms are positive? (Update, 11/11/09): Laurent Bartholdi from sci.math.research gave the answer. Setting x = N2 and y = v3 for ease of notation, one gets the constraints, x-3y < 0x2-30xy+9y2 < 0x3+45x2y-81xy2 +27y3 > 0

which define 6 lines in the {x,y} plane delimiting two regions. These regions are, a) 1 < 3v3/N2 < 1+Sec[2π/9], b) 1+Sec[4π/9] < 3v3/N2 < 5+2√6 where the exact values in terms of the secant function was pointed out by Warut Roonguthai from the NMBRTHRY mailing list. Or, v3 is approx. within, a) 0.333N2 < v3 < 0.768N2,b) 2.253N2 < v3 < 3.299N2,

For ex, for N = 163, then 20.7 < v < 27.3, and 39.2 < v < 44.4, and one can choose an infinite number of rational v within those two ranges. (End update.) (Update, 4/19/10): It is easily seen that Ryley's Identity is a 6th-deg polynomial in the constant N. A 3rd-deg polynomial, in fact, is possible. William Ellison cited one example from the book Cubic Forms by Yuri Manin as, (m3-36n9)3 + (-m3+35mn6+36n9)3 + (33m2n3+35mn6)3 = m(32m2n2

+34mn5+36n8)3 for arbitrary n. Note how this has neat coefficients that are only powers of 3. Robert Israel pointed out that a small tweak can reduce the powers as, (27m3-n9)3 + (-27m3+9mn6+n9)3 + (27m2n3+9mn6)3 = m(27m2n2

+9mn5+3n8)3

(End update.) 3b. Form: (a3+b3)(c3+d3) = N Theorem: “Any rational number N is the product of two sums of rational cubes." Proof: ((1+18N-27N2)3 + (-1+18N+27N2)3) ((1+3N)3 + (1-3N)3) = 63N (1+27N2)3 where the RHS is then divided by its cubic factor. Q: Who discovered this identity? Euler? Lebesgue?

4. Form: x4+y4-(z4+t4) = N Norrie's Theorem: “Any rational number N is the sum and difference of four rational fourth powers in an infinite number of ways.” (R. Norrie) ((2a+b)c3d)4 + (2ac4-bd4)4 - (2ac4+bd4)4 - ((2a-b)c3d)4 = a(2bcd)4

where b = c8-d8, for arbitrary {c,d}. 5. Form: x1

5+x25+x3

5+x45+x5

5+x65 = N

(Update, 11/17/09): Choudhry's Fifth Powers Theorem: “Any rational number N is the sum of six rational fifth powers in an infinite number of non-trivial ways.” Proof: (x+a)5 + m5(x+b)5 + m10(x+c)5 - (x+d)5 - m5(x+e)5 - m10(x+f)5 =

-40v35(v30-1)(v10+1)x + 2v30(-v30+12v20+8v10+1)(v30-1)(v10+1) where, {a,b,c,d,e,f,m} = {-v20, v5, -v15-v5-1, v20, -2v15-v5, -v15-v5+1, v3} Since the RHS is only linear in x, one can equate that side to any N, easily solve for x, and express the LHS in terms of N and the free variable v. (End proof.) Note: Method 1 below was when I didn't had access yet to Choudhry's paper, "Representation of every rational number as an algebraic sum of fifth powers of rational numbers", so I came up with my own. As this update is after the one for Seventh Powers, this incorporates some insights from that section. Method 2 is Choudhry's (yielding the identity above) after he gave me a copy of his paper. Method 1. Proof (Piezas): Expand the system, F(x): = (x+a)5+(x+b)5+(mx+mc)5-(x+d)5-(x+e)5-(mx+mf)5 and collecting powers of x, this resolves to, F(x): = 5P1x4 + 10P2x3 + 10P3x2 + 5P4x + P5 where Pk = ak+bk+m5ck-(dk+ek+m5fk), for k = {1,2,3,4,5}. The objective is to eliminate all terms of F(x) other than the linear term x, or to find {a,b,c,d,e,f,m} such that, ak+bk+m5ck = dk+ek+m5fk, for k = 1,2,3,5 (eq.1) which is just an Equal Sums of Like Powers problem. Once found,

since x is arbitrary, let x = 54P44 N. Thus,

F(x): = 55P4

5 N Dividing by the numerical factor, any rational N then is the sum of six rational fifth powers as claimed. The problem, of course, is if there are non-trivial values such that eq.1 can be solved. It turns out this system can be reduced to an elliptic curve, hence N in fact is the sum of six rational fifth powers in an infinite number of ways. To see this, let, {a,b,c,d,e,f,m5} = {p+q, r+s, 1+t, -p+q, -r+s, -1+t, n} where the symmetry of terms simplifies the system considerably. Then let {q, t, r} = {(-rs-nt)/p, (rs-h)/r, -(n+p)} with, s = (-5n2-4np+16p2+5n4+16n3p-4n2p2-60np3-60p4) / (60hn+120hp) and the multi-grade system (eq.1) is solved if {n,p} satisfies the curve, p(n+p)(n2+3np+3p2-1) = 3h2

where n must be the 5th power of a rational. One value I found is n = 25 = 32 with an initial rational point p = 50/9, from which others can then be calculated. This gives an initial set of 9-digit solns to the eqn, ak+bk+32ck = dk+ek+32fk, for k = 1,2,3,5 as {a,b,c,d,e,f} = {-225,478,523; -172,632,729; 112,165,576; -291,996,273; -277,027,261; 100,192,381} thus providing an algebraic identity proving Choudhry's Fifth Powers Theorem. (End proof.) The approach can be extended to higher odd powers, with a corresponding increase of the complexity of the system

to be solved, but Choudhry found an elegant soln for Seventh Powers discussed in the next section. Note: It is not really necessary to eliminate the constant term P5 of F(x) or, equivalently, to solve eq.1 for k = 5. However, since it can be done, then might as well do it anyway. (Update, 12/21/09): Method 2: Choudhry sent me a copy of his paper and it turns out he used a similar approach. After some modification, the method is essentially equivalent to using the form, F(x): = (x+a)5 + m5(x+b)5 + m10(x+c)5 - (x+d)5 - m5(x+e)5 - m10(x+f)5

Collecting powers of x, this is, F(x): = 5P1x4 + 10P2x3 + 10P3x2 + 5P4x + P5 where Pk = ak+m5bk+m10ck-(dk+m5ek+m10fk), for k = {1,2,3,4,5}. To eliminate the three highest terms of F(x), one is to solve, ak+m5bk+m10ck = dk+m5ek+m10fk, for k = 1,2,3 Choudhry found a remarkable identity to do this, given by, {a,b,c,d,e,f,m} = {-v20, v5, -v15-v5-1, v20, -2v15-v5, -v15-v5+1, v3} which reduces F(x) to the simple form, F(x): = -40v35(v30-1)(v10+1)x + 2v30(-v30+12v20+8v10+1)(v30-1)(v10+1) = N Given a constant N, one can easily solve for x. And since v is arbitrary, then N is the sum of six 5th powers in an infinite number of

ways. Also, if N = 0, this provides, after scaling, a concise 36-deg polynomial identity to six 5th powers with a zero sum. (End update.) Q: Can it be reduced to any N as five rational 5th powers in an infinite number of ways? And if the theorem can be reduced for any N, this would imply that x1

5+x25+x3

5+x45+x5

5 = 0 has an infinite number of non-trivial rational solns, since only 3 are known so far. 6a. Form: x1

6+x26+x3

6-(y16+y2

6+y36) = N

(Update, 11/30/09): To prove that any N is the sum and difference of six 6th powers, analogous to Norrie's Theorem for 4th powers, one way is to use a similar method as in 5th powers. Expanding, F(x): = (x+a)6+(mx+mb)6+(nx+nc)6-(x+d)6-(mx+me)6-(nx+nf)6

and collecting powers of x, this resolves to, F(x): = 6P1x5 + 15P2x4 + 20P3x3 + 15P4x2 + 6P5x + P6 where Pk = ak+m6bk+n6ck-(dk+m6ek+n6fk), for k = {1,2,3,4,5,6}. The objective is to eliminate all terms of F(x) other than the linear and constant term, or to find {a,b,c,d,e,f,m,n} such that, ak+m6bk+n6ck = dk+m6ek+n6fk, for k = 1,2,3,4 (Sys.1) Thus, what will be left is, F(x): = 6P5x + P6 = N

where x is then easily solved for any N. However, this author has not been able to find non-trivial {a,b,c,d,e,f} and {m,n}, if any, that solves Sys.1, though it can be shown if one of {m,n} = 1, then the system is trivial. Note: To simplify Sys.1, one can again do the substitution, {a,b,c,d,e,f} = {p+q, r+s, t+u, p-q, r-s, t-u}, though it still yields no non-trivial factor of small degree. However, another approach might prove (or disprove) the conjecture that any N is the sum and difference of six 6th powers. (End update.) 6b. Form: x1

6+x26+x3

6+x46-(y1

6+y26+y3

6+y46) = N

(Update: 12/11/09): Theorem: “Any rational number N is the sum and difference of eight rational 6th powers in an infinite number of ways.” (Piezas) Proof: Using a similar approach as Choudhry's for 7th powers, expand, F(x): = (x+a)6+(mx+mb)6+(x-c)6+(mx-md)6-(x-a)6-(mx-mb)6-(x+c)6-(mx+md)6

and collecting powers of x, this resolves to, F(x): = 12P1x5 + 40P3x3 + 12P5x where Pk = ak+m6bk-ck-m6dk, for k = {1,3,5}. To get rid of the x5

and x3 terms, find {a,b,c,d,m} such that, a+m6b = c+m6d (eq.1) a3+m6b3 = c3+m6d3 (eq.2)

Eq.2 can be completely solved using the modified Binet formula given by, {a,b} = {1-m6n(p-3q), m6n2-(p+3q)}{c,d} = {1-m6n(p+3q), m6n2-(p-3q)} where n = p2+3q2 for arbitrary m,p,q. To solve eq.1 as well, m = 1 must be avoided as the system becomes trivial but {p,q} must be chosen such that p2+3q2 = 1. This is given by, {p,q} = {(u2-3v2 )/(u2+3v2), 2uv/(u2+3v2)} for arbitrary u,v, and m, and avoiding m = 1. After the x5 and x3 terms are eliminated, since x is arbitrary, let x = (12P5)5 N. Thus, F(x): = (12P5)6 N Dividing by the numerical factor, N then is the sum/difference of eight rational 6th powers in an infinite number of ways as claimed. (End proof)

7. Form: x17+x2

7+x37+x4

7+ ... +x87 = N

(Update, 11/10/09): Choudhry's Seventh Powers Theorem: “Any non-zero rational number N is the sum of eight rational 7th powers in an infinite number of non-trivial ways.” Proof: Expand the equation,

F(x): = (x+a)7+(x-a)7+(mx+b)7+(mx-b)7-(x+c)7-(x-c)7-(mx+d)7-(mx-d)7

and collecting powers of x, this resolves to, F(x): = 42(a2+m5b2-c2-m5d2)x5 + 70(a4+m3b4-c4-m3d4)x3 + 14(a6+mb6-c6-md6)x To get rid of the x5 and x3 terms, find {a,b,c,d,m} such that, a2+m5b2 = c2+m5d2

a4+m3b4 = c4+m3d4

and only the linear term is left. Since x is arbitrary, let x = 146

(a6+mb6-c6-md6)6 N. Thus, F(x): = 147(a6+mb6-c6-md6)7 N Dividing by the numerical factor, N then is the sum of eight rational seventh powers as claimed. (End proof) Choudhry chose m = 2 and found {a,b,c,d} with 33-digits! I'm assuming there might be smaller solns, so I posted the problem in sci.math.symbolic to see if someone can find one, for any m > 1. Note 1: It can be shown that the general system, a2+mb2 = c2+md2 (eq.1)a4+nb4 = c4+nd4 (eq.2) can be reduced to solving an elliptic curve. Euler’s complete soln to eq.1 (see Form 10 of Sums of Two Squares) is given by, (pr+mqs)2 + m(ps-qr)2 = (pr-mqs)2 + m(ps+qr)2

If we apply these values for {a,b,c,d} on eq.2 as well, it reduces to the eqn, (mp2-nq2)r2 = (np2-m3q2)s2

which have rational solns if (mp2-nq2)(np2-m3q2) = y2. For the special case relevant to 7th powers when {m,n} = {h5, h3}, the curve is, (h2p2-q2)(p2-h12q2) = y2 (eq.3) For h = 2, Choudhry found the 16-and-17-digit soln {p,q} = {5911167604843137, 12317476831120126} which gives his 33-digit {a,b,c,d}. It might be interesting to know if eq.3 has small solns for some other rational h > 1. Note 2: Update (11/16/09) D. Rusin gave a very thorough analysis here and, as a first step, simplified it to a problem of Equal Sums of Like Powers. By letting {b,d} = {Bm, Dm}, one is to solve, ak+m7Bk = ck+m7Dk, for k = 2,4 (eq.4) Then, by a series of clever transformations, he reduced this system to the rather simple elliptic curve, V2 = U(U+1)(U+q2) where q = (1-m7) / (1+m7) and using Maple's APECS and Magma, eventually gave an explicit soln {U,V}, apparently the smallest, and the same as what Choudhry found. From this initial point, an infinite more can be calculated proving that any rational N is the sum of eight rational 7th powers in an infinite number of ways. For other m such that eq.4 has non-trivial solns, because of its symmetry, it suffices to look within the interval m = {0 to 1}. In addition to m = 2 (or

equivalently m = 1/2), Rusin found possible candidates m = {3/4, 1/5, 2/5, 3/5, 4/5, 1/6, 2/7} though solns, if any, must be large. (End update.) Note 3: When N = 0, the theorem gives a trivial result, hence does not prove there are an infinite number of non-trivial solns to eight 7th powers equal to zero, though many have been found already.

Q: Can this be reduced to seven rational seventh powers?

(Update, 1/30/10): Choudhry

8. Form: x18+x2

8+...+x68 – (y1

8+y28+...+y6

8) = N Choudhry's Eighth Powers Theorem: “Any non-zero rational number N is the sum and difference of twelve rational 8th powers in an infinite number of non-trivial ways." Proof: Choudhry’s method is equivalent to solving the equation, U1 – U2 = M U1:= (ax+y)8 + (p1

8+p38)(x+by)8 + (p2

8+p48)(cx-y)8 + (x-dy)8

U2:= (ax-y)8 + (p18+p3

8)(x-by)8 + (p28+p4

8)(cx+y)8 + (x+dy)8

for variables {a,b,c,d} and {p1, p2, p3, p4} such that, when expanded and collecting {x,y}, all terms vanish except M which is only linear in x. He found a wonderfully simple and symmetric soln given by, {p1, p2, p3, p4} = {t, t3, t5, t7}

{a,b,c,d} = {t12, t12, t4, t20} such that, M = 24 t8(t4-t20-t52+t84+t116-t132) xy7

for arbitrary {t,x,y}. Note how the exponents match up as 1+7 = 3+5 = 8, 12+12 = 4+20 = 24, as well as 4+132 = 20+116 = 52+84 = 136. (Q. What is the reason for the simplicity of the soln, and can this be generalized?) Doing the substitution {x,y} = {24N, t4-t20-t52+t84+t116-t132}, we then get the form, M = (2ty)8 N Dividing by the factor, then any non-zero N is the sum and difference of twelve rational 8th powers in an infinite number of ways as claimed. (Using another approach, it may be possible to reduce this to ten or even eight rational 8th powers.) Of course, if N is itself an 8th power, then this provides a polynomial identity for the 13-term, x1

8+ x28+… x6

8 = y18+ y2

8+… y78

Note: My thanks to Choudhy for providing a copy of his paper "On Sums of Eighth Powers", Journal of Number Theory, Vol 39, Sept 1991. (End update.) 9. Form: x1

11+x211+x3

11+x411+ ... +x12

11 = N Using Choudhry's and Rusin's approach for 7th powers, if there is a non-trivial soln {xi, yi} to the system S11,

ax1k+bx2

k+cx3k = ay1

k+by2k+cy3

k, for k = 2,4,6,8 for some rational {a,b,c} = {p11, q11, r11}, then any rational N is the sum of 12 rational 11th powers. Multi-grade systems valid for k = {2,4,6,8} or even k = {2,4,6,8,10} are known (see the section on Tenth Powers) so it might be possible, with certain assumptions, that S11 can be reduced to an elliptic curve as well. As a numerical experiment, one may test this by using small {p,q,r} < 11, like Choudhry's m = 2, excluding the case p = q = r, and testing values {xi, yi} below some reasonable bound. In general, the conjecture would be that any rational N is the sum of at most 4m rational (4m-1)th powers (with Ryley's Theorem reducing the cubic case to just three cubes), with the problem reducible to a system of equal sums of like powers. (Update, 4/20/10): Enrico Jabara gave a polynomial soln to x1

3+x24+x3

5+x47+x5

8 = z8 as, (288747n4)3 + (292749n)4 + (232717n4)5 + (2874n4)7 + (n4-252728)8 = (n4+252728)8 More generally, as a variant of the easier Waring’s Problem, he proved that any integer N is the sum of consecutive unlike powers, N = ± x1

2 ± x23 ± x3

4 ± x45

N = ± x1

3 ± x24 ± x3

5 ± x46 ± x5

7 ± x68 ± x7

9 ± x810

where the xi are also in the integers. However, for the next in the series that starts with 4th powers,

N = ± x14 ± x2

5 ± … ± (xk-3)k

it is yet unknown what is the ending power k. Source: Representations of numbers as sums and differences of unlike powers, (by Enrico Jabara, Università di Ca’ Foscari).

(Update, 8/7/09): V. Equal Sums of Like Powers The results in this section are by the author. See also the separate section on Equal Sums of Like Powers and those on higher powers. I. Even Powers (Part 1) Theorem 1: Define Ek:= ak+bk+ck-(dk+ek+fk). Let {u,v} =

{a2+b2+c2, a4+b4+c4}. If Ek = 0 for k = 2,4 then, a6+b6+c6-d6-e6-f6 = ± 3(a2-d2)(b2-d2)(c2-d2)3(a8+b8+c8-d8-e8-f8) = 4(a6+b6+c6-d6-e6-f6) (a2+b2+c2)6(a10+b10+c10-d10-e10-f10) = 5(a6+b6+c6-d6-e6-f6) ((a2+b2+c2)2+a4+b4+c4) Note: The d in the RHS of the first eqn can be either {d,e,f}. And the ± sign depends on what side of the eq is chosen to be {a,b,c} or {d,e,f}. Small soln:[6, 23, 25]k = [10, 19, 27]k, for k = 2,4. Corollary: These relations can be woven together. Define n as, a4+b4+c4 = n(a2+b2+c2)2 then,

32(a6+b6+c6-d6-e6-f6)(a10+b10+c10-d10-e10-f10) = 15(n+1)(a8+b8+c8-d8-e8-f8)2 Proof: We can concisely express the last two of the eqns above as, 3 E8 = 4 E6(u) (eq.1)

6 E10 = 5 E6(u2+v) (eq.2) Square eq.1, multiply eq.2 by E6, and then eliminate E6

2 between eq.1 and eq.2 to get, 32 E6E10 = 15 E8

2 (1+v/u2), or,

32 E6E10 = 15 E82 (1+n)

as stated. (End of proof). For the special case when a+b = ± c, since it is true that, a4+b4+(a+b)4 = (1/2)(a2+b2+(a+b)2)2 then n = 1/2 and the above becomes, 64(a6+b6+c6-d6-e6-f6)(a10+b10+c10-d10-e10-f10) = 45(a8+b8+c8-d8-e8-f8)2 which is a concise version of the Ramanujan 6-10-8 Identity. The general case is in fact the first in an infinite family to be given in the following theorems. Theorem 2: Define Fk:= ak+bk+ck+dk-(ek+fk+gk+hk), and qk:= ak+bk

+ck+dk. Let {u,v,w} = {q2, q4, q6}. If Fk = 0 for k = 2,4,6 then,

F8 = ± 4(a2-e2)(b2-e2)(c2-e2)(d2-e2)4 F10 = 5 F8 (u)

8 F12 = 6 F8 (u2+v)

24 F14 = 7 F8 (u3+3uv+2w) Corollary: Define n as a4+b4+c4+d4 = n(a2+b2+c2+d2)2 then, 25(a8+b8+c8+d8-e8-f8-g8-h8) (a12+b12+c12+d12-e12-f12-g12-h12) = 12(n+1)(a10+b10+c10+d10-e10-f10-g10-h10)2 or, equivalently, 25 F8F12 = 12(n+1)(F10)2. Small soln: [2, 16, 21,

25]k = [5, 14, 23, 24]k, k = 2,4,6. Theorem 3: Define Gk:= ak+bk+ck+dk+ek-(fk+gk+hk+ik+jk) and rk:=

ak+bk+ck+dk+ek. Let {u,v,w,x} = {r2, r4, r6, r8}. If Gk = 0 for k = 2,4,6,8 then, G10 = ± 5(a2-f2)(b2-f2)(c2-f2)(d2-f2)(e2-f2)5 G12 = 6 G10 (u)

10 G14 = 7 G10 (u2+v)

30 G16 = 8 G10 (u3+3uv+2w)

120 G18 = 9 G10 (u4+6u2v+3v2+8uw+6x) Corollary: "Define n as a4+b4+c4+d4+e4 = n(a2+b2+c2+d2+e2)2, then 72 G10G14 = 35(n+1)(G12)2." Small soln: [71, 131, 180, 307, 308]k =

[99, 100, 188, 301, 313]k, k = 2,4,6,8.

Theorem 4: Define Hk:= ak+bk+ck+dk+ek+fk-(gk+hk+ik+jk+lk+mk)

and sk:= ak+bk+ck+dk+ek+fk. Let {u,v,w,x,y} = {s2, s4, s6, s8, s10}. If Hk = 0 for k = 2,4,6,8,10, then, H12 = ± 6(a2-g2)(b2-g2)(c2-g2)(d2-g2)(e2-g2)(f2-g2)6 H14 = 7 H12 (u)

12 H16 = 8 H12 (u2+v)

36 H18 = 9 H12 (u3+3uv+2w)

144 H20 = 10 H12 (u4+6u2v+3v2+8uw+6x)720 H22 = 11 H12 (Poly1), where Poly1 is a 5th deg polynomial in terms of u,v,w,x,y. Corollary: "Define n as a4+b4+c4+d4+e4+f4 = n(a2+b2+c2+d2+e2+f2)2, then 49 H12H16 = 24(n+1)(H14)2." Small soln: [22, 61, 86, 127,

140, 151]k = [35, 47, 94, 121, 146, 148]k, k = 2,4,6,8,10. And so on for other k = 2,4,..2n. While an infinite number of solns are known for Theorems 1 to 4, none are known for higher systems. (Q: Anyone with the programming skills and computing power to find a few?) Note 1: The coefficient of the Hi in Theorem 4 as {6, 12, 36, 144, 720} divided by 6 is equal to {1, 2, 6, 24, 120}, or the sequence of factorial n!, so it is easy to extrapolate that the next system involves the sequence 7(n!). Note 2: As an overview, we have the sequence of corollary relations,

32 E6E10 = 15(n+1)(E8)2

50 F8F12 = 24(n+1)(F10)2

72 G10G14 = 35(n+1)(G12)2

98 H12H16 = 48(n+1)(H14)2

This has the general form, 2m2 X2(m-1) X2(m+1) = (m2-1)(n+1)(X2m)2

starting with m = 4,5,6… so one can see the next in the series. However, by taking square roots of the variables, these theorems are also valid for systems with k = 1,2,3,…n, or those satisfying the Prouhet-Tarry-Escott problem. See also the section on Equal Sums of Like Powers. II. Even Powers (Part 2) Theorem 1: "Define, {a,b,c,d} = {m+p, m-p, m+q, m-q} If p2+q2 = m2, then 9(a2+b2-c2-d2)(a4+b4-c4-d4) = 7(a3+b3-c3-d3)2." Theorem 2: "Define, {a,b,c,d} = {m+p, m-p, m+q, m-q}{e,f,g,h} = {m+r, m-r, m+s, m-s} If,

p2+q2 = r2+s2 (eq.1)p2+q2 = 4m2 (eq.2) then 25(a4+b4+c4+d4-e4-f4-g4-h4)(a6+b6+c6+d6-e6-f6-g6-h6) = 21(a5+b5+c5+d5-e5-f5-g5-h5)2." Also, for any m, then ak+bk+ck+dk = ek+fk+gk+hk, for k = 1,2,3. Hence it is a symmetric ideal solution (see Theorem 1 and 2) of the Prouhet-Tarry-Escott problem. Note that eq.1 can be completely solved as {p,q,r,s} = {ux+vy, vx-uy, ux-vy, vx+uy} and substituting this into eq.2 yields, (u2+v2)(x2+y2) = 4m2

which can be solved by using the formula for Pythagorean triples separately on {u,v} and {x,y}. Theorem 3: "Define, {a,b,c,d,e,f} = {m+p, m-p, m+q, m-q, m+r, m-r}{g,h,i,j,k,l} = {m+s, m-s, m+t, m-t, m+u, m-u} If, pk+qk+rk = sk+tk+uk, for k = 2,4 (eq.1)p2+q2+r2 = vm2 (eq.2) then, 49(a6+b6+c6+d6+e6+f6-g6-h6-i6-j6-k6-l6)(a8+b8+c8+d8+e8+f8-g8-h8-i8-j8-k8-l8) = (4/3)(v+21)(a7+b7+c7+d7+e7+f7-g7-h7-i7-j7-k7-l7)2." And, for any m, an+bn+cn+dn+en+fn = gn+hn+in+jn+kn+ln, for n = 1,2,3,4,5. When v = 1, only two primitive solns are known with terms

< 16,000, {p,q,r,s,t,u} = {2, 289, 610, 170, 223, 614} (by "Martin"){p,q,r,s,t,u} = {1575, 11522, 11850, 4647, 8890, 13230} (by James Waldby) after this author posted the problem on the newsgroup sci.math.symbolic. However, when v = 2, there are small solns starting with {p,q,r,s,t,u} = {3,5,8,0,7,7} and it can be shown there are an infinite number. Let {p,q,r,s,t,u} = {z1+z2, -z1+z2, 2z2, z3+z4, -z3+z4, 2z4} and eq.1 and eq.2 become, z1

2+3z22 = z3

2+3z42 (eq.1)

2(z12+3z2

2) = 2m2 (eq.2) Eq. 1 can be completely solved as {z1, z2, z3, z4} = {ux+3vy, vx-uy, ux-3vy, vx+uy} and eq.2 becomes, (u2+3v2)(x2+3y2) = m2

and it is easy to solve this using a variant of the formula for Pythagorean triples separately on {u,v} and {x,y}. The procedure can be extrapolated to higher systems. III. Odd Powers The difficulty with the system, a1

k + a2k + … + am

k = b1k + b2

k + … + bmk, k = 1,3,5,…2n+1 (eq.

1) valid only for odd powers is that terms can be transposed or moved

around to the form, c1

k + c2k + … + cp

k = 0, k = 1,3,5,…2n+1 and it is more challenging to find relations beyond k = 2n+1. The trick then is to “fix the variables in place” by finding a balanced partition such that, a) a1+a2+… +am = b1+b2+… +bm = 0, orb) eq.1 is valid for k = 2 as well and hence terms can no longer be transposed arbitrarily. (For the system k = 1,3,5, the two conditions are equivalent.) a) Condition: a1+a2+… +am = b1+b2+… +bm = 0. Third powers: “If ak+bk+ck = dk+ek+fk, for k = 1,3, where a+b+c = d+e+f = 0, then 9abc(a6+b6+c6-d6-e6-f6) = 2(a9+b9+c9-d9-e9-f9).” Fifth powers: “If ak+bk+ck+dk = ek+fk+gk+hk, for k = 1,3,5, where a+b+c+d = e+f+g+h = 0, (or equivalently, k = 1,2,3,5), then, 7(a4+b4+c4+d4-e4-f4-g4-h4)(a9+b9+c9+d9-e9-f9-g9-h9) = 12(a6+b6+c6+d6-e6-f6-g6-h6)(a7+b7+c7+d7-e7-f7-g7-h7).” More succinctly, "Define Fk = ak+bk+ck+dk -(ek+fk+gk+hk). If Fk = 0 for k = {1,2,3,5}, then 7F4F9 = 12F6F7." Note: As these two systems have complete solns, given here (Form 13) and here (Form 5.3b), respectively, one can substitute those into the given identities and see that they are true. For higher than fifth powers, there are only isolated results with this condition so it harder to come

up with general statements. b) Condition: Valid for k = 2 as well. As Bastien’s Theorem forbids a non-trivial result for the system k = 1,2,3 with only three terms on either side of the eqn, this condition yields a “general” result valid only for those with higher than third powers, starting with, Define Fk: = x1

k+x2k+x3

k+x4k – (y1

k+y2k+y3

k+y4k). If Fk = 0 for k

= 1,2,3,5, then (9F7)(x12+x2

2+x32+x4

2) = 7F9. Define Fk: = x1

k+x2k+x3

k+x4k+x5

k – (y1k+y2

k+y3k+y4

k+y5k). If Fk

= 0 for k = 1,2,3,5,7, then (11F9)(x12+x2

2+x32+x4

2+x52) = 9F11.

Define Fk: = x1

k+x2k+x3

k+x4k+x5

k+x6k – (y1

k+y2k+y3

k+y4k+y5

+y6k). If Fk = 0 for k = 1,2,3,5,7,9, then (13F11)

(x12+x2

2+x32+x4

2+x52+x6

2) = 11F13. Note: All known solns satisfy this and the form is easily extrapolated for higher systems. However, it is proven only for fifth powers and a general proof is needed. (Update, 12/21/09): Jarek Wroblewski observed that the third conjecture can be generalized as (13F11)(x1

2+x22+...+x6

2+y12+y2

+y62)/2 = 11F13, and similarly for the others, even if Fk is not zero for

k = 2. Of course, if it is for k = 2, then it reduces to the form given above.

Part 2. Sums of Squares I. Sums of two squares

1. x2+y2 = zk a. (a+1/a)2 + (b+1/b)2 = c2

b. a2(b2-1)2 + b2(a2-1)2 = c2

c. (a2+b2)2 + (c2+d2)2 = e2

d. a2 + (a+n)2 = b2

e. a2 + b2 = (b+n)2

f. (2ab)2 + (2cd)2 = (a2+b2-c2-d2)2

g. (ac+bd)2 + (ad-bc)2 = (a2+b2)2

2. x2+ny2 = zk; 2b. ax2+by2 = cz2

3. ad-bc = ±14. x2+y2 = z2+15. x2+y2 = z2-16. x2+y2 = z2+nt2

7. x2+y2 = z2+ntk

8. x2+y2 = mz2+nt2

9. c1(x2+ny2) = c2(z2+nt2)

10. mx2+ny2 = mz2+nt2

11. x2+y2 = zk+tk

1. Form: x2+y2 = zk A. Introduction While integers a,b,c that satisfy a2+b2 = c2 are called Pythagorean triples, the ancient Babylonians already knew there were triangles whose sides satisfy that relationship more than a thousand years earlier. The famous tablet Plimpton 322 (pre-1500 BC, now kept in Columbia University) contains pairs of numbers in sexigesimal which

can be seen as part of a Pythagorean triple. The largest pair is (18541, 12709) and a quick calculation shows that the difference of their squares is also a square, 185412-127092 = 135002. Quick for us using a calculator but the size of this example shows the Babylonians must have known of a method to generate solutions other than randomly scribbling figures in the sand. The smallest primitive solns (where a,b,c have no common factor) are: {3, 4, 5}, {5, 12, 13}, {7, 24, 25}, {8, 15, 17}, etc. Theorem: “For primitive triples a2+b2 = c2, exactly, 1) one of a,b is odd, and c is always odd.2) one of a,b is divisible by 3 and/or 4. 3) one of a,b,c is divisible by 5.” which shows the importance of the triple {3,4,5}. Note: It is possible the divisibility by 3,4,5 is contained in just one term, such as in the first term of, (60v)2 + (900v2-1)2 = (900v2+1)2

Theorem: “One soln can lead to another, or if a2+b2 = c2, then (a+2b+2c)2 + (2a+b+2c)2 = (2a+2b+3c)2 is another triple.” Starting with {a, b, c} = {±3, ±4, 5}, one can iteratively generate all primitive Pythagorean triples. (Barning, 1963; Roberts, 1977) Theorem: “All odd numbers > 1, as well as multiples of 4, appear in a primitive Pythagorean triple.” Proof: Use the identities: (2m+1)2 + n2 = (n+1)2, where n = 2(m2+m)(4m)2 + n2 = (n+2)2, where n = 4m2-1

Theorem: The complete non-trivial soln to x12+x2

2 = y12, with a

scaling factor t, is given by, ((a2-b2)t)2 + (2abt)2 = ((a2+b2)t)2 Proof: For any soln xi where x1 ≠ y1, one can always find rational

{a,b,t} using the formulas: {a,b,t} = {x1+y1, x2, (-x1+y1)/(2x22)}. In

general, one can completely solve x12+x2

2+...+ xn2 = y1

2. See Sums of Three Squares. Update (7/2/09): Adam Bailey gave a variant of the complete soln to a2+b2 = c2 as, (p+2)2r2 + (2/p+2)2r2 = (p+2/p+2)2r2 where r is a scaling factor, since for any {a,b,c} > 0, one can always find rational {p,r} using the formulas {p,r} = {-(a+b-c)/(a-c), 2/(a+b-c). Thus, this shows that the complete soln to a2+b2 = c2, as well as Binet's complete one for a3+b3+c3 = d3, is expressible (with a scaling factor) in just n-1 parameters, where n is the degree of the equation. (End note). Pythagorean triples were already known by the classical Greek mathematicians. But the Babylonians and Greeks didn’t know about complex numbers which makes it easy for us to identically solve the more general equation x2+y2 = zk for any positive integer k, though for k > 2, it is not necessarily the complete soln. For example, when k = 3, the method yields, with a rational scaling factor u, (a3-3ab2)2u6 + (3a2b-b3)2u6 = (a2+b2)3u6

but does not always cover the alternative soln with scaling factor v,

(a3+ab2)2v6 + (a2b+b3)2v6 = (a2+b2)3v6

This will be discussed more in Sums of Two Squares, as well as the similar equation a2+b2 = cn+dn. More generally for other k, given the expansion of the complex number (a+bi)k = A+Bi where {A,B} are polynomials in terms of {a,b}, and i = Ö-1, then, A2+B2 = (a2+b2)k Similarly, for the case of four squares equal to an nth power, one can use quaternions. Given the expansion of the quaternion (a+bi+cj+dk)n = A+Bi+Cj+Dk, then, A2+B2+C2+D2 = (a2+b2+c2+d2)n though there is an easier method to be discussed later. One can also solve the equation A2+B2 = (a2+b2)k by factoring over Ö-1 to get, (A+Bi)(A-Bi) = (a+bi)k(a-bi)k

Equating factors yields a system of two equations in two unknowns A,B which, being only linear, can then easily be solved for. This yields, A = (pk+qk)/2, B = -(pk-qk)i/2, where {p,q} = {a+bi, a-bi} and B, after simplification, is a real value. This technique of equating factors will prove useful for similar equations. Update (6/17/09): Pythagorean triples {a,b,c} = {x2-1, 2x, x2+1} can be used in the form,

ak+bk+ck = 2(x8+14x4+1)k/4, for k = 4,8 The expression x8+14x4+1 is no ordinary polynomial. Equated to zero, it is one of the octahedral equations, involved in the projective geometry of the octahedron. It also appears in a formula for the beautiful j-function j(τ) in terms of the Dedekind eta function η(τ) as, j(τ) = 16(x8+14x4+1)3/(x5-x)4

where x = 2η2(τ) η4(4τ) / η6(2τ) If you have Mathematica, you can easily verify this. For ex, choosing the particular value d = (1+Sqrt[-163])/2 which conveniently yields an integer, j(d) = N[16(x8+14x4+1)3/(x5-x)4 /. x → 2*DedekindEta[d]2 DedekindEta[4d]4 / DedekindEta[2d]6 /. d → (1+Sqrt[-163])/2, 100] = -6403203

Furthermore, using -Numerator+1728Denominator of the j-function formula above (note that 1728 = 123), one gets another octahedral equation. Expressed in terms of the same Pythagorean triples {a,b,c}, this is, (a4+b4+c4)3 - 54(abc)4 = 8(x12-33x8-33x4+1)2

Why Pythagorean triples, when expressed in a constrained manner, spell out the two octahedral equations, I do not know. (End note.) Update (10/11/09): Pythagorean triples also appear in the equation x2 = -1 where x, instead of the imaginary unit, is to be a quaternion. To recall, given the expansion of (a+bi+cj+dk)n = A+Bi+Cj+Dk, then, A2+B2+C2+D2 = (a2+b2+c2+d2)n

In a Mathematica add-on package, this object is given as Quaternion[a,b,c,d] and one can use NonCommutativeMultiply to multiply them, and for the case n = 2, we get, Quaternion[a,b,c,d] ** Quaternion[a,b,c,d] = Quaternion[a2-b2-c2-d2, 2ab, 2ac, 2ad] (eq.1) One can see two immediate consequences. First, as expected, (a2-b2-c2-d2)2 + (2ab)2 + (2ac)2 + (2ad)2 = (a2+b2+c2+d2)2. Second, while the Fundamental Theorem of Algebra states that an nth degree univariate equation with complex coefficients has n complex roots, this does not apply when the roots are quaternions. For example, the nth roots of plus/minus unity or, xn = ±1, have n complex solns. But if x is to be a quaternion, then there can be an infinite number of them. Let n = 2 and take the negative case: x2 = -1. Thus, we wish (eq.1) to become, Quaternion[a2-b2-c2-d2, 2ab, 2ac, 2ad] = Quaternion[-1,0,0,0]. Since {a,b,c,d} are to be real, then a = 0, and {b,c,d} should satisfy, b2+c2+d2 = 1 which have an infinite number of rational solns, a subset of which is d = 0 and {b,c} as rationalized Pythagorean triples. (If c = d = 0, then the only soln is b = ±1 in which case the quaternion reduces to the imaginary unit.) This will also be discussed in the section on Sums of Three Squares. Of course, the most general setting for solving x2 = -1 would be x as the octonions, where the coefficients of the non-real part would have to satisfy the sum of seven squares, y1

2+y22+y3

2+y42+y5

2+y62+y7

(End update.) B. Special Forms 1. (a+1/a)2 + (b+1/b)2 = c2

2. a2(b2-1)2 + b2(a2-1)2 = c2

3. (a2+b2)2 + (c2+d2)2 = e2

4. a2 + (a+n)2 = b2

5. a2 + b2 = (b+n)2

6. (2ab)2 + (2cd)2 = (a2+b2-c2-d2)2

7. (ac+bd)2 + (ad-bc)2 = (a2+b2)2 If you know of any other form and its soln, pls submit them. Form 1: (a+1/a)2 + (b+1/b)2 = c2

Euler (x+1/x)2 + (y+1/y)2 = z2

{x,y} = {4p/(p2-1), (3p2+1)/(p3+3p)} Form 2: a2(b2-1)2 + b2(a2-1)2 = c2

Euler x2(y2-1)2 + y2(x2-1)2 = z2

{x,y} = {4p/(p2+1), (3p2-1)/(p3-3p)}

with a very similar soln to the previous and which is useful in making {x2+y2, x2+z2, y2+z2} all squares, or the Euler Brick Problem, to be discussed more in the section on Simultaneous Polynomials. Form 3: (a2+b2)2 + (c2+d2)2 = e2 Euler (a2+b2)2 + (b2+d2)2 = (b2+8c2+d2)2, where d = (b2+3c2)/(2c), and a2-b2 = 10c2

The conditional eqn can be solved as {a,b,c} = {u2+10v2, u2-10v2, 2uv}. This can be generalized as, let {p,q} = {2n2, n2-1}, then, (a2+b2)2 + (b2+d2)2 = (b2+pc2+d2)2, where d = (b2+qc2)/(2c), and a2-(n-1)b2 = n(n2+1)c2 for any n, with Euler’s as the case n = 2. Another approach is to use the parametrization for Pythagorean triples in the form, (a2+b2)2 + (c2+d2)2 = (p2+q2)2 where {a,b,p,q}= {u2-v2-w2, 2uv, u2+v2+w2, 2uw}, and conditional eqn, c2+d2 = 4uw(u2+v2+w2) . One soln to the conditional eqn is to let {v,w} = {e/(2u), u}, and it becomes c2+d2 = e2+8u4 which can easily solved as discussed in the Section on x2+y2 = z2+ntk. Q: Any others for (a2+b2)2 + (c2+d2)2 = e2? Update (6/20/09): After this author posted the question in sci.math, Dave Rusin gave a very elegant five-parameter soln. He

notes that one way to approach the problem is to treat the legs (each of which are sums of two squares) as the product of two sums of two squares since we know that, (x1x3+x2x4)2 + (x1x4-x2x3)2 = (x1

2+x22)(x3

2+x42)

Thus we get, ((x1

2+x22)(x3

2+x42))2 + (2(y1

2+y22)(y3

2+y42))2 = ((y1

2+y22)2 +

(y32+y4

and, after some clever manipulation of some eqns describing a 5-dimensional projective variety over Q, gave the soln as, {x1, x2, x3, x4} = {2d, 1+z, 1-z, 2(a2+b2-c2)}

{y1, y2, y3, y4} = {4ac, 4bc, 1-z, 2(a2+b2+c2)} where z = (a2+b2)2+6(a2+b2)c2+c4-d2 for four free parameters {a,b,c,d}. Note: Incidentally, the related form, ((u1

2-u22)(u3

2-u42))2 + (2(v1

2-v22)(v3

2-v42))2 = ((v1

2-v22)2 + (v3

v42)2)2

has, {u1, u2, u3, u4} = {2d, 1-z2, 1+z2, 2(a2-b2-c2)}

{v1, v2, v3, v4} = {4ac, 4bc, 1+z2, 2(a2-b2+c2)} where z2 = (a2-b2)2+6(a2-b2)c2+c4-d2 with parameters {a,b,c,d}. (End update.)

Form 4: a2 + (a+n)2 = b2

Given a triangle with legs as consecutive integers {p, p+1}, then a second can be found, Theorem: “If p2 + (p+1)2 = r2, then q2 + (q+1)2 = (p+q+r+1)2, where q = 3p+2r+1.” (Fermat) (Update, 6/24/09): Maurice Mischler pointed out that all integer solns to the triple (p, p+1, r) is given by, ((x-1)/2)2 + ((x+1)/2)2 = y2

where {x,y} satisfy the Pell equation x2-2y2 = -1. Since x is always odd, then terms are integers. (End note). (Update, 12/25/09): Given the co-prime triple {a,b,c}, the difference between the hypotenuse and a leg, or c-a = n, can assume any non-zero value n. In contrast, the difference between the two legs, or b-a = n, is only IF the prime factors of n is of form 8m ±1 which is sequence n = {1, 7, 17, 23, 31,...} in the OEIS. These triples can be completely parameterized by the form, ((x-n)/2)2 + ((x+n)/2)2 = y2

where {x,y} are co-prime solns of the Pell-like equation x2-2y2 = -n2. Thus, x2-2y2 = -72 has {x,y} = {1,5}, etc, and it is easily shown that if there is one co-prime soln, then there is an infinite number of them by generalizing Fermat's result as, Theorem: “If p2 + (p+n)2 = r2, then q2 + (q+n)2 = (p+q+r+n)2, where q = 3p+2r+n.” (End update.)

(Update, 12/30/09): Given the first four primitive Pythagorean triples, {3,4,5}, {5,12,13}, {7,24,25}, {8,15,17}, it can be noticed that the sum of their legs, {3+4, 5+12, 7+24, 8+15} = {7, 17, 31, 23} are primes of the form 8n±1. The relevant theorem is this: Theorem: “For every prime p = 8n±1, there is one and only one primitive and positive Pythagorean triple {a,b,c} such that a+b = p.” Thus, no matter how many two-part positive partitions {a,b} of p there may be, there is one and only one such that a2+b2 = c2 for some integer c. Proof: (Andrew Usher) Using the complete parameterization of primitive Pythagorean triples as {m2-n2, 2mn, m2+n2} where m > n > 0, the sum p of its two legs can be expressed as the Pell-like equation, (m2-n2) + (2mn) = (m+n)2 -2n2 = x2-2y2 = p (eq.1) If solvable, eq.1 has an infinite number of integer solns, but it can be shown that, a) For p a prime, eq.1 has only one fundamental soln {x0, y0} from which all others can be derived via a recursion.b) The recursion can show that {x0, y0} will be the only one with m > n > 0. If p is not a prime, then (a) will not be true. For example p = 7(23) = 161 has two, leading to two primitive triples {17, 144, 145} and {44, 117, 125} with the sum of their legs as 17+144 = 44+117 = 7(23) = 161. (End update) Form 5: a2 + b2 = (b+n)2

When a leg and the hypotenuse differ by a constant n, a general formula is, (nx)2 + y2 = (y+n)2 where 2y = n(x2-1), for arbitrary {n,x}. (Of course, for y to be integral, then either n should be even, or x is odd.) (Update, 12/24/09) Form 6: (2ab)2 + (2cd)2 = (a2+b2-c2-d2)2 Euler observed that the form, 2(p2v4+x4+y4+z4) = (pv2+x2+y2+z2)2 (eq.1) (with p inserted by this author) is equivalent to the triplet of simultaneous eqns, p(2vx)2 + (2yz)2 = (pv2+x2-y2-z2)2

p(2vy)2 + (2xz)2 = (pv2-x2+y2-z2)2

p(2vz)2 + (2xy)2 = (pv2-x2-y2+z2)2 When p = ±1, this is a special case of Descartes’ Circle Theorem, and the triplet are Pythagorean triples. Piezas Let p = -1. If a4+b4 = c4+d4, then one soln to (eq.1) is, {v,x,y,z} = {2(ab-cd)(ab+cd), (a2+b2+c2+d2)(a2-b2+c2-d2), 2(ac-bd)(a2+c2), 2(ac-bd)(b2+d2)} This is discussed more in Form 18 here. (End update.) Form 7: (ac+bd)2 + (ad-bc)2 = (a2+b2)2

One can also find solutions to x2+y2 = z2 using the equation p2+q2 = r2+s2 and vice-versa, Fibonacci (ad+be)2 + (ae-bd)2 = (dc)2 + (ec)2, if a2+b2 = c2

Conversely, if a2+b2 = c2+d2, then, P. Volpicelli (ac+bd)2 + (ad-bc)2 = (a2+b2)2 A. Fleck (a2c-b2c+2abd)2 + (a2d-b2d-2abc)2 = (a2+b2)3 The solns by Volpicelli and Fleck suggest a generalization can be found for all positive integer k. Using a variation of the technique introduced earlier, assume, A2+B2 = (a2+b2)g(c2+d2)h and equating factors, A+Bi = (a+bi)g(c+di)h, A-Bi = (a-bi)g(c-di)h

call this as System 1, then solving for A,B can give a soln to, A2+B2 = (a2+b2)g+h

where A,B are polynomials in {a,b,c,d} with the constraint a2+b2 = c2+d2 (call this eq.1). Thus, for k = g+h = 4,

(ac3-3bc2d-3acd2+bd3)2 + (bc3+3ac2d-3bcd2-ad3)2 = (a2+b2)4, if a2+b2 = c2+d2

which continues Volpicelli's and Fleck's identities, and so on for all k. In Volpicelli’s soln, if we let {c,d} = {a, ±b}, then we get either the usual formula for Pythagorean triples or, after removing common factors, a tautology. But if we solve for {a,b,c,d} in eq.1 using Euler’s complete soln, (pr+qs)2 + (ps-qr)2 = (pr-qs)2 + (ps+qr)2

plus the fact that for non-negative integers {g,h} there are k+1 ways to get the sum g+h = k, this gives the more general result: Theorem: “Let F := (p2+q2)(r2+s2). Then Fk for k > 0 is identically the sum of two squares in k+1 ways.” Proof: It is well known Fk for k = 1 is a sum of two squares via the Brahmagupta-Fibonacci Two-Squares Identity. In fact, Euler’s soln for (eq.1) is just this in disguise since, (pr+qs)2 + (ps-qr)2 = (pr-qs)2 + (ps+qr)2 = (p2+q2)(r2+s2) thus easily proving that for k = 1, there are indeed two ways to express Fk as the sum of two squares. However, for k >1, there are not just one, but k+1 corresponding identities. To prove this, since one is to solve, A2+B2 = (a2+b2)g(c2+d2)h where A,B are polynomials in {a,b,c,d}, the solns are dependent on the particular values of the exponents {g,h} chosen. And there are k+1 ways such that g+h = k. Since a2+b2 = c2+d2, then,

A2+B2 = (a2+b2)g+h

By Euler’s soln {a,b} = {pr+qs, ps-qr}, so A2+B2 = ((p2+q2)(r2+s2))g+h = Fk

in k+1 ways. (End proof.) For example, for F3, let exponents {g,h} of the factors be in turn {0,3}, {1,2}, {2,1}, {3,0}. Solving for A,B in System 1, these yield the four identities, {A1, B1} = {c(c2-3d2), d(3c2-d2)}

{A2, B2} = {ac2-2bcd-ad2, bc2+2acd-bd2}

{A3, B3} = {a2c-2abd-b2c, a2d+2abc-b2d}

{A4, B4} = {a(a2-3b2), b(3a2-b2)} where {a,b,c,d} = {pr+qs, ps-qr, pr-qs, ps+qr}. To give a random numerical example, let {p,q,r,s} = {1, 2, -2, 3} so (p2+q2)(r2+s2) = 65. These then yield {Ai, Bi} as {488, 191}, {140, 505}, {320, 415},

{524, 7} such that A2+B2 = 653. And so on for other k, for arbitrary {p,q,r,s}. An interesting variant to Pythagorean triples is x2+y3 = z4: H. Mathieu (q2(p2-2))2 + (2q2)3 = (pq)4, if p2-2q2 = 1 ((p4-p2)/2)2 + p6 = (pq)4, if p2-2q2 = -1 Solns can be given by the nth triangular number (n2+n)/2 that is also a

square and depends on solving the above Pell equation. Note that solving n in (n2+n)/2 = y2 for rational n entails making its discriminant a square 1+8y2 = x2, or equivalently, x2-2(2y)2 = 1. More generally though, Piezas (4q2d4(p2-2))2 + (4q2d3)3 = (2pqd2)4, if p2-dq2 = 1 (4p2d3(p2-1))2 + (2pd)6 = (2pqd2)4, if p2-dq2 = -1 thus it is not just limited to triangular numbers. Another variant is x4+y3 = z2: K. Brown p4 + (q2-1)3 = (q3+3q)2, if p2-3q2 = 1 (See Brown’s “Miscellaneous Diophantine Equations”.) More generally, p4 + (dq2-1)3 = d(dq3+3q)2, if p2-3dq2 = 1 Note: As pointed out by Michael Somos, the above equation is also true if p2+3dq2 = -1. (Even for negative integer d, this is not solvable for integer {p,q}, but is easily solvable in the rationals for any d.)

2. Form: x2+ny2 = zk This is just a generalization of the form x2+y2 = zk.

Euler x2+ny2 = (p2+nq2)k Same technique of equating factors over Ö-n, {x+yÖ-n, x-yÖ-n} = {(p+qÖ-n)k, (p-qÖ-n)k} and then solve for x,y. Example, for k=2, (p2-nq2)2 + n(2pq)2 = (p2+nq2)2 which for n=1 gives the familiar and complete parametrization (after scaling) for Pythagorean triples. For k = 3, we get, (p3-3npq2)2 + n(3p2q-nq3)2 = (p2+nq2)3 However, for k > 2 this method does not generally give all solns (Pepin). For example, for the particular case k = 3 and n = 47, a class of solns not given by the above is, (13u3+60u2v-168uv2-144v3)2 + 47(u3-12u2v-24uv2+16v3)2 = 23

(3u2+2uv+16v2)3 by Pepin. This is also discussed in Quadratic Polynomials as a kth Power.

2b. Form: ax2+by2 = cz2

To solve this form for an infinite number of solns, all it takes is one initial point {x,y,z}. Either of two methods can be used, each with their own strengths.

Method 1: Use the identity by this author, ax2+bxy+cy2+dz2 = (am2+bmn+cn2+dp2)(u2+cdv2)2 where {x,y,z} = {mu2+cdmv2, nu2-2pduv-d(bm+cn)v2, pu2+(bm+2cn)uv-cdpv2} For example, given m2+2n2 = 2p2 we have {a,b,c,d} = {1, 0, 2, -2}, and initial soln {m,n,p} = {4, 1, 3}. Using these, we get the parametrization, x2+2y2 = 2z2

where {x,y,z} = {4(u2-4v2), u2+12uv+4v2, 3u2+4uv+12v2}. Similarly for, x2+y2 = 2z2

where {x,y,z} = {u2-2v2, u2+4uv+2v2, u2+2uv+2v2}, and so on. Method 2: Use the identity, ax1

2+bx22+cx3

2 = (ay12+by2

2+cy32)(az1

2+bz22+cz3

where {x1, x2, x3} = {uy1-vz1, uy2-vz2, uy3-vz3} with {u,v} =

{az12+bz2

2+cz32, 2(ay1z1+by2z2+cy3z3)}

with initial soln {y1, y2, y3} and arbitrary {z1, z2, z3}, thus giving

a three-variable parametrization. Given the same ex, y12+2y2

2y32, we have {a,b,c} = {1, 2, -2} and small soln {y1, y2, y3} =

{4, 1, 3}. Substituting these in the formula for the xi, and letting {z1, z2, z3} = {x,y,z} for convenience, we get the alternative identity, x1

2+2x22 = 2x3

where {x1, x2, x3} = {4(x+2y-2z)(x-y-z), x2-8xy-2y2+12yz-2z2,

3x2+6y2-8xz-4yz+6z2}. The advantage of the first method is that it yields simple forms and can be used on eqns with cross terms (i.e. where b is non-zero). However, the second method is useful in that it can be generalized into n diagonal forms: Theorem: "In general, given one solution to a1y1

2+a2y22+…+

anyn2 = 0, then an infinite more can be found."

Proof: We simply generalize the soln given in Method 2. For four addends, ax1

2+bx22+cx3

2+dx42 = (ay1

2+by22+cy3

2+dy42)

(az12+bz2

2+cz32+dz4

where {x1, x2, x3, x4} = {uy1-vz1, uy2-vz2, uy3-vz3, uy4-vz4}

with {u,v} = {az12+bz2

2+cz32+dz4

2, 2(ay1z1+by2z2+cy3z3+dy4z4)} with initial {y1, y2, y3, y4} for arbitrary {z1, z2, z3, z4}, thus now giving a four-variable parametrization. And so on for any n

addends as the pattern for the xi and {u,v} are easily seen. (These methods are also discussed in Section 006.) 3. Form: ad-bc = ±1 This is inserted here because it is important to the next two forms, x2+y2 = z2±1. One can solve ad-bc = ±1 in the integers using the transformation, {a,b,c} = {q2, p+q, p-q} such that it becomes the Pell equation, p2-(d+1)q2 = ±1 Alternatively, let {a,b,c,d} = {p+s, -q+r, q+r, p-s} to transform it to, p2+q2 ±1 = r2+s2

simple solns of which are, E.Fauquembergue: (a+1)2 + (2a)2 + 1 = (2a+1)2 + (a-1)2

(2a2-b2+1)2 + (2ab)2 + 1 = (2a2+1)2 + (b2-1)2

Piezas (a2-c2+a+1)2 + (2c)2 + 1 = (a2-c2+a-1)2 + (2a+1)2

This form is discussed more in the section on Sums of Three Squares.

Q: Any more solns to p2+q2 ± 1 = r2+s2?

4. Form: x2+y2 = z2+1 In general, given an initial soln to x2+y2 = z2+h, one can find a quadratic parametrization (by this author) in the form, (pn+a)2 + (pn2+2an+b)2 = (pn2+2an+c)2 + h, if a2+b2 = c2+h where p = 2(-b+c) for arbitrary {n,h}. Choudhry (p2q+p-q)2 + (2pq+1)2 = (p2q+p+q)2 + 1 A. Gerardin (dx)2 + (d2y2-1)2 = (d2y2+d)2 + 1, where x2-2(d+1)y2 = 1 More generally, one can convert this to a Pell equation for a broader class. Piezas u2 + ((d-v2)/2)2 = ((d+v2)/2)2 ± 1, where u2-dv2 = ±1. (With d chosen such that terms are integers.) Another way would be to use Euler’s complete soln for x2+y2 = z2+t2, (ac+bd)2 + (ad-bc)2 = (ac-bd)2 + (ad+bc)2

and set one term, say, ad-bc = ±1, a form discussed previously. Alternatively, using the simple identity, (2m)2 + (2m2-n)2 = (2m2-n+1)2 + (2n-1) set n = 1 or 0 to get, (2m)2 + (2m2-1)2 = (2m2)2 + 1(2m)2 + (2m2)2 = (2m2+1)2 - 1 the second of which is for the next section.

5. Form: x2+y2 = z2-1 For this form, starting with an initial identity, one can generate an infinite sequence of identities. Proof (Piezas):

If a2+b2 = c2 ± d2, then, (4ac-2bc+2t)2 + (2ac-4bc+2t)2 = (4ac-4bc+3t)2 - d4, where t = a2+b2+c2

which implies one soln to u2+v2 = w2-x2 leads to another. Furthermore, iterative use leads to u2+v2 = w2-xk with k a power of two, and x = 1 a special case. For example, the identity given in the previous section, (2m)2 + (2m2)2 = (2m2+1)2 - 1

yields, (8m4+16m3+12m2+8m+2)2 + (8m3+8m2+4m+2)2 = (8m4+16m3+16m2+8m+3)2 - 1

which, in turn, yields another, ad infinitum. Other solns are,

E. Grigorief ((a2-b2+c2-d2)/2)2 + (ab+cd)2 = ((a2+b2+c2+d2)/2)2 - 1, where ad-bc = ±1 One can also use Genocchi’s and Pepin’s method, (2pq(r2-2s2))2 + (4pqrs)2 = (p2+(r4+4s4)q2)2 - 1, if p2-(r4+4s4)q2 = ±1 which for the case r = 1 has a parametric soln to be discussed in Simultaneous Polynomials. (This also discusses the complete soln, by Genocchi and Pepin, such that {x2+y2-1, x2-y2-1} are both squares and also involves solving a Pell eqn.) As an overview, it turns out that, x2+y2 = z2±1x3+y3 = z3±1x4+y4 = z2+1 can be solved using Pell equations, the higher powers to be discussed in their respective sections.

6: Form x2+y2 = z2+nt2

This generalizes the two forms above. For the special case when n = 1, this is discussed in mx2+ny2 = mz2+nt2, the last section below. A.Gerardin (1-ma)2 + (1-mb)2 = (mc)2 + 2, if m = 2(a+b)/(a2+b2-c2) Piezas (2a+bn)2 + (2b+an)2 = (2c-cn)2 + n(2a+2b)2, if a2+b2 = c2

for some arbitrary constant n which was derived by modifying Gerardin’s. For the special case t = 1, or Pythagorean-like triples x2+y2 = z2+h, it can be proven that for any h, there is an infinity of solns given by, (2m+1)2 + (2m2+2m-n)2 = (2m2+2m-n+1)2 + 2n (2m)2 + (2m2-n)2 = (2m2-n+1)2 + (2n-1) and the general quadratic parametrization given earlier. Q: Any other simple formulas that work for all h not derived from these two? Note: For the case h = 0, or the Pythagorean triples x2+y2 = z2, it is known that the ratio N/B where N is the number of primitive solns with z below a bound B asymptotically approaches 1/(2π) ≈ 0.159154, a result established by Lehmer. If we generalize this to x2+y2 = z2+h and define the ratio R(h) = N/B for any h, it seems for h < 0, there might be asymptotics which involve 1/√-h. For example, for h = -1, given N with increasing bound B = 10, 102, 103, and so on gives the sequence of ratios as, R(-1) = 0.2, 0.14, 0.126, 0.1238, 0.1251, 0.12497, 0.12499, …

which seem to be converging to 0.125 = 1/8. For small square-free h at bound B = 106 and R(h) rounded to five decimal places, paired with its conjectured asymptotic value: R(-2) = 0.176791/(4√2) = 0.17677 R(-3) = 0.288681/(2√3) = 0.28867 R(-5) = 0.223661/(2√5) = 0.22360 R(-6) = 0.203961/(2√6) = 0.20412 R(-7) = 0.37799 1/√7 = 0.37796 R(-10) = 0.158141/(2√10) = 0.15811 is highly suggestive considering the simplicity of the form. Interestingly, for h > 0, the R(h) do not seem to have simple ones involving √h. (Note: Computing R(h) is time-consuming, and I’m grateful to the Mathematica code provided by Daniel Lichtblau of Wolfram Research, as well as help provided by Andrzej Kozlowski and James Waldby. If someone can prove the exact values are in fact the correct asymptotics, pls let me know.)

7. Form: x2+y2 = z2+ntk J. Rose

(4n2)2 + (4n3)2 = (4n2(n-1))2 + (2n)5

Mehmed-Nadir x2 + y2 = z2 + (a2+b2)5 {x,y,z} = {(a2+b2)(a2-b2)b, ((a2+1)(a2+b2)2+4b4)/2, ((a2-1)(a2+b2)2-4b4)/2} E. Barisien ((n+2)(n2-2n-2))2 + (4n(n+1))2 = (2(n+1)(n+2))2 + n6

These can be generalized by the identity given previously by this author for Pythagorean-like triples where the constant term 2n or 2n-1 can be equated to any kth power pk, and n then easily solved for. Q: Can Mehmed-Nadir’s identity be generalized to x2 + y2 = z2 + (a2+b2)k for other k not using the identity for Pythagorean-like triples? See also sum of three squares x2 + y2 + z2 = (a2+b2)k, one of which is also by Mehmed-Nadir.

8. Form: x2+y2 = mz2+nt2 The parallelogram law states that given a parallelogram's four sides {t,t,z,z} (since it necessarily has opposite sides equal) and two diagonals {x,y}, then, x2+y2 = 2z2+2t2 (eq.1) If the diagonals are equal as well, or x = y, then the parallelogram law reduces to the well-known Pythagorean theorem, a2+b2 = c2. (See

also the quadrilateral law, on the form a2+b2+c2+d2 = x2+y2+z2.) Solns to eq.1 are, Euler {x,y,z,t} = {2(pr-qs), 2(ps+qr), (p+q)r-(p-q)s, (p-q)r+(p+q)s} M. Gruber {x,y,z,t} = {4(p-q)q-2s, 4(p-q)r, s, 4p(p-q)-s}, where s = 2p2-q2-r2 Paul Cheffers {x,y} = {z+t, z-t} More completely, by using the transformation {2z, 2t} = {p+q, p-q} on (eq.1), it suffices to solve the form x2+y2 = p2+q2, which can be done completely. Alternatively, if n is a constant that is the sum of two squares n = a2+b2, x2+y2 = n(z2+t2) then this has soln {x,y} = {az+bt, bz-at} for constant {a,b} and arbitrary {z,t} since this yields, x2+y2 = (a2+b2)(z2+t2) For example, x2+y2 = 5(z2+t2) has {x,y} = {2z+t, z-2t}, with Cheffer's as the case {a,b} = {1,1}, and so on. Q: Any soln to the more general form x2+y2 = mz2+nt2?

9. Form: c1(x2+ny2) = c2(z2+nt2) In a manner, this generalizes the previous section. For constants m1, m2, given an equation of form, Case 1: m1(a2+b2) = (r2+s2)(c2+d2)

Case 2: m1(a2+b2) = m2(c2+d2) these can be solved if m1, m2 are sums of two squares using the general identity, (a2+b2)(p2+q2) = (c2+d2)(r2+s2) where {a,b,c,d} = {ru+sv, su-rv, pu+qv, qu-pv}. For Case 1, it is constant {p,q} with arbitrary {r,s,u,v}; for Case 2, constant {p,q,r,s} with arbitrary {u,v}. Example 1: 5(a2+b2) = (r2+s2)(c2+d2), then {a,b,c,d} = {ru+sv, su-rv, u+2v, 2u-v}. Example 2: 5(a2+b2) = 13(c2+d2), then {a,b,c,d} = {3u+2v, 2u-3v, u+2v, 2u-v}.

10. Form: mx2+ny2 = mz2+nt2

Theorem: “There is a complete integral solution to ap2+bpq+cq2 = ar2+brs+cs2.” First. we transform this to diagonal form by the transformation {p,q,r,s} = {u-bv, 2av, x-by, 2ay} to get, u2-Dv2 = x2-Dy2

where D = b2-4ac is the discriminant. This can be solved in an even more general form. Theorem: “The equation mu2+nv2 = mx2+ny2 has the complete solution, m(ac+bdn)2 + n(bc-adm)2 = m(ac-bdn)2 + n(bc+adm)2, for arbitrary {a,b,c,d}.” The identity is by Euler. Note that either side is equal to (ma2+nb2)(mc2+nd2). Proof (Piezas): Given any non-trivial soln {u,v,x,y}, one can always find rational {a,b,c,d} using the formulas, {a,b,c} = {(y-v)/(2dm), (u-x)/(2dn), dn(v+y)/(u-x)} where it can be set d=1 without loss of generality. For example, substituting these into u = ac+bdn one finds it is true only if mu2+nv2 = mx2+ny2, and similarly for the other variables. The idea was to solve for {a,b,c,d} and express them in terms of u,v,x,y. These were found using Mathematica’s solve command, Solve[{u,v,x} = {ac+bdn, bc-adm, ac-bdn}, {a,b,c}] which finds {a,b,c} and where the resulting radical solns can then be simplified using the relation mu2+nv2 = mx2+ny2. Incidentally, if we

set the fourth variable y = bc+adm = 0, Euler’s identity reduces to, m(ma2-nb2)2 + n(2abm)2 = m(ma2+nb2)2 which for m=n=1, not surprisingly, is the well-known formula for Pythagorean triples. There are also other versions. S. Realis (complete) u2+nv2 = x2+ny2

{u,v,x,y} = {a2-n(a-b)2+n(a-c)2, b2-(a-b)2+n(b-c)2, a2+nb2-nc2, c2-(a-c)2-n(b-c)2} V. Lebesgue (p+q+r-s)2 + (p+q-r+s)2 = (p-q+r+s)2 + (p-q-r-s)2, if pq = rs or, alternatively, (p+q)2 + (r-s)2 = (r+s)2 + (p-q)2, if pq = rs A.Desboves x2+y2 = (m2+n2)z2

{x,y,z} = {(u2-dv2)m, nu2-2duv+dnv2, u2-2nuv+dv2}, where d = m2+n2

11. Form: x2+y2 = zk+tk (Update, 11/9/09): W. Lewis asked if results were known for the eqn

x2+y2 = z5+t5. Since I didn't know of any relevant identity, my first thought was to use Mathematica's search function to look for small solns and see if a parametrization could be found. One can use two approaches: Method 1: Use the Bramagupta-Fibonacci Two-Square Identity for the problem in its general case. This is given by, (pr+qs)2 + (ps-qr)2 = (p2+q2)(r2+s2) (eq.1) There are 4 free variables {p,q,r,s} to dispose of. Select {p,q} = {xk, yk}. The RHS, after distributing, becomes, (pr+qs)2 + (ps-qr)2 = x2k(r2+s2) + y2k(r2+s2) Then solve for r2+s2 = tk which was discussed in the previous page and is easy to do. Eq. 1 then becomes, (pr+qs)2 + (ps-qr)2 = (tx2)k + (ty2)k For k = 3, this is satistied by {p,q,r,s,t} = {x3, y3, u3-3uv2, 3u2v-v3, u2+v2} with 4 free variables. For ex, let {u,v,x,y} = {2,1,3,4}, then, 7582 + 1692 = (5*9)3 + (5*16)3

Method 2: For co-prime terms, the small solns given by Mathematica were enough to provide a clue for a parametrization to the eqn, x2+y2 = zk+1 A. For all ODD k in terms of powers and near-powers of 2. This is given by, (2m+1)2 + (22m+1-1)2 = (24m+2)1 + 1

(23m+2)2 + (26m+3-1)2 = (24m+2)3 + 1 (25m+3)2 + (210m+5-1)2 = (24m+2)5 + 1 etc. The pattern is easily seen and one can notice how z has a constant form regardless of k. (Update, 2/19/11): Oliver Couto gave an EVEN k counterpart, (nk/2(nk-2))2 + (2nk-1)2 = (n3)k + 1 hence, (n(n2-2))2 + (2n2-1)2 = (n3)2 + 1 (n2(n4-2))2 + (2n4-1)2 = (n3)4 + 1 (n3(n6-2))2 + (2n6-1)2 = (n3)6 + 1 Piezas For EVEN k = 4m+2: (n3-2n)2 + (2n2-1)2 = n6 + 1 (n5-2n3+2n)2 + (2n4-2n2+1)2 = n10 + 1 (n7-2n5+2n3-2n)2 + (2n6-2n4+2n2-1)2 = n14 + 1 and so on.

II. Sum / Sums of three squares

1. x2+y2+z2 = tk

2. x2+y2+z2 = u2+v2

3. (x2-1)(y2-1) = (z2-1)2

4. x2+y2+z2 = u2+v2+w2

5. x2+y2+z2 = (u2+v2+w2)Poly(t)6. x2+y2+z2 = 3xyz

1. Form: x2+y2+z2 = tk Geometrically, the case k = 2 of Form 1 can be seen as a cuboid with sides {x,y,z} and space diagonal t,

or, for t = r = 1, a sphere of unit radius,

By analogy to Pythagorean triples, the numbers {x,y,z,t} is also known as a Pythorean quadruple. The simplest integer soln is, n2 + (n+1)2 + (n2+n)2 = (n2+n+1)2

which easily proves that, unlike in Pythagorean triples, every integer appears at least once in an integer Pythagorean quadruple. Similarly for quintuples, n2 + (n-2)2 + (2n+1)2 + (3n2+2)2 = (3n2+3)2

and so on. Theorem: The complete soln to x1

2+x22+x3

2 = y12 is given by,

((a2-b2-c2)t)2 + (2abt)2 + (2act)2 = ((a2+b2+c2)t)2 where t is just a scaling factor. Proof (Piezas): For any soln xi where x1 ≠ y1, one can always find rational {a,b,c,t} using the

formulas: {a,b,c,t} = {x1+y1, x2, x3, (-x1+y1)/(2(x22+x3

Similarly, Theorem: The complete non-trivial soln to x1

2+x22+x3

2+x42 = y1

2 is given by the scaled Euler-Aida Ammei identity,

((a2-b2-c2-d2)t)2 + (2abt)2 + (2act)2 + (2adt)2= ((a2+b2+c2+d2)t)2 Proof: For any soln xi where x1 ≠ y1, one can always find rational {a,b,c,d,t} using the formulas: {a,b,c,d,t} = {x1+y1, x2, x3, x4, (-

x1+y1)/(2(x22+x3

2+x42))}.

And so on. One can easily see the pattern for other n sums of squares. Update (10/11/09): The eqn x2+y2+z2 = 1, which also defines a sphere of unit radius centered at the origin, also appears in the eqn x2 = -1 where x, instead of the imaginary unit, is to be a quaternion. To recall, given the expansion of (a+bi+cj+dk)n = A+Bi+Cj+Dk, then, A2+B2+C2+D2 = (a2+b2+c2+d2)n In a Mathematica add-on package, this object is given as Quaternion[a,b,c,d] and one can use NonCommutativeMultiply to multiply them, and for the case n = 2, we get, Quaternion[a,b,c,d] ** Quaternion[a,b,c,d] = Quaternion[a2-b2-c2-d2, 2ab, 2ac, 2ad] (eq.1) One can note two immediate consequences. First, as expected, (a2-b2-c2-d2)2 + (2ab)2 + (2ac)2 + (2ad)2 = (a2+b2+c2+d2)2. Second, while the Fundamental Theorem of Algebra states that an nth degree univariate equation with complex coefficients has n complex roots, this does not apply when the roots are quaternions. For example, the nth

roots of plus/minus unity or, xn = ±1 have n complex solns. But if x is to be a quaternion, then there can be an infinite number of them. Let n = 2 and take the negative case: x2 = -1. Thus, we wish (eq.1) to have the product: Quaternion[-1,0,0,0]. Since the {a,b,c,d} are to be real, then a = 0, and {b,c,d} should satisfy, b2+c2+d2 = 1 which have an infinite number of real and rational solns, a subset of which are rationalized Pythagorean triples. The rational soln with the smallest denominator is {b,c,d} = ±{1,0,0}, in which case the quaternion reduces to the imaginary unit. The next smallest is ±{1,2,2}/3. Of course, the most general setting for solving x2 = -1 would be x as the octonions, where the coefficients of the non-real part would have to satisfy the sum of seven squares, y1

2+y22+y3

2+y42+y5

2+y62+y7

2 = 1 and, if all yi are to be non-zero, the soln with the smallest denominator is ±{1,1,1,1,2,2,2}/4. (End update.) Theorem (H. Monck): If a2+b2+c2 = d2, then (a+b+d)2 + (a+c+d)2 + (b+c+d)2 = (a+b+c+2d)2. Thus, one soln can generate others (with sign changes in a,b,c,d yielding several distinct ones), just like for Pythagorean triples. Q: Any similar identity for four squares or more? Other versions for sums of three squares are, A. Desboves (complete soln) a2+b2+c2 = d2

{at, bt, ct, dt} = {2(p2-q2+r2), 2(p-q)2-2r2-2p(q-r), p2-(q-r)2-4r(p-q),

3(p2+q2+r2)-2q(2p+r)} for arbitrary p,q,r and where t is just a scaling factor. (V. Lebesgue earlier gave a more complicated 6-parameter soln.) Proof (Piezas): For any soln a,b,c,d, one can always find rational p,q,r,t using the formulas, {p,q,r,t} = {2a+2b, 3a-2d, a-2c, -8(2a-2b+c-3d)} Piezas (complete) t2(2u-v+2w)2 + t2(-u+2v-2w)2 + t2(-1+w)2 = t2(2u-2v+3w)2, where w = (u2+v2+1)/2 and t is a scaling factor. Since the eqn is homogenous, after some changes one can also use w = u2+v2+x2 though for simplicity it can be set x=1 without loss of generality. Proof: For any particular soln a,b,c,d, one can always find rational u,v,t given by, {u,v} = {(2a-b-2d)/t, (-a+2b+2d)/t}, where t = -(2a-2b+c-3d) Note that the scaling factor t is essentially the same as the one for Desboves. Partial solns in terms of binary quadratic forms can also be given, A. Gerardin (6x2-14xy+6y2)2 + (3x2-3y2)2 + (2x2-2y2)2 = (7x2-12xy+7y2)2 Notice that 62+32+22 = 72. This can be generalized as, Piezas (ax2-2dxy+ay2)2 + (bx2-by2)2 + (cx2-cy2)2 = (dx2-2axy+dy2)2, if a2+b2+c2 = d2

Euler, Lebesgue (a2-b2-c2+d2)2 + (2ab-2cd)2 + (2ac+2bd)2 = (a2+b2+c2+d2)2 which is also known as the Lebesgue Three-Square Identity. The square (a2+b2+c2+ab+ac+bc)2 can be identically expressed as the sum of 3 or 4 squares, Catalan ((a+b)(a+c))2 + ((a+b)(b+c))2 + (ab-ac-bc-c2)2 = (a2+b2+c2+ab+ac+bc)2

J. Neuberg: a2(a+b+c)2 + b2(a+b+c)2 + c2(a+b+c)2 + (ab+ac+bc)2 = (a2+b2+c2+ab+ac+bc)2

Likewise, (a2+b2+2c2)2 as 3 squares (in two ways), or 4 squares, E. Catalan: (2(a+b)c)2 + (a2-b2)2 + (2ab-2c2)2 = (a2+b2+2c2)2 (2(a+b)c)2 + (2(a-b)c)2 + (a2+b2-2c2)2 = (a2+b2+2c2)2 (a2-c2)2 + ((a+c)(b+c))2 + ((a+c)(b-c))2 + (2ac-b2-c2)2 = (a2+b2+2c2)2 (Update, 6/25/10): Alain Verghote gave a similar identity, (2(a+b)c)2 + (2(a-b)c)2 + (2a2+2b2-c2)2 = (2a2+2b2+c2)2

though this can be derived from Catalan's by letting c → c'/2. J. Euler ((a2+c2)(b2-c2))2 + (2bc(a2-c2))2 + (4abc2)2 = ((a2+c2)(b2+c2))2 Just like one can solve the sum of two squares equal to a kth power in the form, A2+B2 = (a2+b2)k it can also be done for three squares, A2+B2+C2 = (a2+b2+c2)k for any positive integer k. In fact, it can be done for n squares and the general result is that: Theorem 1: “For all n and k, then x1

2+x22+… +xn

2 = (y12+y2

+yn2)k, or the kth power of the sum of n squares is identically the sum

of n squares.”Theorem 2: “For all n > 1 and k > 2, then x1

2+x22+… +xn+1

(y12+y2

2+… +yn2)k, or the kth power of the sum of n squares is

identically the sum of n+1 squares.” Note 1: The case k = 2 is discussed in Sums of n Squares.Note 2: The cubic version A3+B3+C3-3ABC = (a3+b3+c3-3xyz)k is discussed as Form 21 in Cubic Polynomial as a kth Power. Proof for Theorem 1: (Piezas) The proof is simple. Let the expansion of the complex number (a±bi)k be,

U+Vi = (a+bi)k; U-Vi = (a-bi)k

where U,V are expressions in the arbitrary a,b. It is easy to see their product, or norm, is, U2+V2 = (a2+b2)k Note that b can be factored out in V, or V = V1b, so if we let {a,b} = {p1,√p0}, then, U2 + p0V1

2 = (p12+p0)k

And since p0 is arbitrary, one can choose it to be the sum of squares p0

= p22+p3

2+…+pn2 giving,

U2 + V1

2p22 + … + V1

2pn2 = (p1

2+ p22+…+pn

Thus proving the kth power of n squares is itself the sum of n squares. (End proof.) For k = 2, this is, (p1

2-p0)2 + p0(2p1)2 = (p12+p0)2

so if p0 = p2

2, this gives the classic formula for Pythagorean triples.

But if p0 = p22+p3

2, and changing variables, yields an analogue for three squares, (a2-b2-c2)2 + (2ab)2 + (2ac)2 = (a2+b2+c2)2 which, after scaling, is the complete soln. For k = 3,

(a3-3ab2-3ac2)2 + b2(-3a2+b2+c2)2 + c2(-3a2+b2+c2)2 = (a2+b2+c2)3 and so on for all positive integer k and n. (Using identities by Mehmed Nadir and Matsunago, we will see that the kth power of the sum of n squares is also identically the sum of n+1 squares for k = 3m or 4m. Update: It can be done for all k > 2. See further below.) Mehmed Nadir (also known as Mehmet Nadir) a2(a2+b2)2 + (2ab2)2 + (a2-b2)2b2 = (a2+b2)3 Matsunago: (a4-b4)2 + (4a2b2)2 + (2ab(a2-b2))2 = (a2+b2)4 Q: Is there one for k = 5? (See update below.) While we already know that (a2+b2)k is expressible as the sum of two squares, for what k is it identically the sum of three non-zero squares? By solving a2+b2 = (p2+q2)m, one can see there are solns when k = 3m, 4m for all integer m > 0. Letting d = b2, we can also express these as, a2(a2+d)2 + (2ad)2 + (a2-d)2d = (a2+d)3

(a4-d2)2 + (4a2d)2 + (2a(a2-d))2d = (a2+d)4

where d can then be set as the sum of n squares d = x2

2+x32+…+xn

2 thus proving, “ ...The kth power of the sum of n squares is the sum of n+1 squares for k = 3m or 4m.” But for what other k is there? One limited form with b = 1, is A. Gerardin (y-1)2 + y2 + (y+1)2 = x2 + 1, if x2-3y2 = 1

Thus the kth power of (x2+1)k, for all solns x to x2-3y2 = 1, should be expressible as the sum of three squares for any positive integer k. For ex, substituting {p,q,r} = {y-1, y, y+1} into the general identity, then for k = 2 this is, (y2+4y)2 + (2y2-2y)2 +(2y2-2)2 = (x2+1)2, if x2-3y2 = 1 and so on for all k. A closer one is, (a-2b)2 + (2a-b)2 + (2a+2b)2 = 9(a2+b2) though this does not give Mehmed-Nadir’s or Matsunago’s elegant forms. Update (6/18/09, 10/12/09) Theorem: "For all k > 2, the expression (a2+b2)k is the sum of three squares as polynomials in {a,b} with integral coefficients." Proof: Given the tautology, a2+b2 = a2+b2

Multiply both sides by (a2+b2)2n, (a2+b2) (a2+b2)2n = (a2+b2)2n+1

Distribute, a2(a2+b2)2n + b2(a2+b2)2n = (a2+b2)2n+1

and it is easy to express either of the squares on the LHS as the sum of two squares. And since the eqn x1

2+x22+… +xn

(y12+y2

2+… +yn2)m can be solved for any positive integer m

(Theorem 1 discussed at the start of this Section), then one can also solve,

a2(a2+b2)2n + b2(a2+b2)2n = (p2+q2)(2n+1)m where {a,b} are polynomials in {p,q}. Together with the identity for k = 4, which is effectively for all k = 4m, then it is possible to express (a2+b2)k as the sum of three squares for all k > 2. (End proof.) For k = 5, this gives, a2(a2+b2)4 + (4a3b2-4ab4)2 + (a4b-6a2b3+b5)2 = (a2+b2)5 One can then always find expressions {a,b} such that the sum is (p2+q2)5m. And so on for other prime k > 2. However, more generally, Theorem 2: “For all n > 1 and k > 2, then x1

2+x22+… +xn+1

(y12+y2

2+… +yn2)k, or the kth power of the sum of n squares is

identically the sum of n+1 squares.” Proof: As before, one simply expresses b2 = d, set d as any non-zero number of squares, distribute the third term appropriately, then use Theorem 1. For example, for k = 5, this is,

a2(a2+d)4 + (4a3d-4ad2)2 + (a4-6a2d+d2)2d = (a2+d)5

After setting, say, d = b2+c2, a2(a2+b2+c2)4 + (4a3(b2+c2)-4a(b2+c2)2)2 + b2(a4-6a2(b2+c2)+(b2+c2)2)2 + c2(a4-6a2(b2+c2)+(b2+c2)2)2 = (a2+b2+c2)5 or 4 squares equal to the 5th power of 3 squares, one can then use Theorem 1 to find {a,b,c} such a2+b2+c2 = (p2+q2+r2)m, for any m. For m = 2, this would be {a,b,c} = {p2-q2-r2, 2pq, 2pr}. And so on for any n > 1 and k > 2. (End update.)

For (a2+b2)k as the sum of four squares, Barisien gave several examples for small k. These are easy to find since solns to u2+v2 = zm and x2+y2 = zn immediately imply, (u2+v2)(x2+y2) = (ux)2+(vx)2+(uy)2+(vy)2 = zm+n. E. Barisien: a2(a2-b2-c2)2 + b2(a2+b2-3c2)2 + c2(a2-3b2+c2)2 + (2a2b)2 + (2a2c)2 + (4abc)2 = (a2+b2+c2)3 Q: How about the kth power of three squares (a2+b2+c2)k identically the sum of n squares with n > 4? Any other prime k? F. Hrodmadko n2 + (n+1)2 + (nx)2 = (nx+1)2, if x = n+1. There is a similar identity for third powers, by Jandacek, n3 + (x-n-1)3 + (nx)3 = (nx+1)3, if x = 3n2+3n+2. Proving that any integer n appears at least once as a soln to ak+bk+ck = (c+1)k for k = 2 or 3. Finally, the system of eqns, x1

k+x2k+…+ xn

k = y1k+y2

k+…+ ynk, for k = 1,2

by Bastien’s theorem (to be discussed more in the section on Equal Sums of Like Powers) has non-trivial solns only for n > 2. It is possible two terms on one side are equal to zero and hence reduces to the minimal form,

ak+bk+ck = dk, the complete soln of which obeys the relationship ab+ac+bc = 0 and is given by, (p2+pq)k + (pq+q2)k + (-pq)k = (p2+pq+q2)k, for k = 1,2 implying at least one term is always negative. There are analogous cases for multi-grade third, fourth, fifth powers to be given later.

2. Form: x2+y2+z2 = u2+v2 One can set any of its terms equal to 1, but depending on which side of the equation it is chosen, it may need slightly different Pell equations. To set this form either as p2+q2+1 = s2+t2, or p2+q2+r2 = s2+1, can involve, x2-(d+1)y2 = ±1; or x2-(d2+1)y2 = ±1 respectively. To find the complete soln of this form, use the more general one for x1

2+x22+x3

2 = y12+y2

2+y32,

(a-b)2 + (c+d)2 + (e-f)2 = (a+b)2 + (c-d)2 + (e+f)2

where ab-cd+ef = 0. It is easy to make one of the terms vanish. (a-b)2 + (c+d)2 + 4n = (a+b)2 + (c-d)2

where n = ab-cd. Let {a,b,c} = {x+y, x-y, y2} and this condition becomes, x2-(d+1)y2 = n

where one can then set n = ±1. If d,x,y are odd, then all the terms above are even and can be reduced. Alternative versions are, Catalan (complete) (ad+qr)2 + (bd-pr)2 + w2 = (a+bd-pr)2 + (ad-b+qr)2, where r = (a2+b2-w2)/2, ap+bq = 1. If we set p = q = 0, this reduces to, (ad)2 + (bd)2 + w2 = (a+bd)2 + (ad-b)2, if a2+b2 = w2

or, explicitly, (u2+v2)2 + (2duv)2 + (du2-dv2)2 = (du2-2uv-dv2)2 + (u2+2duv-v2)2 Integral solns to p2+q2+r2 = s2+1 can then be obtained by setting the last term u2+2duv-v2 = ±1. Let {u,v} = {x-dy, y} and this transforms into the Pell equation x2-(d2+1)y2 = ±1. Other solns are, E.Fauquembergue: (a+1)2 + (2a)2 + 1 = (2a+1)2 + (a-1)2

(2a2-b2+1)2 + (2ab)2 + 1 = (2a2+1)2 + (b2-1)2

Piezas (a2-c2+a+1)2 + (2c)2 + 1 = (a2-c2+a-1)2 + (2a+1)2

(a2+c2+a-1)2 + (2c)2 + (2a+1)2 = (a2+c2+a+1)2 + 1

which were derived using the complete soln of x12+x2

2+x32 =

y12+y2

2+y32. For the special case c = 0, these two become the same

non-trivial identity, (a2+a-1)2 + (2a+1)2 = (a2+a+1)2 + 1 while Fauquebergue’s for either {a,b} = 0 reduces to a tautology. The ff has been discussed previously, A. Gerardin (y-1)2 + y2 + (y+1)2 = x2 + 1, if x2-3y2 = 1. Q: Does this generalize to an identity needing Pell equations distinct from the one derived from Catalan’s? E. Lucas (4m)2 + (8m+1)2 + (8m+2)2 = (12m+2)2 + 1

3. Form: (x2-1)(y2-1) = (z2-1)2 This eqn was studied by Schinzel and Sierpinski. Expanded out, this is just a special case of the previous form being, x2 + y2 + (z2-1)2 = x2y2 +1 This has an infinite number of integral solns. A pair is given by, {x,y} = {z+t, z-t}, if t2-2z2 = 2. (Smallest soln is {t,z} = {10,7}.){x,y} = {z+2t, z-2t}, if z2-2t2 = -1

Q: Any others? As with Gerardin’s and others, this can also be solved as x2 + y2 + (z2-1)2 = (x2y2 +1)k for any k > 0.

4. Form: x2+y2+z2 = u2+v2+w2 This form is important since certain equal sums of like powers for k = 4,5,6 may have this as a “side” condition. The complete soln already has been given earlier, (a+b)2 + (c+d)2 + (e+f)2 = (a-b)2 + (c-d)2 + (e-f)2

where ab+cd+ef = 0. It is easy to give other conditions such that one of the terms vanish or has special forms like equal to unity, etc. The complete soln for k = 1,2 is, L. Dickson (ad+e)k + (bc+e)k + (ac+bd+e)k = (ac+e)k + (bd+e)k + (ad+bc+e)k, for k = 1,2 Note: This can be true for k = 4 as well if (a+b)(c+d) = -3e. (Update, 12/29/09): Alain Verghote Define {p,q} = {x2+2y2, 2x+xy2}, then, p2 = (x2)2 + (2xy)2 + (2y2)2

q2 = (2x)2 + (2xy)2 + (xy2)2 Since there is a common middle term, then,

(2x)2 + (xy2)2 + (x2+2y2)2 = (2x+xy2)2 + (x2)2 + (2y2)2

This belongs to the general form, ak + bk + (c+d)k = (a+b)k + ck + dk, which is already for k = 1, and will also be true for k = 2 if ab-cd. (End update.)

C. Goldbach p2 + q2 + (p+q+3r)2 = (p+2r)2 + (q+2r)2 + (p+q+r)2

Note: Can be simultaneous for k = 2,3 if 6(pq+pr+qr)+5r2 = 0. Alternatively, this can be more simply expressed as, (p-r)k + (q-r)k + (p+q+r)k = (p+r)k + (q+r)k + (p+q-r)k

which is already for k = 2, but is also valid for k = 3 if 6pq = r2. Euler: a2 + b2 + c2 = (a+2d)2 + (b+2d)2 + (c+2d)2, if a+b+c+3d = 0. In general, a2 + b2 + c2 = (a+pm)2 + (b+qm)2 + (c+rm)2, if m = -2(ap+bq+cr)/(p2+q2+r2) R. Davis

a2+b2+c2 = (ap+bq+cr)2 + (aq+br+cp)2 + (ar+bp+cq)2, if p+q+r =1 and 1/p+1/q+1/r = 0. Q: This seems a particularly elegant soln. In what context does it appear? L. Lander x1

2+x22+x3

2 = y12+y2

{x1, x2, x3} = {b(ac+d), (a-c)(ac-d), ab(a+c)}{y1, y2, y3} = {b(ac-d), (a+c)(ac+d), ab(a-c)} where d = b2-c2. Any soln that also satisfies x1x2x3 = y1y2y3 (which is the case for this example) can then generate solns to, z1

k+z2k+z3

k+z4k = z5

k+z6k+z7

k+z8k, for k = 1,3,5

A. Choudhry (ay+b)2 + (cy+ad)2 + (-ay+b+cd)2 = (ay-b)2 + (cy-ad)2 + (-ay-b-cd)2

for any variable. For certain values a,b,c,d,y, which require solving a particular elliptic curve, this also becomes true for k = 2,3,4. A. Choudhry x1

k+x2k+x3

k = y1k+y2

{x1, x2, x3} = {-2am+(b+c)n, (a+b)m+(a+c)n, (a-b)m-(a-c)n}{y1, y2, y3} = { 2am+(b+c)n, -(a+b)m+(a+c)n, -(a-b)m-(a-c)n}

for k = 1,2. This can be extended for k = 6, but a quartic polynomial in c must be made a square as will be discussed in the section on Sixth Powers. Q: Any other soln for k=2 that can be used for k=6?

5. Form: x2+y2+z2 = (u2+v2+w2)Poly(t) H. Burhenne x1

2+x22+x3

2 = (p2+q2+r2)(x2+y2+z2)2

{x1, x2, x3} = {ps-2tx, qs-2ty, rs-2tz}, where {s,t} = {x2+y2+z2, px+qy+rz} for arbitrary p,q,r and x,y,z. One can also set {p,q,r} = {y,z,x} to solve x1

2+x22+x3

2 = (x2+y2+z2)3. This was generalized as, S. Realis ax1

2+bx22+cx3

2 = (ap2+bq2+cr2)(ax2+by2+cz2)2

{x1, x2, x3} = {ps-2tx, qs-2ty, rs-2tz} where {s,t} = {ax2+by2+cz2, apx+bqy+crz} where Burhenne’s is simply the case a=b=c=1. (This can be generalized to n addends as discussed in Section 003.) The next one is also essentially the same. G. Malfatti

p2+q2+r2 = (a2+b2+c2)(d2+e2+f2)2 {p,q,r} = {(d2+e2-f2)a+2bdf+2cef, (-d2+e2+f2)b+2adf-2cde, (d2-e2+f2)c+2aef-2bde} H. Mathieu x2+y2+z2 = (a2+b2+c2)(c2m2+c2n2+(am+bn)2) {x,y,z} = {abm+(b2+c2)n, abn+(a2+c2)m, acn-bcm}

6. Form: x2+y2+z2 = 3xyz Known as the Markov equation, Frobenius proved that x2+y2+z2 = nxyz has positive integer solns only for n=1,3 with the former reducible to the latter by scaling variables. Markov gave various parametrizations (?) to this equation though this author hasn’t been able to find them yet. (The similar equation x3+y3+z3 = 3xyz will be discussed later in the section on third powers.) Define {x,y,z} as a Markov triple, with the first few ones as {1,1,1}, {2,1,1}, {5,1,2}, {13,1,5}, etc. It can be shown that one soln leads to another, Theorem: “If x2+y2+z2 = 3xyz, then x2+y2+(3xy-z)2 = 3xy(3xy-z)." More generally, if we let z = (3xy ± p)/2, it transforms to: p2-(9x2-4)y2 = -4x2 (eq.1) which is a Pell-like eqn. This approach is essentially solving the Markov equation as a quadratic in z and making its discriminant a square. The problem now is to find an x such that this has a soln {p,y}. If for a particular x there is one, then there is in fact an

infinity. For the particular case x = 1, let {x,y,z} = {1, y, (3y ± p)/2}, then eq.1 is, p2 -5y2 = -4 which guarantees a bisection (every other term) of the Fibonacci numbers, namely y1 = {1, 2, 5, 13, etc.}, will appear in a Markov triple,. Since {p,y} are either both odd or even, then z will always be an integer. For x = 2, let {x,y,z} = {2, y, (3y ± 2r)}, then eq.1 is, r2 -2y2 = -1 and a bisection of the Pell numbers as y2 = {1, 5, 29, 169, etc} will also be in a triple. So what other x are there? There are x that are neither Fibonacci nor Pell, the smallest of which are x = 194, 433, 1325, etc. Since the Markov eqn is symmetric and the terms {x,y,z} can be interchanged, the elements of the bisections naturally can also be used as an x. For example, if we use x = 5, eq.1 is, p2-221y2 = -100 This has three fundamental y, namely y = {1, 2, 13} but a non-fundamental one is y = 194 which explains this smallest non-Fibonacci, non-Pell, Markov number and others. In turn, we can then use this y as an x. In general, if a number appears in a Markov triple, then it will appear in an infinite number of triples. Q. Any parametric soln?

III. Sums of four or more squares

1. a2+b2+c2+d2 = ek

2. a2+b2+c2+d2 = e2+f2

3. a2+b2+c2+d2 = e2+f2+g2

4. a2+b2+c2+d2 = e2+f2+g2+h2

5. a2+b2+c2+d2+e2 = f2

6. 2(a2+b2+c2+d2) = (a+b+c+d)2

1. Form: a2+b2+c2+d2 = ek Theorem: “The complete soln to a2+b2+c2+d2 = e2 is given by the scaled Euler-Aida Ammei identity, ((p2-q2-r2-s2)t)2 + (2pqt)2 + (2prt)2 + (2pst)2= ((p2+q2+r2+s2)t)2 where t is a scaling factor.” Proof (Piezas): For any given soln {a,b,c,d}, one can always find rational {p,q,r,s,t} using the formulas, {p,q,r,s,t} = {a+e, b, c, d, (e-a)/(2(b2+c2+d2)} (End proof.) Other solns are: J. Neuberg: a2(a+b+c)2 + b2(a+b+c)2 + c2(a+b+c)2 + (ab+ac+bc)2 = (a2+b2+c2+ab+ac+bc)2

E. Catalan: (p2-q2)2 + ((p+q)(p+r))2 + ((p+q)(p-r))2 + (2pq-q2-r2)2 = (p2+2q2+r2)2 A. Martin

(4pr+s)2 + (4pr-s)2 + (4qr+s)2 + (4qr-s)2 = (4p2+4q2+2r2)2, if s = 2p2+2q2-r2 G. Metrod x2 + (x+2pq)2 + (x+4pq)2 + (x+6pq)2 = (p2+5q2)2, if x = (p2-6pq-5q2)/2 The last is a special case of (x+ay)2 + (x+by)2 + (x+cy)2 + (x+dy)2 = z2. This can be solved as a quadratic in x and making its discriminant a square entails solving an expression of the form ry2+16z2 = t2 which can easily be solved. For k > 2, just like the other sums of n squares, the equation a2+b2+c2+d2 = ek has an identically true soln for all positive integer k. In fact, by Lagrange’s Four-Square Theorem, all positive integers can be expressed as the sum of at least four squares. Other results are, Euler {ap+bq+cr+ds, ar-bs-cp+dq, -as-br+cq+dp, aq-bp+cs-dr}{-aq+bp+cs-dr, as+br+cq+dp, ar-bs+cp-dq, ap+bq-cr-ds}{ar+bs-cp-dq, -ap+bq-cr+ds, aq+bp+cs+dr, as-br-cq+dp}{-as+br-cq+dp, -aq-bp+cs+dr, -ap+bq+cr-ds, ar+bs+cp+dq} The sum of the squares of each row or column is (a2+b2+c2+d2)(p2+q2+r2+s2). Note that each term has two negative signs other than those on the main diagonal slanting towards the right which have none. By letting {p,q,r,s} = {b,c,d,a} the common sum becomes (a2+b2+c2+d2)2 thus giving a 4x4 semi-magic square of squares. (If the two main diagonals also have the same sum, then it is known as a full magic square.) Another form is {p,q,r,s} = {d,b,a,c}. However, let {p,q,r,s} = {c,d,a,b},

{2ac+2bd, a2-b2-c2+d2, -2ab+2cd, 0}{2bc-2ad, 2ab+2cd, a2-b2+c2-d2, 0}{a2+b2-c2-d2, -2ac+2bd, 2bc+2ad, 0}{0, 0, 0, (a2+b2+c2+d2)} or a 3x3 semi-magic square of squares and which gives equivalent forms of the Lebesgue Three-Square Identity, (2ac+2bd)2 + (a2-b2-c2+d2)2 + (2ab-2cd)2 = (a2+b2+c2+d2)2 (There is probably an octonion 8x8 version of Euler’s 4x4 square though this author has not been able to find it yet.) A more general version was given by Lagrange as, x1

2+mx22+nx3

2+mnx42 = (a2+mb2+nc2+mnd2) (p2+mq2+nr2+mns2)

where {x1, x2, x3, x4} = {ap-mbq-ncr+mnds, aq+bp-ncs-ndr, ar+mbs+cp+mdq, as-br+cq-dp}

2. Form: a2+b2+c2+d2 = e2+f2 The smallest positive primitive solns are, 12 + 12 + 32 + 32 = 22 + 42 = 2022 + 22 + 32 + 32 = 12 + 52 = 2612 + 12 + 42 + 42 = 32 + 52 = 3422 + 22 + 22 + 52 = 12 + 62 = 37 If a = b = z, and c = d = t, then we get the special case,

e2+f2 = 2(z2+t2) (eq.2) which can be given a complete soln. See also Form 8 of "Sums of Two Squares". Q: Any simple identity for eq.1 not a derivation of the others in the subsequent section?

3. Form: a2+b2+c2+d2 = e2+f2+g2 Euler: ak + bk + ck + (a+b+c)k = (a+b)k + (a+c)k + (b+c)k, for k = 1,2 E. Catalan (-a+d)k + (-b+d)k + (-c+d)k + dk = ak + bk + ck, if a+b+c = 2d Birck (-a+b+c)k + (a-b+c)k + (a+b-c)k + (a+b+c)k = (2a)k + (2b)k + (2c)k, for k = 1,2 These three are different forms of the same identity. Birck's form appears in the 8th power Birck-Sinha Identity as well as for Sinha’s 2-4-6 Identity to be discussed later.

4. Form: a2+b2+c2+d2 = e2+f2+g2+h2 J. Zehfuss (2a)k + (2b)k + (2c)k + (2d)k = (-a+b+c+d)k + (a-b+c+d)k + (a+b-c+d)

k + (a+b+c-d)k, for k = 1,2 Note that for d = 0, this reduces to Birck’s version. This can also be expressed, after minor changes in signs as, M. Hirschhorn (4p+1)2 + (4q+1)2 + (4r+1)2 + (4s+1)2 = 4(p+q+r+s+1)2 + 4(p-q-r+s)2 + 4(p+q-r-s)2 + 4(p-q+r-s)2

or, Hirschhorn's Odd-Even Identity, proving that the sum of four distinct odd squares is the sum of four distinct even ones. Proof: This can easily be shown true by equating the terms on the LHS of both forms, solving for {p,q,r,s}, and substituting into the RHS of the second. A third equivalent form, more symmetrical in one sense, is, J. Wroblewski (t+x+y+z)k + (t-x-y+z)k + (t-x+y-z)k + (t+x-y-z)k = (t-x-y-z)k + (t+x+y-z)k + (t+x-y+z)k + (t-x+y+z)k, for k = 1,2 where {x,y,z} in the LHS is just negated in the RHS. (Update, 2/2/6/10): Alain Verghote gave a generalization as two (2.5.6) identities, Form 1: (-a+b+c+d+e)k + (a-b+c+d+e)k + (a+b-c+d+e)k + (a+b+c-d+e)k + (a+b+c+d-e)k = (2a)k + (2b)k + (2c)k + (2d)k + (2e)k + (a+b+c+d+e)k, for k = 1,2 Form 2:(-a+b+c+d+e)2 + (a-b+c+d+e)2 + (a+b-c+d+e)2 + (a+b+c-d+e)2 + (a+b

+c+d-e)2 = (-a+b+c+d)2 + (a-b+c+d)2 + (a+b-c+d)2 + (a+b+c-d)2 + (a+b+c+d+e)2 + (2e)2

Note how for e = 0, Form 1 reduces to Zehfuss', while Form 2 reverts into a tautology. These can be further generalized as, (-a+b+c+d+e+f)k + (a-b+c+d+e+f)k + (a+b-c+d+e+f)k + (a+b+c-d+e+f)k + (a+b+c+d-e+f)k + (a+b+c+d+e-f)k = (2a)k + (2b)k + (2c)k + (2d)k + (2e)k + (2f)k + 2(a+b+c+d+e+f)k, for k = 1,2 and so on. Setting f = 0 reduces to Form 1. The significance of Birck's and Zehfuss' symmetric versions is that they can give rise to the multi-grade identity,

(-a+b+c+d)k + (a-b+c+d)k + (a+b-c+d)k + (a+b+c-d)k + (2e)k + (2f)k + (2g)k + (2h)k =(-e+f+g+h)k + (e-f+g+h)k + (e+f-g+h)k + (e+f+g-h)k + (2a)k + (2b)k + (2c)k + (2d)k, for k = 1,2,4,6,8 if abcd = efgh and an+bn+cn+dn = en+fn+gn+hn, for n = 2,4. This has an infinite number of solns and one can also let d = h = 0 for a simpler version. See Eighth Powers for details. However, to create an analogous multi-grade using Verghote’s identity would entail a more complicated set of conditions on ten variables. (End update.)

5. Form: a2+b2+c2+d2+e2 = f2 The complete soln is given by the scaled Euler-Aida Ammei identity, ((a2-b2-c2-d2-e2)t)2 + (2abt)2 + (2act)2 + (2adt)2 + (2aet)2 =

((a2+b2+c2+d2+e2)t)2 Other solns are, Fauquembergue n-Squares Identity (a2+b2-c2-d2+x)2 + (2ac+2bd)2 + (2ad-2bc)2 + 4x(c2+d2) = (a2+b2+c2+d2+x)2

where it can be set as x = e2 (or any n squares). It can also be shown that the square of six squares is identically the sum of five squares in two ways: Piezas (a2+b2+c2-d2-e2-f2)2 + 4(ad+be+cf)2 + 4(ae-bd)2 + 4(af-cd)2 + 4(bf-ce)2 = (a2+b2+c2+d2+e2+f2)2 (a2+b2-c2-d2+e2+f2)2 + 4(ac+bd)2 + 4(ad-bc)2 + 4(ce+df)2 + 4(cf-de)2 = (a2+b2+c2+d2+e2+f2)2 Q: Any other square of n distinct squares expressible as the sum of n-1 squares (or less) other than this and the Lebesgue Three-Square? (See update in next page.) E. Barisien (x6)2 + (4x2y4)2 + (4xy5)2 + (2y6)2 + (2xy(2x2+y2)(x2+2y2))2 = (x6+8x4y2+8x2y4+2y6)2 Q: Any other similar polynomial whose square is expressible as the sum of 5 or less squares?

6. Form: 2(a2+b2+c2+d2) = (a+b+c+d)2 (Update, 6/26/11) Contributed by Christoph Soland. A configuration of four mutually tangent circles is known as Soddy circles. Configurations come in dual pairs with the same tangency points. Let {e1, e2, e3, e4} be the curvatures (1/radius) of the four circles and {f1, f2, f3, f4} those of the dual circles. Then, 2(e1

2+e22+e3

2+e42) = (e1+e2+e3+e4)2 (eq.1)

and similarly for the fi. For solutions to eq.1, let x2+y2+z2 = t2. Then, {e1, e2, e3, e4} = {t+x+y+z, t+x-y-z, t-x+y-z, t-x-y+z} {f1, f2, f3, f4} = {t-x-y-z, t-x+y+z, t+x-y+z, t+x+y-z} Ref.: Coxeter, H.S.M. Introduction to Geometry, 2ed. Note: Parametric solutions to eq.1 where all ei are squares 2

(v4+x4+y4+z4) = (v2+x2+y2+z2)2 are also known. See Form 18 here.

IV. Some Identities of Squares

1. Euler-Aida Ammei Identity2. Brahmagupta-Fibonacci Two-Square Identity3. Euler Four-Square Identity4. Degen-Graves-Cayley Eight-Squares Identity5. V. Arnold’s Perfect Forms6. Lagrange’s Identity

7. Difference of Two Squares Identity(Update, 10/26/09): The Lebesgue Polynomial Identity is given by, (a2+b2-c2-d2)2 + (2ac+2bd)2 + (2ad-2bc)2 = (a2+b2+c2+d2)2 We can generalize this to the form, x1

2+x22+… +xm

2 = (y12+y2

2+… +yn2)2

It can be proven there are polynomial identities with integer coefficients for all m = n (discussed in Sums of Three Squares). But what about when m < n? It turns out we can generalize the one by Lebesgue by looking at its underlying structure. Note that this can be expressed as, (p1

2+p22+q1

2+q22)2 - (p1

2+p22-q1

2-q22)2 = 4(p1

2+p22)

(q12+q2

2) (eq.1) Let {a,b} = {p1

2+p22, q1

2+q22} and this reduces to the basic

Difference of Two Squares Identity, (a+b)2 - (a-b)2 = 4ab a special case of Boutin's Theorem also given at the end of this section which generalizes it to a sum and difference of kth powers. Since (p1

2+p22)(q1

2+q22) = r1

2+r22 by the Bramagupta-Fibonacci

Two-Square Identity, then the RHS of eq.1 can be expressed as the sum of two squares, which explains the Lebesgue Identity. But there are the higher Euler Four-Square Identity and Degen Eight-Square Identity. Thus, let {a,b} = {p1

2+p22+p3

2+p42, q1

2+q22+q3

2+q42}

and we get, (p1

2+p22+p3

2+p42+q1

2+q22+q3

2+q42)2 - (p1

2+p22+p3

2+p42-q1

2-q22-

q32-q4

2)2 = 4(p12+p2

2+p32+p4

2)(q12+q2

2+q32+q4

2) Since the RHS can be expressed as four squares, this gives an identity for the square of 8 squares as the sum of five squares. Using the Degen Eight-Square, this gives one for the square of 16 squares as nine squares. In general, we can have a Theorem 3 to complement the first two in Sums of Three Squares. Theorem 3: Given x1

2+x22+… +xm

2 = (y12+y2

2+… +yn2)2, where

m ≤ n, one can identically express the square of n squares as m squares for: I. All m = n. II. For m < n, with n even (v > 1):a) {m,n} = {2v±1, 2v}b) {m,n} = {4v-3, 4v}c) {m,n} = {8v-7, 8v} III. For m < n, with n odd (v > 1):a) For n = 4v-1, then m = 4v-3.b) For n = {8v-1, 8v-3, 8v-5}, then m = 8v-7. Thus, for example, (y1

2+y22+… +y16

2)2 can be identically expressed as m squares for m = {1, 9, 13, 15, 16}. (Q. Is this the most number of m when m ≤ n?) However, when m < n, and after eliminating the possibility of (x1

2+x22+x3

2)2 expressed as two non-zero squares, two cases are still unresolved: n = 5 and n = 8v-7. Whether or not there are identities for these remains to be seen. Proof: One simply uses the difference of two squares identity: (a+b)2 + (a-b)2 = 4ab (eq.1)

Let a = (p12+p2

2+p32+p4

2+ ... +p2u2). Disregarding the square

numerical factor, the RHS of eq.1 becomes (by distributing in pairs) as, b(p1

2+p22) + b(p3

2+p42) + ...

Let b = (q1

2+q22). Since by the Bramagupta-Fibonacci Identity

(x12+x2

2)(y12+y2

2) = z12+z2

2, then the RHS can be expressed as 2u squares. Thus, {m,n} = {2u+1, 2u+2} or, equivalently, {2v-1, 2v}. However, if the identity is not used on one pair which is instead expressed as four squares, then the RHS has 2u+2 squares. Thus {m,n} = {2u+3, 2u+2} or, {2v+1, 2v}, proving part (Ia) of Theorem 3. Similarly, let a = (p1

2+p22+p3

2+p42+ ... +p4u

2). Disregarding again the square numerical factor, the RHS (by distributing fourwise) is, b(p1

2+p22+p3

2+p42) + b(p5

2+p62+p7

2+p82) + ...

Let b = (q1

2+q22+q3

2+q42). Since by the Euler Four-Square Identity

(x12+x2

2+x32+x4

2)(y12+y2

2+y32+y4

2) = z12+z2

2+z32+z4

2, then the RHS can be expressed as 4u squares. Thus, {m,n} = {4u+1, 4u+4}, or {4v-3, 4v}, proving part (Ib). The last part is proven using the last such identity (by the Hurwitz Theorem) namely, the Degen Eight-Square Identity. Finally, part II of the Theorem can be proven by simply setting the appropriate number of variables yi as equal to zero. For example, we can reduce the {5,8} Identity to a {5,7} by letting one of the eight squares as zero to get, (a2+b2+c2+d2-e2-f2-g2)2 + 4(ae-bf-cg)2 + 4(af+be-dg)2 + 4(ag+ce+df)2 + 4(bg-cf+de)2 = (a2+b2+c2+d2+e2+f2+g2)2

and so on. (End update.) 1. Euler-Aida Ammei Identity Theorem: “The square of the sum of n squares is itself a sum of n squares.” (x1

2-x22-…-xn)2 + å(2x1xn)2 = (x1

2+x22+…+xn

Examples: (a2-b2)2 + (2ab)2 = (a2+b2)2 (a2-b2-c2)2 + (2ab)2 + (2ac)2 = (a2+b2+c2)2 (a2-b2-c2-d2)2 + (2ab)2 + (2ac)2 + (2ad)2 = (a2+b2+c2+d2)2 and so on. Note that these can be alternatively expressed as the basic identity, (x1

2-x0)2 + (2x1)2x0 = (x1

2+x0)2

for arbitrary x1 and x0, where for the theorem it was set x0 = x2

2+x32+

…+xn2. By letting x0 = x2

2, one can see this basic identity is essentially the formula for Pythagorean triples. A stronger result by M. Moureaux is that, “The kth power of the sum of n squares, for k a power of 2, is itself a sum of n squares.” After some time with Mathematica, I observed that there seems to be this beautifully consistent pattern in the identities based on a certain algebraic form. Let the summation å be from m = {2 to n}, then,

(a)2 + å(2x1xn)2 = (x12+x2

2+…+xn2)2

(b)2 + å(4ax1xn)2 = (x1

2+x22+…xn

2)4, (c)2 + å(8abx1xn)2 = (x1

2+x22+…xn

2)8, (d)2 + å(16abcx1xn)2 = (x1

2+x22+…xn

2)16, (e)2 + å(32abcdx1xn)2 = (x1

2+x22+…xn

2)32, and so on, where a = -(x1

2-x22-…-xn) and b = (4x1

4+4ax12-a2), c =

(4a4+4a2b-b2), d = (4b4+4b2c-c2), e = (4c4+4c2d-d2), etc. It seems there is this “recurrence relation” involving the algebraic form 4u4+4u2v-v2. I have no proof this in fact goes on, though it would be odd if the pattern stops. (And the algebraic form factors over √2, which is only appropriate since we are dealing with powers of two.) To generalize the Euler-Aida Ammei identity: For what k is the kth power of n diagonal quadratic forms identically a sum of like form? Or, c1w1

2 + c2w22 + …+ cnwn

2 = (c1x12 + c2x2

2 + …+ cnxn2)k

It turns out for the monic case, or when c1 = 1, the answer is for all positive integer k. In a previous section, it was proven that, “the kth power of the sum of n squares (x1

2+x22+… xn

2)k is itself the sum of n squares”. It takes only a very small modification of the proof to generalize this. Theorem 1. “The expression (c1x1

2 + c2x22 + …+ cnxn

2)k, for c1 =

1, is identically a sum of n squares of like form for all positive integer n and k.” Proof: (Piezas) To recall, let the expansion of the complex number (a±bi)k be, U+Vi = (a+bi)k; U-Vi = (a-bi)k

where U,V are expressions in the arbitrary a,b. Their product, or norm, is, U2+V2 = (a2+b2)k Since b can be factored out in V, or V = V1b, if we let {a,b} = {p1,√p0}, then, U2 + p0V1

2 = (p12+p0)k

All these are familiar. The slightly different step is that since p0 is arbitrary, one can choose it to be the sum of squares of form p0 =

c2p22+c3p3

2+…+cnpn2 and distributing terms, we get,

U2 + c2p2

2V12 + … + cnpn

2V12 = (p1

2 + c2p22 +…+ cnpn

thus proving the kth power of the right hand side of the eqn is a sum of squares of like form. (End proof.) For k = 2 and all ci = 1, this is of course the Euler-Aida Ammei identity and the inspiration for the proof. For k = 3, this is, (p1

3-3p0p1)2 + p0(3p12-p0)2 = (p1

2+p0)3

for p0 = c2p22 +…+ cnpn

2 and so on. For the more general non-monic

case (c1y12 + …+ cnyn

2)k, some particular identities are also known for the case k=3, J. Neuberg ax1

2+bx22+cx3

2 = (ap2+bq2+cr2)3

{x1, x2, x3} = {p(y-2z), q(y-2z), ry}, if y = 4(ap2+bq2)-z, z =

ap2+bq2+cr2

G. de Longchamps ax1

2+bx22+cx3

2+dx42 = (ap2+bq2+cr2+ds2)3

{x1, x2, x3, x4} = {p(y-2z), q(y-2z), ry, sy}, if y = 4(ap2+bq2)-z, z

= ap2+bq2+cr2+ds2

This author observed that this can be generalized as, ax1

2+bx22+cx3

2+dx42+ex5

2 = (ap2+bq2+cr2+ds2+et2)3

{x1, x2, x3, x4, x5} = {p(y-2z), q(y-2z), ry, sy, ty}, if y = 4

(ap2+bq2)-z, z = ap2+bq2+cr2+ds2+et2 and so on for n variables by simply modifying z. A more systematic approach is given below. Theorem 2. “The non-monic expression (c1 x1

2 + c2x22 + …+ cnxn

2)k is identically a sum of n squares of like form for all positive integer n

and odd k.” Proof: The proof is a variation of the one above. One simply solves the equation, U2+c1V2 = (x2+c1y2)k, by equating its linear factors, U+V√-c1 = (x+y√-c1)k, U-V√-c1 = (x-y√-c1)k, and easily solving for U,V as expressions in x,y. Since, for odd k, x can be factored out in U, or U = U1x, if we let {x,y} = {√p0, p1} then, p0U1

2 + c1V2 = (p0+c1p12)k,

where p0 can then be set as the sum of squares p0 = c2p2

2 +…+ cnpn2,

giving, c1V2 + (c2p2

2 +… + cnpn2)U1

2 = (c1p12 + c2p2

2 +…+ cnpn2)k

proving that, for odd k, the right hand side is identically the sum of squares of like form. (End of proof.) For k = 3, this is, c1(c1p1

3-3p0p1)2 + p0(3c1p12-p0)2 = (c1p1

2+p0)3

for p0 = c2p2

2 +…+ cnpn2, and so on for all odd k.

2. Brahmagupta-Fibonacci Two-Square Identity

(ac+bd)2 + (ad-bc)2 = (a2+b2)(c2+d2) This can be generalized as, (ac+nbd)2 + n(ad-bc)2 = (a2+nb2)(c2+nd2) From the Two-Square we can derive the Euler-Lebesgue Three-Square, (a2+b2-c2-d2)2 + (2ac+2bd)2 + (2ad-2bc)2 = (a2+b2+c2+d2)2 This can be generalized by the Fauquembergue n-Squares Identity. It is a bit difficult to convey with limited notation but in one form can be seen as, (a2+b2-c2-d2+x)2 + (2ac+2bd)2 + (2ad-2bc)2 + 4x(c2+d2) = (a2+b2+c2+d2+x)2

where x is arbitrary and can be chosen as any sum of n squares. Note that for x = 0 this reduces to the Euler-Lebesgue. For the case x as a single square this gives, after minor changes in variables, (a2+b2+c2-d2-e2)2 + (2ad+2ce)2 + (2ae-2cd)2 + (2bd)2 + (2be)2 = (a2+b2+c2+d2+e2)2 distinct from the Euler-Aida Ammei identity for n = 5 which is given by, (a2-b2-c2-d2-e2)2 + (2ab)2 + (2ac)2 + (2ad)2 + (2ae)2 = (a2+b2+c2+d2+e2)2 For x = e2+f2, it results in seven squares whose sum is the square of six squares:

(a2+b2+c2+d2-e2-f2)2 + (2ae+2df)2 + (2af-2de)2 + (2be)2 + (2bf)2 + (2ce)2 + (2cf)2 = (a2+b2+c2+d2+e2+f2)2 and so on for other x. 3. Euler Four-Square Identity (a2+b2+c2+d2) (e2+f2+g2+h2) = u1

2 + u22 + u3

2 + u42

u1 = ae-bf-cg-dhu2 = af+be+ch-dgu3 = ag-bh+ce+dfu4 = ah+bg-cf+de Note that a cubic version, in fact, is possible, (x1

3+x23+x3

3+x43)

(y13+y2

3+y33+y4

3) = z13+ z2

3+z33+ z4

3, to be discussed later. Also, by Lagrange's Identity discussed below, the product can be expressed as the sum of seven squares, (a2+b2+c2+d2) (e2+f2+g2+h2) = (ae+bf+cg+dh)2 + (af-be)2 + (ag-ce)2 + (ah-de)2 + (bg-cf)2 + (bh-df)2 + (ch-dg)2

A more general version for squares was also given by Lagrange as, (a2+mb2+nc2+mnd2) (p2+mq2+nr2+mns2) = x1

2+mx22+nx3

2+mnx42

x1 = ap-mbq-ncr+mnds, x2 = aq+bp-ncs-ndr,

x3 = ar+mbs+cp+mdq, x4 = as-br+cq-dp In analogy to the Three-Square, we can also find a Five-Square (by yours truly), (a2+b2+c2+d2-e2-f2-g2-h2)2 + (2u1)2 + (2u2)2 + (2u3)2 + (2u4)2 =

(a2+b2+c2+d2+e2+f2+g2+h2)2 with the ui as defined above. Hence this is another case of a square of n squares expressed in less than n squares. And, in analogy to Fauquembergue’s n squares, another kind of n squares identity can be derived from the Five-Square as, (a2+b2+c2+d2-e2-f2-g2-h2+x)2 + (2u1)2 + (2u2)2 + (2u3)2 + (2u4)2 +

4x(e2+f2+g2+h2) = (a2+b2+c2+d2+e2+f2+g2+h2+x)2

where x again can be any number of squares. For x a square, this can give a 9-square identity. 4. Degen-Graves-Cayley Eight-Squares Identity (DGC) (a2+b2+c2+d2+e2+f2+g2+h2) (m2+n2+o2+p2+q2+r2+s2+t2) = v1

2+v22+v3

2+v42+v5

2+v62+v7

v1 = am-bn-co-dp-eq-fr-gs-htv2 = bm+an+do-cp+fq-er-hs+gtv3 = cm-dn+ao+bp+gq+hr-es-ftv4 = dm+cn-bo+ap+hq-gr+fs-etv5 = em-fn-go-hp+aq+br+cs+dt

v6 = fm+en-ho+gp-bq+ar-ds+ctv7 = gm+hn+eo-fp-cq+dr+as-btv8 = hm-gn+fo+ep-dq-cr+bs+at For convenience, let {a, b,…t} = {a1, a2,…a16}. This can also give a Nine-Square Identity (distinct from the one in the previous section) as, (a1

2+…+ a82- a9

2-…- a162)2 + (2v1)2 + …+ (2v8)2 = (a1

2+ a22 +… +

a162)2

and the DGC n-Squares identity, (a1

2+…+ a82- a9

2-…- a162 + x)2 + (2v1)2 + …+ (2v8)2 + 4x(a9

2 +…

+ a162) = (a1

2+ a22 +… + a16

2 + x)2

Since the DGC is the last bilinear n-squares identity, these two should also be the last of their kind. (Update, 10/26/09): Just like the Two-Square and Four-Square, the Eight-Square Identity can be generalized. For arbitrary {u, v}, (a2+ ub2+c2+ud2+ve2+uvf2+vg2+uvh2) (m2+un2+o2+up2+vq2+uvr2+vs2+uvt2) = x1

2+ux22+x3

2+ux42+vx5

2+uvx62+vx7

2+uvx82

x1 = am-bnu-co-dpu-eqv-fruv-gsv-htuvx2 = bm+an+do-cp+fqv-erv-hsv+gtvx3 = cm-dnu+ao+bpu+gqv+hruv-esv-ftuvx4 = dm+cn-bo+ap+hqv-grv+fsv-etvx5 = em-fnu-go-hpu+aq+bru+cs+dtu

x6 = fm+en-ho+gp-bq+ar-ds+ctx7 = gm+hnu+eo-fpu-cq+dru+as-btux8 = hm-gn+fo+ep-dq-cr+bs+at (End update) 5. V. Arnold’s Perfect Forms Let {a,b,c} be in the integers. Given the equation (au2+buv+cv2)(ax2+bxy+cy2) = az1

2+bz1z2+cz22. If for any integral integral

{u,v,x,y} one can always find integral {z1, z2}, then the binary quadratic form F(a,b,c) is defined as a perfect form. Theorem 1: “The product of three binary quadratic forms F(a,b,c) is of like form.” Proof: (an1

2+bn1n2+cn22)(au2+buv+cv2)(ax2+bxy+cy2) =

az12+bz1z2+cz2

where, z1 = u(n3x+cn2y)+cv(n2x-n1y) z2 = v(an1x+n4y)-au(n2x-n1y) and {n3, n4 } = {an1+bn2, bn1+cn2} from which immediately follows, Corollary:“If there are integers {n1, n2} such that F(a,b,c) = 1, then F(a,b,c) is a perfect form.” Note that if F(a,b,c) is monic, the soln {x,y} = {1,0} immediately

implies this form is perfect. But by dividing z1, z2 with c, a, respectively, and modifying the expressions for {n3, n4} will result in a second theorem, Theorem 2: “If there are integers {n1, n2, n3, n4} such

that an12+bn1n2+cn2

2 = ac, n3 = (an1+bn2)/c, n4 = (bn1+cn2)/a, then F(a,b,c) is a perfect form.” Proof: (an1

2+bn1n2+cn22)(au2+buv+cv2)(ax2+bxy+cy2) =

(az12+bz1z2+cz2

2)(ac) where, z1 = v(n1x+n4y)-u(n2x-n1y) z2 = u(n3x+n2y)+v(n2x-n1y) and {n3, n4 } = {(an1+bn2)/c, (bn1+cn2)/a}. The expressions are essentially the same as in Theorem 1 but have been divided by c,a. This second class is relevant to quadratic discriminants d with class number h(d) = 3m. For imaginary fields with h(-d) = 3, there are sixteen fundamental d, all of which have its associated F(a,b,c) as perfect forms. For brevity, only the first three will be given and in the format {a,b,c}, {n1, n2, n3, n4}: d = 23; {2,1,3}, {1,1,1,2}d = 31; {2,1,4}, {2,0,1,1}d = 59; {3,1,5}, {-2,1,-1,1} For real fields with h(d) = 3, there are forty-two d in the Online Encyclopedia of Integer Sequences, all of which also have F(a,b,c) as perfect. However, while most of the ni for negative d with h(d) = 3,6

were only single digits, for positive d these can get quite large. For ex, d = 2857; {2,51,-32}, {3326866, -127404, -4879, 86873547} Q: Any other theorems regarding the product of two or three binary quadratic forms? We can generalize this somewhat and go to diagonal n-nary quadratic forms (one without cross terms), F(a1,a2,…an):= a1x1

2 + a2x22 +…+ anxn

If we consider the equation, (a1x1

2 + a2x22 +…+ anxn

2)(a1y12 + a2y2

2 +…+ anyn2) = a1z1

a2z22 +…+ anzn

then for what constants {a1, a2,…an} is there such that the product of two diagonal n-nary quadratic forms is of like form? Most of the results have been limited to the special case of all ai = 1 and n = 2,4,8, namely the Brahmagupta-Fibonacci, Euler, and Degen-Graves identities discussed above. The first can be generalized to the form {1,p}, the second to {1, p, q, pq} by Lagrange, and the third to {1,1, p,p, q,q, pq, pq}. Ramanujan in turn generalized Lagrange’s Four-Square Theorem and found 54 {a,b,c,d} such that ax1

2+bx22+cx3

2+dx42 can represent all positive integers, namely,

{1,1,1,v}; v = 1-7{1,1,2,v}; v = 2-14{1,1,3,v}; v = 3-6{1,2,2,v}; v = 2-7{1,2,3,v}; v = 3-10{1,2,4,v}; v = 4-14{1,2,5,v}; v = 6-10

which is the complete list. (Note: Incidentally, it would have been expected that the last would be for v = 5-10. What is the smallest positive integer not expressible by the form {1,2,5,5}?) All 54 are then perfect quaternary forms since, needless to say, the product of two positive integers is always a positive integer. For the first case {1,1,1,1} this is just Euler’s four-square identity and the zi have a bilinear expression in terms of the xi and yi. It might be interesting to know if the other 53 {a,b,c,d} have similar formulas for their zi. In general, for ai not all equal to unity, what other results are there for the product of two n-nary quadratic forms, especially for n not a power of two? Update, 9/21/09: Turns out the form {1,2,5,5} cannot express the number 15. See "The 15 and 290 Theorems" by Conway, Schneeberger, and Bhargava. 6. Lagrange’s Identity A faintly similar identity to the sum-product of n squares given previously is, (x1y1 + … + xnyn )

2 + S (xkyj – xjyk)2 = (x12

+ … + xn2) (y1

2 + … +

) for 1£ k<j£n. For n=3, this has 4 addends, (x1y1+x2y2+x3y3 )

2 + (x1y2 – x2y1 )2 +(x1y3 – x3y1 )

2 + (x2y3 –

x3y2 )2 = (x1

2+x22+x3

2) (y12+y2

2+y32)

while n=4 already involves 7 addends, and is an alternative way to express Euler's Four-Square Identity. A general identity was found by

A. Cauchy and J. Young. The case n=3 was also rediscovered by T.Weddle while studying the semi-axes of an ellipsoid. 7. Difference of Two Squares Identity A difference-product identity on the other hand is given by, (x1

2+ … +xn2 + (y1

2+…+yn))2 – (x12+ … +xn

2 – (y12+…+yn))2 = 4

(x12+ … +xn

2)(y12+ … +yn

Proof: Let a = x1

2+ … +xn2, b = y1

2+ … +yn2, then this is just the

basic identity, (a+b)2-(a-b)2 = 4ab. (End proof.) If the right hand side of the identity is non-trivially expressible as the sum of n squares, as is the case for n = 2,4,8, this automatically implies a square of 2n squares expressible as the sum of n+1 squares, thus explaining the Three, Five, Nine-Square Identities above. (The case n = 1 just gives the formula for Pythagorean triples.) Let, (a2+b2+c2+d2+e2+f2)2 - (a2+b2+c2-d2-e2-f2)2 = 4(a2+b2+c2)(d2+e2+f2)(a2+b2+c2+d2+e2+f2+g2+h2)2 - (a2+b2+c2+d2-e2-f2-g2-h2)2 = 4(a2+b2+c2+d2)(e2+f2+g2+h2) and together with Lagrange’s Identity for n = 3,4 applied on the RHS respectively, this can be used to prove that the square of 6 squares can be expressed as the sum of five squares, (a2+b2+c2-d2-e2-f2)2 + 4(ad+be+cf)2 + 4(ae-bd)2 + 4(af-cd)2 + 4(bf-ce)2 = (a2+b2+c2+d2+e2+f2)2 and provide an alternative soln to expressing the square of 8 squares as a sum of 8 squares,

(a2+b2+c2+d2-e2-f2-g2-h2)2 + 4((ae+bf+cg+dh)2 + (af-be)2 + (ag-ce)2 + (ah-de)2 + (bg-cf)2 + (bh-df)2 + (ch-dg)2) = (a2+b2+c2+d2+e2+f2+g2+h2)2 since the Euler-Aida Ammei gives it as, (a2-b2-c2-d2-e2-f2-g2-h2)2 + 4a2(b2+c2+d2+e2+f2+g2+h2) = (a2+b2+c2+d2+e2+f2+g2+h2)2 Note 1: By not using the Euler-Aida Ammei identity, are there always alternative solns to the square of n squares as the sum of n squares?Note 2: For what n is the square of n squares identically the sum of less than n squares? (It can be for n = 4, 8, 16 (as 3, 5, 9 squares, respectively). In fact, by setting appropriate variables equal to zero, it is the case for all n ≤ 16 other than n = 3, 5, 9.) (See update at start of this section.) A. Boutin Boutin’s Identity: S ± (x1 ± x2 ±…± xk)k = k! 2k-1x1x2…xk where the exterior sign is the product of the interior signs. (Or, the term is negative if there is an odd number of negative interior signs.) The case k=2 gives the well-known, (a+b)2 - (a-b)2 = 4ab while for k = 3,4, (a+b+c)3 - (a-b+c)3 - (a+b-c)3 + (a-b-c)3 = 24abc (a+b+c+d)4 - (a-b+c+d)4 - (a+b-c+d)4 - (a+b+c-d)4 + (a-b-c+d)4 + (a-b

+c-d)4 + (a+b-c-d)4 - (a-b-c-d)4 = 192abcd and so on for other kth powers. The case k=3 then implies that, Piezas (x1

3+x23+…+xn

3) (y13+y2

3+…+yn3) = z1

3+ z23+z3

3+ z43

or, “The product of two sums of n cubes is the sum of four cubes.” Proof: Simply let a = x1

3+x23+…+xn

3, b = y13+y2

3+…+yn3, and c =

9 in Boutin's Identity for k = 3.

A. Univariate: ax2+bx+c = z2

B. Bivariate: ax2+bxy+cy2 = zk

1. x2+cy2 = zk

2. ax2+cy2 = zk, k odd3. x2+2bxy+cy2 = zk

4. ax2+2bxy+cy2 = zk, k oddC. Bivariate: ax2+bxy+cy2 = dz2

1. ax2+bxy+cy2 = dz2

2. ax2+by2+cz2+dxy+exz+fyz = 03. ax2+cy2 = dzk, k > 2

Given an initial solution, the problem of finding an indefinite number of subsequent rational solns to P(x) = z2 where P(x) is the general univariate polynomial of degree k=2,3,4 was solved by Fermat using a simple method. The general principle can be illustrated by the quadratic case.

A. Univariate form: ax2+bx+c = z2 If P(x) has an initial rational soln, we can assume we are dealing with the form, ax2+bx+c2 = z2

To illustrate, given P(x):= px2+qx+r. Change variables from x to v and let x = v+n (for some indefinite n). Expanding, we get, P(v): = pv2 + (2pn+q)v + pn2+qn+r Thus, if n is a soln to pn2+qn+r = y2, then equivalently, P(v): = pv2 + (2pn+q)v + y2

which is the desired form. In general, given a polynomial P(x) of any degree, if n is a soln to P(x) = y2 then the substitution x = v+n will yield a new polynomial P(v) with the constant term y2. Following Fermat, we then assume that, ax2+bx+c2 = (p/qx-c)2

Subtracting one side from the other and collecting the variable x, (p2-aq2)x2 - (2cpq+bq2)x = 0 One can then easily solve for x as, x = (2cpq+bq2)/(p2-aq2) for arbitrary p,q. (The method then is to remove enough terms so

solving for x involves only a linear equation. This can easily be extended to other powers as shall be seen later.) If integral x is desired, one can set p2-aq2 = ±1 and solve this Pell equation. For non-square, positive integer a, this then provides an infinite number of integral solns x. For the related quadratic form, (x+u)(x+v) = dz2

{x, z} = {p2-u, pq}, if p2-dq2 = u-v{x, z} = {p2-v, pq}, if p2-dq2 = -(u-v) involves the solving the Pell-like equation p2-dq2 = k. (Euler) B. Bivariate form: ax2+bxy+cy2 = zk The problem of making the general binary quadratic form into a kth power can be divided into two classes: the monic form a=1 where generic solns are known for all k and the non-monic form where generic solns are known only for odd k.

1. Form: x2+cy2 = zk Euler: x2+cy2 = (p2+cq2)k Same technique of equating factors {x+yÖ-c, x-yÖ-c} = {(p+qÖ-c)k, (p-qÖ-c)k} and easily solving for x,y. Example, for k=2,

(p2-cq2)2 + c(2pq)2 = (p2+cq2)2 which for c=1 gives the familiar Pythagorean triples, and so on for other k, though it should be pointed out that for k > 2 this does not generally give all solns (Pepin). For example, for the particular case c = 47, all relatively prime solns with odd z are given by the method. But a class of even z is given by, Pepin: (13u3+60u2v-168uv2-144v3)2 + 47(u3-12u2v-24uv2+16v3)2 = 23

(3u2+2uv+16v2)3 Q: How then to find the complete soln of x2+cy2 = zk for k>2? For the negative case, or c = -d, one can use a variation of the method above by employing a Pell equation and equate, x2-dy2 = (p2-dq2)k(r2-ds2) where r2-ds2 = 1 and solve for x,y using x+yÖd = (p+qÖd)k(r+sÖd)x-yÖd = (p-qÖd)k(r-sÖd)

2. Form: ax2+cy2 = zk, k odd To solve ax2+cy2 = (ap2+cq2)k, as before, equate factors, xÖa + yÖ-c = (pÖa + qÖ-c)k

xÖa - yÖ-c = (pÖa - qÖ-c)k

then solve for x,y. x = (ak+bk)/(2Öa), y = (ak-bk)/(2Ö-c), where a = (pÖa + qÖ-c), b = (pÖa - qÖ-c). Example for k = 3, {x,y} = {p(ap2-3cq2), q(3ap2-cq2) } and so on for all odd k. For even k there is the problem of {x,y} containing the radical Öa, though it disappears in the monic case a=1.

3. Form: x2+2bxy+cy2 = zk Euler, Lagrange: x2+2bxy+cy2 = (p2+2bpq+cq2)k Equating factors, x+(b+d)y = (p+(b+d)q)k

x+(b-d)y = (p+(b-d)q)k

where d = Ö(b2-c). Then solve for {x,y} giving, x = ((-b+d)ak+(b+d)bk)/(2d) y = (ak-bk)/(2d), where a = (p+(b+d)q), b = (p+(b-d)q)

4. Form: ax2+2bxy+cy2 = zk, k odd Fauquembergue:

ax2+2bxy+cy2 = ak(x2+2bxy+acy2)k Equating factors, x+((b+d)/a)y = am(p+(b+d)q)k

x+((b-d)/a)y = am(p+(b-d)q)k

Solving for {x,y}, x = am((-b+d)ak+(b+d)bk)/(2d) y = am+1(ak-bk)/(2d), where a = (p+(b+d)q), b = (p+(b-d)q), d = Ö(b2-ac), and k = 2m+1. (Note that for a=1, this reduces to the monic case discussed by Euler and Lagrange.) Q: There are certain {a,b,c} such that ax2+2bxy+cy2 = zk for even k has no soln for rational x,y,z. For what {a,b,c}, with a ¹ 1, can we find integral polynomial solutions for all k? C. Bivariate form: ax2+bxy+cy2 = dz2

5. Form: ax2+bxy+cy2 = dz2 This form is merely a special case of the general theorem given by Desboves in the next section. A.Gerardin ax2+bxy+cy2-dz2 = (am2+bmn+cn2-dp2)(au2+buv+cv2)2 {x,y,z} = {-(am+bn)u2-2cnuv+cmv2, anu2-2amuv-(bm+cn)v2, p(au2+buv+cv2)}, for arbitrary u,v.

Piezas ax2+bxy+cy2-dz2 = (am2+bmn+cn2-dp2)(u2-cdv2)2 {x,y,z} = {mu2-cdmv2, nu2+2pduv+d(bm+cn)v2, pu2+(bm+2cn)uv+cdpv2} For the particular case a=d=1, the complete soln as established by Desboves is x2+bxy+cy2 = z2

{x,y,z} = {u2-cv2, 2uv+bv2, u2+buv+cv2} which can be derived by using the initial soln {m,n,p} = {1,0,1} on either of the two previous general identities. If b=0, after some modification this becomes, x2+mny2 = z2

{x,y,z} = {mu2-nv2, 2uv, mu2+nv2}

6. Form: ax2+by2+cz2+dxy+exz+fyz = 0 S. Realis (complete soln) Given one soln to ax2+by2+cz2 = 0 then an infinite more can be found. aX2+bY2+cZ2 = (ax2+by2+cz2)(ap2+bq2+cr2)2 X = x(-ap2+bq2+cr2) – 2p(bqy+crz),

Y = y(ap2-bq2+cr2) – 2q(apx+crz),Z = z(ap2+bq2-cr2) – 2r(apx+bqy) for arbitrary {p,q,r} (or x,y,z since if the equation is equal to zero, either factor is of the same form). Or alternatively, to show its “internal structure” (after a small exchange of variables), ax1

2+bx22+cx3

2 = (ap2+bq2+cr2)(ax2+by2+cz2)2

{x1, x2, x3} = {pv1-2xv2, qv1-2yv2, rv1-2zv2} where {v1, v2} =

{ax2+by2+cz2, apx+bqy+crz} for arbitrary {x,y,z}. In fact, a more general statement can be made. Theorem: "Given one solution to a1y1

2+a2y22+…+ anyn

2 = 0, then an infinite more can be found." Proof: We simply generalize the soln. For four addends, ax1

2+bx22+cx3

2+dx42 = (ap2+bq2+cr2+ds2)(ax2+by2+cz2+dt2)2

{x1, x2, x3, x4} = {pv1-2xv2, qv1-2yv2, rv1-2zv2, sv1-2tv2} where

{v1, v2} = {ax2+by2+cz2+dt2, apx+bqy+crz+dst} for arbitrary {x,y,z,t}. And so on for any n addends. This was previously discussed in Section 003. A similar general identity exists for cubes while there is more limited one for fourth powers of the form x1

4+x24 = x3

4+ x44 such that an initial soln leads to subsequent ones.

A. Desboves (complete soln)

Theorem: “Given one integral soln to ax2+by2+cz2+dxy+exz+fyz = 0, then an infinite more can be given by a polynomial identity.” aX2+bY2+cZ2+dXY+eXZ+fYZ = (ax2+by2+cz2+dxy+exz+fyz)(bp2+fpq+cq2)2 X = -x(bp2+fpq+cq2)Y = (dx+by+fz)p2 + (ex+2cz)pq - cyq2

Z = -bzp2 + (dx+2by)pq + (ex+fy+cz)q2

for arbitrary p,q.

7. Form: ax2 + cy2 = dzk, k > 2 Pepin 3(pr+3qs)2-(pr+9qs)2 = 2(r2-3s2)3, and 3(3pr-15qs)2-(5pr-27qs)2 = 2(r2-3s2)3, where {p,q} = {r2+9s2, r2+s2}. Note: Pepin’s result also solves a2+mb2 = c2+md2, but the complete soln of this is, (pr+mqs)2 + m(ps-qr)2 = (pr-mqs)2 + m(ps+qr)2

for all {m,p,q,r,s}. However, Pepin’s variant form is, (pr+m2qs)2 + m(mpr-mnqs)2 = (npr-m3qs)2 + m(pr+mqs)2

and is true for all {p,q,r,s} only if {m,n} satisfies the elliptic curve, m3-m+1 = n2, one soln of which is {m,n} = {3, 5}. Q: Other poly solns to ax2 + cy2 = dzk for d >1, k >2?

I. Two variables1. {x2+axy+by2, x2+cxy+dy2}2. {x2-ny2, x2+ny2} (Congruent numbers)3. {x2+y, x+y2}4. {x2+y2-1, x2-y2-1}5. {x2+y2+1, x2-y2+1}

II. Three variables6. {x ± y, x ± z, y ± z} (Mengoli's Six-Square Problem)

7. {x2-y2, x2-z2, y2-z2}8. {x2+y2, x2+z2, y2+z2} (Euler Brick Problem)9. {x2+y2+z2, x2y2+x2z2+y2z2}10. {-x2+y2+z2, x2-y2+z2, x2+y2-z2}11. {2x2+y2+z2, x2+2y2+z2, x2+y2+2z2}12. {2x2+2y2-z2, 2x2-y2+2z2, -x2+2y2+2z2}13. {x2+yz, y2+xz, z2+xy}14. {x2+y2+xy, x2+z2+xy, y2+z2+xy}15. {x2-xy+y2, x2-xz+z2, y2-yz+z2}

III. Four variables16. {x2+axy+y2, x2+bxz+z2, y2+cyz+z2} 17. {a2+b2+c2, a2+b2+d2, a2+c2+d2, b2+c2+d2}18. {a2b2+c2d2, a2d2+b2c2}19. {a2b2+c2d2, a2c2+b2d2, a2d2+b2c2}20. {1+abc, 1+abd, 1+acd, 1+bcd}

Most of these were solved by Euler. If you have additional polynomial solutions, pls send them. Or if you have poly solns to a set of simultaneous poly not in the list, those are also most welcome. I. Two variables

Form 1: {x2+axy+by2, x2+cxy+dy2} Euler {x,y} = {(a-c)4-8(b+d)(a-c)2+16(b-d)2, 8(a-c)(a2-4b-c2+4d)} though for the special case when {a,b} = {-c,d}, this implies discriminants are a2-4b = c2-4d, hence yields trivial y = 0. In general, for integral {a,b,c,d} there is an infinite number of non-trivial integral {x,y} such that the two polynomials are squares. To see this, consider the two eqns, {x1

2+ax1+b, x22+cx2+d} = {t1

2, t22}

which have the soln, {x1, x2} = {(p2-b)/(a-2p), (q2-d)/(c-2q)} for free variables {p,q}. To make x1 = x2, it is easy to equate the two formulas to form a quadratic equation in {p,q}. (p2-b)/(a-2p) = (q2-d)/(c-2q) Solving for q, this has a rational root iff the discriminant, a quartic

polynomial in p, is made a square and is given by, 4p4-8cp3-4(2b-ac-4d)p2+8(bc-2ad)p+4(b2-abc+a2d) = z2

Since the polynomial is monic, this is easily attained with one soln as, 4p = (a2+4b-2ac+c2-4d)/(a-c) Treating the quartic as an “elliptic curve”, from this initial point, in general, one can then compute an infinite number of other points.

Form 2: {x2-ny2, x2+ny2} These simultaneous equations involve what is known as congruent numbers: 1. Lucas {x,n,y} = {p2+q2, p3q-pq3, 2} 2. A.Gerardin {x,n,y} = {16p8+24p4q4+q8, 4p4+q4, 4pq(4p4-q4)} {x,n,y} = {p8+6p4q4+q8, 2(p4+q4), 2pq(p4-q4)} Q: Any other n as a polynomial with small degree? 3. J.Maurin, A. Cunningham The ff identity proves that one soln leads to another. {u2-nv2, u2+nv2} = {(x4-2nx2y2-n2y4)2, (x4+2nx2y2-n2y4)2},

where {u,v} = {x4+n2y4, 2xyzt}, if (x2-ny2)(x2+ny2) = t2z2. Any poly soln to the more general concordant forms {x2+my2, x2+ny2} = {t2, z2}, esp to the case when m ≠ 1?

Form 3: {x2+y, x+y2} Simply equate {x2+y, x+y2} = {(x+p)2, (y+q)2} and solve for x,y. (Euler)

Form 4: {x2+y2-1, x2-y2-1} 1. Bhaskara {x,y} = {y2/2+1, (8a2-1)/(2a)} {x,y} = {8a4+1, 8a3} with the last one also found by E. Clere. It may be the case that this has a series of higher degree parametrizations, like the recursive one for x3+y3+z3 = 1 found by D. Lehmer. (See update below.) 2. A.Genocchi (complete rational soln in p,q,r,s) {x,y} = {(2r2-t)/t, 4pqrs/t}, if t = r2 - (p4+4q4)s2

For integral solns, it suffices to solve the Pell equation r2 - (p4+4q4)s2

= ±1. The following identity is essentially the same then,

3. T. Pepin If p2-(r4+4s4)q2 = ±1, (p2+(r4+4s4)q2)2 + (4pqrs)2 - 1 = (2pq(r2+2s2))2, (p2+(r4+4s4)q2)2 - (4pqrs)2 - 1 = (2pq(r2-2s2))2, Update: After a closer inspection, it turns out Bhaskara’s second soln is just the first in a family. Note that if r = 1 in Pepin's identity, then the Pell equation, p2-(1+4s4)q2 = -1 has fundamental polynomial soln {p,q} = {2s2, 1} and from this we can then generate an infinite sequence of polynomial solns. The first yields, (8s4+1)2 ± (8s3)2 = (8s4±4s2)2 + 1 which is Bhaskara’s, while the next is, (2t2-1)2 ± (16s3t)2 = (8s4±4s2)2(2t)2 + 1, where t = 8s4+1 and so on. However, if in Pepin's identity we let s=1, p2-(r4+4)q2 = -1 this also has a parametric soln given by {p,q} = {r2(r4+3)/2, (r4+1)/2} which is integral for odd r. Thus this gives rise to a second family with the first member, ((r4+2)(t-2)/2)2 ± (r3t)2 = (r2(r2±2)t/2)2 +1, if t = (r4+1)(r4+3)

and so on. In summary, the two families deal with the Pell equation x2-dy2 = ±1 with discriminant d of form m2+1 and n2+4 (where the latter is restricted only to odd n since it reduces to the former for even n). Interestingly, as will be seen later, a similar polynomial family exists for third powers x3+y3 = z3±1 which also depends on a parametric soln to a Pell equation, though now this involves a discriminant of form 3(4m3-1).

Form 5: {x2+y2+1, x2-y2+1} J. Drummond {x,y} = {2n2, 2n} Q: Does this pair have a complete soln similar to the one found by Genocchi-Pepin? This is the only soln known so far. Update (7/3/09): This author checked all {x,y} < 1000 and found that, other than the one given by Drummond, the only other solns were x = y = v such that v satisfies the Pell eqn u2-2v2 = 1. II. Three variables Form 6: {x ± y, x ± z, y ± z} 1. Petrus {x,y,z} = {2(pr+qs)2+2(pq+rs)2, 2(pr-qs)2+2(pq-rs)2, 2(pr+qs)2-2(pq+rs)2} where the expressions {pqrs, (p2+s2)(q2+r2)} are to be made squares.

One particular soln is {p,q,r,s} = {112, 12, 35, 15}. 2. Euler {x,y,z} = {a2d2+b2c2+2t2, 2(abcd+t2), 2(abcd-t2)}, where t2 = (a2-b2)(c2-d2)/4 It remains to make {abcd, (a2-b2)(c2-d2)} squares, a very similar condition to the one above. One particular soln is {a,b,c,d} = {9, 4, 81, 49}. By squaring variables, this reduces to the single condition (a4-b4)(c4-d4) = n2. There are in fact polynomial solns to the general problem, one of deg 16 and another of deg 20, given below: 3. Euler: (16-deg) {x,y,z} = {(81+14n4+n8)2+z, 16n2(27+3n2+n4+n6)2-z, 32n4

(81+30n4+n8)} 4. M. Rolle: (20-deg) {x,y,z} = {1+21a4-6a8-6a12+21a16+a20, 10a2-24a6+60a10-24a14+10a18, 6a2+24a6-92a10+24a14+6a18} Any poly soln of deg < 16? Note that if {x ± y, x ± z, y ± z} are squares, this immediately implies that {x2-y2, x2-z2, y2-z2} are also squares.

Form 7: {x2-y2, x2-z2, y2-z2} W. Lenhart

{x,y,z} = {(p2+q2)/(p2-q2), (r2+s2)/(r2-s2), 1} Two expressions are already squares. To make it all three, {p,q,r,s} = {8n, n2+9, 8n(n2-9), n4+2n2+81}

Form 8: {a2+b2, a2+c2, b2+c2} Also known as the Euler Brick Problem: Find {a,b,c,}, or the length, width, height of a brick such that the diagonals {x,y,z} are rational, given by, a2 + b2 = x2

a2 + c2 = y2

b2 + c2 = z2

The smallest integer soln is {44, 117, 240}, 442 + 1172 = 1252

442 + 2402 = 2442

1172 + 2402 = 2672

Olson has proved that an integer Euler brick has the product abcxyz as divisible by 344452. A perfect cuboid would have the space diagonal a square as well, or a2+b2+c2 = d2 for some rational d, but none are known with odd side less than 100 billion (1011). See Durango Bill's The Integer Brick Problem, or Oliver Knill's Hunting for the Perfect Cuboid. Theorem (A. Bremner): “Given the Euler brick {a,b,c,x,y,z}, then {ab, bc, ac, by, az, cx} is another Euler brick.”

Some polynomial solns are: 1. N. Saunderson (in 1740. Later re-discovered by Euler, C. Chabanel, J. Neuberg) {a,b,c} = {(4p2-r2)q, 4pqr, (4q2-r2)p} where {p,q,r} is the Pythagorean triple, {m2-n2, 2mn, m2+n2}. Explicitly, {a,b,c} = {2mn(3m4-10m2n2+3n4), 8mn(m4-n4), m6-15m4n2+15m2n4-n6} though these are factorable polynomials. This soln has the smallest degree known, and it is unknown if there others with deg ≤ 6. (Note also that the middle diagonal a2+c2 = y2 is a perfect 6th power.) 2. C. Kunze {a,b,c} = {2pq, pq2-p, p2q-q} The first two diagonals become squares and it remains to make b2+c2 one, p2q2(p2+q2-4) + p2+q2 = z2

which is a square if p2+q2 = 4, so {p,q} = {2(r2-s2)/(r2+s2), 4rs/(r2+s2)}. This yields the Saunderson-Euler soln given in [1]. 3. Euler {a,b,c} = {(p2-1)/(2p), (q2-1)/(2q), 1}, where {p,q} = {4t/(t2+1), (3t2-1)/(t3-3t)}

The last two diagonals become squares while a2+b2 is, p2(q2-1)2 + q2(p2-1)2 = x2

with the given {p,q} satisfying this condition. This yields an 8th deg polynomial. However, it can also be derived from Saunderson’s soln using the theorem cited above. 4. W. Lenhart {a,b,c} = {(p2-1)/(2p), 2q/(q2-1), 1} a very similar transformation to Euler's. Likewise, these values make the last two diagonal as squares but one still has to solve a2+b2 = x2, or, (p2-1)2(q2-1)2 + 16p2q2 = x2

This belongs to the more general problem of, (p2-m)2(q2-m)2 + 4mnp2q2 = x2

for some constants m,n which has the soln x = (p2-m)(q2-m) + 2mn, if p2+q2 = m+n, hence is reduced to this simpler problem. For Lenhart’s case, with {m,n} = {1, 4), one needs to solve p2+q2 = 5 which has soln, {p,q} = {(s2+4s-1)/(s2+1), 2(s2-s-1)/(s2+1)}. 5. R. Rignaux {a,b,c} = {2pqrs, rs(p2-q2), pq(r2-s2)}, where p2q2(r2-s2)2 + r2s2(p2-q2)2 = t2 (eq.1)

After equating some variables as equal to unity, eq.1 is essentially the same condition given by Euler, and can be solved using Euler’s soln. Note: This author checked the equation a2+b2+c2 = d2 for all given polynomial solns and found that identities (1), (2), after minor change of variables, yield the 8-deg, v8+68v6-122v4+68v2+1 = d2

while (3),(4),(5) is a 16-deg poly to be made a square. As the parametrizations are not complete, a negative result to this hyper-elliptic curve certainly does not imply the non-existence of a perfect Euler brick. (Update, 7/2/10): Bremner’s paper, The rational cuboid and a quartic surface gives more parametrizations of 8th deg, and points out that evidence suggests these can be found for every even deg ≥ 6.

Form 9: {x2+y2+z2, x2y2+x2z2+y2z2} 1. Euler {x,y,z} = {p2+q2-r2, 2pr, 2qr} where {p,q,r} = {8n(n2-1), n4-10n2+5, n(n2+1)(n2-3)} 2. J.Euler {x,y,z} = {(p4-6p2+1)(p2+1), 4p(p2-1)2, 8p2(p2-1)} 3. J.Euler

{x2+y2+z2, x2y2+x2z2+y2z2} = {(a2+b2)2c2, (2ab)2(a4+b4)2} where {x,y,z} = {2a2b, 2ab2, (a2-b2)c}, if a2+b2 = c2

The next three forms can be grouped together.

Form 10: {-x2+y2+z2, x2-y2+z2, x2+y2-z2} Euler {x,y,z} = {a(a+b)p-2c2q, a(a+b)p+2c2q, acp+2cq2}, where {p,q} = {3a+4b, a+2b} First two expressions are already squares. To make it all three, {a,b,c} = {m2+2n2, 2mn, m2-2n2}

Form 11: {2x2+y2+z2, x2+2y2+z2, x2+y2+2z2} Legendre {x,y,z} = {u2-11v2, u2+4uv+11v2, u2-4uv+11v2}, if u4+8u2v2+121v4 = t2. with small solns {u,v} = {2, 1} and {11, 2}. More generally, Piezas For constant m of form m = n2-2, for the three expressions, {mx2+y2+z2, x2+my2+z2, x2+y2+mz2}

where {x,y,z} = {u2-bv2, u2+auv+bv2, u2-auv+bv2} and {a,b} = {2n, 5n2-9}, the last two are squares. The first is also a square if, n2u2-2(5n4-33n2+36)u2v2+n2(5n2-9)2v4 = t2 with Legendre’s as the case n = 2.

Form 12: {2x2+2y2-z2, 2x2-y2+2z2, -x2+2y2+2z2} E. Grebe: {x,y,z} = {p-q, p+2q, 2p+q} These three forms belong to the more general class {ax2+ay2+bz2, ax2+by2+az2, bx2+ay2+az2}. Q: Any solns for other a,b?

Form 13: {x2+yz, y2+xz, z2+xy} Euler If z = 4(x+y) the first two become squares. It remains to make the last expression 16(x+y)2+xy a square. Two of his solns are: {x,y} = {s2+8st, -8st+t2}{x,y} = {5s2+8st, 8st+13t2}

Form 14: {x2+y2+xy, x2+z2+xy, y2+z2+xy} S. Ryley {x,y,z} = {p(p2-3q2)(3p2-q2), -p(3p2-5q2)(p2+q2), q(5p4-10p2q2+q2)}

Form 15: {x2-xy+y2, x2-xz+z2, y2-yz+z2} J. Cunliffe {x,y,z} = {(3-5n+n2)(-1-n+3n2) /(-2+2n+n2), n(4-5n), 4-8n+3n2}

Form 16: {x2+axy+y2, x2+bxz+z2, y2+cyz+z2} A.Martin {x,y,z} = {n(2m+an), m2-n2, n(2m+cn)} This makes two of the expressions squares. To make it all three, one should solve for m,n in {2m+an, 2m+cn} = {p2-q2, 2pq+bq2} for arbitrary p,q. This involves the denominator (a-c) so the method runs into a complication when a=b=c. Cunliffe’s soln is for the case a=b=c=-1. Any soln for a=b=c=1? In general, E.Turriere proved that if {ax2+bxy+cy2, dx2+exz+fz2, gy2+hyz+iz2} are all squares with rational variables then a polynomial soln involves finding rational points on a certain sextic surface. III. Four variables

Form 17: {a2+b2+c2, a2+b2+d2, a2+c2+d2, b2+c2+d2} Euler {a,b,c,d} = {4n(-1+n4), 2n(1-6n2+n4), -1+7n2-7n4+n6, 4n(-1+n4)} Note that a2+b2+c2 = (1+n2)6. Also, {a,b,c,d} = {4pqr, (p-q)(p+3q)r, 2(p2+3q2)s, (p-3q)(p+q)s} where, {r,s} = {(p-3q)(p+q)(p2+3q2), 2pq(p-q)(p+3q)} Again, a2+b2+c2 is a sixth power equal to (p2+3q2)6.

Form 18: {a2b2+c2d2, a2d2+b2c2} Euler {a,b,c,d} = {3pq, q(2p2+q2), 2(p2-q2), p(p2+2q2)}

Form 19: {a2b2+c2d2, a2c2+b2d2, a2d2+b2c2} Euler {a,b,c,d} = {4pq, 2(p2+3q2), (p-q)(p+3q), (p-3q)(p+q)} Note that if their squares are u2, v2, w2, then they can also be

expressed as, {u,v,w} = {m2+7n2, 2(m2-mn+2n2), 2(m2+mn+2n2)} where {m,n} = {p2-3q2, 2pq}. Form 20: {1+abc, 1+abd, 1+acd, 1+bcd} W. Wright Let d = 4p(ac+p)(bc+p)/(abc2-p2)2, where p = (ab-ac-bc)/2. The last three expressions become squares. One can then easily solve the first as 1+abc = q2 for any of the three free variables a,b,c. PART 5. Pell Equations I. Complete SolutionII. TransformationsIII. Polynomial ParametrizationsIV. Diophantine Equations needing Pell Eqns I. Complete Solution C. Hermite Given an initial soln {p,q} to p2-dq2 = ±1, one can find an infinite number of solns {x,y} by equating, x2-dy2 = (p2-dq2)k Using the same technique of equating factors,

x+yÖd = (p+qÖd)k, x-yÖd = (p-qÖd)k, we easily solve it as {x, y} = { (ak +a-k)/2, (ak -a-k)/(2Öd) }, where a = p+qÖd, and for the negative case p2-dq2 = -1 only odd powers k should be used. Note that the contribution of a-k diminishes as k increases so effectively are approximated by {x,y} ≈ {ak/2, ak /(2Öd)}. P.Paoli, S. Realis (all rational solns) Given an initial soln {p,q} to p2-nq2 = c for some constant c, then, x2-ny2 = c where x = (pu2+2nquv+npv2)/(u2-nv2), y = (qu2+2puv+nqv2)/(u2-nv2) for arbitrary {u,v}. However, if integral x,y is desired, then one can choose {u,v} to solve the Pell eqn u2-nv2 = ±1, with two solns given by (-p,q) and (p,q). Since there is an infinite number of {u,v}, then so for the x,y. The formulas for x,y without their denominators satisfy, x2-ny2 = (p2-nq2)(u2-nv2)2 Euler More generally, given an initial soln {p,q} to mp2-nq2 = c, then mx2-ny2 = c where, x = (pu2+2nquv+mnpv2)/(u2-mnv2), y = (qu2+2mpuv+mnqv2)/(u2-mnv2)

and an infinite number of integral x,y can be found by solving u2-mnv2 = ±1. The Paoli-Realis soln is just the special case m=1. Again, without the denominators, the x,y satisfy, mx2-ny2 = (mp2-nq2)(u2-mnv2)2 Solns for special forms are given by, S.Realis (n+2)x2-ny2 = (n+2)p2-nq2

{x, y} = {(n+1)p+nq, (n+2)p+(n+1)q} x2+nxy-ny2 = p2+npq-nq2

{x, y} = {(n+1)p-nq, (n+2)p-(n+1)q} II. Transformations 1. W. Brouncker, J. Wallis Brouncker-Wallis theorem: “If x2-dy2 = -1 (eq.1), then (2x2+1)2-d(2xy)2 = 1.” Proof: Square both sides of eq.1, (x2-dy2)2 = (-1)2

(x2+dy2)2 - d(2xy)2 = 1

Since dy2 = x2+1 from eq.1, then, (2x2+1)2-d(2xy)2 = 1 Example: x2-13y2 = -1; {x,y} = {18, 5),p2-13q2 = 1; {p,q}= {2(182)+1, 2(18)(5)} = {649, 180} This shows that fundamental solns to the negative Pell equation, when they exist, are smaller than those for the positive case. After a challenge by Frenicle de Bessy, Brouncker solved x2-313y2 = 1 by solving the relatively easier negative case. (Note: Letting {x,y} = {u√-1, v√-1} gives the purely positive variant: “If u2-dv2 = 1, then (2u2-1)2-d(2uv)2 = 1”, which implies one soln leads to some of the other solns.) 2. A. Cayley Theorem: “If there is an odd fundamental soln {u,v} to u2-dv2 = ±4, then it leads to the fundamental soln for x2-dy2 = ±1.” Case 1: If u2-dv2 = 4, and {x,y} = {(u2-3)u/2, (u2-1)v/2}, then x2-dy2 = 1. Ex. Let d = 5. Then {u,v} = {3,1}, and {x,y} = {9, 4}. Case 2: If u2-dv2 = -4, and {x,y} = {(u2+3)u/2, (u2+1)v/2}, then x2-dy2 = -1. Ex. Let d = 5. Then {u,v} = {1,1} and {x,y} = {2, 1}. In contrast, let d = 37. Since the fundamental soln of the negative

case is even {u,v} = {12, 2}, using the formula does not lead to fundamental {x,y} = {6,1}. Note: The first case can easily be transformed into the second using the same trick, {u’,v’} = {u√-1, v√-1}. 3. Euler Theorem: “If there is an odd fundamental soln {u,v} to u2-dv2 = -4, then it leads to the fundamental soln for x2-dy2 = 1 using the transformation {x, y} = {(u4+4u2+1)(u2+2)/2, (u2+3)(u2+1)uv/2}.” Ex. Let d = 61. Then {u,v} = {39, 5}, and {x,y} = {1766319049, 226153980}. This was found by combining the Brouncker-Wallis and Cayley theorems, and shows that fundamental solns of u2-dv2 = -4, when odd, are much smaller than those for x2-dy2 = 1 since x is a sixth degree polynomial in u. Historically, Euler used a variant of the identity above (this author modified to make it more aesthetic) to find x2-61y2 = 1 which has the largest solns for d < 100. Parametric fundamental solns to u2-dv2 = -4 with odd {u,v} are, (-1+n)2 - (5-2n+n2) (1)2 = -4((1+n)(1+2n+n3))2 - (5+6n+3n2+2n3+n4) (n2+1)2 = -4 for even n = 2m. (The second, for n = 2, yields d = 61.) Q: Any others? Note 1: For x2-dy2 = ±4 to have a solution in odd integers a necessary (but not sufficient) condition is that d is odd number of form 8n+5. (Though there are exceptions like d = 37, 101, etc.)Note 2: For odd d, if x2-dy2 = -4 is solvable, then so is x2-dy2 = -1.

But for even d, this does not necessarily apply. For example, x2-20y2 = -4 has the soln {4,1}, but x2-20y2 = -1 has none. 4. A. Cunningham, R. Christie By doing the transformations a) {x√2, y√2} = {p, q}, and b) {x√-2, y√-2} = {p, q}, on the Brouncker-Wallis theorem, we get the variants, (p2-1)2 - d(pq)2 = 1, if p2-dq2 = 2(p2+1)2 - d(pq)2 = 1, if p2-dq2 = -2 If d = 8n+r is prime, then the ff always have solns: d = 8n+1 for u2-dv2 = -1; d = 8n+3 for p2-dq2 = -2; d = 8n+5 for u2-dv2 = -1; d = 8n+7 for p2-dq2 = 2. Theorem: “Primes of form d = 4n+1 solve x2-dy2 = -1, while those of form d = 4n-1 solve one case of x2-dy2 = ±2 (Legendre). For such d, the fundamental soln to x2-dy2 = ±2 leads to the one for u2-dv2 = 1.” 5. F. Arndt (2pr2-1)2 - pq(2rs)2 = 1, if pr2-qs2 = 1(pr2+qs2)2 - pq(2rs)2 = 1, if pr2-qs2 = ±1(pr2-1)2 - pq(rs)2 = 1, if pr2-qs2 = 2(pr2+1)2 - pq(rs)2 = 1, if pr2-qs2 = -2 which are generalizations of the identities discussed previously. Q: Any other transformations?

(Update, 12/14/09): Given the fundamental soln {p,q} to x2-dy2 = 1, Gerry Martens defined the following functions: 2G[n] := (p+q√d)n + (p-q√d)n

A[i,j] := (G[i]-1) (G[i+2j+1]+1)/4 B[i,j] := (G[i]+1) (G[i+2j+1]-1)/4 2C[i,j] := (√A[i,j] - √B[i,j] )2 + 1 2D[i,j] := (√A[i,j] + √B[i,j] )2 + 1 then {A,B} and/or {C,D} are squares for certain d. 1. For {A,B} as squares, d = {2, 5, 8, 10, 13, 17, 20, etc}. 2. For {C,D} as squares, d = {2, 7, 8, 14, 23, 31, 34, etc}.3. For {A,B,C,D} as squares, d = {2, 8, 50, 98, 200, etc}. Note: Apparently, a necessary but not sufficient condition is that d = u2+v2 for [1], while d = 2m2 for [3]. (End update.) (Update, 12/21/09): (Piezas) This arose in trying to solve in the rationals, y4 + 2(u2+auv+bv2)y2 + (u2+v2)4 = z2

Note that this can be made a square if u2+auv+bv2 = (u2+v2)2. Given the quadratic equation, x2+ax+b = 0, and its roots {2x1, 2x2} = {-a

+√d, -a-√d}, where d = a2-4b, define the sequence, Pn = 1/d (x1n-x2

n). Theorem: “If b = -1, then an infinite number of solns to the Diophantine equation,

(u2+v2)2 = (u2+auv+bv2)w2

can be given by, {u,v,w} = {P2m+1, P2m, P4m+1}, and,{u,v,w} = {P2m+1, -P2m+2, P4m+3}.” This can be verified in Mathematica after a little manual tweaking. Example 1. Given x2-x-1 = 0, where {x1, x2} = {(1+√5)/2, (1-√5)/2},

then P(n): = 1/√5 (x1n-x2

n) yields the Fibonacci sequence, P(n): = {1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,…} so the eqn, (u2+v2)2 = (u2-uv-v2)w2, has solns {u,v,w} = {2,1,5}, {2,-3, 13}, {5, 3, 34}, {5, -8, 89}, etc. Example 2. Given x2-2x-1 = 0, where {x1, x2} = {1+√2, 1-√2}, then

P(n): = 1/√2 (x1n-x2

n), after removing the common factor 2, gives the Pell numbers, P(n): = {1, 2, 5, 12, 29, 70, 169,…} so (u2+v2)2 = (u2-2uv-v2)w2, has {u,v,w} = {5, 2, 29}, {5,-12, 169}, etc. (End update.) (Update, 12/23/09): I realized my solns had u2+v2 = w, or (P2m)2 +

(P2m+1)2 = P4m+1, which is a known attribute of Fibonacci numbers. Thus, in general, the Diophantine equation,

(u2+v2)2 = (u2+auv+bv2)w2

can be solved by, {u,v,w} = {p-aq, 2q, u2+v2}, where {p,q} are solns of the Pell equation, p2-(a2-4b)q2 = 1 generalizing the theorem given in the previous update. (End update.) III. Polynomial Parametrizations The degree of a polynomial soln to x2-dy2 = ±1 will be given by the P(n) defining the variable x. A. Degrees 1, 2, 3, 4, 6 Deg 1 (y2n+x)2 – (y2n2+2xn+d)y2 = x2-dy2

Deg 2 (an2+2ny+x)2 – (n2+2bn+d) (an+y)2 = z2, where a = (x-z)/d, b = (x+z)/y, and x2-dy2 = z2. Deg 3 (2an3+3abn2+6ny+x)2 – (4n2+4bn+d) (an2+abn+y)2 = x2-dy2

where a = 2(3x-by)/(bd) and b is a root of the cubic b3y-3b2x+3bdy-dx = 0. Deg 4 (Avanzi and Zannier) (a2-2b(b2+2b-c))2 – (a2+8bc+16c) (b2+c)2 = -16c3, if a = b2+2b+c Appropriate choice of b and ±c will result in conventional Pell equations. For {b,c} = {2n, 4} this yields, after removing common factors, ((1+n)(1+2n+n3))2 – (5+6n+3n2+2n3+n4) (n2+1)2 = -4 The case n=0 gives the fundamental soln to x2-5y2 = -4 while n=2 gives Euler’s soln to x2-61y2 = -4. In general, if n is even then x’ is odd, so using Euler’s transformation this gives a parametric fundamental soln to x2-dy2 = 1 for some d. Alternatively, for {b,c} = {2n-1, 1} yields, (-1+2n+2n2-4n3+4n4)2 – 2(1+2n+2n4) (1-2n+n2)2 = -1 Deg 6 (G. Ricalde) (8(a3+b3)2+1)2 – ((a+b)2+4) (4(a3+b3)(a2+b2))2 = 1, if b = a+1 Avanzi and Zannier (2n(1+2n-2n2+8n3-4n4+5n5))2 - (1+4n+4n2+4n4) (1-2n+6n2-4n3+4n4)2 = -1 B. Degrees 5, 7, 11, 13, 17

For these all known poly solns to x2-dy2 = 1 are of the form, (2pr2-1)2 - pq(2rs)2 = 1, or equivalently, (pr2+qs2)2 - pq(2rs)2 = 1, hence also satisfy pr2-qs2 = 1. Thus, the variables {p,q,r,s} of the entries below are understood to refer to these equations. Deg 5 (Avanzi and Zannier) {p,q,r,s} = {2+3m2+2(-m+m3)n-(1+4m2)n2+2mn3, -1+2mn, -n, 1+mn-n2}{p,q,r,s} = {-2+3m2+2(m+m3)n-(1-4m2)n2+2mn3, -1+2mn, n, 1+mn+n2} and, (2+an2)2 – a(1+n)(n(2-n+n2))2 = 4, if a = 1+3n-n2+n3

(2+an2)2 – a(1+n)(n(-2+n+n2))2 = 4, if a = -7-n+3n2+n3

For any n, the {x,y} of the pair above are even so it reduces to x2-dy2 = 1. Setting a = 1+3n-n2+n3 = 0 involves a radical that can be expressed in terms of the Weber class poly for Ö-11, similar to the one for deg 11 which involves Ö-23. Deg 7 (Avanzi and Zannier) {p,q,r,s} = {(-1+n2+2n3)/2, 2(-3+2n), -1-2n+2n2, (-1-n+2n3)/2} Piezas p = n/4q = -4 + (1+2a+3a2-6a3+a4)n - 2a2(1-2a2+a3)n2 + (-1+a)2a4n3

r = 3+3a-a2 + (-1-3a-4a2+3a3)n + a2(2+2a-3a2)n2 + (-1+a)a4n3

s = (1-(1+2a)n+a2n2

A deg-17 soln will result by setting a = bn for some rational b. Deg 11 (Weber) {p,q,r,s} = {-1+n2+n3, -5-4n+n2+n3, (-2+n)(1+n)(-2+n2)/2, (-1+n)(-2-2n+n3)/2} This was found using a modular equation between two elliptic functions at arguments f(w) and f(23w). Note that p,q have discriminant d=23 and p in fact is the inverse of the Weber class poly for Ö-23. Deg 13 (Piezas) {p,q,r,s} = {(1-n2)n/4, -4(4-4n+4n3-n5+n7), 2+2n-n2+n4+n5, (1-n2-n3)/4} This was derived by using a variant of the 2-variable deg-7 soln and likewise setting a = bn for b = -1.

IV. Diophantine Equations needing Pell Eqns These equations are also found in the various chapters of this work but I thought it would be a good idea to bring some of them together in one section. For the basics, go to Pell Equations. 1. ad-bc = ±1

{a,b,c} = {y2, x+y, x-y}, if x2-(d+1)y2 = ±1 2. x2+y2 = z2

Euler (v2+1)2 + (x2+1)2 = (v2+8y2+1)2, if v = (3y2+1)/(2y) and x2-10y2 = 1 3a. x2+ny3 = z4

H. Mathieu (q2(p2-2))2 + (2q2)3 = (pq)4, if p2-2q2 = 1 (p2(p2-1)/2)2 + p6 = (pq)4, if p2-2q2 = -1 This can be generalized as, Piezas (4q2(p2-2))2 + d(4q2)3 = (2pq)4, if p2-dq2 = 1 (4p2(p2-1))2 + (2p)6 = d2(2pq)4, if p2-dq2 = -1 Just multiply the first eq by d8 and the second by d6. 3b. x4+y3 = nz2

K.Brown p4 + (q2-1)3 = (q3+3q)2, if p2-3q2 = 1

More generally, p4 + (dq2-1)3 = d(dq3+3q)2, if p2-3dq2 = 1 (See Brown’s “Miscellaneous Diophantine Equations”.) 4. x2+y2 = z2-1 E. Grigorief ((a2-b2+c2-d2)/2)2 + (ab+cd)2 = ((a2+b2+c2+d2)/2)2 - 1 {a,b,c} = {y2, x+y, x-y}, if x2-(d+1)y2 = ±1 5. x2+y2 = z2+1 A. Gerardin (cx)2 + (c2y2-1)2 = (c2y2+c)2 + 1, where x2-2(c+1)y2 = 1 6. x2+y2 = z2+2k Piezas Let w = 2uv-2v2, then (u2+2uv-kw)2 + (ku2+2kuv-w)2 = (k-1)2(u2+w)2 + 2k, if u2-2v2 = ±1 7. x2+y2+z2 = t2+1

(u2+v2)2 + (2duv)2 + (du2-dv2)2 = (du2-2uv-dv2)2 + (u2+2duv-v2)2 Set u2+2duv-v2 = ±1 by letting {u,v} = {x-dy, y} to get the Pell eqn x2-(d2+1)y2 = ±1. A. Gerardin (y-1)2 + y2 + (y+1)2 = x2 + 1, where x2-3y2 = 1 Any Pell equation for x2+y2+z2 = t2-1? 8. ax2+bx+c2 = z2

Fermat x = (2cpq+bq2)/(p2-aq2), where for integral solns let p2-aq2 = ±1. 9. (n+u)(n+v) = dz2

Euler: {n, z} = {p2-u, pq}, if p2-dq2 = u-v{n, z} = {p2-v, pq}, if p2-dq2 = -(u-v) 10. mx2-ny2 = k Euler Given an initial soln mp2-nq2 = k for some constant k, then,

x = (pu2+2nquv+mnpv2)/(u2-mnv2), y = (qu2+2mpuv+mnqv2)/(u2-mnv2) and an infinite more can be found by solving u2-mnv2 = ±1. 11. {x2+y2-1, x2-y2-1} both squaresA.Genocchi (complete rational soln in p,q,r,s) {x,y} = {(2r2-t)/t, 4pqrs/t}, if t = r2 – (p4+4q4)s2 . For integral solns, it suffices to solve the Pell equation r2 – (p4+4q4)s2 = ±1. Similarly, T. Pepin {x2+y2-1, x2-y2-1} = {(2pq(r2+2s2))2, (2pq(r2-2s2))2} {x,y} = {p2+q2(r4+4s4), 4pqrs}, if p2-(r4+4s4)q2 = ±1. For the particular case {r,s} = {3,1}, this yields, {x2+y2-1, x2-y2-1} = {(22pq)2, (14pq)2} {x,y} = {p2+85q2, 12pq}, if p2-85q2 = ±1. (The discriminant of this Pell equation appears in two other forms given below.) 12. x3+y3+z3 = 1 Piezas

1. (1-ac+bc)3 + (a+c2-ac3)3 + (ac3-b-c2)3 = 1 {a,b,c} = {12qrt, 3(q-r)(3q+r)t, 3s2t2} 2. (1-ac-bc)3 + (a+c2-ac3)3 + (ac3+b-c2)3 = 1 {a,b,c} = {12qrt, 3(q+r)(3q-r)t, 3s2t2} where, for both, r = p-18qs3t3 and the condition p2-3(108s6t6-1)q2 = s. There is a complete soln when s = 1. Piezas (9+44pq-404q2)3 + (10+45pq-417q2)3 = (12+56pq-518q2)3 + 1, if p2-85q2 = -4 (9+44pq+404q2)3 + (10+45pq+417q2)3 = (12+56pq+518q2)3 + 1, if p2-85q2 = 4 Using another initial soln 63 + 83 = 93 – 1, we get, (6+222pq+4014q2)3 + (8+270pq+4806q2)3 = (9+312pq+5616q2)3 - 1, if p2-321q2 = 1 Given one initial soln to x3+y3+z3 = 1 it is possible to find identities like these as discussed in the section on "Third Powers". 13. x3+y3+z3+x+y+z = 0 L.Aubry

{x,y,z} = {u+v, -u+v, -2(3n+1)v}, if u2-(1+12n+36n2+36n3)v2 = n with n some constant. (For n=1, this entails solving u2-85v2 = 1.) In general, for square n one can always find integral u,v as the conditional equation has the form u2 = av2+n2 for some constant n and which has the soln, courtesy of Fermat as, v = 2nxy/(x2-ay2) where if integral v is desired one has to simply solve the Pell equation x2-ay2 = ±1. 14. x4+bx2y2+y4 = z2

Euler x4+bx2y2+y4 = (x3+y2)2, where b = nx2+2x, and x2-ny2 = 1 More generally, let b = nx2+2v, x4+bx2y2+y4 = (vx2+y2)2, if v2-ny2 = 1 15. x4+y4 = z2+1 E. Fauquembergue (17p2-12pq-13q2)4 + (17p2+12pq-13q2)4 = (289p4+14p2q2-239q4)2 + (17p2-q2)4 where q2-17p2 = ±1. It was proven by Fermat that x4+y4 = z2 has no non-trivial solns so this is the next best thing.

P.S. This is the list I have come up so far. Surely there are others? If you know of one, pls send it.

Chapter 6: Third Powers I. Sum / Sums of cubes (Part 1, in blue)

1. x3+y3 = z3

2. x3+y3+z3+t3 = 03. x3+y3+z3 = 14. x3+y3+z3 = 25. x3+y3+z3 = (z+m)3

6. p(p2+bq2) = r(r2+bs2)7. (x+c1y)(x2+c2xy+c3y2)k = (z+c1t)(z2+c2zt+c3t2)k

8. x3+y3+z3 = at3

9. x3+y3 = 2(z3+t3)10. w3+x3+y3+z3 = nt3

11. x3+y3+z3 = t2

12. ak+bk+ck = {x2, y2, z3}, k = 1,2,313. xk+yk+zk = tk+uk+vk, k = 1,314. xk+yk+zk = tk+uk+vk, k = 2,315. x3+y3+z3 = 3t3-t16. x3+y3+z3 = m(x+y+z)17. x1

k+x2k+x3

k+x4k = y1

k+y2k, k = 1,2,3

18. x1k+x2

k+x3k+x4

k = y1k+y2

k+y3k+y4

k, k = 1,2,3

19. ax3+by3+cz3 = N20. ax3+by3+cz3+dxyz = 021. x3+y3+z3-3xyz = tk

1. Sum of two cubes: x3+y3 = z3 P.Tait (x3+z3)y3 + (x3-y3)z3 = (y3+z3)x3, if x3+y3 = z3

Or course, by Fermat’s Last Theorem, x3+y3 = z3 has no non-trivial integer solns.

2. Form: x3+y3+z3+t3 = 0 F. Vieta (a4-2ab3)3 + (a3b+b4)3 + (2a3b-b4)3 = (a4+ab3)3 P. Sondat (ap-q)3 + (bp-q)3 + (cp+q)3 + (dp+q)3 = (a3+b3+c3+d3)(a+b+c+d)3

{p,q} = {a+b+c+d, a2+b2-c2-d2} Thus, given an initial soln, subsequent ones can be found. A similar identity exists for a3+b3+2(c3+d3) = 0. Euler (complete) (p+q)3 + (p-q)3 = (r+s)3 + (r-s)3

{p,q} = {3(bc-ad)(c2+3d2), (a2+3b2)2 - (ac+3bd)(c2+3d2)}{r,s} = {3(bc-ad)(a2+3b2), -(c2+3d2)2 + (ac+3bd)(a2+3b2)}

For details on how Euler derived this, see “Euler’s Extended Conjecture and ak+bk+ck = dk for k>4” by this author. J. Binet (complete) a3+b3 + n(c3+d3) = 0 a = (1 - mn(p-3q))rb = (-1 + mn(p+3q))rc = (m2n - (p+3q))rd = (-m2n + (p-3q))r where m = p2+3q2 and r is just a scaling factor. Proof (Piezas): For any rational soln a,b,c,d, one can always find rational p,q,r, to get those particular values using the formulas, {p,q,r} = {(bc+ad-2(bd+ac))/(2v), (bc-ad)/(2v), -v(c+d)/(3(bc-ad))} where v = a2-ab+b2. For example, for n=1, to get the soln {3,5,4,-6}, the formulas give {p,q,r} = {1,1,1/3}. In fact, by permuting the {a,b,c,d}, one finds there can be six distinct {p,q,r} up to sign. Binet’s original soln only discussed the case n=1 but with a little tweaking by this author, one can generalize it and a derivation will be given later. It should be pointed out that the algebraic form a2+ab+b2, or equivalently x2+3y2, appears a lot when dealing with third and fourth powers. It may be of interest that while x2+y2 = 1 defines a circle, x2+xy+y2 = 1 yields an ellipse. J. Steggall (complete) (s4-rs(p+3q))3 + (-s4+rs(p-3q))3 = (r2-s3(p+3q))3 + (-r2+s3(p-3q))3

if r = p2+3q2. For s=1 reduces to Binet’s (with n=1). Note that Euler, Binet, and Stegall’s solns involve quartic polynomials. Noam Elkies' complete soln uses only cubic polynomials, N. Elkies (complete) a3+b3+c3+d3=0 av = s3-m-2r3+(r2-s2)t+(s-2r)t2

bv = -s3+m-r3+nt-(s+r)t2

cv = t3+m+r3+nt-(s+r)t2

dv = -t3+2rs2-r2s+2r3-nt+(s+r)t2 where {m,n} = {rs2-2r2s, s2+2r2} and v is just a scaling factor. A.Werebrusow (complete) If 3w2xy = a2+ab+b2, then, (-ax+wy2)3 + (ay-wx2)3 = (-bx+wy2)3 + (by-wx2)3 = ((a+b)x+wy2)3 + (-(a+b)y-wx2)3, and, (ax-wy2)3 + (-bx+wy2)3 = (ay-wx2)3 + (-by+wx2)3 For the second formula, set {a,b,w,x,y} = {(p-3q)r, (p+3q)r, r, 1, p2+3q2} to get Binet’s soln. Variations of this were found by K. Schwering and Ramanujan. A.Werebrusow:

(-a2d+bc2)3 + (ad2-b2c)3 = c3(ab-cd)3 + d3(ab-cd)3, if a3+b3 = c3+d3

Thus, this is another identity such that one soln to a3+b3+c3+d3 = 0 leads to a second. Related to the identity below, A. Desboves (b3-d3)(a2d-bc2)3 + (a3-c3)(ad2-b2c)3 = (ab-cd)3(-b3c3+a3d3) C. Hermite ((a+2b)c-1)3 + (-a-2b+c2)3 = ((a-b)c-1)3 + (-a+b+c2)3, if c = a2+ab+b2

S. Baba (pq)3 + (p+6q3)3 = (p-6q3)3 + ((p+12)q)3, if p = q6-4 J. Young (a2+6ab3-3c)3 – (a2-6ab3-3c)3 = b3(a2+6a+3c)3 – b3(a2-6a+3c)3, if c = b6-1 The above generalizes Baba’s. Also useful for finding Martin triples. There are also quadratic parametrizations though these are no longer complete. J. Young (p2+16pq-21q2)3 + (-p2+16pq+21q2)3 + (2p2-4pq+42q2)3 = (2p2+4pq+42q2)3

Ramanujan (3x2+5xy-5y2)3 + (4x2-4xy+6y2)3 + (5x2-5xy-3y2)3 = (6x2-4xy+4y2)3

Other binary quadratic form solns have been found by other authors such as Gerardin, Womack, etc. A generalization has been found by this author as, Piezas (ax2-v1xy+bwy2)3 + (bx2+v1xy+awy2)3 + (cx2+v2xy+dwy2)3 + (dx2-

v2xy+cwy2)3 = (a3+b3+c3+d3)(x2+wy2)3

where {v1, v2, w} = {c2-d2, a2-b2, (a+b)(c+d)} thus for any soln to a3+b3+c3+d3 = N where N is zero or any number of cubes, then one can always find a quadratic parametrization. It is also the case that, (ax-v1y)k + (bx+v1y)k + (cx+v2y)k + (dx-v2y)k = (ax+v1y)k + (bx-

v1y)k + (cx-v2y)k + (dx+v2y)k, for k =1,3 Piezas If a+b = c+d. Let {u1, u2} = {c-d, a-b}, then, (ax2+u1xy+by2)3 + (bx2-u1xy+ay2)3 + (cx2-u2xy+dy2)3 + (dx2+u2xy

+cy2)3 = (a3+b3+c3+d3)(x2+y2)3 There are many particular cubic equations with this property, one of

which is 93+133+193+233 = 283, (9+23 = 13+19) as well as those in a nice arithmetic progression like, 113+123+133+143 = 203

313+333+353+373+393+413 = 663

the latter found by D. Rusin in the context of a certain elliptic curve. There is also a quadratic identity, similar to the general one, J. Nicholson (au+pv)3 + (-au+qv)3 + (bu+nv)3 + (-bu+mv)3 = ((m+n)b2+(p+q)a2)3 (m3+n3+p3+q3) where {u, v} = {(m2-n2)b-(p2+q2)a, (m+n)b2+(p+q)a2}

3. Form: x3+y3+z3 = 1 It turns out solving this completely in the integers may involve a certain elliptic curve. The complete integral soln to this equation is given by Werebrusow’s identity with one term set equal to 1. (1-ac+bc)3 + (a+c2-ac3)3 + (ac3-b-c2)3 = 1, where a2+ab+b2 = 3c(ac-1)2

Proof (Piezas): Given any non-trivial soln {x,y,z}, one can always find rational {a,b,c} = {(y-c2)/(1-c3), (-c2x-z)/(1-c3), (1-x)/(y+z)}. The conditional equation above is a quadratic in b and to make its discriminant a square, one has to solve, y2 = 3(4a2c3-8ac2+4c-a2)

which is an elliptic curve in c, or equivalently, y2 = 3(4c3-1)a2-24c2a+12c a quadratic curve in a. This is easier made a square if its constant term is also a square. Let c = 3v2, y2 = 3(108v6-1)a2-216v4a+36v2

This is of the form y2 = ax2+bx+c2 where an infinite number of integral solns can be found by solving the Pell equation p2-aq2 = ±1. (See the relevant section in Part 2, Quadratic Polynomial as a kth power.) Sparing the reader some algebra, this gives, Piezas 1. (1-ac+bc)3 + (a+c2-ac3)3 + (ac3-b-c2)3 = 1 {a,b,c} = {12qrt, 3(q-r)(3q+r)t, 3s2t2} 2. (1-ac-bc)3 + (a+c2-ac3)3 + (ac3+b-c2)3 = 1 {a,b,c} = {12qrt, 3(q+r)(3q-r)t, 3s2t2} where, for both, r = p-18qs3t3 and the condition p2-3(108s6t6-1)q2 = s. By setting s = ±1, this is of course a Pell equation. Note that the second addend is unchanged, implying a pair of solns which share a term thus solving the system, x1

3+x23 = x3

3+x43 = x5

3+1 Example, for the case s = t = 1, this gives p2-321q2 = 1, with

fundamental {p,q} = {215, 12} yielding the pair, {x,y,z} = {4528, 3753, -5262}{x,y,z} = {-3230, 3753, -2676} It turns out that for s = 1, the Pell equation, p2-3(108t6-1)q2 = 1 has a parametrization for its fundamental soln given by {p1, q1} =

{216t6-1, ±12t3}. From this one can then generate an infinite family. Using trivial {p0, q0} = {1, 0} on the expressions for x,y,z yields the known 4-deg identity, (1-9t3)3 + (9t4)3 + (3t-9t4)3 = 1 while {p1, ±q1} with one sign yields a 10-deg, (1+9t3+648t6-3888t9)3 + (-135t4+3888t10)3 + (-3t-81t4+1296t7-3888t10)3 = 1 while the other sign gives a 16-deg. The next, {p2, ±q2} gives a 22 and 28-deg, {p3, ±q3} a 34 and 40-deg, and so on for an infinite sequence with degree 6n+4. One of the terms is always a polynomial with only even powers, unchanged for ±t, so a term is shared by two distinct solns to x3+y3+z3 = 1. (This in fact is the same infinite family that can be derived by a recursion found by D.H.Lehmer.) Note: Aside from s = 1, I do not know of any other non-trivial integral s such that p2-3(108s6t6-1)q2 = s is solvable in the integers {p,q} for a given t.

One can also use the binary quadratic forms together with Pell equations to find an infinite number of integral solns if an initial one is known. To recall, (ax2-v1xy+bwy2)3 + (bx2+v1xy+awy2)3 + (cx2+v2xy+dwy2)3 + (dx2-

v2xy+cwy2)3 = (a3+b3+c3+d3)3(x2+wy2)3

{v1, v2, w} = {c2-d2, a2-b2, (a+b)(c+d)} so it suffices to find one soln to a3+b3+c3+1 = 0. It is easy to set {x,y} = {p+v2q, 2q} to transform the fourth addend x2-v2xy+cwy2 to the

form p2-nq2. (The discriminant n is a function of a,b,c with 3! = 6 possible values and it is the experience of this author that at least one has n > 0). Assume d = ±1 and one can then solve the Pell equation p2-nq2 = ±1. For example, using the famous taxicab number 1728 = 93+103 = 123+13, and after a little modification, (9+44pq-404q2)3 + (10+45pq-417q2)3 = (12+56pq-518q2)3 + 1, if p2-85q2 = -4 (9+44pq+404q2)3 + (10+45pq+417q2)3 = (12+56pq+518q2)3 + 1, if p2-85q2 = 4 Using another initial soln, 63 + 83 = 93 – 1, we get, (6+222pq+4014q2)3 + (8+270pq+4806q2)3 = (9+312pq+5616q2)3 - 1, if p2-321q2 = 1 and so on. A. Gerardin

(p4+9pq3)3 + (3q2)6 = (3p3q+9q4)3 + p12

For p=1 this gives a parametrization to a3+b3+c3 = 1.

4. Form: x3+y3+z3 = 2 (1+ax3)3 + (1-ax3)3 = 6a2x6 + 2 It suffices to make a = 6 to satisfy the equation, with x arbitrary so there are parametrizations for x1

3+x23+x3

3 = n for n = 1,2. (Update, 12/14/09): Alain Verghote gave a simpler derivation of the above. Given the basic identity, (1+n)3 + (1-n)3 = 6n2 + 2 one then simply lets n = 6x3. (End update.) There is a fifth power version, x1

5+x25+x3

5+x45+x5

5+x65+x7

5 = n, also for n =1,2, with n=1 to be discussed in a later chapter. For n=2 by Seiji Tomita, surprisingly this is dependent on Pythagorean triples and is analogous to the cubic case, (1+ax5)5 + (1-ax5)5 + (1+bx5)5 + (1-bx5)5 + (-1+cx5)5 + (-1-cx5)5 + (dx4)5 = 2 where {a,b,c,d} should satisfy the two conditions a2+b2 = c2 and 20a2b2 = d5, one soln of which is {a,b,c,d} = {270, 360, 450, 180}.

5. Form: x3+y3+z3 = (z+m)3 J. Jandasek (m =1) n3 + (3n2+2n+1)3 + (3n3+3n2+2n)3 = (3n3+3n2+2n+1)3

This implies that any integer n appears in a cubic quadruple at least once. A similar identity exists for second powers as was already discussed. G. Ampon (m =1) (3n2)3 + (6n2+3n+1)3 + (9n3+6n2+3n)3 = (9n3+6n2+3n+1)3

Q: Any soln for some other constant m?

6. Form: p(p2+bq2) = r(r2+bs2) Derivation: To derive Binet’s version, for convenience, set, (p+q)3 + (p-q)3 = (r+s)3 + (r-s)3

Expanding, we get p(p2+3q2) = r(r2+3s2). This can generalized to, p(p2+bq2) = r(r2+bs2) (eq.1) Assume that, r2+bs2 = (p2+bq2)(u2+bv2) (eq.2)

Factor over √-b and equate factors, r+s√-b = (p+q√-b)(u+v√-b)r-s√-b = (p-q√-b)(u-v√-b) Solve for {r,s}, r = pu-bqv (eq.3)s = pv+qu (eq.4) Substitute eq.2 into eq.1 (by eliminating p2+bq2), p = r(u2+bv2) (eq.5) Substitute r from eq.3 into eq.5, p = (pu-bqv)(u2+bv2) Collect {p,q}, p(-1+u(u2+bv2)) = qbv(u2+bv2) And the equation is true if, p = bv(u2+bv2), q = -1+u(u2+bv2) Substitute these into eq.3 and eq.4 to get, r = bv, s = -u+(u2+bv2)2 and we have all unknowns {p,q,r,s} for any b. But this can be generalized even further.

7. Form: (x+c1y)(x2+c2xy+c3y2)k = (z+c1t)(z2+c2zt

+c3t2)k

Since by using the transformation {x,y} = {p-c1q, q} on the left hand side (and a similar one for the right) will transform this into the form p(p2+apq+bq2)k, one can simply assume c1 = 0 without loss of generality, p(p2+apq+bq2)k = r(r2+ars+bs2)k Piezas {p,q,r,s} = {bvwk, -1+uwk, bv, -(u+av)+wk+1}, if w = u2+auv+bv2

This is easily be proven for k=1,2. For k=1 and a=0, this reduces to the formula for third powers given previously. For k=2, this is relevant to equal sums of fifth powers to be discussed later. Using computer algebra one can see it is also true for other k>2, but I have no proof it is the case for all positive integer k.

8. Form: x3+y3+z3 = mt3 A. Werebrusow (p2r3+qs3)3 + (p2r3-qs3)3 + (-6rs2)3 = 2(p2r3)3, if pq = ±6 Other poly solns for m ≥ 2? Using Ryley’s Identity below (also given in Part 1), it can be shown there are integral {x,y,z,t} for any constant m,

S. Ryley (p3+qr)3 + (-p3+pr)3 + (-qr)3 = m(6mnp2)3, {p,q,r}= {m2+3n3, m2-3n3, 36m2n3} with arbitrary n. Note that the first term is a 6-deg poly in m. Can smaller deg poly be found that works for any m? (Update, 4/19/10): A 3rd-deg polynomial in m is possible. William Ellison cited one example from the book Cubic Forms by Yuri Manin as, (m3-36n9)3 + (-m3+35mn6+36n9)3 + (33m2n3+35mn6)3 = m(32m2n2

+34mn5+36n8)3 for arbitrary n. Note how this has simple coefficients that are only powers of 3. Robert Israel pointed out that a small tweak can reduce the powers as, (27m3-n9)3 + (-27m3+9mn6+n9)3 + (27m2n3+9mn6)3 = m(27m2n2

+9mn5+3n8)3 See also Ryley's Theorem. (End update.)

9. Form: x3+y3 = 2(z3+t3) Given one soln to the above equation, a second one can be found as, A. Gerardin x3+y3+2(z3+t3) = (p3+q3+2(r3+s3)) (p3-2r3)3

{x,y,z,t} = {p(v+3r3), q(v-3r3), -2rv, s(v-3r3)}, if v = p3+r3

As we’ll later see, the more general result by Desboves is that given an initial soln to the sum of cubes ax1

3+a2x23+...+ anxn

3 = 0, then one

can find a second. While the complete soln to a3+b3 = n(c3+d3) is given by the modified Binet formula, some nice partial solns for n=2 are, A. Gerardin (a3+3b3)3 + (a3-3b3)3 = 2a9 + 2(3ab2)3 (a2+4ab-b2)3 + (-a2+4ab+b2)3 = 2(a+b)6 - 2(a-b)6

9b. Form: x3+y3 = 2(z3+Poly) (Update, 12/14/09): Alain Verghote 4(ac+bd+e)3+4(ad+bc-e)3 = x3+3xy2

where {x,y} = {(a+b)(c+d), (a-b)(c-d)+2e}. Note that the polynomial x2+3y2 appears often when dealing with third and fourth powers.

10. Form: w3+x3+y3+z3 = ntk (Update, 12/14/09): Gerhard Paseman

(a-1)3 - (a+1)3 + (c+1)3 - (c-1)3 = (6b)2, where a2 + 6b2 = c2

(Update, 12/15/09): Alain Verghote ((s+1)2)3 + (2-(s+1)2)3 + (-(s-1)2)3 + (-2+(s-1)2)3 = 64x6, where s = 3x2

for arbitrary x. One can choose x = y2 to make the sum a 4th power. (End update) A. Martin (also the case k = 3 of Boutin's Identity) (-a+b+c)3 + (a-b+c)3 + (a+b-c)3 + d3 = (a+b+c)3, if 24abc = d3

Piezas This is a special case of a more general identity. Using the nice initial soln 113+123+133+143 = 203, then, (11x2+xy+14y2)3 + (12x2-3xy+13y2)3 + (13x2+3xy+12y2)3 + (14x2-xy+11y2)3 = 203(x2+y2)3 Binary quadratic form solns to x1

3+x23+…+xn

3 = t3 can be found for any n > 2 given an initial soln and the identity found by this author discussed previously. Piezas (a2+ab)k + (a2-ab)k + (b2+ab)k + (b2-ab)k = 2(a2+b2)k, for k = 1,2,3. By finding expressions {a,b} such that a2+b2 = cm (which is easily done), the above equation provides a template for solving,

x13 + x2

3 + x33 + x4

3 = 2t3m

for any m. A similar identity for fourth powers was found by Ramanujan, x1

4 + x24 + x3

4 = 2t2m

to be discussed later and a fifth power version was found by this author as well.

I. Sum / Sums of cubes (Part 2, in blue)

1. x3+y3 = z3

2. x3+y3+z3+t3 = 03. x3+y3+z3 = 14. x3+y3+z3 = 25. x3+y3+z3 = (z+m)3

6. p(p2+bq2) = r(r2+bs2)7. (x+c1y)(x2+c2xy+c3y2)k = (z+c1t)(z2+c2zt+c3t2)k

8. x3+y3+z3 = at3

9. x3+y3 = 2(z3+t3)10. w3+x3+y3+z3 = nt3

11. x3+y3+z3 = t2

12. ak+bk+ck = {x2, y2, z3}, k =1,2,3 (Martin triples)13. xk+yk+zk = tk+uk+vk, k = 1,314. xk+yk+zk = tk+uk+vk, k = 2,315. x3+y3+z3 = 3t3-t16. x3+y3+z3 = m(x+y+z)17. x1

k+x2k+x3

k+x4k = y1

k+y2k, k = 1,2,3

18. x1k+x2

k+x3k+x4

k = y1k+y2

k+y3k+y4

k, k = 1,2,3

19. ax3+by3+cz3 = N20. ax3+by3+cz3+dxyz = 021. x3+y3+z3-3xyz = tk

11. Form: x3+y3+z3 = t2 V. Bouniakowsky: (n3+1)3 + (-n3+2)3 + (3n)3 = (3(n3+1))2

E. Catalan (a4+2ab3)3 + (b4+2a3b)3 + (3a2b2)3 = (a6+7a3b3+b6)2 A. Gerardin (9a4+8ab3)3 + (4ab3)3 + (4b4)3 = (27a6+36a3b3+8b6)2 Any other solns with addends as polynomials of small degree? The ff identity proves one soln to x3+y3+z3 = t2 leads to another, A. Werebrusow If a3+b3+c3 = d2, then, (a-px)3 + (b+px)3 + (c-x)3 = (d-qx)2, where x = 3(a+b)p2+3c-q2, q = (3p(a2-b2)+3c2)/(2d), for arbitrary p.

12. Form: ak+bk+ck = {x2, y2, z3} for k =1,2,3 (aka Martin triples) A Martin triple is a triple of integers {a,b,c} such that, a + b + c = x2 (eq.1)a2 + b2 + c2 = y2 (eq.2)a3 + b3 + c3 = z3 (eq.3) The smallest in positive integers is M1 = {1498, 2461, 7490}, so, 1498 + 2461 + 7490 = 1072

14982 + 24612 + 74902 = 80252

14983 + 24613 + 74903 = 75973

There are also an infinite number of distinct Martin triples unscaled by a square factor, and this can be proved via a polynomial identity or an elliptic curve. Originally, A. Martin solved only eq.2 and eq.3, and gave a polynomial identity of high degree (see below), but any soln can be made valid for eq.1 as well. Each term is simply multiplied with a scaling factor m which is the square-free part of the sum a+b+c = mn2. For ex, the above was derived from, 142 + 232 + 702 = 752

143 + 233 + 703 = 713

and since a+b+c = 14+23+70 = 107, this yields M1 after multiplying all terms by m = 107. These "pre-triples" are quite rare. Given a3+b3+c3 = z3 (eq.3), of around 267000 positive and primitive solns with z < 100,000 in J. Wroblewski’s database here (which is up to z = 106), only ten can be transformed into Martin triples, with the next three being,

{3, 34, 114}{18, 349, 426}{145, 198, 714} However, if positive and negative solns are allowed there is a parametrization to this, A. Martin (using Young’s soln) (a2+6ab3-n)k + (-a2+6ab3+n)k + (b(a2-6a+n))k

where n = 3(b6-1), for k = 3 already sums up to a cube q3. Expanding for k = 2, if it is to be a square p2, as a polynomial in a this has the form, (b2+2)a4+c3a3+c2a2+c1a+c0

2(b2+2) = p2

where the ci are polynomials in b. This quartic is easily made a square

if b2+2 = y2. If so, two small solns are, a = (b8+8b6+12b4-b2+4)/(2b2+4) a = 6(b8+2b6-b2-2)/(b8+8b6+12b4-b2+4) which can then be used to compute further rational points on this curve. It still remains to make b2+2 = y2 which is easily done as b = (u2-2v2)/(2uv). Considering this results in high degree polynomials it is not surprising the smallest positive soln found by Martin had 17 digits. In general, it can be proven that finding these triples may involve an elliptic curve. Given an initial soln to a3+b3+c3 = d3, one can always generate quadratic form parametrizations using an identity found by this author. For ex, using the smallest positive pre-triple {14,

23, 70}, define the function M(x) as, M(x):= (-8x2+36x+14)k + (9x2-25x+23)k + (-6x2-8x+70)k For k = 3 this is already the perfect cube (x2-9x+71)3. For k = 2, this yields the quartic, M(x):= 181x4-930x3+1335x2+1262x+752 = y2

which is to be made a square. One soln, not surprisingly, is x = 0 with another as x = 68/75, and so on. It turns out an analogous situation can be extended to fourth powers as, ak+bk+ck+dk = {p2, q4} for k = 2,4 with the only known soln, 159352+270222+579102+592602 = 885972

159354+270224+579104+592604 = 701214

found by this author using Wroblewski’s database for fourth powers. Any others?

13. Form: ak+bk+ck = dk+ek+fk, k =1,3 Just like for k = 1,2, the eqn ak+bk+ck = dk+ek+fk, (eq.1) also has a complete soln for k = 1,3. We can solve this in two ways. First, using the form by Lander,

(u+x)k + zk + (v+y)k = (v+x)k + yk + (u+z)k

call this L0, a complete soln for k = 1 which he used for k = 1,5. Expanding L0 for k = 3, u2x-v2x+ux2-vx2+v2y+vy2-u2z-uz2 = 0 As a quadratic in z, this has discriminant D, D:= -4(x-y)uv2-4(x2-y2)uv+u2(u+2x)2

which must be made a square. Since this discriminant as a polynomial in v is a quadratic with a square constant term, it is easily made a square using Fermat’s method. A special case is when one of the terms of L0 is zero, say letting y = 0, since this gives the complete symmetric ideal soln of degree four, (t+a)k+(t+b)k+(t+c)k+(t+d)k+(t+e)k = (t-a)k+(t-b)k+(t-c)k+(t-d)k+(t-e)k

for k = 1,2,3,4 which is true for any t iff ak+bk+ck+dk+ek = 0 for k = 1,3. Another approach to solve the system k = 1,3 which turns out to be simpler is to use another form, (a+bp+q)k + (b-bp+q)k + (c+ap+q)k = (a+cp+q)k + (b+ap+q)k + (c-cp+q)k

call this L1 since this author found it while studying Lander’s work on L0 for k = 1,5. This is a subset of L0, yet still complete for multi-grades with one k >1 (the easy proof will be given in Fifth Powers). For k = 1,3, expanding L1 at k = 3 gives just a linear condition in q, -a2+b2+c2+ab+ac+bc+2(b+c)q = 0

Solving for q then gives the complete soln of eq.1 in four variables {a,b,c,p}. (And just like in the previous approach, any of the terms of L1 can be set equal to zero, say a+bp+q = 0, solve for q, and the linear condition will now be in the variable p.) One can see the approach can easily be generalized for k = 1,4 and higher. So L1 for k = 1,4, as a polynomial in p after removing the trivial factor p(p-1), is, (Poly10)p2+(Poly20)p+(Poly30) = 0 with discriminant D that is a quadratic polynomial in q, while for k = 1,5, (Poly11)p2+(Poly21)p+(Poly31) = 0 has a D that is a quartic in q, and for k = 1,2,6, (Poly12)p2+(Poly22)p+(Poly32) = 0 with D that is again a quartic in q, so the problem can be treated as finding rational points on the curve D = y2. Thus, L1 is a very versatile form because when expanded for higher powers, the variable p remains of relatively low degree and for either of the systems k = 1,5 or k = 1,2,6, the complete soln involves solving only a quadratic eqn and an elliptic curve. These will be discussed in more detail later. Another special case of eq.1 also has abc = def. One way to solve this is to use the form, (ap)k + (bq)k + (cr)k = (bp)k + (cq)k + (ar)k

which already satisfies the condition. Expanding for k = 1,3 and solving the two eqns yields,

{p,q,r} = {ab-c2, -a2+bc, -b2+ac} There is a second distinct form though, (ap)k + (bq)k + (cr)k = (-bp)k + (-cq)k + (ar)k

with {p,q,r} = {ab+c2, -a2-bc, b2-ac}. In fact, it can easily be proven that if eq.1 has abc = def, then a+b+c = d+e+f = 0. Proof: The system ak+bk+ck = dk+ek+fk for k = 1,3 where abc = def, by eliminating {c,f} from this system depends on the final eqn, ab(a+b) = de(d+e) call this E1. However, letting {c,f} = {-a-b, -d-e} and substituting into the system, this also satisfies k = 1,3 given E1, proving the assertion. (End proof) An earlier proof was given by Choudhry, (Update, 3/23/10): Choudhry (2001): "If ak+bk+ck = dk+ek+fk, for k = 1,3, (eq.1 & 2) and abc = def (eq.3), then a+b+c = d+e+f = 0." Proof: One can use the identity, (a+b+c)3 + 2(a3+b3+c3)-6abc = 3(a+b+c)(a2+b2+c2) and a similar one in the variables {d,e,f}. If eqs.1,2,3 hold, then (a+b+c)(a2+b2+c2) = (d+e+f)(d2+e2+f2) (eq.4) For a+b+c = d+e+f ≠ 0, this implies a2+b2+c2 = d2+e2+f2. But Bastien’s Theorem excludes a non-trivial soln to,

ak+bk+ck = dk+ek+fk, for k = 1,2,3, Thus, for eq.4 be non-trivially true, it must be the case that a+b+c = d+e+f = 0. Eqs 1,2,3 are then satisfied if, ak+bk+(-a-b)k = dk+ek+(-d-e)k, for k = 1,3 where ab(a+b) = de(d+e), parametric solns of which are easily found. (End proof.) Source: Triads of cubes with equal sums and equal products, The Mathematics Student, Vol. 70, 2001. (End update) Theorem (Piezas): If ak+bk+ck = dk+ek+fk for k = 1,3 where a+b+c = d+e+f = 0 (or equivalently, abc = def), then, 9abc(a6+b6+c6-d6-e6-f6) = 2(a9+b9+c9-d9-e9-f9) (eq.1) Proof: Simply substitute {c,f} = {-a-b, -d-e} into eq.1 and this also has E1:= ab(a+b)-de(d+e) as a factor. (Update, 6/22/10): The system, x1x2x3 = y1y2y3

x13+x2

3+x33 = y1

3+y23+y3

has several solns: 1. Gerardin (a2p)3 + (b2q)3 + (abr)3 = (a2q)3 + (-b2r)3 + (-abp)3

where {p,q,r} = {a3-2b3, a3+b3, 2a3-b3}. 2. Choudhry gave a 3-parameter version,

(ap)3 + (bq)3 + (cr)3 = (aq)3 + (-br)3 + (-cp)3

where {p,q,r} = {a3-2b3-c3, a3+b3+2c3, 2a3-b3+c3}. 3. Piezas (ap)3 + (bq)3 + (cr)3 = (ar)3 + (bp)3 + (cq)3

where {p,q,r} = {ab-c2, -a2+bc, -b2+ac}. The last identity, discussed in the previous paragraph, is also valid for x1+x2+x3 = y1+y2+y3 = 0, as well as, 9x1x2x3 (x1

6+x26+x3

6-y16-y2

6-y36) = 2(x1

9+x29+x3

9-y19-y2

9-y39)

14. Form: xk+yk+zk = tk+uk+vk, k = 2,3 Identities that are good for squares can be extended for particular values to cubes as well, just as this one, C. Goldbach p2 + q2 + (p+q+3r)2 = (p+2r)2 + (q+2r)2 + (p+q+r)2

Alternatively, this can be more symmetrically expressed as, (p-r)k + (q-r)k + (p+q+r)k = (p+r)k + (q+r)k + (p+q-r)k

which is already for k = 2, but is also valid for k = 3 if 6pq = r2.

15. Form: x3+y3+z3 = 3t3-t M. Noble Given a3+b3+c3 = 3d3-d, then, {a,b,c,d} = {(4p-4r)/(5q), (p+4r)/(5q), (3p)/(5q), (4p)/(5q)} where {p,q,r} = {15u2+4v2, 15u2+9uv-4v2, 10uv+3v2} This soln has the special property such that by defining the expressions, {x,y,z} = {a3-d3, b3-d3, c3-d3} then {x-(x+y+z)3, y-(x+y+z)3, z-(x+y+z)3} are perfect cubes. (This is a particular case of a more general identity by Noble.)

16. Form: x3+y3+z3 = m(x+y+z) L.Aubry x3+y3+z3 ± (x+y+z) = 0 {x,y,z} = {u+v, -u+v, -2v(3n+1)}, if u2-(1+12n+36n2+36n3)v2 = ±n with n some constant. For example, for n=1 this entails solving either u2-85v2 = ±1 depending on which sign is involved. In general, if x,y,z need not be co-prime, then, {x,y,z} = {n(u+v), n(-u+v), -2nv(3n2+1)}, if u2-(1+12n

+36n2+36n3)v2 = ±1.

17. Form: x1k+x2

k+x3k+x4

k = y1k+y2

k, k = 1,2,3 By Bastien’s Theorem, the system of eqns, x1

k+x2k+…+xn

k = y1k+y2

k+…+ynk, k = 1,2,3

has a non-trivial soln only for n>3. However, one can always set one term x1 = 0. A simple soln is, J. Nicholson ak + bk + (2a+4b)k + (3a+3b)k = (a+3b)k + (2a+b)k + (3a+4b)k, k = 1,2,3 Piezas But it is also possible two terms on one side are zero. The complete soln for that case is, ak + bk + ck + dk = ek + fk, {d,e,f} = {(-abc)/x, ((a+b)(a+c)(b+c)+y)/(2x), ((a+b)(a+c)(b+c)-y)/(2x)} where x = ab+ac+bc and a,b,c satisfy (a2-x)(b2-x)(c2-x) = y2. Two easy solns can be found. First, let b = c, and, {a,b,c,y} = {p+q, q, q, 2(p+q)qr}, where p2 = 2q2 + r2. Second, let a+b = c,

{a,b,c,y} = {-2p+q, 2p+q, 2q, 4(2p-q)(2p+q)r}, where p2 = q2 + r2. Any more?

18a. Form: x1k+x2

k+x3k+x4

k = y1k+y2

k+y3k+y4

k, k = 1,3 (Update, 8/6/09): Tony Rizzo gave the soln, (p+1)k + (q+1)k + (r-1)k + (s-1)k = (p-1)k + (q-1)k + (r+1)k + (s+1)k, for k = 1,3, where p2+q2 = r2+s2 and solved the condition as, (a-d)2+(b+c)2 = (a+d)2+(b-c)2 where a/b = c/d. One can make this valid for k = 1,2,3 by re-arranging terms as, (p+t)k + (-p+t)k + (q+t)k + (-q+t)k = (r+t)k + (-r+t)k + (s+t)k + (-s+t)k, for k = 1,2,3, for any t and hence is a special case of Theorem 5 discussed in the section on Equal Sums of Like Powers. (End update.)

18b. Form: x1k+x2

k+x3k+x4

k = y1k+y2

k+y3k+y4

k, k = 1,2,3 The complete soln to this system is already known. For multigrades valid for the consecutive powers k = 1,2,…m by Frolov’s Theorem it is always possible to add a constant to each term such that,

x1+x2+… +xn = y1+y2+…+yn = 0 (eq.1) so we can make use of this property. One approach to completely solving k = 1,2,3 is then to use the Chernick-Lander form, (a+b+c)k + (a-b-c)k + (-a-b+c)k + (-a+b-c)k = (d+e+f)k + (d-e-f)k + (-d-e+f)k + (-d+e-f)k

which satisfies eq.1. The solution to this, which is in binary quadratic forms, also solves k = 5 (by Gloden’s Theorem) so will be discussed in the section on Fifth Powers. Necessarily some of the terms will be negative so this form does not parametricize the even system, x1

k+x2k+x3

k+x4k = y1

k+y2k+y3

for k = 2,4,6. To include this, we have to solve k = 1,2,3 without appealing to eq.1. Use the general form F2, (a+bh)k + (c+dh)k + (e+fh)k + (g+h)k = (a-bh)k + (c-dh)k + (e-fh)k + (g-h)k

To satisfy k = 1,2, set {g,f} = {-(ab+cd+ef), -(1+b+d)}. Expanding for k = 3, we get a quadratic in h, (1+b)(1+d)(b+d)h2-(Poly1) = 0 where Poly1 is a quadratic in the variables a,b,c,d,e. To find rational h, one must make its discriminant a square, (1+b)(1+d)(b+d)(Poly1) = y2 (eq.2) which is easily done since the expression is just a quadratic in {a,c,e}.

It is well-known that given an initial soln to F(x) = y2 where F(x) is a quadratic polynomial, then one can find its complete soln. (One way is to use Fermat’s method.) As a polynomial in e, an initial soln to eq.2 is, e = (a+ab-c+cd)/(b+d) from which one can then find the complete soln. Using this, one can prove that, Theorem: Given ak+bk+ck+dk = ek+fk+gk+hk for k = 1,2,3. Define n as a2+b2+c2+d2 = n(a+b+c+d)2, then, 25(a4+b4+c4+d4-e4-f4-g4-h4)(a6+b6+c6+d6-e6-f6-g6-h6) = 12(n+1)(a5+b5+c5+d5-e5-f5-g5-h5)2 This 4-6-5 Identity, by squaring all variables, can become a 8-12-10 Identity good for the even system k = 2,4,6 and analogous to Ramanujan’s 6-10-8 which depends on a k = 2,4. This will be discussed more in Fourth Powers.

18c. Form: x13+x2

3+...+xn3 = Poly

(Update, 12/15/09): Tony Rizzo gave the ff cubic and quintic identities, (ax2+y)3 + (ax2-y)3 + (a-2)(ax2)3 = (a2x3)2 + (axy)2 + (6-a)ax2y2 (eq.1) (ax2+y)5 + (ax2-y)5 + (a-2)(ax2)5 = (a3x5+10xy2)2 + 10(a-10)x2y4 (eq.2) for arbitrary {a,x,y}. For eq.1, when a = 6, the RHS is a sum of two

squares. This itself is a square if {36x3, 6xy} = {2m, m2-1}, thus, (6x2+y)3 + (6x2-y)3 + 4(6x2)3 = (6xy+2)2, if 6xy = 324x6-1 For eq.2, when a = 10, the RHS becomes Bouniakowsky's quintic-square eqn, (10x2+y)5 + (10x2-y)5 + 8(10x2)5 = 100x2(100x4+y2)2 Note: The relationship between the two can be appreciated more by considering the basic identities, (a+y)3 + (a-y)3 + 4a3 = 6a(a2 +y2) (Id.1) (a+y)5 + (a-y)5 + 8a5 = 10a(a2 +y2)2 (Id.2) It is easy to make the sum of Id.1 a square, while that of Id.2 a fourth power. Piezas (a+y)3 + (a-y)3 + 4a3 = (6m)2(2n2-y)2 where {a,y} = {6n, n2-9}, and n = m2, for arbitrary m. (a+y)5 + (a-y)5 + 8a5 = (10m)4(2n2-y)4 where {a,y} = {10n, n2-25}, and n = 100m4, for arbitrary m. (End update.)

19. Form: ax3+by3+cz3 = N

Theorem 1: If ax3+by3+cz3 = N has a solution, then one can generally find a second. (A. Desboves). Proof: ap3+bq3+cr3 = (ax3+by3+cz3)(ax3-by3)3 {p,q,r} = {(ax4+2bxy3), -(2ax3y+by4), z(ax3-by3)} Special cases a=b=1 were found by J.Prestet and A. Legendre while a different version was given by Cauchy. (Desboves also found a rather similar identity for fourth powers.) Special cases of the general form are given by, S. Realis (quadratic) (2x2-4xy+9yz-9z2)3 + (2y2-xy+9xz-18z2)3 = 9(2x2-4xz-yz+y2)3, if x3+y3 = 9z3

Piezas Let q = p3+1, then (px2-p2xy+qyz-qz2)3 + (py2-xy+qxz-pqz2)3 = q(px2-p2xz-yz+y2)3, if x3+y3 = qz3

(Realis’s was the case p=2 of this identity.) E.Lucas (nonic) p3+q3+cr3 = 27 (x3+y3+cz3) (x7y+x4y4+xy7)3 {p,q,r} = {x9+6x6y3+3x3y6-y9, -x9+3x6y3+6x3y6+y9, 3xyz(x6+x3y3+y6)}

More generally, Theorem 2: If ax1

3+a2x23+...+ amxm

3 = N has a solution, then this generally leads to a second. Proof: Desboves’ identity in Theorem 1 is essentially the same as the difference of cubes, a(ax4+2bxy3)3 - b(2ax3y+by4)3 = (ax3+by3)(ax3-by3)3 If ax3+by3 is the sum of cubes c1z1

3+c2z23+…+ cmzm

3 = N, then, a(ax4+2bxy3)3 - b(2ax3y+by4)3 = (c1z1

3+c2z23+…+ cmzm

3)(ax3-by3)3 = 0 so, by distribution, one gets a new identity of the form ax1

3+a2x23+...+

amxm3 = 0 if there are initial solns x,y, and zi. (End proof)

Form p3+q3 = nr3

Expressing a given number n as the sum of two rational cubes is simply the case a = b = 1, and c = 0, of Desboves' identity above. The fact that an initial soln leads to more can be seen in a better light by using the birational transformation, {p,q,r} = {36n-y, 36n+y, 6x}, to reduce the eqn to the elliptic curve, y2 = x3-432n2

and can be reversed as, {x,y} = {12n/(p+q), -36n(p-q)/(p+q)}

More generally though, Theorem 3: If au3+bv3+cw3 = 0 has a solution, then so does y2 = x3-432a2b2c2. Proof: y2-x3+432a2b2c2 = (au3+bv3+cw3)(au3-bv3-cw3)(432b2c2/u6) {x,y} = {4(p2+q)/r2, 4(2p3+3pq)/r3}, where {p,q,r} = {bv3-cw3, 3bcv3w3, uvw} Theorem 4: If au3+bv3+cw3 = 0 has a solution, then so does x3+y3+abcz3 = 0. Proof: x3+y3+abcz3 = 27bcv3w3(au3+bv3+cw3)(p2+q)3

{x,y,z} = {p3+3bqv3, -p3+3cqw3, 3(p2+q)r}, where {p,q,r} as in theorem 3. A generalization was found by J.Sylvester. (Who found theorems 3 and 4?). }

20. Form: ax3+by3+cz3 = dxyz S. Realis (complete, other than the case x = y = z) x3+y3+z3 = 3xyz {x,y,z} = {(a-b)3+(a-c)3, (b-a)3+(b-c)3, (c-a)3+(c-b)3} All positive integers N other than those div by 3 but not by 9 are representable as N = x3+y3+z3-3xyz with integral x,y,z => 0. The

primes (other than 3) are representable in this manner in one and only one way (Carmichael). Theorem 1: If ax3+by3+cz3+dxyz = 0 has a solution, then one can generally find a second. (A. Cauchy) Proof: ap3+bq3+cr3+dpqr = (ax3+by3+cz3+dxyz)(pqr)/(xyz) {p,q,r} = {x(by3-cz3), y(-ax3+cz3), z(ax3-by3)} This generalizes Desboves’ result in a certain way. Cauchy original statement was that given an initial soln {x,y,z}, a second soln {p,q,r} is, p/(v1x) = q/(v2y) = r/(v3z) {v1, v2, v3} = {by3-cz3, -ax3+cz3, ax3-by3} After a little algebraic manipulation, one can get the above identity. E. Lucas Similarly, given an initial soln {x,y,z}, a second {p,q,r} satisfies the elegant relations p/x+q/y+r/z = 0 and apx2+bqy2+ crz2 = 0. One can then solve for {p,q} which gives essentially the same result as Cauchy’s. Theorem 2: If ax3+by3+cz3+dxyz = 0 has a soln, then so does p3+q3+abcr3+dpqr = 0. (J. Sylvester) Proof: {p,q,r} = {f2g+g2h+h2f-3fgh, fg2+gh2+hf2-3fgh, xyz(f2+g2+h2-fg-fh-gh)}, where {f,g,h} = {ax3, by3, cz3}.

Note that this generalizes theorem 4 of the previous section which is just the case d=0. For a=b=1, d=0, and after some simplification this also gives the nonic parametrization of Lucas. C.Souillart, E. Mathieu x3+ny3+n2z3-3nxyz = (p3+nq3+n2r3-3npqr)(u3+nv3+n2w3-3nuvw) {x,y,z} = {pu+n(qw+rv), pv+qu+nrw, pw+qv+ru} which proves that, like the quadratic form x2+y2, the product of two forms x3+ny3+n2z3-3nxyz is of like form. In general, the result extends to cyclic determinants of order m.

21. Form: x3+y3+z3-3xyz = tk Krafft, Lagrange To recall, the algebraic form we can designate as f2:= x2-dy2 factors

over the square root extension d1/2 (which may involve the complex unit). For its cubic analogue, use the general form f3:=

x3+ny3+n2z3-3nxyz which factors linearly over a complex cube root of unity w as, f3:= x3+ny3+n2z3-3nxyz = (x+my+m2z)(x+mwy+m2w2z)(x+mw2y

+m2wz) where w3-1 = 0 and m = n1/3 for convenience. Or alternatively, if one has Mathematica, f3:= Resultant[x+wy+w2z, w3-n, w]

Because it has three linear factors and variables, one can then easily solve the equation, x3+ny3+n2z3-3nxyz = (p3+nq3+n2r3-3npqr)k

for any positive integer k (just like for f2) by factoring both sides. In fact, for the special case f3 = ±1, this is sometimes known as the cubic Pell equation, an analogue to the Pell equation f2 = ±1 since starting with one initial soln {p,q,r}, one can then find an infinite number of solutions {x,y,z}. Specifically, one solves the system, x+my+m2z = (p+mq+m2r)k

x+mwy+m2w2z = (p+mwq+m2w2r)k

x+mw2y+m2wz = (p+mw2q+m2wr)k for x,y,z and where m = n1/3. For k=2, this gives, {x,y,z} = {p2+2nqr, nr2+2pq, q2+2pr} and so on for other k.

II. Cubic Polynomials as kth powers A. Univariate: ax3+bx2+cx+d2 = tk

1. ax3+bx2+cx+d2 = t2

2. ax3+bx2+cx+d2 = t3

B. Bivariate: ax3+bx2y+cxy2+dy3 = tk

1. x3+y3 = t2

2. ax3+by3 = t2

3. x3+y3 = nz2

4. x3+ax2y+bxy2+cy3 = t2

5. x3+ax2y+bxy2+cy3 = t3 Given an initial soln to the curve F(v):= av3+bv2+cv+d = t2, it is easy to do a small transformation F(v) → F(x) such that F(x) has a square constant term, a method discussed in the section on Quadratic Polynomials as kth powers. Thus, for convenience, in the univariate case we will already assume that d is a perfect square. A. Univariate: ax3+bx2+cx+d2 = tk 1. ax3+bx2+cx+d2 = t2

The general method is given by Fermat. One can solve this in two ways as: ax3+bx2+cx+d2 = (px+d)2, or, ax3+bx2+cx+d2 = (px2+qx+d)2

where p,q are free variables. Expanding and subtracting one side from the other yields, ax3 + (b-p2)x2 + (c-2dp)x = 0 (eq.1) p2x4 + (-a+2pq)x3 + (-b+2dp+q2)x2 + (-c+2dq)x = 0 (eq.2) For (eq.1), p can eliminate the x1 term. For (eq.2), p,q can do so for the x1 and x2 terms. For both, one can then solve for x giving, x = (c2-4bd2)/(4ad2)

x = 8d2(c3-4bcd2+8ad4)/(c2-4bd2)2 as two solns to ax3+bx2+cx+d2 = z2. 2. ax3+bx2+cx+d2 = t3

For the monic case a = 1, what Fermat did was equate, x3+3bx2+3cx+d = (x+b)3, then solved for x as 3x = (-b3+d)/(b2-c) though there are b,c,d such that the method is inapplicable. B. Bivariate: ax3+bx2y+cxy2+dy3 = tk

1. Form: x3+y3 = t2 Euler (also by R. Hoppe) (3m4+6m2n2-n4)3 + (-3m4+6m2n2+n4)3 = (6mn(3m4+n4))2

2. Form: ax3+by3 = t2 Using the approach by Krafft, Lagrange discussed in Form 21 above, we can set one variable as zero, say, z = q2+2pr = 0, and solving for r, we can solve, x3+ny3 = t2

as {x,y} = {4p(p3-nq3), q(8p3+nq3)}, or more generally, ax3+by3 = t2 {x,y} = {4p(ap3-bq3), q(8ap3+bq3)} However, for tk with k>2, it already involves polynomials in p,q,r of degree >1 so it is not so easy to set z=0 and find a general integral soln to ax3+by3 = tk for k>2. Is there such a soln for other k? An alternative soln to k=2, for a=b=1 and where x,y are relatively prime integers is, R. Hoppe (also by Euler) (p4+6p2q2-3q4)3 + (-p4+6p2q2+3q4)3 = (6pq)2(p4+3q4)2 This can be modified to solve x3+y3 = nz2.

3. Form: x3+y3 = nz2 E. Fauquembergue (p2+6pq-3q2)3 + (-p2+6pq+3q2)3 = pq (6(p2+3q2))2 so it suffices to find the factorization of n as n = pq. Notice this is essentially the Euler-Hoppe identity. Finally, one soln implies a second, A.Gerardin (x3+4y3)3 + (-3x2y)3 = nz2(x3-8y3)2, if x3+y3 = nz2

4. Form: x3+ax2y+bxy2+cy3 = t2 Lagrange, Legendre This is the more general equation and has the beautiful soln, {x,y} = {u4-2bu2v2-8cuv3+(b2-4ac)v4, 4v(u3+au2v+buv2+cv3)} for arbitrary variables u,v and where t is a sixth degree polynomial. Derivation: This can be found by generalizing f3. To recall, f3:= Resultant[x+wy+w2z, w3-n, w] However, if we extend w3-n to the more general cubic v3-av2+bv-c = 0, this results in a long trivariate polynomial, or f31, f31: = Resultant[x+vy+v2z, v3-av2+bv-c, v] rather tedious to explicitly write down. For a=b=0 and c=n, this just reduces to f3. Lagrange noted that if the variable z = 0, the equation

f31 = tk naturally reduces to the bivariate form x3+ax2y+bxy2+cy3 =

tk. So using the system, x+v1y+v1

2z = (p+v1q+v12r)k

x+v2y+v22z = (p+v2q+v2

x+v3y+v32z = (p+v3q+v3

with the vi as the three roots of the cubic v3-av2+bv-c = 0, one first

solves for x,y,z. For k=2, a soln to f31 = tk, after much simplication, is given by, {x,y,z} = {p2+cr(2q+ar), 2pq-2bqr-(ab-c)r2, q2+2(p+aq)r+(a2-b)r2} One can then set z = 0, solve for p to get a soln purely in {x,y}. By using a small transformation, Legendre found a more aesthetic version (given at the start of this section) as, {x,y} = {u4-2bu2v2-8cuv3+(b2-4ac)v4, 4v(u3+au2v+buv2+cv3)} which solves x3+ax2y+bxy2+cy3 = t2.

5. Form: x3+ax2y+bxy2+cy3 = t3 For k>2, unfortunately it involves polynomials in p,q,r of degree >1 so it is harder to set z=0. Using another method, one can still find k=3. Theorem: There is generally an infinite family of polynomial solutions {x,y} to x3+ax2y+bxy2+cy3 = t3. For simplicity’s sake, we can transform u3+au2v+buv2+cv3 = t3 into one with a=0 using the transformation u = x-ay, v = 3y to get, x3 - 3(b2-3c)xy2 + (2b3-9bc+27d)y3 = t3 or simply x3+pxy2+qy3 = t3. To solve this, using a birational transformation, we get, P. von Schaewen, J. von Sz.Nagy x3+pxy2+qy3 = z3

{x,y,z} = {pm2-3n2±t, 6mn, pm2+3n2±t}, if {m,n,t} satisfy the elliptic curve E, E: = p2m4+12qm3n-6pm2n2-3n4 = t2 One soln of which is, {m,n,t} = {u2-3q2v, p2qv, pu(u3-3q2v2)}, where {u,v} = {p3+9q2, 3(p3+6q2)}. Since we can use ±t, this yields two solns x,y,z. Using the negative case, after removing common factors this gives, {x,y} = {-q(u2-6q2v), p(u2-3q2v)} with {u,v} as defined above, and so on for the positive case. From this initial {m,n}, one can then come up with an infinite number of polynomial solns. As a last point, we can extrapolate Lagrange’s method to quartics (and higher). Forming, f41:= Resultant[x+vy+v2z+v3w, v4-av3+bv2-cv+d, v] one can easily solve f41 = tk for any positive integer k by solving the system, x+viy+vi

2z+vi3w = (p+viq+vi

2r+vi3s)k

for {x,y,z,w} and where the vi are the four roots of v4-av3+bv2-cv+d = 0. To reduce f41 to the bivariate form by setting two variables z = w = 0 unfortunately entails solving an equation of degree >1 even for just k=2, hence to find a bivariate or univariate polynomial soln to x4+ax3y

+bx2y2+cxy3+dy4 = t2 is a more difficult problem than for the cubic case. A limited soln is possible though, as we’ll see in the section on fourth powers.

PART 7. Fourth Powers I. Sum / Sums of biquadrates (This page covers eqs 1-9.)

1. a4+b4 = c4+d4

2. pq(p2+q2) = rs(r2+s2)3. pq(p2-q2) = rs(r2-s2)4. pq(p2+hq2) = rs(r2+hs2)5. x4+y4 = z4+nt2

6. x4+y4 = z4+nt4

7. u4+nv4 = (p4+nq4)w2

8. u4+nv4 = x4+y4+nz4

9. u4+v4 = x4+y4+nz4

10. ak+bk+ck = dk+ek+fk, k = 2,411. ak+bk+ck = 2dk+ek, k = 2,412. ak+bk+ck = dk+ek+fk, k = 2,3,413. x4+y4+z4 = 2(x2y2+x2z2+y2z2)-t2

14. x4+y4+z4 = t4

15. x4+y4+z4 = ntk

16. v4+x4+y4+z4 = ntk

17. vk+xk+yk+zk = ak+bk+ck+dk, k = 2,418. 2(v4+x4+y4+z4) = (v2+x2+y2+z2)2

19. x1k+x2

k+x3k+x4

k+x5k = y1

k+y2k+y3

k, k = 1,2,3,4

20. x1k+x2

k+x3k+x4

k+x5k = y1

k+y2k+y3

k+y4k+y5

k, k = 1,2,3,4

21. x14+x2

4+…xn4, n > 4

II. Quartic Polynomials as kth Powers1. ax4+by4 = cz2

2. ax4+bx2y2+cy4 = dz2

3. au4+bu2v2+cv4 = ax4+bx2y2+cy4

4. ax4+bx3y+cx2y2+dxy3+ey4 = z2

I. Sum / Sums of biquadrates Fourth powers are rather interesting since quite a number of situations seem to stop at this degree. For example, it is already known that solving P(x) = 0 in the radicals is generally solvable only for degree k ≤ 4. Likewise, using Fermat’s method it seems that solving P(x) = y2 in the rationals and where P(x) has a square constant term is generally solvable only for k ≤ 4. Also, in the context of Diophantine equations, solns are known to, ak+bk+ck = dk, and ak+bk = ck+dk

for k ≤ 4 and it is conjectured (a special case of Euler’s Extended Conjecture, or EEC) that these have no non-trivial integer solns for k > 4. Whether in fact this is the case remains to be seen. 1. Form: a4+b4 = c4+d4 For a database of solns, see Jarek Wroblewski's results here. An infinite sequence of polynomial identities can be found using an elliptic "curve" discussed below, with the smallest having 7th deg. The beautiful version given by Gerardin has small coefficients {1,2,3}, (a+3a2-2a3+a5+a7)4 + (1+a2-2a4-3a5+a6)4 =(a-3a2-2a3+a5+a7)4 + (1+a2-2a4+3a5+a6)4

Choudhry gave a 5th power version, (a-a3-2a5+a9)5 + (1+a2-2a6+2a7+a8)5 + (2a3+2a4-2a7)5 =(a+3a3-2a5+a9)5 + (1+a2-2a6-2a7+a8)5 + (-2a3+2a4+2a7)5 which in fact is good for k = 1,5. While there are identities for k = 6, none are known with such small coefficients. (Anyone can find one?) J. Steggall (a3+b)4 + (1+ad)4 = (a3+d)4 + (1+ab)4, where d = 2ac-b and {a,b,c} must satisfy a quadratic in b, a4+3a2c-b2c+2abc2-2a2c3+1 = 0. Solving this in the rationals entails making its discriminant a square, and involves the elliptic curve, z2 = (1+a4)c+3a2c2-a2c4

If we do the substitution a = (n+1)/(n-1), one soln is c = 9n4/(2n8-2n6+n4-2n2+2) from which we can compute other rational points. Note that the resulting square, after reversing the substitution, has a factor in common with the square in Euler’s second alternative method to solve this eqn. (Update, 2/22/10): Theorem: “The eqn a4+b4 = c6+d6 has an infinite number of unscaled, integral solns.” (Piezas) Proof: The first known example by Giovanni Resta had the form, (m2n)4(x4+y4) = (mn)6(u6+v6) (eq.1) with primitive soln {m,n} = {5, 73}; and {u,v,x,y} = {3, 4, 18, 31}. Eq.1 can be simplified as, m2(x4+y4) = n2(u6+v6)

Hence the situation is reduced to the easier problem of finding, (x4+y4)(u6+v6) = w2 (eq.2) where gcd(x,y) = 1 and gcd(u,v) = 1. However, the second factor can also be treated as a constant, so this becomes, 25(193)(x4+y4) = w2

which is an elliptic curve. Starting with the known {x,y} = {18, 31}, an infinite number of other rational points can then be computed. Hence, another soln to eq.1 is: {m,n} = {5, 228420709564570748140960777}; and {u,v,x,y} = {3, 4, 56328091105014, 7382523294847} and so on, ad infinitum. It seems quite interesting that the Pythagorean triple {3, 4, 5} appears in this problem. Note: A computer search can probably find smaller co-prime solns to eq.2, though the relation u3y2-v3x2 = 0 should be avoided as it is trivial. (End update.) 2. Form: pq(p2+q2) = rs(r2+s2) It is easily seen that a4+b4 = c4+d4 and pq(p2+q2) = rs(r2+s2) are equivalent forms. Given two pairs of Pythagorean triples, (p2-q2)2 + (2pq)2 = (p2+q2)2, and (r2-s2)2 + (2rs)2 = (r2+s2)2 which define two right triangles, the related problems of finding: 1) equal products of a leg and hypotenuse, or 2) equal products of two legs,

(2pq)(p2+q2) = (2rs)(r2+s2) (eq.1) (2pq)(p2-q2) = (2rs)(r2-s2) (eq.2) is essentially solving the equation, (p+hq)4 + (r-hs)4 = (r+hs)4 + (p-hq)4

where h = 1 for the first, h = Ö-1 for the second and which can be proven by expanding this equation. Euler’s soln for the first case involves solving the simple elliptic curve y2 = (a3-b)(b3-a). To derive this, given, pq(p2+q2) = rs(r2+s2) (eq.1) let {p,q,r,s} = {x, ay, bx, y} to get (a-b3)x2 + (a3-b)y2 = 0 or equivalently, y2/x2 = (b3-a)/(a3-b) 1. First method: One can find a,b such that the above becomes a square. Euler, after a series of ingenious algebraic manipulation, gave, b = a(1+v), where v = 3(1-a2)3/(1+10a2+a4+4a6) which will give a 7th deg soln in the variables {p,q,r,s}. 2. Second method: Assume that x = (a3-b). One then has to find, y2 = (a3-b)(b3-a)

The case a = b is trivial. But this curve has infinite number of rational points, another one given by, b = a - 8a(-1+18a2-18a6+a8)/(1+100a2+190a4-44a6+9a8) this yields a 13th deg poly soln to the problem and is related to Stegall’s. From this, other rational points can be found and parametrizations are known for degrees 7, 13, 19, etc. More will be said about this later. Other solns are, A. Gerardin {p,q,r,s} = {a7+a5-2a3+a, 3a2, a6-2a4+a2+1, 3a5} equivalent to, but simpler than, Euler’s 7th deg. (Incidentally, this was also derived by Swinnerton-Dyer in his early papers at the tender age of 16.) T. Hayashi {p,q,r,s} = {2v3(u4+2v4), u4vw, 2uv6, uw(u4+2v4)}, if u4+3v4 = w2

The simplest non-trivial soln is {u,v,w} = {1,2,7} which yields {x1,

x2, x3, x4} = {542, 103, 514, 359} so that x14+x2

4 = x34+x4

4. Note

that given an initial soln to u4+3v4 = w2, subsequent ones can be found as, {u,v,w} = {a4-3b4, 2abc, 12a4b4+c4}, if a4+3b4 = c2

which is just a special case of an identity by Lagrange. Q. Any polynomial soln to u4+3v4 = w2?

3. Form: pq(p2-q2) = rs(r2-s2) Euler If s = p, the equation reduces to solving p2 = q2-qr+r2, two solns of which are, {p,q,r} = {u2+3v2, u2+2uv-3v2, 4uv}{p,q,r} = {u2+3v2, u2-2uv-3v2, u2+2uv-3v2} The form x2+xy+y2, and its equivalent u2+3v2, appears quite often when dealing with third powers. Turns out it also plays a significant role for 4th powers. L.J.Lander {p,q,r,s} = {v5-2v, v5+v, -2v4+1, v4+1} Using this initial soln, more can be derived using a general identity in the ff section. 4. Form: pq(p2+hq2) = rs(r2+hs2) To recall, via a substitution this is equivalent to the form, a4 + b4 = c4 + d4

and this also non-trivially solves, a4 + b4 + c4 = d4 + e4 + f4

J. Chernick If pq(p2+hq2) = rs(r2+hs2), then, (p2-hq2)k + (r2+hs2)k + (-h)(k/2)(2rs)k = (p2+hq2)k + (r2-hs2)k + (-h)(k/

2)(2pq)k, for k = 2,4 though Chernick's identity was set at h = -1. For h = 1, this implies that the system, a2 + b2 - c2 = d2 + e2 - f2

a4 + b4 + c4 = d4 + e4 + f4

is solvable, one of which is {a,b,c,d,e,f} = {7847, 21460, 3504, 21172, 10585, 7104} derived from the smallest soln, 594 + 1584 = 1334 + 1344. See also Form 18 on how this general form appears in the context of Descartes' Circle Theorem. Piezas Theorem: "Given one soln {p,q,r,s} to (p+hq)4 + (r-hs)4 = (r+hs)4 + (p-hq)4 or equivalently, pq(p2+hq2) = rs(r2+hs2), then subsequent ones can be found." Proof: (ab(a2+hb2)-cd(c2+hd2)) = (pq(p2+hq2)-rs(r2+hs2)) (m3q2-n3s2)4 a = m3pq2-3m2nqrs+2n3ps2

b = q(m3q2-n3s2)c = -2m3q2r+3mn2pqs-n3rs2

d = s(m3q2-n3s2)

where {m, n} = {3p2+hq2, 3r2+hs2}. Thus if the RHS = 0, then so will be the LHS. The form x2+3y2 appears again. This identity is implicit in a method used by L.J. Lander. As was already mentioned, many polynomial solns for the case h = 1 were already known and Sinha observed that these were of degree d = 7, 13, 19, 31, 37, or d = 6n+1, and asked if it occurs only if d was prime. In “Unsolved Problems In Number Theory”, R. Guy mentions it was not necessarily the case, as there is a d = 49 though there was none known for d = 25. Turns out there is. This author found that using the above identity and initial soln, {p,q,r,s} = {-2(1+10v2+v4+4v6), v-17v3-17v5+v7, -2v(4+v2+10v4+v6), 1-17v2-17v4+v6} and after removing common factors, it yields a 25-deg poly. In fact, by judicious permutation of the {p,q,r,s}, one can find a 42-deg soln which is definitely not of form d = 6n+1 and doesn’t seem to be a composition of 6 and 7-deg polynomials. 5. Form: x4+y4 = z4+nt2 Desboves (a2+b2)4 + (a2-b2)4 = (2ab)4 + 2(a4-b4)2 Since the equation a2+b2 = ck is easily solved, this shows that x4k+y4 = z4+2t2 has an infinite no. of solns. (Or x4+y4 = z4k+2t2 since it is trivial to solve 2ab = ck.) E. Fauquembergue (also for x4+y4 = 1+z2)

(17p2-12pq-13q2)4 + (17p2+12pq-13q2)4 = (17p2-q2)4 + (289p4+14p2q2-239q4)2 As equal sums of two biquadrates, it remains to make the last term a square. With the trivial p = q to be avoided, one small soln is {p,q} = {11, 3}. Fauquembergue's identity depends on the unique integral soln of u2-2v4 = -1 with {u,v} > 1 and which is {u,v} = {239, 13}. Whether there are other identities similar in form to this remains to be seen. Incidentally, by making the third term as q2-17p2 = ±1, this also solves x4+y4 = z2+1. Since it was proven by Fermat that x4+y4 = z2 has no non-trivial integral soln, then this is the closest near-miss (just like there are integer solns to x3+y3 = z3±1 which are also near-misses of Fermat’s last theorm). Note: This author checked x4+y4 = z2±1 and found a lot of solns for the positive case, with the smallest being {x,y,z} = {5, 7, 55}. In contrast, there were none at all for the negative case with {x,y} < 103. Why the asymmetry? To compare, there are roughly an equal number of positive integer solns to x3+y3 = z3±1 below a bound z. (Update, Jan 3, 2011): By exhaustive computer search, Jim Cullen showed that there is no soln to x4+y4 = z2-1 with {x,y} < 106. If any exists, by modular arguments it can be shown to have the form (10u)4+(10v)4 = z2-1. A table of solutions to the form (10x)4+(100y)2 = z2-1 is also given in the link provided. 6. Form: x4+y4 = z4+nt4 N. Elkies (192m8-24m4-1)4 + (192m7)4 = (192m8+24m4-1)4 + 192m4

or equivalently, (192m8-24m4-1)4 + (192m7)4 = (192m8+24m4-1)4 + 12(2m)4

This was found while considering near-misses to the Fermat curve, x4+y4 = z4. Three of the terms are polynomials of high degree while the “excess” is only a quartic thus providing very good approximations to the curve. To illustrate, for m = 2, this gives, 245764 + 487674 ≈ (49535.0000000000063...)4

so the fourth root of the sum is very close to an integer. 7. Form: u4+nv4 = (p4+nq4)w2 R. Carmichael Define {r,s} = {p4-nq4, p4+nq4}. Then, (pm-2prs)4 + n(qm+2qrs)4 = (p4+nq4)(m2+16np4q4r2)2, with m = 4np4q4+s2do See also Sophie Germain's Identity. 8. Form: u4+nv4 = x4+y4+nz4 A. Gerardin Let r = p8-4n2, (6pn2+pr)4 + n(3p4n-r)4 = (6pn2)4 + (pr)4 + n(3p4n+r)4

(Note that if n=0, the identity is trivial and doesn’t provide a counter-example to FLT for k = 4.) 9. Form: u4+v4 = x4+y4+nz4 R. Norrie Define r = p8-q8. Then for any constant n, ((2n+r)p3q)4 + (2np4-q4r)4 = ((2n-r)p3q)4 + (2np4+q4r)4 + n(2pqr)4

for arbitrary p,q, so any number is the sum/difference of four rational fourth powers as was already mentioned in Part. Again, for n=0 the identity is trivial. It may be interesting to ask for what n is there a rational poly soln to x4+y4+z4 = t4+n?

10. Form: ak+bk+ck = dk+ek+fk (k = 2,4) Bastien’s Theorem also given below states that the above eqn is trivial if simultaneously true for k = 1,2,3,4. However, this does have solns when k = 2,4; k = 1,2,4, or even k = 2,3,4 and, of course, when only for single exponents. When valid for both k = 2,4, this quadratic-quartic system will be called throughout this work as Q1, and this form is significant because it may appear as a side condition for higher powers such as for the 5th, 6th, 7th, and 8th. First, we have these general theorems, T. Sinha

Theorem: If ak+bk+ck = dk+ek+fk, k = 2,4, then (a2-f2)(b2-f2) = (c2-d2)(c2-e2). The eqn (a2-f2)(b2-f2) = (c2-d2)(c2-e2) is quite easy to satisfy and can be reduced to solving an eqn of form x2+ny2 = z2. However, it is not enough to imply Q1. More generally, any two of the equations in the system implies the third, but it is not the case that one implies the other two. Tarry, Escott Theorem: If ak+bk+ck = dk+ek+fk, k = 2,4, then (a+t)k + (-a+t)k + (b+t)k + (-b+t)k + (c+t)k + (-c+t)k = (d+t)k + (-d+t)k + (e+t)k + (-e+t)k + (f+t)k + (-f+t)k

k = 1,2,3,4,5 for any t. This is a special case of a theorem of the Prouhet-Tarry-Escott Problem which I’m assuming is due to Tarry or Escott since they established related theorems for this field. Note that appropriate pairs have the common sum 2t and is known as a symmetric solution. Any multi-grade k = 1-5 of this form is then reducible to Q1. By Bastien's Theorem, the system, x1

k+x2k+…+ xm

k = y1k+y2

k+…+ ymk

for k = 1,2,…n has non-trivial solns only if m>n. Thus, one side must have at least n+1 terms. For the case k = 1,2,3,4,5, this implies an eqn with 6 terms on either side. However, it is possible some terms are equal to zero. For example, given the special case of Q1,

ak+bk+ck = 2dk+ek, k = 2,4, then, (a+d)k + (-a+d)k + (b+d)k + (-b+d)k + (c+d)k + (-c+d)k = 2(2d)k + (e+d)k + (-e+d)k, k = 1,2,3,4,5 which has only 4 terms on the RHS. Chernick Theorem: If (a+d)k + (-a+d)k + bk = (c+d)k + (-c+d)k + dk, k = 2,4, then (b-d)k + (-b-d)k + (2d)k = (a+2d)k + (-a+2d)k + (-c-2d)k + (c-2d)k, k = 1,2,3,5 which is one of two identities. Note that if the terms of Q1 are expressed as xi and yi, this also satisfies x1+x2 = y1+y2. More generally,Piezas Theorem: If (p-q+r)k + (p+q-r)k + xk = (p+q+r)k + (p-q-r)k + yk, k = 2,4, then, (2p+q+r)k + (2p-q-r)k + (-2p-q+r)k + (-2p+q-r)k = (p+x)k + (p-x)k + (-p+y)k + (-p-y)k, k = 1,2,3,5, and where one of the terms of the latter eqn can be set to zero to get either of Chernick’s two solns. Chernick, Escott Theorem: If 1k + 2k + 3k = uk + vk + (u+v)k, k = 2,4, then,

(u-7)k + (u-2v+1)k + (3u+1)k + (3u+2v+1)k = (u+7)k + (u-2v-1)k + (3u-1)k + (3u+2v-1)k, k = 2,4,6. Sinha Theorem: If (a+3c)k + (b+3c)k + (a+b-2c)k = (c+d)k + (c+e)k + (-2c+d+e)k, k = 2,4, then, ak + bk + (a+2c)k + (b+2c)k + (-c+d+e)k = (a+b-c)k + (a+b+c)k + dk + ek + (3c)k, k = 1,3,5,7 excluding the trivial case c = 0. This was derived using the next theorem. Birck, Sinha Theorem: If ak+bk+ck = dk+ek+fk, k = 2,4 where a+b ≠ c; d+e ≠ f, then, (a+b+c)k + (-a+b+c)k + (a-b+c)k + (a+b-c)k + (2d)k + (2e)k + (2f)k =(d+e+f)k + (-d+e+f)k + (d-e+f)k + (d+e-f)k + (2a)k + (2b)k + (2c)k

for k = 1,2,4,6,8. Piezas Theorem: If ak+bk+ck = dk+ek+fk for k = 2,4. Define {x,y} = {a2+b2+c2, a4+b4+c4}. Then, a6+b6+c6-d6-e6-f6 = 3(a2-d2)(b2-d2)(c2-d2) = 3(a2b2c2-d2e2f2)3(a8+b8+c8-d8-e8-f8) = 4(a6+b6+c6-d6-e6-f6)(x)

6(a10+b10+c10-d10-e10-f10) = 5(a6+b6+c6-d6-e6-f6)(x2+y) If we define Fk:= ak+bk+ck-dk-ek-fk, then for even k > 4, Fk is of the form F6Poly1 and also a highly composite number. More significantly, the identities can be combined and for a,b,c,d,e,f with appropriate properties, these can be woven together into a very neat form. For example, define n as, a4+b4+c4 = n(a2+b2+c2)2 and Q1 implies the general identity, 32(a6+b6+c6-d6-e6-f6)(a10+b10+c10-d10-e10-f10) = 15(n+1)(a8+b8+c8-d8-e8-f8)2 For the special case when a+b = ±c, since, a4+b4+(a+b)4 = (1/2)(a2+b2+(a+b)2)2 then n = 1/2, and the above becomes, 64(a6+b6+c6-d6-e6-f6)(a10+b10+c10-d10-e10-f10) = 45(a8+b8+c8-d8-e8-f8)2 which is a concise version of Ramanujan’s 6-10-8 Identity discussed in the next section. (In the general case, this turns out to be just the first in a family of identities.) (Update, 7/31/09): This can be extended to the "solution chain" a1

k+a2k+a3

k = a4k+a5

k+a6k = a7

k+a8k+a9

k = a10k

+a11k+a12

k = ..., k = 2,4, for a chain divisible into pairs. For the special case when {a1+a2, a4+a5, a7+a8, a10+a11} = {a3, a6, a9, a12}, then,

6+a26+a3

6+a46+a5

6+a66-a7

6-a86-a9

6-a106-a11

6-a126)

(a110+a2

10+a310+a4

10+a510+a6

10-a710-a8

10-a910-a10

10-a1110-a12

= 45(a18+a2

8+a38+a4

8+a58+a6

8-a78-a8

8-a98-a10

8-a118-a12

8)2, One can test this by the soln chain [28, 175, 203] = [77, 140, 217] = [107, 113, 220] = [5, 188, 193] given here, found by Gloden in the 1940's. (End update.) It seems then the system Q1 is quite important and worth a second look. To solve, ak+bk+ck = dk+ek+fk

for k = 2,4, one trick is to reduce the degree of the two eqns by using the substitution {a,b,c,d,e,f} = {p+q, r+s, t+u, t-u, r-s, p-q} to get, pq+rs+tu = 0p3q+pq3+r3s+rs3+t3u+tu3 = 0 By eliminating u between them, one ends up with a quartic in t, (pq+rs)t4-(p3q+pq3+r3s+rs3)t2+(pq+rs)3 = 0 or a cubic in either p,q,r,s. However, it is possible to use another substitution that can reduce the degree even further. Theorem: "The system ak+bk+ck = dk+ek+fk for k = 2,4, call it Q1, can be reduced to solving a quadratic. Its discriminant D is a quartic polynomial in one variable and the problem of making D a square can be treated as an elliptic curve." (In fact, as was already mentioned, similar results can be given for the three systems k = 1,4, or k = 1,5, or k = 1,2,6. Another substitution will be used for these which is slightly more efficient since the first

system ends up with a discriminant D that is only a quadratic polynomial.) Proof: We simply use a different substitution, call this form F1, {a,b,c,d,e,f} = {p+qu, r+su, t+u, t-u, r-su, p-qu} to get, pq+rs+t = 0p3q+r3s+t3+(pq3+rs3+t)u2 = 0 where the degree of the variable u has been reduced to merely a quadratic. Eliminating t between them yields, (Poly1)u2-(Poly2) = 0 where, Poly1:= p(q-q3)+r(s-s3)Poly2:= p3(q-q3)-3pqrs(pq+rs)+r3(s-s3) and the problem is reduced to making its discriminant a square, y2 = (Poly1)(Poly2) which is just a quartic in p (or r), thus proving the theorem. This quartic in fact has a square leading and constant term so is easily made a square and from that initial point, other rational ones may be computed. To prove that this is the complete characterization of non-trivial solns, solve for p,q… in terms of the original variables to get, {p,q,r,s,t,u} = {(a+f)/2, (a-f)/(c-d), (b+e)/2, (b-e)/(c-d), (c+d)/2, (c-d)/2} and even if there is division by a variable, when c = d the system ak

+bk = ek+fk for k = 2,4 has only trivial solns hence this exceptional case is of no interest and does not affect the generality of the theorem. While it has been proven that Q1 can be reduced to solving a quadratic, a third constraint may be useful to explore various solns. This constraint is an appropriate linear or quadratic relation between the terms such that the final eqn remains a quadratic and involve a simpler elliptic curve. This author found nine relations, six linear and three quadratic,

10.1 a+b = nc; d+e = nf10.2 a+b ≠ c; d+e ≠ f10.3 a+b±c = n(d+e±f)10.4 na+b+c = d+e+nf10.5 na+b = e+nf10.6 a+d = n(c+f)10.7 (a2-f2)c2 = -(b2-e2)d2

10.8 a2+nab+b2 = d2+nde+e2

10.9 a2+nad+d2 = b2+nbe+e2

for any n. All of these have a complete soln in terms of an elliptic curve though may simplify into quadratic forms for particular values of n or be trivial when n = 0. The linear relations will be discussed in the next section.

10.1. Constraint: a+b = nc, d+e = nf The complete soln for the case n = 1 is given by, ak + bk + (a+b)k = ck + dk + (c+d)k, k = 2,4 where a2+ab+b2 = c2+cd+d2. By modifying it slightly to the form,

ak + bk + (-a-b)k = ck + dk + (-c-d)k

makes it valid for k = 1 as well. One soln is, define {m,n} = {p-3q, p+3q}, then for k = 1,2,4, (2p-mr)k + (2pr-n)k + (-m-nr)k = (2p-nr)k + (2pr-m)k + (-n-mr)k

for arbitrary {p,q,r}. (Update, 1/25/10): Wroblewski Given, (a2)k + (b2)k + (a2+b2)k = (c2)k + (d2)k + (c2+d2)k, for k = 2,4 (eq.1) or equivalently, a4+a2b2+b4 = c4+c2d2+d4

then we get the [k.8.8] identity, [ap+cq, -ap+cq, bp+dq, -bp+dq, dp+aq, dp-aq, cp+bq, cp-bq]k = [ap+dq, -ap+dq, bp+cq, -bp+cq, cp+aq, cp-aq, dp+bq, dp-bq]k, for k = 1,2,4,8 (Notice how {c,d} of the LHS is replaced by {d,c} in the RHS.) for arbitrary {p,q} and some constants {a,b,c,d}, hence giving an identity in terms of linear forms. An example is {a,b,c,d} = {1, 26, 17, 22}, though Choudhry has given a 7th deg identity for, a4+ma2b2+b4 = c4+mc2d2+d4

for arbitrary m. See Form 3 here. To find a [8.7.7], since {p,q} are arbitrary, one can equate appropriate xp = yq so a pair of terms cancel out. Note: Eq. 1, or more generally its unsquared version, is at the core of the Ramanujan 6-10-8 Identity. (End update.) (Update, 2/3/10): Choudhry Gave a soln to ak+bk+ck = dk+ek+fk, k = 1,2,4, for integer {a,b,c,d,e,f} with the interesting property that one term is equal to 1. (-p+v)k + (-q+v)k + (p+q-2v)k = (-p-q+2uv)k + (p+q-2uv-1)k + 1k where {p,q} = {2u2v+u-v, 4uv+3v+1}. Source: On Representing 1 as the sum or difference of kth powers of integers, The Mathematics Student, Vol. 70, 2001. Note: Since this also has a+b+c = d+e+f = 0, its terms obey the beautful Ramanujan-Hirschhorn Theorem given below. (End update.) Ramanujan (a+b+c)k + (b+c+d)k + (a-d)k = (c+d+a)k + (d+a+b)k + (b-c)k

if ad = bc, for k = 2,4. This form appears in the beautiful 6-10-8 and 3-7-5 Identities, Ramanujan 64[(a+b+c)6+(b+c+d)6+(a-d)6-(c+d+a)6-(d+a+b)6-(b-c)6][(a+b+c)10+(b+c+d)10+(a-d)10-(c+d+a)10-(d+a+b)10-(b-c)10] = 45[(a+b+c)8+(b+c+d)8+(a-d)8-(c+d+a)8-(d+a+b)8-(b-c)8]2

M. Hirschhorn 25[(a+b+c)3-(b+c+d)3-(a-d)3+(c+d+a)3-(d+a+b)3+(b-c)3][(a+b+c)7-(b+c+d)7-(a-d)7+ (c+d+a)7-(d+a+b)7+(b-c)7] = 21[(a+b+c)5-(b+c+d)5-(a-d)5+(c+d+a)5-(d+a+b)5+(b-c)5]2 where for both, ad = bc, with the discovery of the latter inspired by the former. These can be generalized as, Ramanujan-Hirshhorn Theorem 1a: If ak+bk+ck = dk+ek+fk, k = 2,4 (eq.1) where a+b+c = d+e+f = 0, then, 64(a6+b6+c6-d6-e6-f6)(a10+b10+c10-d10-e10-f10) = 45(a8+b8+c8-d8-e8-f8)2 (eq.2) 25(a3+b3+c3-d3-e3-f3)(a7+b7+c7-d7-e7-f7) = 21(a5+b5+c5-d5-e5-f5)2 (eq.3) Proof: By doing the substitution {a,b,c,d,e,f} = {p, q, -p-q, r, s, -r-s}, all three equations have the common factor p2+pq+q2-(r2+rs+s2) = 0 so the problem is reduced to finding expressions {p,q,r,s} that satisfy this. (End proof) It has a complete soln. Let {p,q,r,s} = {-u+v, u+v, -x+y, x+y}, and the condition becomes, u2+3v2 = x2+3y2

with the complete soln to u2+nv2 = x2+ny2 as, (ac+nbd)2 + n(bc-ad)2 = (ac-nbd)2 + n(bc+ad)2. There are also interesting relations using odd powers k = 1,3, Piezas

Theorem 1b: If ak+bk+ck = dk+ek+fk for k = 1,3 (eq.1) where a+b+c = d+e+f = 0, then, 9abc(a6+b6+c6-d6-e6-f6) = 2(a9+b9+c9-d9-e9-f9) (eq.2) Proof: Doing the same substitution {a,b,c,d,e,f} = {p, q, -p-q, r, s, -r-s}, the two equations have the common factor pq(p+q)-rs(r+s) = 0. (End proof) This is also discussed in Form 13 of Third Powers. Another uses k = 1,3,5, Piezas Theorem 2: If ak+bk+ck+dk = ek+fk+gk+hk, k = 1,3,5 where a+b+c+d = e+f+g+h = 0, then, 7(a4+b4+c4+d4-e4-f4-g4-h4)(a9+b9+c9+d9-e9-f9-g9-h9) = 12(a6+b6+c6+d6-e6-f6-g6-h6)(a7+b7+c7+d7-e7-f7-g7-h7) which again has a complete soln. For higher powers, Theorem 3: If ak+bk+ck+dk = ek+fk+gk+hk, k = 2,4,6, define n as, a4+b4+c4+d4 = n(a2+b2+c2+d2)2 then, 25(a8+b8+c8+d8-e8-f8-g8-h8)(a12+b12+c12+d12-e12-f12-g12-h12) = 12(n+1)(a10+b10+c10+d10-e10-f10-g10-h10)2 Theorem 4: If ak+bk+ck+dk+ek = fk+gk+hk+ik+jk, k = 2,4,6,8 define n as a4+b4+c4+d4+e4 = n(a2+b2+c2+d2+e2)2 then, 72(a10+b10+c10+d10+e10-f10-g10-h10-i10-j10)(a14+b14+c14+d14+e14-f14-g14-h14-i14-j14) = 35(n+1)(a12+b12+c12+d12+e12-f12-g12-h12-i12-j12)2

and so on based on higher systems k = 2,4,…2m. Also, buried in the musty pages of an old journal, there is a similarly beautiful identity, Sinha's 2-4-6 Identity, Sinha 5[a2+b2+c2]*[(a+b+c)4 + (a+b-c)4 + (a-b+c)4 + (-a+b+c)4 - (2a)4 - (2b)4 - (2c)4] = [(a+b+c)6 + (a+b-c)6 + (a-b+c)6 + (-a+b+c)6 - (2a)6 - (2b)6 - (2c)6] which this author realized could be extended (and no further) as, 5[a2+b2+c2+d2]*[(a+b+c-d)4 + (a+b-c+d)4 + (a-b+c+d)4 + (-a+b+c+d)4 - (2a)4 - (2b)4 - (2c)4 - (2d)4] =[(a+b+c-d)6 + (a+b-c+d)6 + (a-b+c+d)6 + (-a+b+c+d)6 - (2a)6 - (2b)6 - (2c)6 - (2d)6] One can see that Sinha’s is just the case d=0. How this was derived will be discussed in the section on Eighth Powers. V. Tariste (23x-57y)k + (40x+6y)k + (17x+63y)k = (23x+57y)k + (40x-6y)k + (17x-63y)k

The identity is in fact valid for k =2,4. There is nothing really special about these particular values as these can be generalized as, Piezas Let {u1, u2, u3, c} = {a+2b, 2a+b, a-b, a+b}, then for k = 2,4,

(ax+u1y)k + (bx-u2y)k + (cx-u3y)k = (ax-u1y)k + (bx+u2y)k + (cx

+u3y)k

(ax+u2y)k + (bx-u3y)k + (cx+u1y)k = (ax+u3y)k + (bx+u1y)k + (cx

+u2y)k

And as sums of one side, (ax+u1y)k + (bx-u2y)k + (cx-u3y)k = (ak+bk+ck)(x2+3y2)k/2

(ax+u2y)k + (bx-u3y)k + (cx+u1y)k = (ak+bk+ck)(x2+3xy+3y2)k/2

Furthermore, (ax2+2u1xy-3ay2)k + (bx2-2u2xy-3by2)k + (cx2-2u3xy-3cy2)k = (ak

+bk+ck)(x2+3y2)k where the last is a template identity that can generate quadratic parametrizations using an initial soln, as long as it satisfies the condition a+b = c. For example, the smallest soln, 24+24+44+34+44 = 54, was used by Ramanujan. Another, found by this author, is 504+504+1004+44+154 = 1034 which yields, (50x2+300xy-150y2)4 + (50x2-300xy-150y2)4 + (100x2-300y2)4 + (4x2+12y2)4 + (15x2+45y2)4 = 1034(x2+3y2)4 and so on for others involving a+b = c. Note that the expressions {a+2b, 2a+b, a-b, a+b} are intimately connected to the form x2+3y2 since,

(a+2b)2 + 3a2 = (2a+b)2 + 3b2 = (a-b)2 + 3(a+b)2 = 4(a2+ab+b2) However, for general n, given the system, a+b = nc; d+e = nfa2+b2+c2 = d2+e2+f2

a4+b4+c4 = d4+e4+f4

and the substitution {a,b,c,d,e,f} = {p+q,r+s, t+u, t-u, r-s, p-q}, where set t = 1 without loss of generality, solve for {p,s,u} and substitute these into the last eqn. This will have a quadratic factor in r with discriminant D, D:= (1+n2)2q4 + 8n(2+n2)q3 - 2(-17-6n2+n4)q2 - 8n(2+n2)q + (1+n2)2 which must be made a square. Since this quartic polynomial has a square leading and constant term, this is easily done. Two small solns are, q1 = -2/n, q2 = (n-1)/(n+1) Though these lead to a trivial soln, they can be used to compute other rational points on the curve. A non-trivial point found using Fermat’s method is, q3 = (n6+2n4+n2-4)/(4n(n2+1)2) 10.2. Constraint: a+b ≠ c; d+e ≠ f Birck, Sinha Theorem: If ak+bk+ck = dk+ek+fk, k = 2,4

where a+b ≠ ±c; d+e ≠ ±f, then, (a+b+c)k + (-a+b+c)k + (a-b+c)k + (a+b-c)k + (2e)k + (2f)k + (2g)k =(d+e+f)k + (-d+e+f)k + (d-e+f)k + (d+e-f)k + (2a)k + (2b)k + (2c)k

for k = 1,2,4,6,8. This author found an extension of this to tenth powers and will be discussed later. The proof for this particular theorem of Birck will be discussed on the section on Eighth Powers. From this, Sinha derived a version applicable to odd exponents, T. Sinha Theorem: If ak+bk+ck = dk+ek+fk, k = 2,4 (or Q1) where a+b-c = 2(d+e-f) and a+b ≠ c, d+e ≠ f, then, (2a-3h)k + (2a-h)k + (2b-3h)k + (2b-h)k + (2f+h)k = (2c+h)k + (2c+3h)k + (2d-h)k + (2e-h)k + (3h)k

for k = 1,3,5,7, where 2h = d+e-f. To solve Q1 and the side conditions, Sinha gave a system of linear relations (modified by this author to be consistent with the others), call it S, a+b-c = 2(d+e-f)(a+b)-(d-e) = 0; 13a+13b+11c+4f = 0 These can then define three of the variables. For example, solving for {a,c,f}, and substituting into Q1 for either k = 2 or 4, this forces the

remaining three variables, say they are {x,y,z}, into a quadratic eqn of form, a1x2 + a2y2 + a3z2 + a4xy + a5xz + a6yz = 0 One can solve this by using an identity of Desboves’ (found in the section on Second Powers) that needs an initial soln {x,y,z}. Another approach, producing more aesthetic results, is that since this is a quadratic in any of the variables, say z, for rational solns its discriminant must be made a square, b1x2+b2xy+b3y2 = t2

where the bi are in terms of the ai. Thus, if one can find a system of useful linear relations between the {a,b,c,d,e,f}, the problem will be reduced to solving this eqn. As this author found out, there are at least four systems S,

1. a+b-c = 2(d+e-f); (a+b) ± (d-e) = 0; 13a+13b+11c+4f = 02. a+b-c = 2(d+e-f); (a-b) ± (d+e) = 0; 4c+7d+7e-f = 0

counting positive and negative cases of the second eqn as distinct. These then yield the polynomial solns to Sinha's problem, 1. (-3x+2y+z)k + (-3x+2y-z)k + (2x-4y)k = (8x-y)k + (2x+3y)k + (14x-2y)k

where 121x2-10xy-5y2 = z2, D = 70. 2. (5x+2y+z)k + (5x+2y-z)k + (-14x-4y)k = (14x+3y)k + (4x-y)k + (6x-2y)k

where x2-50xy-5y2 = z2, D = 70. (Sinha’s)

3. (6x+3y)k + (4x+9y)k + (2x-12y)k = (-x+3y+3z)k + (-x+3y-3z)k + (-6x-6y)k

where x2+10y2 = z2, D = -10. 4. (2x+3y)k + (-4x+y)k + (-10x-4y)k = (3x+y+z)k + (3x+y-z)k + (2x-2y)k

where 49x2+40xy+10y2 = z2, D = -10. where D is the discriminant of the conditional eqn and some manipulation was done to attain these simple forms. Since all have a square leading coefficient, these can be transformed to the form u2+dv2 = w2 which has a complete soln. Note 1: Any other linear relations or S that is not merely a permutation of variables and change of signs? For ex, the first pair (a+b)±(d-e) = 0 is equivalent to either of the ff: (a+b)±(d+f), (a-c)±(d-e), (a-c)±(d+f), while the second pair (a-b) ± (d+e) = 0 is the same as (a-b)±(d-f), (a+c)±(d+e), (a+c)±(d-f). I do not know if the above four solns are the only ones in terms of binary quadratic forms. Note 2: Sinha’s theorem can be expressed more concisely as given, (a+3c)k + (b+3c)k + (a+b-2c)k = (c+d)k + (c+e)k + (-2c+d+e)k, for k = 2,4, then, ak + bk + (a+2c)k + (b+2c)k + (-c+d+e)k = (a+b-c)k + (a+b+c)k + dk + ek + (3c)k

for k = 1,3,5,7, (excluding the trivial case c = 0).

Piezas There can be more solns to Q1 with a+b-c = 2(d+e-f) though now an elliptic curve has to be solved. The linear relation used is either case of (a+b) = ±(d+e). The general form is, (-y+t)k + (y+t)k + pk = (-z±t)k + (z±t)k + qk, where t is chosen so the eqn satisfies a+b-c = 2(d+e-f). For the positive case, (-y+t)k + (y+t)k + (48-2t)k = (-12x+t)k + (12x+t)k + (24)k, where t = 13-x2, and y2 = -2x4+100x2+46 This elliptic curve has trivial solns x = {1,3,5,7} but from these we can find non-trivial ones such as 4307x = 2367, and so on. For the negative case, (-y+t)k + (y+t)k + (2u+6v)k = (-z-t)k + (z-t)k + (-2u)k, where t = u+v, and u,v must satisfy the simultaneous eqns u2+6v2 = y2 and u2+12uv+24v2 = z2. Solving the first as {u,v} = {p2-6, 2p}, the second becomes, p4+24p3+84p2-144p4+36 = z2

with integral solns p = {-6,-4,-2,0,1,3,5} from which rational ones can be derived though this has to be checked for triviality. Ramanujan 34 + (2v4-1)4 + (4v5+v)4 = (4v4+1)4 + (6v4-3)4 + (4v5-5v)4

This has the nice form of involving a constant term and can be generalized as, Piezas For any constant n, (2n)k + (n+p)k + (n-p)k = (2r)k + (q+r)k + (q-r)k, k = 2,4 {p,q,r} = {mx2+nx+3m, mx2+nx-3m, 2mx+n}, and three free variables {m,n,x}. Example, let n = 1, x = 2, so, 2k + (7m+3)k + (7m+1)k = (3m-1)k + (5m+3)k + (8m+2)k

though for any {m,n,x} this family belongs to the case a+b = c, d+e = f so can’t be used for Sinha’s problem. R. Norrie (a4-2)4b4 + a4(b4+2)4 + (2a3b)4 = (a4+2)4b4 + a4(b4-2)4 + (2ab3)4 10.3. Constraint: a+b±c = n(d+e±f) Piezas It turns out the system, a+b±c = n(d+e±f) (eq.1)a2+b2+c2 = d2+e2+f2 (eq.2)a4+b4+c4 = d4+e4+f4 (eq.3) also has a soln for any n of which Sinha’s problem was just the special

case n = 2. The general problem is equivalent to finding rational points on a certain elliptic curve (though this curve may be degenerate for some n). To show this, first express the variables in the form, {a,b,c,d,e,f} = {p+q, r+s, t+u, t-u, r-s, p-q} reducing the degrees of eq.2 and 3, a trick which will be also useful when dealing with sixth powers. As the explicit expressions are a bit messy, only the steps will be given:

1. Solve for p,u using eqs.1,2. 2. Substituting these into eq.3, it has a quadratic factor in r with

discriminant D.3. D is a quartic in s. Hence this must be made a square with two

small solns for any n as s = q±t. By treating D = y2 as an elliptic curve, from an initial soln subsequent ones can then be found.

Example: Using Sinha’s problem n = 2, and the negative case a+b-c = 2(d+e-f), doing the substitutions, we get, 3p-q-r+3q-3t+u = 0 (eq.1)pq+rs+tu = 0 (eq.2)p3q+pq3+r3s+rs3+t3u+tu3 = 0 (eq.3) Solve for {p,u} and substitute into eq.3. One factor will be a quadratic in r. The square-free discriminant D of this is of form, D: = c1s4+c2s3+c3s3+c4s+c5

2 = y2

with a square constant term and where the ci are polynomials in {q,t}. Four small solns are: s = {q, q-t, q+t, (q-t)/7}. These may then be used to generate subsequent ones. For general n, using s = q+t with t = 1, a quadratic form soln can be found for the + case of eq.1 as,

{a,b,c} = {1+2nq+r, n(1+2q), -1+n-r}{d,e,f} = {n+q+nq+r, 1-n+q-nq-r, 1+2q} where r = q+(n+1)q2, for arbitrary {n,q}. This also satisfies {a-b+c, d+e-f} = {0,0} though one can also find parametric ones such that this is not the case. 10.4. Constraint: na+b+c = d+e+nf Piezas Just like the previous, this has a soln for any n. The inspiration for this form is the smallest polynomial soln to the sixth power multi-grade ak

+bk+ck = dk+ek+fk, k = 2,6 which, among others, surprisingly obeys the side condition 3a+b+c = d+e+3f. (Other n for sixth powers obeying na+b+c = d+e+nf are now known.) For the quartic version, for general n again this can be reduced to finding rational points on an elliptic curve. Let, na+b+c = d+e+nf (eq.1)a2+b2+c2 = d2+e2+f2 (eq.2)a4+b4+c4 = d4+e4+f4 (eq.3) and likewise {a,b,c,d,e,f} = {p+q, r+s, t+u, t-u, r-s, p-q},

1. Solve for r,u in eq.1,2. When substituted into eq.3, this becomes a quadratic in q.

2. Discriminant D of eq.3 is a quartic polynomial in s (but a sextic in p,t).

3. For general n, four small solns are s = {p-nt, p-t, p+t, p}. Using these and others, subsequent ones can then be found. Thus, any soln to eq.1,2,3 must solve D = y2.

Example: Using s = p+t with t = 1 yields, {a,b,c} = {2n-3p+np, -n-n2+3p+np, -n(-1+n-2p)}{d,e,f} = {n(1+n-2p), -n+n2+3p-np, -2n+3p+np} for arbitrary {n,p}, with this particular soln also satisfying {a+b-c, d-e+f} = {0,0}. 10.5. Constraint: na+b = e+nf Piezas For n = 1, this can be seen as the case n = 0 of a+b+nc = nd+e+f of the system above, and can be completely solved as polynomials thus providing a four-parameter soln to ak+bk+ck = dk+ek+fk, for k = 2,4. Interestingly, this ultimately involves solving the simple equation x2+my2 = z2. Furthermore, any soln implies ak+bk+ck+dk = ek+fk+gk

+hk for k = 1,2,3,5. Theorem: If (p+u)k + (p-u)k + (2q)k = (p+v)k + (p-v)k + (2r)k, for k = 2,4, then, (p+2q)k + (p-2q)k + (-p+2r)k + (-p-2r)k = (-2p+u)k + (-2p-u)k + (2p+v)k + (2p-v)k, for k = 1,2,3,5 The complete soln is given by the simultaneous eqns, -3p2+q2+3r2 = u2

-3p2+3q2+r2 = v2

Proof: Just solve for {u,v} in terms of {p,q,r}. This is simpler than it looks since the complete parametrization to both is in linear forms.

Alternatively, one can just directly solve the form, ak+bk+ck = dk+ek+fk for k = 2,4 (or Q1) with a+b = d+e, eliminate e,f using resultants, giving the final eqn, 2a2+5ab+2b2-2c2-3ad-3bd+3d2 = 0 Solving this for a, its discriminant must be made a square and after suitable transformations, turns out to involve the eqn, x1

2+nx22 = x3

2+nx42

which has a complete soln. Sparing the reader some algebra, the final eqn, with e = a+b-d, is solved as, {a,b,c,d,e} = {-3pr-4qr+2ps+qs, 5qr-ps, 3pr+qs, 3qr+ps, -3pr-2qr+qs} It remains to find f. Substituting these values into Q1, one ends up with a quadratic function to be made a square. To simplify matters, let, {f, r,s} = {(4p+2q)z, (2p+q)2y, 2x+(6p+q)(p+2q)y} and this final condition becomes, x2-6q(3p+q)(p2-q2)y2 = z2

This is an eqn of form x2+ny2 = z2 which can be completely solved as {x,y,z} = {u2-nv2, 2uv, u2+nv2} so we now have a four-parameter soln in the variables {p,q,u,v}. For a specific example, let {p,q} = {0,1} and we get the nice identity,

(5y)k + (x-2y)k + (x+2y)k = xk + (3y)k + zk, if x2+24y2 = z2. (And so on for all p,q.) An alternative formulation can be given as, Theorem: If (p-q+r)k + (p+q-r)k + xk = (p+q+r)k + (p-q-r)k + yk, for k = 2,4, then, (2p+q+r)k + (2p-q-r)k + (-2p-q+r)k + (-2p+q-r)k = (p+x)k + (p-x)k + (-p+y)k + (-p-y)k, k = 1,2,3,5 Note that the k = 2,4 system still has x1+x2 = y1+y2. The complete soln is given by, {r,p} = {(x2-y2)/(8q), s/(24q)}, where s2 = -192q4 +96(x2+y2)q2-3(x2-y2)2 so the problem is to find {q,x,y} which makes the quartic a square. However, we already know that this variant of Q1 has a polynomial soln so after a little reverse algebra, we find, {q,s,x,y} = {vw/2, 6vw(2t+vw), uw+v, uw-v}, with w = (3tv+z)/(2u2-2v2), if 4(u2-v2)2 + 3(4u2-v2)t2 = z2

Just like before, the last condition is of the form x2+dy2 = z2 and is easily solved completely as, t = 4mn(u2-v2)/(m2-3(4u2-v2)n2) for free variables {m,n,u,v}. Given the constraint na+b = e+nf on Q1, we have discussed the case n = 1. For general n, na+b = e+nf (eq.1)

a2+b2+c2 = d2+e2+f2 (eq.2)a4+b4+c4 = d4+e4+f4 (eq.3)nb+c = -(nd+f) (eq.4) where again {a,b,c,d,e,f} = {p+q, r+s, t+u, t-u, r-s, p-q} and the fourth eqn is optional. Solve for {u,s} in eq.1,2. Substituted into eq.3 this becomes the quadratic in q, v1q2 = v2t2, where v1 = (p-nr)3-(p-n3r)t2, v2 = (p3-nr3)-(p-nr)t2 Thus the objective is to find v1v2 = y2 which is only a quartic in t. Four small solns are: t = {p+r, p-r, (p-nr)/(n+1), (p-nr)/(n-1)}. Non-trivial soln can be computed from these trivial ones. If the fourth eqn is also to be used, solve for q and eq.3 becomes a quadratic in t with a discriminant D that is only a quartic in p,r. Without loss of generality, let r =1 and this is, D:= (4+n2)2p4 - 4n(4+5n2)p3 + 2(4+3n2+4n4)p2 - 4n(5+4n2)p + (1+4n2)2 Since this has a square leading and constant term, this is easily solved with three small solns: p = {n, 1/n, -1/n}. 10.6. Constraint: a+d = n(c+f) Lastly, we have the system, a+d = n(c+f) (eq.1)a2+b2+c2 = d2+e2+f2 (eq.2)a4+b4+c4 = d4+e4+f4 (eq.3)

Let {a,b,c,d,e,f} = {p+q, r+s, t+u, t-u, r-s, p-q}. Solve for u,p. Substituted into eq.3, this has a quadratic factor in r. The square-free discriminant D of this is a quartic in s. Small solns to finding rational points on the curve D = y2 are: s = {q+t, q-t, (1+n)q+(1-n)t}. Note 1: Any other system consisting of Q1 and one linear relation between the terms (or two, as long as both are new) which would suffice to reduce the problem to solving a quadratic or an elliptic curve?Note 2: There is another system but it involves a quadratic relation between the terms, J. Chernick If pq(p2-q2) = rs(r2-s2), then (p2+q2)k + (r2-s2)k + (2rs)k = (2pq)k + (r2+s2)k + (p2-q2)k, for k = 2,4. This can be re-phrased as, If a2+b2 = f2, and c2 = d2+e2, and ab = de, then ak+bk+ck = dk+ek

+fk, for k = 2,4. A soln was given by Gloden as, {a,b,c,d,e,f} = {p2-q2, p2+r2, 2pq, p2+q2, p2-r2, 2pr}, where p2 = q2+qr+r2, and can be solved by {p,q,r} = {u2+uv+v2, u2-v2, 2uv+v2}. This form is useful for solving a special case of Descartes' Circle Theorem when all variables are squares, 2(w4+x4+y4+z4) = (w2+x2+y2+z2)2 discussed in the next page, Form 18.

(The identities discussed here are in blue.)

1. a4+b4 = c4+d4

2. pq(p2+q2) = rs(r2+s2)3. pq(p2-q2) = rs(r2-s2)4. pq(p2+hq2) = rs(r2+hs2)5. x4+y4 = z4+nt2

6. x4+y4 = z4+nt4

7. u4+nv4 = (p4+nq4)w2

8. u4+nv4 = x4+y4+nz4

9. u4+v4 = x4+y4+nz4

10. ak+bk+ck = dk+ek+fk, k = 2,411. ak+bk+ck = 2dk+ek, k = 2,412. ak+bk+ck = dk+ek+fk, k = 2,3,413. x4+y4+z4 = 2(x2y2+x2z2+y2z2)-t2

14. x4+y4+z4 = t4

15. x4+y4+z4 = ntk

16. v4+x4+y4+z4 = ntk

17. vk+xk+yk+zk = ak+bk+ck+dk, k = 2,418. 2(v4+x4+y4+z4) = (v2+x2+y2+z2)2

19. x1k+x2

k+x3k+x4

k+x5k = y1

k+y2k+y3

k, k = 1,2,3,4

20. x1k+x2

k+x3k+x4

k+x5k = y1

k+y2k+y3

k+y4k+y5

k, k = 1,2,3,4

21. x14+x2

4+…xn4, n > 4

11. Form: ak+bk+ck = 2dk+ek (k = 2,4) The case e = 0 was studied by Ramanujan. To recall, he gave the formulas for k = 2,4, ak + bk + ck = 2(ab+ac+bc)k/2

ak(b-c)k + bk(a-c)k + ck(a-b)k = 2(ab+ac+bc)2k/2

(a2b+b2c+c2a)k + (ab2+bc2+ca2)k + (3abc)k = 2(ab+ac+bc)3k/2

(and so on) with a+b+c = 0. As was already mentioned, this form gives rise to a 5th power multi-grade with just 10 terms, (a+d)k + (-a+d)k + (b+d)k + (-b+d)k + (c+d)k + (-c+d)k = (2d)k + (2d)k + (e+d)k + (-e+d)k, k = 1,2,3,4,5. For the complete radical soln to the system x1

k+x2k+x3

k = 2x4k+ x5

k, k = 2,4, express it in the form, (2a)k + (2b)k + (p+q)k = 2(2c)k + (p-q)k

which gives, a2+b2-2c2+pq = 02a4+2b4-4c4+p3q+pq3 = 0 a useful trick to reduce the degree. Combining the two by eliminating q then gives the beautifully simple condition, (a2+b2-2c2)p4 - 2(a4+b4-2c4)p2 + (a2+b2-2c2)3 = 0 Unfortunately, the discriminant of this is a sextic in a,b (and an octic in c) so cannot be treated as an elliptic curve. However, it does have solns one of which, implied in Ramanujan’s case when last term is p-q = 0, is {a,b,c,p} = {u2-v2, 2uv+v2, u2+uv+v2, u2+2uv}

which makes the discriminant zero. (There are other solns as well.) Piezas Parametric solns to ak+bk+ck = 2dk+ek have been found in two cases: I. When a+b = c.

1. If e = 0, then a2+ab+b2 = d2. (Ramanujan’s)2. If e = 2d, then a2+ab+b2 = 3d2.

II. When a+b = 2d. Let, (p+r)k + (-p+r)k + qk = 2rk + sk, then p2+2q2 = 6r2, and 2p2+q2 = s2. The first eqn can be solved as {p,q,r} = {2u2-4uv-4v2, u2+4uv-2v2, u2+2v2} and substituting this into the second gives the elliptic curve, (3u2-4uv-6v2)2 + 32u2v2 = s2

with one non-trivial soln being {u,v} = {12, 5}, and so on. Note: Any other case?

12. Form: ak+bk+ck = dk+ek+fk (k = 2,3,4) A. Choudhry (ay+b)k + (cy+ad)k + (-ay+b+cd)k = (ay-b)k + (cy-ad)k + (-ay-b-cd)k

This is already identically true when k = 2. But let, {a,b,c,d} = {x2+3, (x+3)3(x-1), -2(x2-3), 2(x2+2x+3)},

where x satisfies the elliptic curve, y2 = (x2-4x+3)2-48x2

and this also becomes valid for k = 3,4. Some integer points on this curve are given by x = {-3, -1, 0, 15} while one non-trivial soln is x = 1/5, and so on. Piezas Rational solns to ak+bk+ck = dk+ek+fk, k = 2,3,4 imply a rational soln t to the quadratic eqn, (a2+b2-e2-f2)t2 - 2(a3+b3-e3-f3)t + (a4+b4-e4-f4) = 0 Proof: The above is equivalent to the identity, (c2-d2)t2 - 2(c3-d3)t + (c4-d4) = 0, where t = c+d. 13. Form: x4+y4+z4 = 2(x2y2+x2z2+y2z2)-t2 This is equivalent to (x+y-z)(x-y+z)(-x+y+z)(x+y+z) = t2 hence is essentially the problem of finding a triangle with rational area using Heron’s formula. One approach is to reduce it to the equation p2+q2+r2 = s2 of which the complete soln by Desboves is known. Equate the first three factors as, {(x+y-z), (x-y+z), (-x+y+z)} = {p2, q2, r2}, and by solving for x,y,z, and substituting into the last factor as x+y+z = s2, one then gets the afore-mentioned condition. Other ways are,

1. Brahmagupta {x,y,z} = {a2/b+b, a2/c+c, a2/b+a2/c-b-c} 2. Kurushima (Euler would later give a similar identity.) {x,y,z} = {ac(b2+d2), bd(a2+c2), (bc+ad)(ab-cd) } 3. Gauss {x,y,z} = {abcd(a2+b2), ab(c+d)(a2c-b2d), ab(a2c2+b2d2) } 4. A. Martin Let {m,n} = {a2+b2, a2-b2}, then, {x,y,z} = {(c2-d2)m, 2cdm+2d2n, (c2+d2)m+2cdn} 14. Form: x4+y4+z4 = t4 This equation is significant since Euler made the rather reasonable conjecture that it would take at least k kth powers to sum up to a kth power which turned out to be erroneous. The first counter-example was found by Lander, Parkin as, 275 + 845 + 1105 + 1335 = 1445

via computer search but there is no systematic method to find other primitive solns for fifth powers. For fourth powers, a counter-example was later found as well as a method to generate further solns, albeit they can quickly become enormous.

(Update, 2/8/10): Piezas If p4+q4+r4 = 1, then (p+q)4 - r4 + 1 = 2(p2+pq+q2)2 also holds. (Update, 2/8/10): Demjanenko, Elkies The complete soln to p4+q4+r4 = 1 (eq.0) can be given in the form, (x+y)4 + (x-y)4 + z4 = 1 where, ay2 = (8mn-3a)x2-2bx-2mn (eq.1)±az2 = 4bx2+8mnx-b (eq.2) and {a,b} = {2m2+n2, 2m2-n2}, for some constants {m,n} hence is a problem of making two quadratics in x as squares and, for appropriate {m,n}, reduces to solving an elliptic curve. One can solve {y,z} in the radicals to see that it holds and it is easily proven that this is the complete soln. Proof: Let {x+y, x-y, z} = {p,q,r}. As eq.1 and eq.2 are also quadratics in {m,n}, solve for m, assume eq.0 as true to rationalize the discriminant, and take the positive sign of the square root to get, m1 = n((p+q)2+r2-1) / (2(p2+pq+q2+p+q))

m2 = n(p2+pq+q2-p-q) / ((p+q)2-r2-1)

(One can use the identity given by the author to rationalize the disciminant of m2.) It will then be seen that m1 = m2 if eq.0 is true, hence one can always find rational {m,n} in terms of {p,q,r}. For example, using the smallest {p,q,r} = {217519/s, 414560/s, 95800/s} and s = 422481, we get {m,n} = {6007, 30080}. (End proof) However, one can also start with {m,n} as it is possible they are relatively small. Elkies’ first soln used only {m,n} = {8,-5}, and (eq.1) becomes, 153y2 = -779x2-206x+80 which has initial soln {x1,y1} = {3/14, 1/42}. This yields a parametrization for x as, x = (51p2-34p-5221)/(238p2+10906) which, when substituted into (eq.2), becomes the problem of finding a value such that the quartic polynomial in p is a square. Clever transformations may then reduce the size of the coefficients. This can be treated as an elliptic curve and since x1 = 3/14 has p = -3779/17, from this initial point one can calculate more, proving that eq.0 has an infinite number of distinct solns. (End update.) Elkies (85v2+484v-313)4 + (68v2-586v+10)4 + (2u)4 = (357v2-204v+363)4

if u2 = 22030+28849v-56158v2+36941v3-31790v4, a soln of which is v = -31/467. This gave the first rational soln to x4+y4+z4 = t4 (eq.0) and is just the case {m,n} = {8,-5} of the

previous identity. This elliptic curve in fact has an infinite number of rational points, thus providing an infinite number of solns to eq.0. 15. Form: x4+y4+z4 = ntk E. Fauquembergue (ab)4 + (ac)4 + (bc)4 = (a4+a2b2+b4)2, if a2+b2 = c2

This beautifully simple soln has a counterpart p4+q4+r4+s4 = t2 (eq.1) which also depends on Pythagorean triples and again found by Fauquembergue, (2a2bc3)4 + (2ab2c3)4 + (2ab(a4+b4))4 + ((a2-b2)c4)4 = (4a2b2(a4+b4)2-c12)2, if a2+b2 = c2

However, since eq.1 has a lot of small solns, there is probably a parametrization where {p,q,r,s} are binary quadratic forms, and t is a quartic form (which can then be specialized to become a square). F. Proth (and others) ak + bk + (a+b)k = 2(a2+ab+b2)k/2, for k =2,4 As was already pointed out, this is quite a ubiquitous algebraic form. Its first even kth powers can be expressed as a sum of three squares, (ab)2 + (ab+b2)2 + (a2+ab)2 = (a2+ab+b2)2, and a square-biquadratic sum by E. Escott, (ab)4 + (ab+b2)4 + ((a2+ab)(a2+ab+2b2))2 = (a2+ab+b2)4,

These imply that the form (a2+ab+b2)2k is expressible as the sum of three squares using the method discussed below. Ramanujan Let a+b+c = 0, then, a4 + b4 + c4 = 2(ab+ac+bc)2

a4(b-c)4 + b4(a-c)4 + c4(a-b)4 = 2(ab+ac+bc)4

(a2b+b2c+c2a)4 + (ab2+bc2+ca2)4 + (3abc)4 = 2(ab+ac+bc)6

In his Notebooks, Ramanujan gave these as well as one for k=8 and wrote “…and so on”. The rest can be found by noting that the first is equivalent to Proth’s so what one has to do is to find expressions a,b such that a2+ab+b2 = (p2+pq+q2)k which can be done by factoring over a complex cube root of unity ω, (a-bω) (a-bω2) = (p-qω)k (p-qω2)k and by using the reliable method of equating factors, one can then easily solve for {a,b}. It should be pointed out Ramanujan may have used another method to derive his more elegant expressions which are conditionally dependent on a+b+c = 0. (Incidentally, this form has a generalization for third and fifth powers.) S. Realis x4+y4+z4= 3t2 {x,y,z} = {5a4+4a3+9a2+10a+5, 5a4+10a3+9a2+4a+5, 5a4+16a3+27a2+16a+5}

Any other solns to x4+y4+z4= ntk for n>2? E. Fauquembergue (2p2-2q2)4 + (p2-4q2)4 + (3pq)4 = (p2+2q2)4 + (4p4-11p2q2+16q4)2

16. Form: v4+x4+y4+z4 = ntk There is yet no polynomial identity for a4+b4+c4+d4 = t4 so this form is the closest so far. E. Fauquembergue (2a2bc3)4 + (2ab2c3)4 + (2ab(a4+b4))4 + ((a2-b2)c4)4 = (4a2b2(a4+b4)2-c12)2, if a2+b2 = c2

Any other polynomial soln to a4+b4+c4+d4 = ntk? (Note: Yes, there is for n = 18, 63, etc. Add the ones by Haldemann.) (Update, 2/9/10): Lee Jacobi, Daniel Madden The complete soln to a4+b4 = c4+d4 and a4+b4+c4 = d4 can be reduced to solving an elliptic curve. It turns out that for a special case of four 4th powers equal to a 4th power namely, a4+b4+c4+d4 = (a+b+c+d)4 (eq.0) with variables as ±, then one can do so as well, a small example of which is,

9554+17704+(-2634)4+54004 = (955+1770-2634+5400)4 = 54914

Jacobi's and Madden’s clever soln depended on the identity, a4+b4+(a+b)4 = 2(a2+ab+b2)2 an identity also useful for solving a4+b4+c4+d4+e4 = f4 as quadratic forms. (See Form 21 below.) Their method starts by adding (a+b)4 + (c+d)4 to both sides of eq.0, a4+b4+(a+b)4+c4+d4+(c+d)4 = (a+b)4+(c+d)4+(a+b+c+d)4

and, using the identity, it is seen that eq.0 is equivalent to the special Pythagorean triple, (a2+ab+b2)2 + (c2+cd+d2)2 = ((a+b)2 + (a+b)(c+d) + (c+d)2)2 Sparing the reader the intermediate algebraic manipulations, using the transformation, {a,b,c,d} = {p+r, p-r, q+s, q-s}, the eqn, (p+r)4+(p-r)4+(q+s)4+(q-s)4 = (2p+2q)4 can be solved if {p,q,r,s} satisfy two quadratics to be made squares, (m2-7)p2+4(m2-1)pq+4(m2-1)q2 = (m2+1)r2 (eq.1)8mp2+8mpq-(3m2-8m+3)q2 = (m2+1)s2 (eq.2) for some constant m. This is easily proven by solving {r,s} in radicals and substituting it into the eqn to see that it holds. These two quadratic conditions define an elliptic curve. One can then try to find a suitable m and find rational {p,q,r,s}. Or, given a known {p,q,r,s}, a value m can be derived as,

m = (3q2+s2) / ((p+2q)2-r2) Example: Given {a,b,c,d} = {5400, 1770, -2634, 955}, then {p,q,r,s} = {3585, -1679/2, 1815, -3589/2}, and m = 961/61. This initial soln set can be used to find an infinite more. Since eq.1 and eq.2 are homogeneous, we can assume q = 1 without loss of generality. Starting with a soln to eq.1 as p1 = -2(3585)/1679, a complete parameterization is, p = (30/1679) (4291863+112196282u+110805419u2)/(448737-463621u2) which makes eq.1 a square for any u. But substituting this to eq.2, and still with q = 1, produces a rational 4th deg polynomial in u that still has to be made a square. To find an initial rational point u, equating p-p1 = 0 yields the factor 55770003+56098141u = 0, hence this is one suitable value. Treating the quartic polynomial as an elliptic curve, an infinite more rational u can be computed, thus proving that the eqn, a4+b4+c4+d4 = (a+b+c+d)4

has an infinite number of distinct, rational solns. (End update.) 17. Form: vk+xk+yk+zk = ak+bk+ck+dk, k = 2,4 Piezas Theorem: If ak+bk+ck+dk = ek+fk+gk+hk, for k = 2,4, and abcd = efgh. Define n as a4+b4+c4+d4 = n(a2+b2+c2+d2)2 then, 32(a6+b6+c6+d6-e6-f6-g6-h6)(a10+b10+c10+d10-e10-f10-g10-h10) = 15(n+1)(a8+b8+c8+d8-e8-f8-g8-h8)2

Note if d = h = 0, the condition abcd = efgh disappears and this reduces to the theorem given earlier of which Ramanujan’s 6-10-8 Identity is a special case. Proof: Given, a2+b2+c2+d2 = e2+f2+g2+h2 (eq.1)a4+b4+c4+d4 = e4+f4+g4+h4 (eq.2) Eliminate h between the two to get a quadratic in {a,b,c,d} and solve for any, say d. Solve for h in eq.1 and we get the complete radical soln of eq.1 and eq.2. Substitute these two values into the 6-10-8 identity and it has a factor, call this P, which is the same polynomial that results when the values are substituted into abcd-efgh = 0. Thus if the latter is satisfied so is the former. (End proof.) A solution will be given below. Theorem. If ak+bk+ck+dk = ek+fk+gk+hk, for k = 2,4,6, where abcd = efgh, and a+b ≠ ±(c+d); e+f ≠ ±(g+h), (call the entire system V1) then, (a+b+c-d)k + (a+b-c+d)k + (a-b+c+d)k + (-a+b+c+d)k + (2e)k + (2f)k + (2g)k + (2h)k =(e+f+g-h)k + (e+f-g+h)k + (e-f+g+h)k + (-e+f+g+h)k + (2a)k + (2b)k + (2c)k + (2d)k

for k = 1,2,4,6,8,10. Again, for d = h = 0, the condition abcd = efgh disappears so reduces to Birck’s Theorem (given later). This is then an extension to tenth powers and the proof will be discussed in that section. Even for non-zero variables it is possible one or two of the addends will vanish, as we shall see later. This author hasn’t been able to find a parametric soln to V1 but was able to do so for the simpler case when valid only for k = 2,4. Given,

k + x2k + x3

k + x4k = y1

k + y2k + y3

k + y4k, k = 2,4

assume it to have the form, call it F1, ak(p+q)k + bk(r-s)k + ck(r+s)k + dk(p-q)k = ak(p-q)k + bk(r+s)k + ck(r-s)k + dk(p+q)k

which has x1x4 = y1y4 and x2x3 = y2y3. This sufficient but not necessary constraint is completely parameterized by F1. To see this, use the Mathematica command, Solve[{a(p+q), d(p-q), a(p-q)} = {x1, x4, y1}, {a,p,d}] to express {a,d,p} in terms of {x1, x4, y1}. Substituting these into d(p+q) = y4, we get, x1x4/y1 = y4 which, if true, yields the appropriate {a,d,p}, with q a free variable, and similarly for {b,c,r}. For example, given, 16k + 48k + 58k + 99k = 22k + 29k + 96k + 72k, k = 2,4 which is the smallest in distinct integers with NO sign changes such that x1+x2+x3+x4 = y1+y2+y3+y4 = 0. This has 16(99) = 22(72); 48(58) = 29(96), then, {a,d,p,q} = {1, 9/2, 19, -3}{b,c,r,s} = {1/2, 1, 77, -19}

with {q,s} arbitrary. F1 takes care of the necessary requirement that x1x2x3x4 = y1y2y3y4. To simplify matters, assume further that x1+x2 = y1+y2 (not obeyed by the example above). After some algebra, we get the soln, {p,q,r,s} = {ab2-ac2, be, a2b-bd2, ae}, where a2(3b2-c2) + e2 = (b2+c2)d2. This conditional eqn is easily solved parametrically. First, let the LHS be a2n+e2 = y2 and find {a,e}, then the RHS as b2+c2 = z2 and find {b,c}, to get the form y2 = z2d2 which then gives d. For example, let {b,c} = {3,4} entails solving the conditional equation, 11a2+e2 = (5d)2

which is easily done. (Note: If the particular form F1 is extended to k = 6, it yields only trivial solns so another form has to be used.) 18. Form: 2(v4+x4+y4+z4) = (v2+x2+y2+z2)2 The equation, 2(a4+b4+c4) = (a2+b2+c2)2 has the simple soln a+b = c, which is central to Ramanujan's 6-10-8 Identity. Turns out when there are four terms, 2(v4+x4+y4+z4) = (v2+x2+y2+z2)2 (eq.1) it has a parametric soln as well. Note that this is a special case of

Descartes' Circle Theorem, 2(x1

2+x22+x3

2+x42) = (x1+x2+x3+x4)2

where the xi are squares. Euler’s clever soln uses the fact that eq.1 is equivalent to either, (2vx)2 + (2yz)2 = (v2+x2-y2-z2)2

(2vy)2 + (2xz)2 = (v2-x2+y2-z2)2

(2vz)2 + (2xy)2 = (v2-x2-y2+z2)2 so is a special case of three simultaneous Pythagorean triples. More generally, for any d, we are to solve, 2(d2v4+x4+y4+z4) = (dv2+x2+y2+z2)2 (eq.2) and Euler's soln can be given as {v,x,y,z} = {2pqrs(a2-t2), t(a2-t2), t(abrs-cpqt), a(abrs-cpqt)}, where {a,b,c} = {pq(r2-ds2), p2+dq2, r2+ds2}, and {p,q,r,s,t} satisfies the conditional eqn, pqrs(p2-dq2)(r2-ds2) = t2 For d = 1, one soln is {p,q,r,s} = {2m2-n2, m2+n2, m2+n2, m2-2n2}. Note: Strictly speaking, Euler’s analysis was focused on the case d = ±1 and this author inserted the d to generalize it. Any soln for other d? (Update, 9/13/09): I noticed we can have a better form if we assume t = rs(r2-ds2) and the conditional eqn becomes the symmetric, pq(p2-dq2) = rs(r2-ds2) This has appeared already in Section 013 and has many solns and is connected to the two distinct eqns,

a4+b4 = c4+d4

ak+bk+ck = dk+ek+fk, k = 2,4 Thus, we can also solve 2(n2v4+x4+y4+z4) = (nv2+x2+y2+z2)2 as: I. When n = -1: {v,x,y,z} = {2(ab-cd)(ab+cd), (a2+b2+c2+d2)(a2-b2+c2-d2), 2(ac-bd)(a2+c2), 2(ac-bd)(b2+d2)}, where a4+b4 = c4+d4. II. When n = 1: Let ak+bk+ck = dk+ek+fk, for k = 2,4, where a2+b2 = f2, d2+e2 = c2, and ab = de, then, {v,x,y,z} = {ae-bd, (a+d)(a-d), a(c+f), d(c+f)} A soln given by Gloden is {a,b,c,d,e,f} = {p2-q2, p2+r2, 2pq, p2+q2, p2-r2, 2pr}, where p2 = q2+qr+r2. This can be solved by {p,q,r} = {u2+uv+v2, u2-v2, 2uv+v2}. (End update.) 19. Form: x1

k+x2k+x3

k+x4k+x5

k = y1k+y2

k+y3k, k =

1,2,3,4 The system of eqns, x1

k+x1k+…+ xn

k = y1k+y1

k+…+ ynk, k = 1,2,3,4

has non-trivial solns only for n > 4. However, it is possible two terms on one side are equal to zero. Piezas

Given ak+bk+ck+dk+ek = fk+gk+hk, for the special case when h = a+b+c, then, {d,e,f,g} = {(p+u)/(2q), (p-u)/(2q), (p+v)/(2q), (p-v)/(2q)} where p = (a+b)(a+c)(b+c), q = (ab+ac+bc), and {a,b,c} satisfies the two quadratic conditions,p(p+4abc) = u2

p(p+4abc) - 4q3 = v2

An example of a particular soln is {a,b,c} = {-21, -4, 12} which yields the eqn, (-63)k + (-12)k + 36k + (-35)k + 10k = (-62)k + 37k + (-39)k

Q: Any general polynomial identity for this? 20. Form: x1

k+x2k+x3

k+x4k+x5

k = y1k+y2

k+y3k+y4

+y5k, k = 1,2,3,4

The complete soln to this system, call this E4, is unknown. However, for the case when x2+x3 = y2+y3 it can be shown that this involves making a quartic polynomial a square. Use the form, (a+bj)k+(c+dj)k+(e+fj)k+(g+hj)k+(i+j)k = (a-bj)k+(c-dj)k+(e-fj)k+(g-hj)k+(i-j)k Let f = -d to satisfy the constraint and i = -(ab+cd+ef+gh), h = -(1+b+d+f) for k = 1,2. Expand this for k = 3,4, to get the auxiliary resultants,

(Poly10)j2+(Poly20) = 0 (eq.1)(Poly11)j4+(Poly21)j2+(Poly31) = 0 (eq.2) where the Polyi are in {a,b,c,d,e,g}. Eliminate j between eq.1 and 2 (set j = √v for convenience) and one gets a final resultant, call this R, which is only a quadratic in a,c,e,g. Solving for g, this has a discriminant D which is only a quadratic in e so to solve D = y2, one uses a quadratic form e = Q(x) for some variable x. After a rational g is found, to find j, solve, -(Poly10)(Poly20) = z2

which is a quadratic in e. Substituting the quadratic form e = Q(x) into this, it becomes, Poly(x) = z2 (eq.3) which is a quartic in x, hence any soln to E4 with the constraint x2+x3 = y2+y3 must satisfy eq.3, a quartic polynomial to be made a square. Note: Since the final resultant R involves six variables {a,b,c,d,e,g}, the explicit soln is a mess. There might be a way to clean this up though using a more suitable form than the one given without losing its generality. But using the simpler form, (a+b)k+(c+d)k+(e+f)k+(g+h)k+(i+j)k = (a-b)k+(c-d)k+(e-f)k+(g-h)k+(i-j)k for example, is messier.

21. Form: x14+x2

4+…xn4, n > 4

E. Fauquembergue (4x4-y4)4 + 2(4x3y)4 + 2(2xy3)4 = (4x4+y4)4 C.Haldeman (2x2+12xy-6y2)4 + (2x2-12xy-6y2)4 + (4x2-12y2)4 + (3x2+9y2)4 + (4x2+12y2)4 = 54(x2+3y2)4 This gives as a first instance 24+24+44+34+44 = 54. Similar identities were found by A. Martin and Ramanujan. A generalization has been found by this author. Piezas (ax2+2v1xy-3ay2)k + (bx2-2v2xy-3by2)k + (cx2-2v3xy-3cy2)k = (ak

+bk+ck)(x2+3y2)k where {v1, v2, v3, c} = {a+2b, 2a+b, a-b, a+b}, for k = 2,4 Thus, it suffices to decompose a sum which starts as a4+b4+(a+b)4+… into a sum and difference of any number of biquadrates and apply it to the identity. For example, using the eqn 504+504+1004+44+154 = 1034, this yields, (50x2+300xy-150y2)4 + (50x2-300xy-150y2)4 + (100x2-300y2)4 + (4x2+12y2)4 + (15x2+45y2)4 = 1034(x2+3y2)4 R. Norrie

(x+y)4 + (-x+y)4 + (2y)4 + (u2-v2)4 + (2uv)4 = (u2+v2)4, if 2uv(u2-v2) = x2+3y2

Note that {u2-v2, 2uv} are the legs of a Pythagorean triple. Thus, if the product of the legs can be expressed as x2+3y2, it then gives a soln to the above equation, the smallest of which is the particular equation given earlier, {2,2,4,3,4; 5}. For the sum of five or six biquadrates, Sophie Germain Sophie Germain’s Identity is given by x4+4y4 = ((x+y)2+y2) ((x-y)2+y2) = (x2+2xy+2y2)(x2-2xy+2y2) The quartic form can be generalized as, x4+(-a2+2b)x2y2+b2y4 = (x2+axy+by2)(x2-axy+by2) where Germain's was the case {a,b} = {2,2}. Another interesting case is {a,b} = {2,-1}, x4-6x2y2+y4 = (x2+2xy-y2)(x2-2xy-y2) = (x2-y2)2 - (2xy)2 which sometimes appear when dealing with Pythagorean triples. (Update, 2/3/10): Choudhry Given the eqn, a4+4b4 = c4+4d4 (eq.1) use the transformation, {a,b,c,d} = {(x2+2x-2)z+xy, 2xz+y, (x2+2x+2)z+xy, y}

so that, substituted into eq.1, it has a linear factor in y. This first soln can be used to compute an infinite sequence of polynomial solns, as eq.1 is an elliptic curve in disguise. The transformations, {a,b,c,d} = {p+q, r-s, p-q, r+s}, then {p,q,r,s} = {u, mv, nu, v} reduce eq.1 to the form, (m-4n3)u2+(m3-4n)v2 = 0 so one is to solve, -(m-4n3)(m3-4n) = z2

One soln is, {p,q,r,s} = {8x+2x3, -4+2x2, 4+4x2, -2x+x3} from which others can then be computed. Source: The Diophantine Equation A4+4B4 = C4+4D4, Indian Journal of Pure and Applied Mathematics, Vol. 29, 1998. (End update.) R.Carmichael (a4-2b4)4 + (2a3b)4 + 4(2ab3)4 = (a4+2b4)4 Note that the form (a4+2b4)4 can also be expressed as a sum of both squares and biquadrates, E. Fauquembergue (a4-2b4)4 + (2a3b)4 + (8a2b6)2 = (a4+2b4)4 (2a2b2)4 + (2a3b)4 + (a8-4a4b4-4b8)2 = (a4+2b4)4

II. Quartic Polynomials as kth Powers

1. ax4+by4 = cz2

2. ax4+bx2y2+cy4 = dz2

3. au4+bu2v2+cv4 = ax4+bx2y2+cy4

4. ax4+bx3y+cx2y2+dxy3+ey4 = z2

1. Form: ax4+by4 = cz2 Theorem: "If the form ax4-y4 = z2 has a solution, then so does the form x4-ay4 = z2." Proof: 1. If ax4-y4 = z2 has a soln, then so does x4+4ay4 = z2, since, z4 + 4a(xy)4 = (ax4+y4)2, if ax4-y4 = z2. 2. And if in turn x4+4ay4 = z2 has a soln, then so does x4-ay4 = z2, since, z4 - a(2xy)4 = (x4-4ay4)2, if x4+4ay4 = z2. Combining the two, (ax4+y4)4 - a(2xyz)4 = (4ax4y4-z4)2, if ax4-y4 = z2. For example, the eqn 2x4-y4 = z2 has smallest soln {x,y,z} = {13, 1, 239}. The theorem implies that p4-2q4 = r2 has solns as well, yielding {p,q} = {57123, 6214}, which is not necessarily the smallest. Furthermore, given a non-trivial integral soln to ax4+bx2y2+cy4 = dz2, one can generally find an infinite more as proven by an identity of

Desboves. For the special case when a = c = 1, we have Lagrange x4+by4 = z2

{x,y,z} = {p4-bq4, 2pqr, 4bp4q4+r4}, if p4+bq4 = r2

Desboves More generally, if ap4+bq4 = cr2, then, ax4+by4 = cz2

{x,y,z} = {p(4m2-3(m+n)2), q(4n2-3(m+n)2), r(4(m+n)4-3(m-n)4)}, where {m,n} = {ap4, bq4} re-arranged by this author to make it more aesthetic. (Note how the algebraic form 4u2-3v2 appears). This can be generalized even more to cover the form ax4+bx2y2+cy4 = dz2. The identity involves polynomials of high degree, 9th deg in the {p,q}, but special cases like a+b = c need only cubics, Desboves If ap4+bq4 = (a+b)r2, then, ax4+by4 = (a+b)z2

where {x,y,z} = {-d2p2+4bcq2, (2c2-d2)pq±2cdr, 4bdpq(4acp2+d2q2)±(2c2-d2)(d2p2+4bcq2)r}, where {c,d} = {a+b, a-b}. with z as a cubic. And since this involves the ± sign, for this particular

form there are solns that share the same x variable. For ex, the eqn 2x4-y4 = z2 has small solns {x,y} = {13, 1} and {1525, 1343}. Using Desboves’ identity, the second yields a pair of {x,y} as, {28145221, 2165017}{28145221, 30838067} Euler, Lagrange Let {u,v} be the legs of a triangle. The problem of making the sum of the legs and its squares as a square and a fourth power, respectively, or u+v = y2,u2+v2 = x4

can be reduced to a single condition. These simultaneous equations are true for, {u,v} = {(y2+z)/2, (y2-z)/2}, if 2x4-y4 = z2

A soln given by Euler is, {x,y,z} = {p3+2pq2-qr, p3-4pq2+qr, p6+p4q2-6p3qr+24p2q4-8q6}, if p4+8q4 = r2

Piezas 2x4-y4 = z2

{x,y} = {12(p+q)3-(3p+2q)v, 12(p+q)3-(4p+3q)v}, if 2p4-q4 = r2

where v = 2p2+8pq+5q2+r, and z is a sixth deg polynomial. Note that with the sign of p held constant, the other two variables q,r can come in four combinations: {+,+}, {-,+}, {+,-}, {-,-} generally giving four

distinct values {x,y}. S.Realis, A.Gerardin In general, one soln to ax4+by4 = cz2 can lead to subsequent ones. Let n,x be unknowns, a(u+px)4 + b(v+qx)4 = c(w-nx+rx2)2 Collecting powers of x, (ap4+bq4-cr2)x4 + (Poly1)x3 + (Poly2)x2 + (Poly3)x + (au4+bv4-cw2) = 0 Assume that (ap4+bq4-cr2) = (au4+bv4-cw2) = 0. Since n is only linear in Poly1 (or Poly3), equate Poly1 = 0, then solve for n. The whole equation then reduces to the form, (Poly2)x2 + (Poly3)x = 0 which, after factoring, is simply a linear eqn in x so we have both our unknowns {n,x} in terms of {p,q,r} and {u,v,w}. These generally lead to addends that are cubic polynomials in terms of p,q,r as in the explicit example given by this author. 2. Form: ax4+bx2y2+cy4 = dz2 Theorem 1: Given an initial non-trivial solution to ap4+bp2q2+cq4 = dr2, one can generally find an infinite number of solutions. (Desboves) (See also no. 4 below) Proof:

ax4+bx2y2+cy4+dz2 = (ap4+bp2q2+cq4+dr2)(z/r)2 {x,y,z} = {p(u2-4cq4w), q(u2-4ap4w), r((p4q4v-w2)2-4p4q4u2v)}, where {u,v,w} = {ap4-cq4, b2-4ac, ap4+bp2q2+cq4} Alternatively, to show its affinity to the diagonal form (when b = 0), {x,y,z} = {p(4m2-3(m+n)2-nt), q(4n2-3(m+n)2-mt), r(4(m+n)4-3(m-n)4+mnt(4m+4n+t)+t(m+n)3)}, where {m,t,n} = {ap4, 4bp2q2, cq4}. Note how when b = t = 0, the identity reduces to the one given in the previous section. Less general solns are also known, 1. J. Lagrange x4+cy4 = z2

{x,y,z} = {p4-cq4, 2pqr, 4cp4q4+r4}, if p4+cq4 = r2

This was already given, but is a special case of the next one, 2. V. Lebesgue x4+bx2y2+cy4 = z2

{x,y,z} = {p4-cq4, 2pqr, (b2-4c)p4q4-r4}, if p4+bp2q2+cq4 = r2

For b = 0, this reduces to the previous. This, in turn, in generalized by, 3. A. Desboves x4+bdx2y2+acd2y4 = z2

{x,y,z} = {ap4-cq4, 2pqr, (b2-4ac)p4q4-d2r4}, if ap4+bp2q2+cq4 = dr2

which for a=d=1 again reduces to the previous. 4. A. Desboves ax4+bx2y2+cy4 = dz2

{x,y,z} = {p(u2-4cdq4r2), q(u2-4adp4r2), r(4p4q4u2v-(p4q4v-d2r4)2)} where {u,v} = {ap4-cq4, b2-4ac}, if ap4+bp2q2+cq4 = dr2

This is basically the identity discussed by Theorem 1 in its original form but takes too long to factor symbolically on a computer. Can the degree of the polynomials be reduced yet still apply to general {a,b,c,d}? Euler Let b = nx2+2x, x4+bx2y2+y4 = (x3+y2)2, if x2-ny2 = 1 One can then use solns of Pell equations to solve the above form for certain b. More generally, let b = nx2+2v, x4+bx2y2+y4 = (vx2+y2)2, if v2-ny2 = 1 3. Form: au4+bu2v2+cv4 = ax4+bx2y2+cy4 Piezas

(√p+√q)k + (√p-√q)k = (√r+√s)k + (√r-√s)k (eq.1) Poly solns to eq.1 have been found this author for k=5,6,8, with k=7 found by D.Rusin. For k=8, this becomes, p4+28p3q+70p2q2+28pq3+q4 = r4+28r3s+70r2s2+28rs3+s4

with one soln, {p,q,r,s} = {(n-1)(n2-n-1), (n+1)(n2+n-1), n3-n-1, n3-n+1} Since the above is a symmetric polynomial, a second transformation {p,q,r,s} = {u/2+v, u/2-v, x/2+y, x/2-y} gives it the simpler form, u4-4u2v2+2v4 = x4-4x2y2+2y4

Any other soln for this as well as for the general case? Also, any non-trivial soln to eq.1 for k>8? (Update, 11/10/09): In "Parametric Solutions of the Quartic Diophantine Equation f(x,y) = f(u,v)", Choudhry discusses how, if an initial soln is known, then one can find another soln in general. He also provides an explicit identity: Choudhry x1

4+2nx12x2

2+x24 = y1

4+2ny12y2

{x1, x2} = {1+n2t+st2-rt3+qt4+3t5+pt6+nt7, -n+pt-3t2+qt3+rt4+st5-

n2t6+t7}{y1, y2} = {1-n2t+st2+rt3+qt4-3t5+pt6-nt7, n+pt+3t2+qt3-

rt4+st5+n2t6+t7}

where {p,q,r,s} = {1-n+n2, -2+4n+n2, -4n+n3, 1+n2+n3}. Note how, for {x1,y1} the signs are different with odd powers t, while for {x2,y2}, they are different for even powers t. For the case n = 0, this reduces to the formula found by Gerardin, while n = 1 is trivial. (End update.) 4. Form: ax4+bx3y+cx2y2+dxy3+ey4 = z2 For the 4th power univariate case to be made a square, Fermat's method still applies. If it has an initial soln, we can use the form with the square constant term, ax4+bx3+cx2+dx+e2 = z2

and assume it as equal to, ax4+bx3+cx2+dx+e2 = (px2+qx+e)2

Using the same approach, the unknowns {p,q} are enough to knock out enough terms such that one can solve for x, given by, x = (be-dp)/(e(p2-a)), where p = (4ce2-d2)/(2e)3

Unfortunately, for the quintic function f(x) = z2, not enough terms can be removed so the method stops here. Interestingly, this is similar to the situation of solving f(x) = 0 where x is in radicals since there are general formulas only for degree 2,3,4 but none for higher. Has it been explicitly proven there is no general formula to find rational and non-zero x that solves f(x) = z2 where f(x) is a quintic or higher degree with rational coefficients and a square constant term?

PART 8. Equals Sums of Like Powers and the PTE Problem I. Some General Conjectures and ProblemsII. Some TheoremsIII. Fifth Powers

5.1 Four terms5.2 Six terms5.3 Seven terms5.4 Eight terms5.5 Ten terms5.6 Twelve terms

Most of the identities I found for kth powers involve multigrade "Equal Sums of Like Powers (ESLP)" (multigrade being the case valid for several exponents). For numerical examples that solve the "Prouhet-Tarry-Escott Problem (PTE)" in certain cases, see Chen Shuwen's site. For the definitive site on the present status of ESLP, check out Jean-Charles Meyrignac's "Computing Minimal Equal Sums of Like Powers". I. Some General Conjectures and Problems 1. Euler’s Extended Conjecture (EEC) “The equation x1

k + x2k +… + xm

k = y1k + y2

k +…+ ynk, (designated

a [k,m,n]), has no solution in the integers for k > m+n, other than the trivial case when all xi = yi.” This was formalized by R. Ekl’s 1998 “New Results in Equal Sums of Like Powers” which, as the title suggests, improved the known results on this problem. Note that for [k,m,n] = [3,2,1], this partially reduces

to Fermat's Last Theorem and is true as proven by A. Wiles, though this is the only proven result. EEC is silent when k = m+n and whether this case has a solution or not is tackled by the stronger version, 2. Lander-Parkin-Selfridge Conjecture (LPS) “The equation x1

k + x2k +… + xm

k = y1k + y2

k +…+ ynk has a non-

trivial solution in the integers for k > 3 if and only if m+n ≥ k.” formalized by J. Meyrignac when he started EulerNet in 1999. Thus this specifically demands that, for example, (4,1,3), (5,1,4), (6,1,5) have a soln. While it is indeed true for the first two, for the third, no soln is known even for (6,1,6) so only time will tell if the conjecture will hold. 3. Prouhet-Tarry-Escott Problem (PTE) “Does the system of equations x1

k + x2k +… + xm

k = y1k + y2

k +…+

ymk, for k = 1,2,…, n, have a non-trivial solution in the integers for

any {m,n} where m > n?” II. Some Theorems (See also the section on Equal Sums of Like Powers in Assorted Identities.) For simplicity, we will designate the multi-grade system a1

k + a2k + …

+ amk = b1

k + b2k + …+ bm

k, valid for k =1,2,...,n, (call this system M) as, [a1, a2,…, am] = [b1, b2,…, bm], k = 1,2,...n where m is its size and n is the degree.

Theorem 1. Bastien’s Theorem (L. Bastien, Sphinx-Oedipe, vol. 8, 1913, pp. 171-172.) "If [a1, a2,…, am] = [b1, b2,…, bm] for k = 1,2,…n has a non-trivial soln, then m > n." In others words, the size of the system is always greater than the degree. To solve the system M up to exponent n, one has to use at least n+1 terms on one side of the equation (since the next theorem allows the other side to have just n terms). The case m = n+1 is called an ideal solution and are now known for all degrees n ≤ 11 other than n = 10. The next theorems allow transformations to be done to these equations. Theorem 2. Frolov’s Theorem (M. Frolov, Bulletin de la Societe Mathematique de France, vol. 17, 1888-9, pp 69-83; vol. 20, 1892, pp.69-84.) "If [a1, a2,…, am] = [b1, b2,…, bm], for k = 1,2,...n, then for any value x, it is true that [x+a1, x+a2,…, x+am] = [x+b1, x+b2,…, x+bm], for k = 1,2,...n." For example, using the solution [1, 5, 6] = [2, 3, 7], k = 1,2, then for any x, (x+1)k + (x+5)k + (x+6)k = (x+2)k + (x+3)k + (x+7)k, for k = 1,2 Corollary 1: For any M, one can always set the first term a1 = 0 using the special case x = -a1. For the example above, with x = -1 this becomes [0, 4, 5] = [1, 2, 6]. Corollary 2: One can also set the terms ai and bi such that their sum, Σai = Σbi = 0. This is called the standard form by Escott and the

importance of this form will be discussed in Theorem 6. Ex, [1, 5, 8, 12] = [2, 3, 10, 11], k = 1,2,3 since (x+1)+(x+5)+(x+8)+(x+12) = 0 gives x = -26/4, then, [-11, -3, 3, 11] = [-9, -7, 7, 9] after removing common factors. It should be pointed out that not all are symmetric solutions. For ex, [1, 75, 87, 205] = [15, 37, 115, 201], k = 1,2,3, yields, [-91, -17, -5, 113] = [-77, -55, 23, 109], k = 1,2,3,5 Naturally enough, these are called non-symmetric solutions and why this example suddenly became applicable for k = 5 as well will be explained in Theorem 6. The next three transformations generally double the number of terms but also extend the range of exponents k. Theorem 3. Tarry-Escott Theorem (Escott, Quarterly Journal of Mathematics, 1910, pp. 141-167; Tarry, L’Intermediaire des Mathematiciens, vol. 19, 1912, pp. 219-221.) "If [a1, a2,…, am] = [b1, b2,…, bm], for k = 1,2,…n, then for any x, it is [a1,…, am, x+b1,…, x+bm] = [b1,…, bm, x+a1,…, x+am], for k = 1,2,…n+1." Ex. [1, 5, 6] = [2, 3, 7], k = 1,2 gives [1, 5, 6, x+2, x+3, x+7] = [2, 3, 7, x+1, x+5, x+6], k = 1,2,3 As expected, the number of terms doubles but with carefully chosen terms, this doubling can be prevented since some terms may appear on both sides of the equation and just cancel out. For example, this one by

Tarry, [1, 5, 10, 16, 27, 28, 38, 39] = [2, 3, 13, 14, 25, 31, 36, 40], k = 1,2,…6 using the theorem starting with x = 11 found the first ideal soln with degree n=7, [1, 5, 10, 24, 28, 42, 47, 51] = [2, 3, 12, 21, 31, 40, 49, 50], k = 1,2,…7 since eight terms on either side cancel out. This author has found a generalization of this particular example to be discussed on the section on Sixth Powers. Theorem 4. (Starts with Odd Powers) "If [a1, a2,…, am] = [b1, b2,…, bm], for k = 1,3,…2n-1, then for any x, it is true that [x+a1,…, x+am, x-b1,…,x-bm] = [x-a1,…,x-am, x+b1,…, x+bm], for k = 1,2,3…2n." Or alternatively, since we are dealing with odd powers and can move all terms to the LHS, "If [a1, a2,…, ap] = 0, for k = 1,3,…2n-1, then [x+a1, x+a2…, x+ap] = [x-a1, x-a2,…,x-ap], for k = 1,2,3…2n." Ex. [0, 7, 8, -1, -5, -9] = 0, k = 1,3 gives, [x+0, x+7, x+8, x-1, x-5, x-9] = [x-0, x-7, x-8, x+1, x+5, x+9], k = 1,2,3,4 Note that appropriate pairs of terms taken from opposite sides of the equation have the common sum 2x. If the original system has m = n+1

and is the special case a1 = 0, this also yields an ideal soln of degree 2n since the term x ± a1 is just the same (as one can see from the example above). Cases with a1 = 0 are, [0, 3] = [1, 2], k = 1[0, 7, 8] = [1, 5, 9], k = 1,3[0, 24, 33, 51] = [7, 13, 38, 50], k = 1,3,5 [0, 34, 58, 82, 98] = [13, 16, 69, 75, 99], k = 1,3,5,7 (A. Letac, 1942) with the last found pre-computer searching. (Along with a second soln, how did Letac find these only known examples back in 1942?) It is then enough to find a k = 1,3,5,7,9 with a1= 0 to give an ideal soln for the missing degreee n =10 but, almost 70 years after Letac, none are known so far. Theorem 5. (Starts with Even Powers) "If [a1, a2,…, am] = [b1, b2,…, bm], for k = 2,4,…2n, then for any x, it is [x+a1,…, x+am, x-a1,…, x-am] = [x+b1,…,x+bm, x-b1,…, x-bm], for k = 1,2,3…2n+1". Ex. [1, 9, 10] = [5, 6, 11], k = 2,4 gives, [x+1, x+9, x+10, x-1, x-9, x-10] = [x+5, x+6, x+11, x-5, x-6, x-11], k = 1,2,3,4,5 If m = n+1, this automatically gives an ideal soln of degree 2n+1. Note that pairs of terms now from the same side of the equation have the common sum 2x. If x = 0, then odd power sums of terms on either side is Σai = Σbi = 0, a consequence implied by Frolov’s theorem. Some examples with m = n+1 are, [1, 7] = [5, 5], k = 2 [1, 9, 10] = [5, 6, 11], k = 2,4

[2, 16, 21, 25] = [5, 14, 23, 24], k = 2,4,6 (G. Tarry, 1913)[71, 131, 180, 307, 308] = [99, 100, 188, 301, 313], k = 2,4,6,8 (Borwein, Lisonek, Percival)[22, 61, 86, 127, 140, 151] = [35, 47, 94, 121, 146, 148], k = 2,4,6,8,10 (Kuosa, Meyrignac, and Shuwen) For the case k = 2,4,6,8, Letac much earlier gave a method using an elliptic curve though it results in relatively large numbers. For k = 2,4,6,8,10, after Choudhry and Wroblewski recently found also an elliptic curve (in 2008), it is now known that this system has an infinite number of solns. In general, Theorems 4 and 5 yield what are called symmetric solutions since sums of appropriate pairs of terms have the common sum 2t. While the above theorems apply to solns with any number of terms, given the special case that is ideal and non-symmetric, there is another that maintains the number of terms but still extends the range k, Theorem 6. Gloden’s Theorem (A. Gloden, Mehrgradige Gleichungen, Noordhoff, Groningen, 1944) "If [a1, a2,…, am] = [b1, b2,…, bm], k = 1,2,…n, where m = n+1, then [ma1-x, ma2-x,…, mam-x] = [mb1-x, mb2-x,…, mbm-x], k = 1,2,…, n, n+2, where x = a1+a2+…+am." Gloden's theorem is equivalent to reducing ideal solns to its standard form (when the sum of terms on each side is zero). Note that, the sum is m(a1+a2+…+am )-mx and since x = a1+a2+…+am = b1+b2+…+bm, then both sides sum to zero. Of course, one can also apply it to symmetric solns but it just gives a trivial result. Example, using the non-symmetric ideal soln given earlier, [1, 75, 87, 205] = [15, 37, 115, 201], k = 1,2,3 we get,

[-91, -17, -5, 113] = [-77, -55, 23, 109], k = 1,2,3,5 and for no other k, where sum of terms Σai = Σbi = 0. Thus, theorem 6 reduces ideal solns to standard form and, for the non-symmetric case, this reduction has the consequence of non-trivially increasing its range, though technically it is no longer ideal since it skips an exponent. (Update, 8/7/09): For an extensive collection of theorems involving equal sums of like powers, see Part 3 of Assorted Identities.

III. Sum / Sums of Fifth Powers 5.1 Four terms5.2 Six terms5.3 Seven terms5.4 Eight terms5.5 Ten terms5.6 Twelve terms

5.1 Four terms While it is conjectured that x1

k + x2k = y1

k + y2k has no non-trivial

rational solns for k > 4, there are solns as roots of quadratics for k = 5. A. Desboves (a+b)k + (-b+ci)k + (-a+b)k + (-b-ci)k = 0, k = 1,2,5 where {a,b,c} = {x2-y2, xy√2, x2+y2}, and i = √-1. Piezas (√p+√q)k + (√p-√q)k = (√r+√s)k + (√r-√s)k, k = 5

{p,q,r,s} = {5vw2, -1+uw2, 5v, -(u+10v)+w3}, where w = u2+10uv+5v2. One can then set v = 5t2 so that {p,r} are squares. Whether {q,s} can then be made non-trivial squares is another matter. More generally, the above equation for k = 5,6,7,8 has a polynomial soln and the first two can be reduced to a common form. For k = 5 this reduces to, √p(p2+10pq+5q2) = √r(r2+10rs+5s2) then {p,q,r,s} = {5(bc-ad)v2, u3-(10bc+w)v2, 5(bc-ad)u2, -v3+(10ad+w)u2} where {u,v,w} = {a2+10ab+5b2, c2+10cd+5d2, ac+5bd} with four free variables {a,b,c,d}. For k = 6, (√p+√q)6 + (√p-√q)6 = (√r+√s)6 + (√r-√s)6 or after expanding, (p+q)(p2+14pq+q2) = (r+s)(r2+14rs+s2) then {p,q,r,s} = {-u2+vw, u2-vx, v2-ux, -v2+uw} where {u,v,w,x} = {a2+14ab+b2, c2+14cd+d2, ac+bc+13ad+bd, ac+13bc+ad+bd} Form: (p+cq)(p2+apq+bq2)k = (r+cs)(r2+ars+bs2)k This generalizes the two forms above and as was already discussed in the section on Third Powers, it takes only a small transformation to

assume c=0 without loss of generality. This author found the soln with c=0 as, {p,q,r,s} = {bvwk, -1+uwk, bv, -(u+av)+wk+1}, where w = u2+auv+bv2

for arbitrary u,v. Note: This can be proven for k=1,2. Using computer algebra, it is easy to see it is also true for other small k > 2. But I have no proof it is the case for all positive integer k. 5.2 Six terms: x1

5+x25+x3

5 = x45+x5

(For a database of solns <more than 5000> within a search radius of 17700, see Duncan Moore's results here.) After Euler’s work on, ak + bk = ck + dk

for k = 3 or 4 and subsequent work by other authors, the next advance was the discovery of polynomial solns to, ak + bk + ck = dk + ek + fk

for k = 5 or 6, with the first one for k = 5 by Sastry and Chowla in 1934 and later for k = 6 by Rao, Brudno, etc. Whether, ak + bk + ck + dk = ek + fk + gk + hk

for k = 7 or 8 will turn out to have polynomial solns as well remains to be seen since particular solns in the integers for both are now known, with more than forty for k = 7 since Ekl found the first one in 1996 and only one so far for k = 8 found by Kuosa in 2006.

(Update: 6/20/09) Special cases: a) x1

5+ x25+ x3

5+ x45+ x5

5 = 0 This is a counter-example to Euler’s sum of powers conjecture and, for fifth powers, there are only three known so far: (1967) [27, 84, 110, 133, -144] = 0 (by L. Lander, T. Parkin)(1996) [5027, 6237, 14068, -220, -14132] = 0 (by R. Scher, E. Seidl)(2004) [55, 3183, 28969, 85282, -85359] = 0 (by J. Frye) Whether this is reducible to solving an elliptic curve like for the fourth power version x1

4+ x24+ x3

4 = x44 still remains to be

seen. b) x1

5+ x25+ x3

5+ x45+ x5

5 = 1 Just like three 3rd powers of signed integers can sum to 1 (a famous example of which involves the taxicab number 1729 = 13

+ 123 = 93 + 103), five 5th powers can also do so. Only eight have been found so far. The smallest was by Lander, Parkin, and Selfridge, while three others were later given by Seiji Tomita, which were independently found by Duncan Moore who also gave four more: [1, 89, 118] = [38, 47, 123][1, 127, 430] = [16, 310, 412][1, 328, 709] = [5, 388, 705]

[1, 588, 772] = [59, 511, 791][1, 561, 1151] = [401, 616, 1146][1, 1073, 2297] = [379, 686, 2306]

[1, 4167, 4283] = [1039, 2601, 4811][1, 4823, 6377] = [1089, 3501, 6611] Note 1: Moore searched within a radius of 17700 and these are the only solns which have a unit term. However, this also includes one with a zero term, found by Bob Scher and Ed Seidl in 1997, [0, 220, 14132] = [5027, 6237, 14068] though another one of this form is unknown. Presumably, the greater the radius, the greater the chances of finding another one. Note 2: The sum and differences of 5 sixth powers can sum to unity and, within a radius of 17800, Moore found five, [1, 132, 133] = [71, 92, 147][1, 173, 294] = [75, 154, 295][1, 55, 330] = [159, 268, 311][1, 500, 515] = [197, 409, 556][1, 1123, 1143] = [573, 815, 1255] One can make a conjecture that, if the radius is great enough, there may be a zero term as well. (Though it is odd there are no more solns with a unit term in the range 2000-17000.) Note 3: Seven 7th powers can also sum to 1, though only two are known so far, and whether they can also sum to zero like for 5th powers is probable but an example is still unknown. See here.) Tomita gave a complete table of [x1, x2, x3] = [x4, x5, x6] for positive

xi < 1000, and Moore made a bigger table with sum almost 18,0005. When all terms are integers with one equal to unity, it is unknown if a parametrization can exist, unlike for third powers where there is an infinite family which recent work by this author has shown involves a Pell equation. No soln is yet known for 1+ x2

5+ x35+ x4

5+x55= z5

with positive xi, as seen in James Waldby's database with z < 10,000.

(For seven 5th powers equal to unity, there are formulas, also dependent on Pell eqns, to be given later.) c) x1

5+ x25+ x3

5+ x45+ x5

5 = y15+ y2

5+ y35+ y4

5+ y55 = z5

The smallest number whose third power is expressible as the sum of three positive third powers in two ways is 41 as, [2, 17, 40] = [6, 32, 33] = [41] For fourth powers as four fourth powers in two ways, Jarek Wroblewski found it to be 31127 (see Wroblewski's database here), [2260, 4870, 17386, 30335] = [2495, 11998, 16430, 30320] = [31127] The first fifth power expressible as the sum of five positive fifth powers in two ways was found by Waldby as 744, [14, 95, 545, 586, 644] = [100, 210, 414, 629, 651] = [744]

(Note: Incidentally, the number 744 figures prominently in the j-function j(τ) since this is the constant term of its series expansion,

j(τ) = 1/q + 744 + 196884q + 21493760q2 + …

which explains why Ramanujan’s constant (and others) exceeds a cube by nearly this amount, eπ√163 = 6403203 + 743.99999999999925…,though this property of 7445 is probably unrelated.)

d) a5+2b5+2c5 = d5

There are solns to a4+2b4+2c4 = d4, the smallest of which is the Pythagorean-like: [2, 2, 3, 4, 4] = [5]. There is also one for fifth powers, the only one so far, given in Waldby’s database as: [526, 526, 1349, 1349, 1355] = [1685]

k+x2k+x3

k = x4k+x5

The smallest soln turns out to be good for both k = 1,5: [24, 28, 67] = [3, 54, 62]. For sixth powers, the smallest is good for both k = 2,6: [3, 19, 22] = [10, 15, 23]. In fact, out of the 198 solns in Tomita's database for [5.3.3] cited above, which is complete for terms < 1000, almost 2/3 (66%) are good for k = 1,5. It should be interesting to check the percentage of [6.3.3] with terms below a bound that is good for k = 2,6. Why they suddenly become good for two powers is food for thought, but may simply be due to the increasing number of terms and the symmetric nature of the equation. For higher powers also with a minimal number of terms, there are solns valid for three powers, like, [344, 902, 1112, 1555] = [479, 662, 1237, 1535], for k = 1,3,7 and others found by A. J. Choudhry whenever a certain multi-variable cubic has a rational root. And also, [51, 253, 412, 600, 624] = [100, 187, 429, 603, 621], for k = 1,3,9

found by Wroblewski and the only one so far. There is only one known for [8.4.4] found by Kuosa, but I’m betting one will eventually be found good for k = 2,8 or, if signed, even good for k = 1,2,8, like some k = 1,2,6. (End note)

(Update, 11/3/09): f) x1k+x2

k+x3k = x4

k+x5k+x6

k, with x1+x2+x3 =

x4+x5+x6 = 0

At the bottom of this webpage, I asked if this system for k = 1,5 has non-trivial solns since the analogous version for 7th powers with eight terms and side-condition x1+x2+x3+x4 = x5+x6+x7+x8

= 0 is solvable, good for k = 1,3,7 and reduced by Choudhry to finding a rational root of a multi-variable cubic. Wroblewski checked his database from 2002 and found four, namely: [105, 1153, -1258] = [455, 582, -1037][31, 47242, -47273] = [5681, 9717, -15398][3914, 51858, - 55772] = [19264, 25403, -44667][12292, 59070, -71362] = [27745, 37983, -65728] The system, in fact, is also reducible to finding a non-trivial rational root of a cubic. Use the form L1, (a+bp+q)k + (b-bp+q)k + (c+ap+q)k = (a+cp+q)k + (b+ap+q)k + (c-cp+q)k

which is already true for k = 1, a general form to be proven later. Let p = q/a+1, and this will equate x1+x2 = x5. Expanding for k = 5, and after removing trivial factors, will result in a cubic in q with coefficients in {a,b,c}. (End update) (Update 6/26/09): In response to an email, Tomita was kind enough to create a complete [6.3.3] table with terms < 1000. True enough, of the 59 solns, almost 3/4 (75%) were good for k = 2,6, a greater percentage than for k = 1,5. For the [7.4.4] in Wroblewski's database (which is not a complete list below a certain bound), and excluding 18 solns to Choudhry's cubic with is tailored to yield k = 1,3,7, there are 40 remaining and almost 1/4 are for k = 1,7. If a complete list with terms < 10,000 can be given, I believe the percentage will be greater.

But why there is this tendency to be multigrade is unknown. (End note) (Update 7/17/09): While browsing the Net, I came across (again) Tom Womack's page. I had read his dissertation years ago and, re-reading it, realized he observed the same point about k = 2,6 being abundant (and Peter Montgomery as well as others before him). In fact, Womack had made a search and found that, out of 207 solns with common sum < 57176, almost 90% were for k = 2,6 (only 22 were just for k = 6), a much higher percentage than the one given by Tomita's smaller soln set. How high will this percentage get for a complete list with a much higher common sum? And what does this imply for eight positive terms valid for k = 1,7 or k = 2,8? (Considering the only one known so far for k = 8 is not multigrade.) (End note.) (Update 10/7/09): Duncan Moore, who has done extensive work on taxicab and cabtaxi numbers, has a bigger databasewhich includes [5.3.3] and [6.3.3]. Below a certain bound, for 5th powers, he found almost 5400 solns, about 59% of which are multigrade for k = 1,5. For 6th powers, there were 405, with 92% multigrade for k = 2,6, an increase from Womack's database. Whether these percentages asymptotically approach a definite value is unknown. (End update.) Various identities for 5th powers have been found, after transposition to symmetric form where {xi, yi} are not necessarily positive, that belong to four classes: 1. x1

5+x25+x3

5 = y15+y2

5+y35, x1+x2 = y1+y2,

2. x15+x2

5+x35 = y1

5+y25+y3

5, (Moessner's, with unknown linear relations between the xi, yi)

3. x1k+x2

k+x3k = y1

k+y2k+y3

k, k = 1,5

4. x1k+x2

k+x3k = y1

k+y2k+y3

k, x1-x2 = y1-y2, k = 1,5

If you know of another, pls send it so it can be added to the collection. The first two classes have beautiful identities. Sastry, Chowla (1934) (u5+25v5)5 + (u5-25v5)5 + (10u3v2)5 = (u5+75v5)5 + (u5-75v5)5 + (-50uv4)5 If {u,v} are chosen such that Abs[u/v] = x is within the range 251/5 < x < 751/5, or roughly 1.904 < x < 2.371, then it yields a (5.1.5) with all terms positive. Outside this range, it becomes a (5.2.4), but there is no {u,v} such that it gives a (5.3.3) with all positive terms. This has the basic form, (u5+av5)5 + (u5-av5)5 - (u5+bv5)5 - (u5-bv5)5 = 20(a2-b2)(u3v2)5 + 10(a4-b4)(uv4)5 If rational {a,b} can be found such that, 20(a2-b2) = mp5 (eq.1)10(a4-b4) = nq5 (eq.2) then it yields an identity for (5,4,m+n). For m = n = 1, Sastry and Chowla found {a,b} = {75, 25} and {p,q} ={10, 50}. Q: Are there others not simply fifth power multiples of these two, or other small m,n? (Update, 11/18/09): Dave Rusin showed that the system can reduced to a certain hyperelliptic curve, thus has only finitely many rational solns, and probably the known soln is the only non-trivial one. He gave the transformation, {a,b,p,q} = {m2y/(5n), m3n, -2mx, 20m2x}, where m = (4x5-5)/(5n2)

and eq.1, eq.2 are true if, -42x10+ 52 = y2 the rational soln of which apparently is only {x,y} = {1,3}. It is hard not to notice this is essentially the smallest Pythagorean triple, -42+52 = 32, and it is interesting that it appears in the context of 5th powers. (End update.) (Update, 12/15/09): Tony Rizzo pointed out that if {a,b} = {75r3, 25r3}, then {m,n} ={r, r2}. For r = 2, this gives the (5,4,6) identity, (u5+600v5)5 + (u5-600v5)5 - (u5+200v5)5 - (u5-200v5)5 = 2(2u3v2)5 + 4(4uv4)5 If {u,v} are chosen such that Abs[u/v] = x is within the range 2001/5 < x < 6001/5, then it yields a (5.1.9) with all terms positive. (End update) Update, 11/23/09): Moessner (1951) t6(a+20t)5 + t3(b-20t2-20t4)5 + (c+20t5)5 = t6(a-20t)5 + t3(b+20t2+20t4)5 + (c-20t5)5 where {a,b,c} = {t6+8t4+12t2-1, t6+8t4-8t2-1, t6-124-8t2-1}. Since t is arbitrary, one can set t = v5 to have equal sums of 5th powers. Note the nice symmetry of terms and how the coefficients, after negation, are palindromic. Surely this is not an isolated result. Anyone else knows of a similarly simple identity, as well as linear relations among the xi, yi, if any, for this? (End update.)

(Update, 2/22/10): Gerardin gave the beautiful 4th power identity with small coefficients {1,2,3}, (a+3a2-2a3+a5+a7)4 + (1+a2-2a4-3a5+a6)4 =(a-3a2-2a3+a5+a7)4 + (1+a2-2a4+3a5+a6)4 and Choudhry gave a 5th power version, (a-a3-2a5+a9)5 + (1+a2-2a6+2a7+a8)5 + (2a3+2a4-2a7)5 =(a+3a3-2a5+a9)5 + (1+a2-2a6-2a7+a8)5 + (-2a3+2a4+2a7)5 which in fact is good for k = 1,5. (Differences between LHS and RHS are highlighted in blue.) While there are identities for k = 6, none are known with such small coefficients. (Anyone can find one?) Other 5th deg identities can be found using the ff. two methods by Choudhry. Given, x1

k+x2k+x3

k+x4k = y1

k+y2k+y3

k+y4k (eq.1)

where, {x1, x2, x3, x4} = {a+b+c, a-b-c, -a+b-c, -a-b+c}{y1, y2, y3, y4} = {d+e+f, -d+e-f, -d-e+f, d-e-f} Expanding one side of the eqn, we get, ∑ xi

∑ xi5 = 80abc(a2+b2+c2)

and similarly for the yi. Eq.1 can be reduced to just 6 terms if x4 = y4. Thus, if,

Method 1: a+b-c = -d+e+f ; ab = de; c(a2+b2+c2) = f(d2+e2+f2)Method 2: a+b-c = -d+e+f ; c = f; ab(a2+b2+c2) = de(d2+e2+f2) then ∑ xi

k = ∑ yik, for k = 1,5.

Choudhry showed that given an initial point, even a trivial one, using appropriate substitutions and solving a system of linear equations, then an infinite number of non-trivial solns can be found. In fact, by directly solving Method 1, one can end up with a quartic equation which has a rational root if a certain quartic polynomial is made a square, hence can be treated as an elliptic curve. Note: Choudhry also used Eq.1 and its substitutions to solve k = 1,3,7. Source: “On Equal Sums of Fifth Powers”, Indian Jour. Pure & Applied Math, Nov. 1997. Update, 11/24/09): Choudhry (1999) has considered the general eqn, ax5+by5+cz5 = au5+bv5+cw5 (eq.1) for the case a+b+c = 0. Excluding the situation when abc(a-b)(a-c)(b-c) = 0, he gave the soln to, ax5+by5+cz5 = t2 (au5+bv5+cw5) as, {x,y,z} = { (p1t2-p1t+q1)r1, (p2t2-p2t+q2)r2, (p3t2-p3t+q3)r3 }

{u,v,w}= { (q1t2-p1t+p1)r1, (q2t2-p2t+p2)r2, (q3t2-p3t+p3)r3 } where.

{p1, p2, p3} = {17a2+14ab+14b2, 14a2+14ab+17b2, 17a2+20ab

+17b2}{q1, q2, q3} = {2a2-ab-b2, -a2-ab+2b2, 2a2+5ab+2b2}{r1, r2, r3} = {a+2b, -2a-b, a-b} given constant {a,b}, for arbitrary t. One can then set t = v5 to satisfy eq.1. As an example, for {a,b,c} = {1,2,-3}, after transposing terms, we get a soln to, x5+2y5+3z5 = u5+2v5+3w5

{x,y,z} = {101t10-101t5-4, 4t12-88t7+88t2, 25t10-25t5+4}{u,v,w}= {-4t12-101t7+101, 88t10-88t5+4, 4t12-25t7+25t2} (End update.) (Update, 2/3/10): Choudhry (-8m6+2mn5)k + (8m5n+n6)k + 2(8m6)k = (8m6+2mn5)k + (8m5n-n6)k + 2(n6)k, for k = 1,5 A soln in positive terms can be acquired if n > m and within the range 22/5 < n/m < 23/5, or approx. 1.32 < n/m < 1.51. With terms transposed to one side, (-8m6+2mn5)k + (8m5n+n6)k - (8m6+2mn5)k - (8m5n-n6)k + 2(8m6)k = 2(n6)k For n = 1, the identity proves that the integer 2 is the sum/difference of six integral 5th powers in infinitely many ways. Source: The

Diophantine Equation x15+x2

5+2x35 = y1

5+y25+2y3

5, Ganita, Vol. 48, No. 2, 1997, 115-116. Note: In one sense, this is a 5th power version of, (-6x3+y3)3 + (6x3+y3)3 - (6x2y)3 = 2(y3)3 where, for y = 1, proves that 2 infinitely is the sum is the sum/difference of three integral 3rd powers. Q: Anyone can give a similar identity expressing 2 as integral 7th powers? (End update.) Lander would later find a three-parameter family which in general would involve 9th degree polynomials. This author found that for special cases, it can be reduced to 7th or 8th deg. Whether there is a 6th deg is unknown. We have already seen that given an initial soln to a1

k+a2k = b1

k+b2k for either k = 2, 3, or 4, one can use this to find

more solns. It turns out there is a quintic version of this, albeit with a small condition, Theorem (Lander): "Given a rational solution to a1

k+a2k+a3

k = b1k

+b2k+b3

k for both k = 1,5, call this system S5, then this can generate subsequent ones." In “Geometric Aspects of Diophantine equations involving Equal Sums of Like Powers”, Lander started out with the form, (u+x)k + zk + (v+y)k = (v+x)k + yk + (u+z)k, (call this L0) a general soln for k = 1. This can also be true for k = 5 using appropriate values. (Note: Incidentally, it should be pointed out that

the eqn, x1k + x2

k + …+ xk+1k = 0 simultaneously valid for three odd

powers k = 1,3,n only has trivial solns when n = 5. Using the form above and Mathematica’s Resultant[] function to eliminate one variable at k = 3 and 5, it will be seen that there are only trivial solns. In contrast, there are non-trivial solns when k = 1,3,7 as found by Choudhry and k = 1,3,9 by Wroblewski. Whether there is a k = 1,3,11 remains to be seen.) Anyway, L0 when expanded out for k = 5 has the nice symmetrical form, (x-z)u4+2(x2-z2)u3+2(x3-z3)u2+(x4-z4)u = (x-y)v4+2(x2-y2)v3+2(x3-y3)v2+(x4-y4)v where all fifth powers vanish and as such is a quartic surface. After a clever geometric analysis, Lander gave a method to derive polynomial expressions for all the variables. Given an initial soln {ai, bi} to L0, this leads to a new one, typically much larger, in two ways. First, 1. Using point P1(x1,y1,z1) Define {u,v,x,y,z} = {-a3+b3, a2-b2, x0+(x1-x0)t , y0+(y1-y0)t, z0+(z1-z0)t}, where, x1 = y1 = z1 = (x0d1+y0d2+z0d3)/(d1+d2+d3) {d1, d2, d3} = {a1

4-b14, a2

4-b24, a3

4-b34}

{x0, y0, z0} = {a1+a3-b3, b2, a3} However, it still remains to find t. When all these expressions are substituted into L0, if {ai, bi} is a soln, then the eqn is a quartic with

four roots {0,0,1,t} so is just a linear eqn in t which is then easily solved. One can also use a second approach, 2. Using point P2(x2,y2,z2) Define {u,v,x,y,z} = {-a3+b3, a2-b2, x0+(x2-x0)t , y0+(y2-y0)t, z0+(z2-z0)t}, where, {x2, y2, z2} = {x1-d0, x1-d0+u, x1-d0+v} x1 = (x0d1+y0d2+z0d3)/(d1+d2+d3) d0 = (d2u+d3v)/(d1+d2+d3) and {d1, d2, d3}, {x0, y0, z0} are defined the same way as in the first point. Using these, eq.0 again reduces to a linear eqn in t. It is also possible to find a polynomial soln using a trivial initial one. Let, {a1, a2, a3, b1, b2, b3} = {p, -p, q, q, r, -r}{a1, a2, a3, b1, b2, b3} = {p, -p, q, r, q, -r} for the first and second points respectively and these will yield 9th degree polynomials. Theorem (Piezas): "The complete soln to S5, or x1

k+x2k+x3

k = y1k

+y2k+y3

k, for k = 1,5 involves solving only a quadratic equation with a discriminant D that is quartic polynomial in one variable. The problem of making D a square can then be treated as an elliptic curve." Proof: It can be shown that Lander’s result can be concisely encoded

in the form L1, (a+bp+q)k + (b-bp+q)k + (c+ap+q)k = (a+cp+q)k + (b+ap+q)k + (c-cp+q)k

which we’ve come across before. This is already true for k = 1. It is a subset of Lander's L0, (u+x)k + zk + (v+y)k = (v+x)k + yk + (u+z)k

excluding the case uv(u-v) = 0. To see this, simply equate the first five terms of L1 with that of L0, and one can always find rational {a,b,c,p,q} using, {a,b,c,p,q} = {(1+t)(st+y-z)/(2t), -(1+t)(st-y+z)/(2t), (1+t)(y+tu-tx+ty-z)/(2t), t/(1+t), (st-y+z+2tz)/(2t)}, where {s,t} = {u+x-y, -(u-v)/u} hence the cases with denominator (1+t)t = uv(u-v) = 0 should be excluded, otherwise there is division by zero. Using these expressions, the sixth terms of L1 and L0 are also seen to be equal. The fact that L1 is less general than L0 does not matter since for the exceptional case, L0 reduces to the four-term form, n1

k + (n2+n3)k = (n1+n2)k + n3k

which, after expanding, is easily seen to have trivial solns for k < 6 other than k = 1, hence is of no interest. With that out of the way, by expanding L1 at k = 5, this reduces to a polynomial in p that is just a quadratic of form, (Poly1)p2+(Poly2)p+(Poly3) = 0 (eq.1)

disregarding the trivial factor p(p-1). The Polyi are in {a,b,c,q} and as polynomials in q are of degree 1,1,3, respectively. (Unfortunately, they are tedious to explicitly write down.) The discriminant D is, D:= (Poly2)2-4(Poly1)(Poly3) of deg 1+3 and so is a quartic in q, thus proving the theorem. To find a polynomial soln, there are two ways: 1st method: Since Poly1 is linear in q, all we have to do is to set Poly1 = 0 and solve for q, with the eqn given by, 2(b+c)(b2+c2)q = (a4-c4)-(a+b)(b+c)(b2+c2) and (eq.1) reduces to the linear eqn (Poly2)p+(Poly3) = 0 where p is easily solved for. Consistent with Lander’s result, this yields ninth deg polynomials in one of the variables {a,b,c}. Incidentally, the sums of the terms as x1+x2, y1+y3 have the simple forms, (a+bp+q) + (b-bp+q) = (a4-c4)/((b+c)(b2+c2))(a+cp+q) + (c-cp+q) = (a4-b4)/((b+c)(b2+c2)) for {p,q} as defined above. This author found out that the degree of the polynomials can be reduced if we assume certain linear relationships between the terms. It’s known that the system for k = 1,5 may have the side condition x1-x2 = y1-y2. Let, (a+bp+q) - (b-bp+q) = (a+cp+q) - (b+ap+q) which is true if c = a+2b. Using this constraint, the polynomials reduce to seventh deg. Since the eqn is homogeneous, it can be set b = 1 without loss of generality and we have the identity, (at+p+q)k + (t-p+q)k + (ct+ap+q)k = (at+cp+q)k + (t+ap+q)k + (ct-cp

+q)k, k = 1,5, c = a+2t = 4a(a-2)(a+3)(a2+4a+5)2

p = (a2+1)(a2+8a+11)(a3+15a2+33a+31)q = -2a(a+1)(a-2)(a2+4a+5)(a3+15a2+33a+31) where the common factor (a+1) of the terms has to be removed. (A. Moessner also found a seventh deg soln in 1951 but I do not know if this is the same since Lander gave the source as an Italian journal.) Alternatively, if c = a-2b is used, they reduce to eighth deg though I haven’t been able to find a linear relationship between the terms. Q: So there are 5th, 7th, 8th, and 9th deg solns. Any for 6th deg? (Note: As was already discussed, the form L1 can also be used for appropriate higher powers k. Later, it will be used to prove that the system k = 1,2,6 can also be reduced, again, to merely solving a quadratic.) (Update 7/13/09) 2nd method: We directly find a q such that the discriminant D is a square. To find a trivial soln, one begins by expanding L1 for k = 5,7 and eliminate p between them. Factoring the resultant, one finds a linear relationship between the {a,b,c,q} given by, 2(b+c)q = a2-b2-c2-ab-ac-bc Using this q, D becomes the simple square, D: = (a2-b2)2(a2-c2)2(b-c)2/(b+c)2

This is a trivial soln but, as was shown in the first method that D is a quartic in q, from this initial point, one can then find an infinite number of non-trivial ones. For simplicity, one can set c = a+2b and the next point yields terms for the xi, yi as 13th-deg polynomials after removing common factors.

Piezas The complete criterion to solve the system x1

k+x2k+x3

k = y1k+y2

+y3k for k = 1,5 has already been given above, with a subset also

satisfying x1-x2 = y1-y2. We can also explore other methods for the sake for the exercise. This subset can be completely determined by either of two nice forms. Let, (a+d)k + (b+d)k + (c-2d)k = (a-d)k + (b-d)k + (c+2d)k

already true for k = 1 and if expanded for k = 5 reduces to the simple eqn, (a4+b4-2c4) + 2(a2+b2-8c2)d2 - 6d4 = 0 One way to solve this is to treat it as a quadratic in d2 and make its discriminant (a quartic polynomial) a square, though is not easily done for this particular quartic. However, we can use the alternative form, (p+r)k + (p-r)k + (2q+s)k = (q+r)k + (q-r)k + (2p+s)k

which, for convenience, after scaling can be assumed s = 1 without loss of generality. By expanding this at k = 5, this is only a quadratic in r2. Solving for r2, r2 = -p2-pq-q2 ± y, where y is the square root of the discriminant, y2 = 4q4+(8+5p)q3+2(4+4p+3p2)q2+(4+8p+8p2+5p3)q+(1+2p+2p2)2 (eq.1)

This is more easily made a square being a polynomial with a square leading and constant term, and illustrates how using the appropriate form can simplify a problem. This quartic is also rather special since its solns are plentiful enough that they can be parametricized linearly in three ways, namely {q1, q2, q3} = {-(80p+40)/(57p+80), -(p+2), -(3p+2)/3}, though the last two are trivial with respect to the original eqn. However, we can always derive non-trivial points. Three parametric solns will then be given, each of which involves a distinct elliptic curve. Thus, (p+r)k + (p-r)k + (2q+1)k = (q+r)k + (q-r)k + (2p+1)k, k = 1,5 has solns, q = – (80p+40)/(57p+80), r = z/(57p+80), 402+10160p+22999p2+14478p3+3249p4 = z2

q = 3(13p2+10p+8)/(9p-16), r = z/(9p-16), 82+656p+1999p2+2922p3+1629p4 = z2

q = (13p2+14p+8)/(3p+16), r = z/(3p+16), 242+1232p+1183p2+542p3+181p4 = z2

where one must solve the elliptic curve in p. The first q was found was found using Fermat’s method. There are trivial p like {-1/2, -80/57, 0}, but these can lead to non-trivial ones, like p = -15/38, etc. The next two q’s were found using Fermat’s method on the two trivial linear solns. These do not completely give all values that make the discriminant a square but it can be shown that the three q's as linear polynomials are the only ones of form q = (ap+b) or q = (ap+b)/(cp+d) for some rational {a,b,c,d}. To show this, given eq.1, substitute q = (ap+b) into the eqn and assume the resulting polynomial to be a square,

u1p4+u2p3+u3p2+u4p+u5 = (v1p2+v2p+v3)2

Since the ui are in terms of {a,b}, then the unknowns are {a,b,v1,v2,v3}. Expanding the above and collecting powers of p, one then has a system of five eqns in five unknowns, and the final eqn in this case has two rational solns. A similar approach for q = (ap+b)/(cp+d) will give a single rational soln. By adding yet another constraint, one can also find a simple form for S5. An example is,

(v+8)k + (u-8)k + (-u+2v+7)k = (-u+2v+8)k + (v-8)k + (u+7)k

This is already true for k = 1. Expanding for k = 5 yields only a quadratic in v and is easily solved. For rational v, one must make its discriminant, a quartic polynomial in u, as a square, y2 = -43281+40110u-4597u2+390u3-31u4

with one small soln as u = 3/2, giving [389, -208, 442] = [474, -123, 272]. From this initial point, one can then compute an infinite number of rational points. This, in fact, is the complete soln to the system, call this M1, x1-x2 = y1-y2x2-nx3 = -ny1+y3

x1k+x2

k+x3k = y1

k+y2k+y3

for k = 1,5 and n = 15. Other than the case n = 1 which is trivial, I haven’t checked yet for what other n this system is solvable. (Update 7/12/09): Another one obeying the system M1 but with n = 2 is,

(1-2q+2r)k + (-3-2q)k + (1+2q)k = (2+q+r)k + (-2+q-r)k + (-1-4q+2r)k

Again, it is already true for k = 1 and, when expanded for k = 5, yields a quadratic in r. Its discriminant is a quartic in q and, if made a square is, y2 = -(384+1112q+1297q2+738q3+171q4) with a small rational point at q = -16/15. Q: Any other simple soln to system M1 for some integer n? (End note.) Note 1: Incidentally, given the monic quartic of form, y2 = x4+(Poly1)x3+(Poly2)x2+(Poly3)x+(Poly4) to be made a square and where the Poly_i are polynomials in a variable p of degree 1,2,3,4 respectively (like eq.1 above), what is the maximum number possible of linear solns of form x = (ap+b) or x = (ap+b)/(cp+d)? Note 2: The complete criterion of x1

k+x2k+x3

k = y1k+y2

k+x3k, k =

1,5, where x1-x2 = y1-y2 has also been given by Bremner though I haven’t seen that work yet. I will update this section to compare methods when I’ve read it. Note 3: It can be proven that k = 1,2,5 yields only trivial rational results. The complete soln to k = 1,2 is by L. Dickson, (Form 4 in Sums of Three Squares) (ad+e)k + (bc+e)k + (ac+bd+e)k = (ac+e)k + (bd+e)k + (ad+bc+e)k

and expanding this at k = 5 gives merely a quadratic polynomial in e with discriminant D,

D = -2(a2-ab+b2)(c2-cd+d2) which must be made a square. But let {a,b,c,d} = {u+v, 2v, x+y, 2y} and it becomes, D = -2(u2+3v2)(x2+3y2) and it is obvious there are no real values such that D becomes a positive real square. The same conclusion was arrived at by Choudhry in his paper, The system of simultaneous equations x1

k+x2k+x3

k = y1k+y2

+y3k, k = 1,2,5 has no non-trivial solns in integers, The Mathematics

Student, Vol 70, 2001. (It is not known, however, if the case k = 2,5 has a soln or not, though it is easily proven using Moore's database that there are none within a radius of 17000.) Note 4: Is a5+b5+c5 = d5+e5+f5 where a+b+c = d+e+f = 0 solvable? (The analogous eqn for seventh powers does have solns.) This can be completely parametricized as, p5+q5+(-p-q)5 = r5+s5+(-r-s)5 which when expanded is, pq(p+q)(p2+pq+q2) = rs(r+s)(r2+rs+s2) though it seems to be unknown if this is non-trivially solvable. (Update: Wroblewski has proven this has non-trivial solns, the smallest of which is {p,q,r,s} = {105, 1153, 455, 582}, yielding, 1055 + 11535 + (-105-1153)5 = 4555 + 5825 + (-455-582)5

though they are sparse in the range 10^5. See update 11/03/09 above.)

5.3 Seven terms As was pointed out by Lander, quite a lot of the small solns found by computer search to, x1

k+x2k+x3

k+x4k = x5

k+x6k+x7

k+x8k (eq.0)

for k = 1,5 have a transposition such that x1+x2+x3+x4 = x5+x6+x7+x8 = 0, call this condition J1. Furthermore, this transposed form is valid for k = 1,2,3,5. This has a complete soln in terms of binary quadratic forms since it can be reduced to solving the beautifully simple equation ax2+by2 = cz2 using Chernick’s approach discussed below. In fact, it can be showed that any soln to eq.0 for k = 1,2,3,5 must satisfy J1. To prove this, use the form F2, (a+bh)k + (c+dh)k + (e+fh)k + (g+h)k = (a-bh)k + (c-dh)k + (e-fh)k + (g-h)k

where it can be set a = 1 without loss of generality. This can be satisfied for k = 1,2 using f = -(1+b+d), g = -(ab+cd+ef). Expanding for k = 3,5 gives, (Poly01)h2 + (Poly02) = 0(Poly11)h4 + (Poly12)h2 + (Poly13) = 0 where the Polyi are in {b,c,d,e}. Eliminating h between the two gives a resultant which completely factors into linear eqns, some trivial but the non-trivial ones all imply J1, thus proving the assertion.

A large class involving eq.0 can be reduced to solving x1k+x2

k+x3k =

y1k+y2

k+y3k for k = 2,4 with the constraint x1+x2 = y1+y2.

Theorem (Piezas): If (p+u)k + (p-u)k + (2q)k = (p+v)k + (p-v)k + (2r)k, for k = 2,4, then, (p+2q)k + (p-2q)k + (-p+2r)k + (-p-2r)k = (-2p+u)k + (-2p-u)k + (2p+v)k + (2p-v)k, for k = 1,2,3,5 The complete soln is given by the simultaneous eqns, -3p2+q2+3r2 = u2

-3p2+3q2+r2 = v2

This, in turn, can be completely solved in terms of binary quadratic forms as was already discussed in Fourth Powers (Part 3). Note that the one for k = 1,2,3,5 have sums of terms on either side equal to zero so, by Theorems 1 and 6 of the Prouhet-Tarry-Escott Problem, is essentially an ideal soln of degree 3 in standard form. The special case of eq.0, x1

k+x2k+x3

k+x4k = x5

k+x6k+x7

when one term is equal to zero entails making a certain quartic polynomial a square (to be discussed later) and has has an infinite family of polynomial solns the smallest of which found by this author is of fifth degree, one by Chernick as ninth degree, though there are others also due to Chernick which need solving an elliptic curve but are notable due to their aesthetic form. Chernick One can derive a soln such that the same variables solve two sets of

systems, for k = 2,4 and k = 1,2,3,5. There are two different pairs, special cases of the more general identity given above with the first being, A. If (2a+c)k + (-2a+c)k + (2b)k = (3c)k + (-c)k + (2d)k, for k = 2,4, then, (2a-2c)k + (-2a-2c)k + (4c)k = (2b+c)k + (-2b+c)k + (-c-2d)k + (-c+2d)k, for k = 1,2,3,5. This can be solved as, 2a2+4b2 = 9c2 (eq.1)2a2+b2 = 2c2+d2 (eq.2) Alternatively, 2a2+4b2 = 9c2 14a2+b2 = 9d2 with the second condition easily found by substituting c from eq.1 into eq.2. To completely solve eq.2, let {a,b,c,d} = {pr+qs, ps-2qr, pr-qs, ps+2qr} and substituting this into eq.1, one has to solve a quadratic. Its discriminant is a quartic polynomial in two variables, say {p,q}, and has to be made a square, so, {r,s} = {4p2-7q2, -3pq ± 2v} where v2 = 7p4-26p2q2+28q4

with one soln as {p,q,r,s} = {3, 1, 29, -47} and an infinite more. Similarly for the second pair, B. If (a+d)k + (-a+d)k + bk = (c+d)k + (-c+d)k + dk, for k = 2,4, then, (b-d)k + (-b-d)k + (2d)k = (a+2d)k + (-a+2d)k + (-c-2d)k + (c-2d)k, for

k = 1,2,3,5. Let, 11a2+4b2 = 9c2 (eq.1)2a2+b2 = 2c2+d2 (eq.2) or alternatively, 11a2+4b2 = 9c2

-4a2+b2 = 9d2 which can be similarly treated as above. These are special cases of the general theorem given above. Recall that, for certain {p,q,r,u,v}, (p+2q)k + (p-2q)k + (-p+2r)k + (-p-2r)k = (-2p+u)k + (-2p-u)k + (2p+v)k + (2p-v)k, for k = 1,2,3,5 Since, after minor sign changes of p, the variables {q, r} are interchangeable (as well as the {u,v}), then there are two ways to set one term equal to zero, either: 1) RHS: 2p-v = 02) LHS: -p+2r = 0 for example, though one can choose any term on one side. Chernick's two identities have, 1) {p,q,r,u,v} = {c, b, d, 2a, 2c}2) {p,q,r,u,v} = {d, b/2, d/2, a, c} which, when substituted into the theorem, explains why one term is equal to zero. J. Chernick, L. Lander

Theorem: "If a2+b2+c2 = d2+e2+f2, and abc = def, then, (a+b+c)k + (a-b-c)k + (-a-b+c)k + (-a+b-c)k = (d+e+f)k + (d-e-f)k + (-d-e+f)k + (-d+e-f)k, for k = 1,2,3,5." For indefinite k, call this system C1. Labeled as, x1

k + x2k + x3

k + x4k = x5

k + x6k + x7

k + x8k, (eq.0)

note that this also satisfies x1+x2+x3+x4 = x5+x6+x7+x8 = 0. The history of C1 is quite interesting. In 1913, Crussol considered a transposed form of C1 to solve k = 1,2,4,6. In 1937, Chernick studied this system in the context of an ideal solution of degree 3 in standard form and gave the complete soln, though he knew by Theorem 6 it could be extended to fifth powers. In the late 1960s, Lander would revisit this system, with a small rearrangement of terms, focusing only on odd powers and solving the cases k = 1,5 and k = 1,3,5. Surprisingly, this system can be extended to solve seventh powers as well as was done by A. Choudhry in the 1990s to find the first numerical solns to k = 1,3,7 though the variables must fulfill a more complicated condition. Note that either side is a special case of Boutin’s Theorem, (a+b+c)3 + (a-b-c)3 - (a+b-c)3 - (a-b+c)3 = 24abc a theorem which generalizes the difference of two squares to other kth powers. Also, in the guise, (-a-b-c)k + (a-b-c)k + (-a-b+c)k + (-a+b-c)k = (-2a)k + (-2b)k + (-2c)k, for k = 1,2,

this appears in the Birck-Sinha Theorem which involves eighth powers. Whether C1 has a soln for k = 1,8 or even k = 1,2,8 still remains to be seen, but perhaps there is. For k = 1,2,3,5, the problem then is to solve the two conditions, a2+b2+c2 = d2+e2+f2 (eq.1)abc = def (eq.2) By adding the condition a±b±c = 0 (for any choice of signs), this implies one of the xi of eq.0 will vanish and hence will involve only seven terms. A. Partial Solutions The section on Sums of n Squares gives a lot of identities for the first eqn such as the nice symmetrical one by Goldbach, (p+r)2 + (q+r)2 + (p+q-r)2 = (p-r)2 + (q-r)2 + (p+q+r)2

and it satisfies eq.2 as well if p2+pq+q2 = r2. Lander also gave the partial soln, {a,b,c} = {q(pr+s), (p-r)(pr-s), pq(p+r)}{d,e,f} = {q(pr-s), (p+r)(pr+s), pq(p-r)} where s = q2-r2. For the special case when one of the terms xi vanishes, let one of the four (d±e±f) be equal to zero. It suffices to consider d+e±f = 0, hence, q(pr-s) + (p+r)(pr+s) ± pq(p-r) = 0 Solving this quadratic in p, its discriminant must be made a square and involves an elliptic curve. Using the positive, then negative case yields,

5q4-8q2r2+4r4 = t1

-3q4+12q3r+4q2r2-8qr3+4r4 = t22

respectively, both of which have small solns, one being {q,r} = {4,1}, etc. B. Complete Solution Completely solving the two eqns, a2+b2+c2 = d2+e2+f2 (eq.1)abc = def (eq.2) just involves binary quadratic forms. Proof: Let, {a,b,c} = {p+qu, r+su, t+u}{d,e,f} = {p-qu, r-su, t-u} where, t = -(pq+rs)p = ((1-q2-s2)r + w) /(2qs) and w satisfies, r2(-1+q-s) (1+q-s) (-1+q+s) (1+q+s) + 4q2s2u2 = w2

Since this is of the form, u2+dv2 = w2

hence is easily completely solved in binary quadratic forms. (End proof.) Chernick used another approach by setting {c,d,e} = {mnf,

ma, nb} which solves eq.2 and transforms eq.1 to, c1a2 + c2b2 = (m2n2-1)f2

where {c1, c2} = {m2-1, n2-1} with its complete soln as, a = c1nu2+2c2uv-c2nv2

b = c1u2-2c1nuv-c2v2

f = c1u2+c2v2

for arbitrary {m,n,u,v}. We can give our own proof that this is complete. Proof: To remove scaling factors, divide the eqn by f2 to get the simpler form, c1p2 + c2q2 = (m2n2-1) (eq.3) where {p,q} = {a/f, b/f}. Given any rational soln {p,q} to eq.3, one can always find rational {u,v} using the formulas, u = (n2-1)(q2-m2)(p+n)v = (m2-1)(p2-1)(q-1) Verifying via Mathematica, one can see that the eqns {p-a/f = 0, q-b/f = 0} using this u,v have eq.3 as a factor, hence will be true if {p,q} make this zero. Thus, Chernick’s soln completely solves the two conditions as binary quadratic forms in four parameters {m,n,u,v}. For the special case when one of the terms xi vanishes, again let one of the four (d±e±f) be equal to zero. They can be easily constructed using {d,e} = {ma, nb} with {a,b,f} as defined above. As was seen, they are binary quadratic forms in {u,v} and interestingly this quartet have the

same discriminant D which if made a square y2, D:= (2m2-1)m2n4-(m4+3m2-2)n2+(2m2-1) = y2

implies there are rational u,v such that one of the terms xi can be chosen to vanish. (The other quartet also have just one discriminant.) This author found two small solns, n = 1/(m+1), or n = (-2m2+5)/(2m2+1) After solving one appropriate xi = 0 for {u,v}, both solns, after removal of common factors, yield xi as just 5th degree polynomials in contrast to the 9th degree ones found by Chernick who used a different method. (Which is ironic since he could have used this one.) Making D a square can then be treated as an elliptic curve and with these initial solns, subsequent ones can then be found.

5.4 Eight terms Several results for eight terms have already been discussed in the previous section. Before discussing the present section, it may be instructive to give an overview of some of the known “side conditions” for, x1

k+x2k+x3

k+x4k = y1

k+y2k+y3

k+y4k, k = 1,3,5, or k = 2,4,6,

x1k+x2

k+x3k+x4

k+x5k = y1

k+y2k+y3

k+y4k+y5

k, k = 1,3,5,7, or k = 2,4,6,8, since there seems to be some general pattern. Fifth powers:

1. x1+x2 = x3+x4 = -(y1+y2)2. x1-x2 = x3-x4 = -(y1-y2)3. x1-x2 = x3-x4 = y1-y24. x1+x2 = y1+y2; x3+x4 = y3+y4

Sixth powers:

1. x1+x2 = y1+y2; x3+x4 = y3+y4 Seventh powers:

1. x1+x2 = x3+x4 = y1+y22. x1-x2 = x3-x4 = y1-y23. x1-x2 = y1-y2; x3-x4 = y3-y4

Eighth powers:

1. x1-x2 = y1-y2; x3-x4 = y3-y4 Of course these are not all of the possible side conditions but small solns tend to follow them hence are the first ones noticed. They are useful since they give more “handles” to deal with the system and, in some cases, reduce it to merely solving a quadratic or making a low-degree polynomial a square. (It is tempting to speculate ninth and tenth powers may obey similar rules though, since only one soln is known for each, very little can be said about their behavior.) (Update: More solns have been found for tenth powers. See update in last section.) Side conditions can also have interesting implications, Piezas Theorem 1: If ak+bk+ck+dk = ek+fk+gk+hk, for k = 1,3,5 where a+b+c+d = e+f+g+h = 0, then the following six relations are also true, [1] 5(a2+b2+c2+d2)(a3+b3+c3+d3) = 6(a5+b5+c5+d5)

[2] 3(a2+b2+c2+d2)(a4+b4+c4+d4-e4-f4-g4-h4) = 4(a6+b6+c6+d6-e6-f6-g6-h6) [3] 9(a2+b2+c2+d2)(a7+b7+c7+d7-e7-f7-g7-h7) = 7(a9+b9+c9+d9-e9-f9-g9-h9) [4] (a3+b3+c3+d3)(a6+b6+c6+d6-e6-f6-g6-h6) = (a9+b9+c9+d9-e9-f9-g9-h9) [5] 7(a3+b3+c3+d3)(a4+b4+c4+d4-e4-f4-g4-h4) = 12(a7+b7+c7+d7-e7-f7-g7-h7) Combining [2] and [3] yields the same relation as combining [4] and [5], namely, [6] 7(a4+b4+c4+d4-e4-f4-g4-h4)(a9+b9+c9+d9-e9-f9-g9-h9) = 12(a6+b6+c6+d6-e6-f6-g6-h6)(a7+b7+c7+d7-e7-f7-g7-h7), and, One can test it with a random example, say, {a,b,c,d} = {21, -13, 9, -17} and {e,f,g,h} = {23, -3, -21, 1}, though it will work in the general case. Note that a+b+c+d = e+f+g+h for this system implies it is also valid for k = 2. [6] has already been given in Fourth Powers but is related to the Chernick-Lander theorem given in the previous section and is analogous to Ramanujan’s 6-10-8 Identity. [3] seems to have analogous versions for any system k = 1,2,3,5,…n for consecutive odd exponents n > 3, while [1] is a general identity that works on four terms a+b+c+d = 0. Proof: Given x1

k+x2k+x3

k+x4k - (y1

k+y2k+y3

k+y4k) = 0 such that

x1+x2+x3+x4 = y1+y2+y3+y4 = 0, this is completely parametricized by the Chernick-Lander form, call this L(k),

L(k): = (a+b+c)k + (a-b-c)k + (-a-b+c)k + (-a+b-c)k – ((d+e+f)k + (d-e-f)k + (-d-e+f)k + (-d+e-f)k) since the three variables {a,b,c} can suffice to express the xi (and likewise {d,e,f} for the yi). Expanding L(k) for k = 1,2,3,5, we get, L(1) = 0, L(2):= a2+b2+c2-d2-e2-f2 , L(3): = abc-def, L(5): = abc(a2+b2+c2)-def(d2-e2-f2), If we wish to set all L(k) = 0, if L(3) = 0, one can see that the eqn L(5) = 0 reduces to just L(2) = 0. Eliminating f between L(2) = 0 and L(3) = abc-def = 0, we get a polynomial, call it P. Substituting the terms of L(k) on any of the statements of Theorem 1 after [1], we get an expression in {a,b,c,d,e,f} and, eliminating f between this and abc-def = 0, we get P again (as a factor). Thus, the roots of P = 0 satisfy both the condition and the theorem, which proves the latter. (For [1], simply substract one side from the other, expand the identity, and one can see it is true if a+b+c+d = 0.) Lander (a+b+c)5 + (a-b-c)5 + (-a-b+c)5 + (-a+b-c)5 = 80abc(a2+b2+c2) (Update, 12/13/09): Tony Rizzo showed that there are non-trivial solns to, 80abc(a2+b2+c2) = d5+e5 (eq.1) thus yielding a (5.3.3) and gave the example {a,b,c,d,e} = {32, 25, 50, 80, 100}. More generally, this gives a (5.3.3) with the side condition x1+x2 = y1+y2. The Sastry-Chowla identity has this constraint, so there is in fact an infinite number of solns to eq.1 given by {a,b,c,d,e} = {u5, 25v5, 50v5, 10u3v2, 50uv4}, for arbitrary {u,v}, with Rizzo's as {u,v} = {2,1}. However, there are other solns as the Sastry-Chowla

identity cannot give a (5.3.3) with all terms positive. Duncan Moore has a database here and out of around 5400 solns, only 12 had the required condition that x1+x2 = y1+y2. Q: Is there any other family that solves eq.1? (End update.) Lander (k = 1,5) (ax2-vxy+w1y2)k + (ax2+vxy+w2y2)k + (x2-vxy+w3y2)k + (x2+vxy

+w4y2)k =

(ax2+vxy+w1y2)k + (ax2-vxy+w2y2)k + (x2+vxy+w3y2)k + (x2-vxy

+w4y2)k

This has the side condition x1+x2 = y1+y2; x3+x4 = y3+y4 which for odd powers is equivalent to x1+x2+x3+x4 = y1+y2+y3+y4 = 0 and was discussed in the previous section. However, this new one is only for k = 1,5 but has a surprising property. Let v = 3a2, and {w1, w2, w3, w4} = {p+q, p-q, r+s, r-s}. Then, the soln is, {p,q,r,s} = {a7+a5-2a3+a, 3a2, a6-2a4+a2+1, 3a5} Note that {p,q,r,s} also solve, (p+q)4 + (r-s)4 = (p-q)4 + (r+s)4

To see this, if the 8-term eqn above was expanded and powers of {x,y} collected, we get a system of four auxiliary eqns in four unknowns, one of which is, w1

4-w24- w3

4+w14 = 0

explaining why it has this property. In fact, by solving this system one

can directly derive the wi, instead of the geometric derivation which Lander used. (Note: Are there other forms of (5,4,4) that also involve the other parametrizations of (4,2,2)?) Kawada, Wooley (h+x)5 + (h-x)5 + (h+y)5 + (h-y)5 + (h+x+y)5 + (h-x-y)5 = 20h(x2+xy+y2+h2)2 - 14h5

(Update, 11/2/09): Tony Rizzo pointed out we can let x2+xy+y2 = z2 for some z. He gave Ramanujan's simple soln to this as {x,y} = {a-d, b+c+d} where a/b = c/d, and b = c. For the special case when y = -x, this then becomes, 2(h+x)5 + 2(h-x)5 + 2h5 = 20h(h2+x2)2 - 14h5

where the whole eqn is now div by 2. Note: I realized Rizzo's observation can be extended to cover some of the identities I found given below. For example, let z = h, and the Kawada-Wooley (KW) identity becomes, (x+z)5 + (-x+z)5 + (y+z)5 + (-y+z)5 + (x+y+z)5 + (-x-y+z)5 = 66z5 where x2+xy+y2 = z2, call this condition C1, the smallest soln of which is {x,y,z} = {3,5,7}. If we choose to express 66 as a sum of powers 2(1)5+2(2)5, this results in a 10-term identity, (x+z)k + (-x+z)k + (y+z)k + (-y+z)k + (x+y+z)k + (-x-y+z)k = 2(z)k + 2(2z)k, with the same conditon C1, though it is now valid for k = 1,2,3,4,5.

This is also discussed in Form 5.5 below. But if we let z2 = nh2, with n = 4 or 28, then the KW identity is,

(h+x)5 + (h-x)5 + (h+y)5 + (h-y)5 + (h+x+y)5 + (h-x-y)5 = mh5 for m = {486, 16806}, where x2+xy+y2 = nh2 for n = {4, 28}, respectively. But if we express m as a sum of powers, then, (h+x)k + (h-x)k + (h+y)k + (h-y)k + (h+x+y)k + (h-x-y)k = 2(3h)k, where x2+xy+y2 = 4h2 (h+x)k + (h-x)k + (h+y)k + (h-y)k + (h+x+y)k + (h-x-y)k = (-h)k + (7h)k, where x2+xy+y2 = 28h2 now valid for k = 1,3,5. These will be discussed in Constraint I immediately after this note, though derived differently using resultants. (End update.) Piezas So far, most of the identities discussed have the condition x1+x2+x3+x4 = y1+y2+y3+y4 = 0. One can avoid this and explore other constraints such as: a) x1+x2 = x3+x4 = x5+x6, as (a+c)k + (-a+c)k + (b+c)k + (-b+c)k + (a

+b+c)k + (-a-b+c)k + uk + vk = 0b) x1-x2 = x3-x4 = x5-x6, as (a+c)k + (a-c)k + (b+c)k + (b-c)k + (a+b

+c)k + (a+b-c)k + uk + vk = 0c) x1-x2 = x3-x4 = -(x5-x6), as (3p+q)k + (p+q)k + (p+r)k + (-p+r)k +

(-3p+q)k + (-p+q)k + uk + vk = 0 I. Constraint: x1+x2 = x3+x4 = x5+x6 We can focus on a special case of this constraint by defining the

polynomial, P(k): = (a+c)k + (-a+c)k + (b+c)k +(-b+c)k + (a+b+c)k + (-a-b+c)k which is one side of the Kawada-Wooley equation. This has the common sum of two terms as 2c. Note this also obeys x1-x2+x3-x4 = x5-x6. If we set the equation, E(k): = P(k) + (r)k + (s)k = 0 to be simultaneously true for k = 1,3,5, then this has a complete parametrization. By eliminating {r,s} from the three equations E(1), E(3), E(5) using resultants, easily done using Mathematica’s Resultant[] function, one gets a final eqn, (a2+ab+b2-28c2) (a2+ab+b2-4c2) = 0 so there are two identities involving P(k), namely, (a+c)k + (-a+c)k + (b+c)k +(-b+c)k + (a+b+c)k + (-a-b+c)k = 2(3c)k, where a2+ab+b2 = 4c2, and, (a+c)k + (-a+c)k + (b+c)k +(-b+c)k + (a+b+c)k + (-a-b+c)k = (-c)k + (7c)k, where a2+ab+b2 = 28c2, both for k = 1,3,5. Also, the polynomial Pk as a sum is quite peculiar since a common factor turns up for k = 1,2,3,4,5. To illustrate, for the first, set {a,b,c} = {2(u2-v2), 2(2uv+v2), (u2+uv+v2)}, then for k = 1,2,3,4,5 we have the sums, Pk = {2(3)(u2+uv+v2), 2(11)(u2+uv+v2)2, 2(33)(u2+uv+v2)3, 2

(83)(u2+uv+v2)4, 2(35)(u2+uv+v2)5}

while for the second, let, {a,b,c} = {2(u2+6uv+2v2), 2(2u2-2uv-3v2), (u2+uv+v2)}, then,

Pk + ck = {7(u2+uv+v2), 7(17)(u2+uv+v2)2, 73(u2+uv+v2)3, 7

(545)(u2+uv+v2)4, 75(u2+uv+v2)5} Thus, in general, for these k and no higher, they have the common factor u2+uv+v2. These identities also have other properties. The first, by solving a2+ab+b2 = t2k, gives a soln to, x1

5 + x25 + x3

5 + x45 + x5

5 + x65 = 2(35y5k)

the quintic analogue of Ramanujan’s family of solns to x1

4 + x24 + x3

= 2(y2k). See Form 11 of Fourth Powers. Also, since the a,b,c are binary quadratic forms, these can provide solns to, x1

5 + x25 + x3

5 + x45 + x5

5 + x65 + x7

5 = 1 by setting one term as equal to ±1 and solving the resulting Pell eqn. For ex, we have, (a+c)k + (b+c)k + (a+b+c)k + (-a-b+c)k + (-a+c)k + 2(-3c)k = ±1 where {a,b,c} = {2(p+q)(p+5q), 8(p+4q)q, p2+8pq+19q2}, for k = 1,3,5, if p2-13q2 = ±1. II. Constraint: x1-x2 = x3-x4 = x5-x6 We can focus first on a special case. Define the eqn, (a+c)k + (a-c)k + (b+c)k + (b-c)k + (a+b+c)k + (a+b-c)k + (u)k + (v)k = 0, for k = 1,3,5, (eq.1)

This has the common difference of the first three pairs of terms as 2c and also has a complete parametrization, but now involves an elliptic curve. Note that this also has the sum of its first four terms as x1+x2+x3+x4 = x5+x6. Let {a,b,u,v} = {p+q, -p+q, d-4q, -d-4q}, then eq.1 is true if, c2+11d2 = 4p2, and 3c2+d2 = 12q2, where the ratio c/d = {1, 1/3} should be avoided. A small soln is {c,d} = {37, 15} and it is easy to convert the two quadratic conditions into an elliptic curve. A special case solvable as quadratic forms is, (a+b+c)k + (a+b-c)k + (-a+b+c)k + (-a+b-c)k + (8b+c)k + (8b-c)k = (7b)k + (13b)k, for k = 1,3,5, (eq.2) where -a2+126b2 = 5c2. The common difference of the first three pairs is also 2c but the sum of its first four terms is now x1+x2+x3+x4 = (x5+x6)/4. From the initial soln {a,b,c} = {1,1,5}, one can get a parametrization. However, for the complete soln to the general case, let,

(a+m)k + (a-m)k + (b+m)k + (b-m)k + (c+m)k + (c-m)k = (d+n)k + (d-n)k

Expanding for k = 1,3,5, we can eliminate d,n and find the resultant in terms of m which is only a quadratic. Explicitly, the complete soln is, d = a+b+c, and {m,n} = {u/(2d), v/(2d)} where {u,v} are, (a+b)4 + (a+b)(4a2+9ab+4b2)c + (3a+2b)(2a+3b)c2 + 4(a+b)c3 + c4 = u2

(a2-b2)2 – 3ab(a+b)c – (2a2+3ab+2b2)c2 + c4 = v2

One must find {a,b,c} such that these two quartic polynomials in c are squares which in general is not easily done. (Surprisingly, for a k = 1,3,5,7 with a similar constraint, it involves only two quadratics to be made squares which is easier.) Without loss of generality, we can set a = 1 and one soln for the first eqn is, c = -(2+3b)/(3+4b) It reduces the second eqn, after removing square factors, to solving, 25+24b-48b2+64b4 = y2

with initial b = 3/16 from which other rational points can then be computed. (Update, 8/13/09): One can use a variant, (p+aq)k + (p+bq)k + (p+cq)k + (d+q)k = (p-aq)k + (p-bq)k + (p-cq)k + (d-q)k, for k = 1,3,5 already satisfying the constraint (after appropriate transposition of terms). Solving this system involves only making a quartic and quadratic polynomial into squares, hence is a slight improvement over the previous method. (End update.) III. Constraint: x1-x2 = x3-x4 = -(x5-x6) This is a generalization of a single numerical result by Tarry. It can be the case that a k = 1,3,5 can give rise to a k = 2,4,6. Given, (3p+q)k + (p+q)k + (p+r)k + (-p+r)k + (-3p+q)k + (-p+q)k + uk + vk = 0, for k = 1,3,5, then, (4p+q)k + (2p+r)k + (p+u)k + (p+v)k = (4p-q)k + (2p-r)k + (p-u)k + (p-v)k, for k = 2,4,6

which has a complete soln in terms of an elliptic curve as, {p,q,r,u,v} = {2x2+x+1, x2+6x+1, -8(x2+2x), 6x2+4x-2+y, 6x2+4x-2-y} and where x,y must satisfy, y2 = 15x4+104x3+223x2+54x+4 though some of the small rational solns are trivial. (Update, 8/4/09): Alternatively, the system can be expressed as, (a-2b+c)k + (a-2b-c)k + (b+3c)k + (b+c)k = (a-d)k + (a+d)k + (-b+3c)k + (-b+c)k, for k = 1,3,5, then, (a+c-d)k + (a+c+d)k + (-a+2b+2c)k + (b-4c)k = (a-c-d)k + (a-c+d)k + (-a+2b-2c)k + (b+4c)k, for k = 1,2,4,6,

where {a,b,c,d} satisfy a2+ab+2b2 = 8c2 and a2-7ab+42b2 = 8d2. This is easily proven by solving for {c,d} in terms of {a,b} and subsituting into the two systems, though the ratio a/b = {-1/3, -1, 2, 5} should be avoided as they are trivial, with a small non-trivial soln as {a,b} = {15,1}. (End update) Note 1: These are the only known classes for k = 1,3,5. Any other class not discussed here?Note 2: Also, in Third Powers, it was seen that [a,b,c] = [d,e,f] for k = 1,3 where abc = def has solns. Is the corresponding [a,b,c,d] = [e,f,g,h] for k = 1,3,5 where abcd = efgh also solvable? 5.5 Ten terms

The system of eqns, x1

k+ x2k+…+ xn

k = y1k+ y2

k+…+ ynk, for k =

1,2,3,4,5 is non-trivial only when n > 5, but two terms on one side can be equal to zero. Piezas Use the form, (a+c)k + (-a+c)k + (b+c)k + (-b+c)k + dk + ek = fk + gk + 0 + (2c)k + 0 + (2c)k

where {d,e,f,g} = {(cw+u)/w, (cw-u)/w, (cw+v)/w, (cw-v)/w}. Note that this is a symmetric ideal solution with pairs of terms having the common sum 2c. The system can be solved if {a,b,c} satisfy the two conditions, (a2+ab+b2-c2)(a2-ab+b2-c2)w-w3 = u2

(a2+ab+b2-c2)(a2-ab+b2-c2)w = v2

where w = a2+b2-2c2. The cases (a2-c2)(b2-c2)(a2+b2-2c2) = 0 must be avoided since it gives either trivial results or there is division by zero. An example of an appropriate soln is {a,b,c} = {-11, 13, -7} which gives {u,v} = {384, 2688}. For the special case a2+ab+b2-c2 = 0, this implies v = 0 and f = g. After some manipulation, we then get a nice soln to the form, (a+c)k + (-a+c)k + (b+c)k + (-b+c)k + (a+b+c)k + (-a-b+c)k = 2(c)k + 2(2c)k, for k = 1,2,3,4,5 where a2+ab+b2 = c2.

5.6 Twelve terms Piezas This is the fifth power counterpart to the third and fourth power identities found by this author. For k = 1,2,3,4,5, (a1x+v1y)k + (a2x-v2y)k + (a3x+v3y)k + (a4x-v3y)k + (a5x+v2y)k +

(a6x-v1y)k =

(a1x-v1y)k + (a2x+v2y)k + (a3x-v3y)k + (a4x+v3y)k + (a5x-v2y)k +

(a6x+v1y)k

and the sum-product identity, (a1x2+2v1xy+3a6y2)k + (a2x2-2v2xy+3a5y2)k + (a3x2+2v3xy+3a4y2)k

+ (a4x2-2v3xy+3a3y2)k + (a5x2+2v2xy+3a2y2)k + (a6x2-2v1xy

+3a1y2)k = (a1k+a2

k+a3k+a4

k+a5k+a6

k)(x2+3y2)k

where {a1, a2, a3, a4, a5, a6} = {a+c, b+c, -a-b+c, a+b+c, -b+c, -a+c} and {v1, v2, v3} = {a+2b, 2a+b, a-b}, for five arbitrary variables a,b,c,x,y. (Note that the vi are the same as the ones for fourth powers for Form 21.) These are symmetric ideal solns, with the common sum 2cx for the linear forms and 2c(x2+3y2) for the quadratic forms. They have the form, (a+n)k + (-a+n)k + (b+n)k + (-b+n)k + (c+n)k + (-c+n)k = (d+n)k + (-d+n)k + (e+n)k + (-e+n)k + (f+n)k + (-f+n)k

where appropriate pairs of terms have the common sum 2n and can be solved by any soln to am+bm+cm = dm+em+fm, m = 2,4, for any n. For

the special case when a+b = d+e, then it is also true that, (a+n)k + (a-n)k + (b+n)k + (b-n)k + (c+n)k + (-c+n)k = (d+n)k + (d-n)k + (e+n)k + (e-n)k + (f+n)k + (-f+n)k, k = 1,2,3,4,5 (Eq.M) when a+b = 2n. Expressed as {xi, yi}, in addition to having the common sum x1+x4 = x2+x3 = x5+x6 = a+b, (likewise for the yi since a+b = d+e), it also has the near-common difference of x1-x4 = x2-x3 = x5+x6 = 2n, and same for the yi. One simple soln to Eq.M is {a,b,c,d,e,f,n} = {5y, x-2y, x+2y, x, 3y, z, (x+3y)/2} based on the simple identity, (5y)k + (x-2y)k + (x+2y)k = xk + (3y)k + zk, for k = 2,4, if x2+24y2 = z2.

PART 9. Sum / Sums of Sixth Powers 6.1 Four terms As was already discussed in a previous section, the equation (√p+√q)6 + (√p-√q)6 = (√r+√s)6 + (√r-√s)6 has a polynomial soln. 6.2 Six Terms: Equal sum of three sixth powers, x1

6+x26+x3

6 = x46+x5

Duncan Moore has done an exhaustive search within a radius of 17,800. Out of about 400 solns, it turns out that 92% (!) are good for k = 2,6. Why that is the case, no one knows, though it may have to do

with the algebraic form x2+xy-y2. (See Bremner and Kuwata's work below.) Data results are here. Identities, whether in terms of a polynomial or an elliptic curve, are known only for multi-grades k = 2,6, or 1,2,6: 1. x1

k+x2k+x3

k = y1k+y2

k+y3k, for k = 2,6, and x1y1(x1

2-y12)+x2y2

(x22-y2

2)+x3y3(x32-y3

2) = 0. Additional side conditions include, a) 3x1+x2+x3 = y1+y2+3y3 b) 2x2 = y2+y3, 2x3 = y2-y3

k+x2k+x3

k = y1k+y2

k+y3k, for k = 1,2,6. Additional conditions

include, a) nx1+x2-x3 = ny1+y2-y3, for n = 12, 15, 21, etc. If you know of another class, pls send it. The second condition of the first class can be expressed as a set of three simultaneous eqns given below. As was mentioned, small solns to (k.3.3) for k = 4 or 6 also turn out to be valid for k = 2. The smallest for k = 6 is, 23k + 10k + 15k = 3k + 19k + 22k

which, in fact, is for k = 2,6. This deceptively simple-looking eqn has a lot of structure. If expressed as, {-23, 10, 15} = {3, 19, -22} labelled {a,b,c,d,e,f} respectively, then, a2+ad-d2 = -(b2-be-e2)b2+be-e2 = -(c2-cf-f2)c2+cf-f2 = -(a2-ad-d2)3a+b+c = d+e+3f

Not surprisingly, this is just the smallest instance of a parametric soln. Cubic and 4th powers have already been discussed and it was seen that a lot of identities involved equivalent forms of F3:= x2+xy

+y2. For 5th and 6th powers, it seems now it is the form F5:= x2+xy-

y2 that is implicit in some identities. For example, recall that, (√p+√q)5 + (√p-√q)5 = (√r+√s)5 + (√r-√s)5

{p,q,r,s} = {5vw2, -1+uw2, 5v, -(u+10v)+w3} where w = u2+10uv+5v2 and it takes only a small change of variables {u,v} = {x-2y, x+2y} to transform this to the form x2+xy-y2. Though F3 and F5 are similar-looking, they are quite different since the former has discriminant d = -3 and hence factors over the imaginary field √-3, while the latter has d = 5 and factors over the real field √5. It has been shown that solns to a2+ab+b2 = c2+cd+d2 also solve ak+bk+(a+b)k = ck+dk+(c+d)k for k = 2,4, and vice versa. For k = 2,6, one now needs a system of three equations, Theorem 1: (Bremner, Kuwata) If there are {a,b,c,d,e,f} such that, call this system S1, a2+ad-d2 + (b2-be-e2) = 0b2+be-e2 + (c2-cf-f2) = 0c2+cf -f2 + (a2-ad-d2) = 0 is true, then so is system S2, ak+bk+ck = dk+ek+fk, for k = 2,6 Proof: Simply add the expressions together to get the identically true

statement, (a2+ad-d2)k + (b2-be-e2)k + (b2+be-e2)k + (c2-cf-f2)k + (c2+cf-f2)k + (a2-ad-d2)k = 2(a2k+b2k+c2k-d2k-e2k-f2k), for k = 1,3 If the LHS is zero, then so is the RHS, which proves the theorem. (Masato Kuwata, Equal Sums of Sixth Powers and Quadratic Line Complexes, 2007. End proof.) (Note: This probably contributes to the reason why more than 90% of solns are multi-grade for k = 2,6 since being valid for one power means being valid for the other. Unfortunately, no similar expressions are yet known for 8th powers!) Corollary 1: (Piezas) Solutions to S1 imply the ff sums are squares, 4(a2+ad-d2) + 5b2 = (b+2e)2, 4(b2+be-e2) + 5c2 = (c+2f)2, 4(c2+cf-f2) + 5a2 = (a+2d)2

4(a2-ad-d2) + 5c2 = (c-2f)2, 4(b2-be-e2) + 5a2 = (a-2d)2, 4(c2-cf-f2) + 5b2 = (b-2e)2

In Unsolved Problems in Number Theory, R. Guy asked if the reverse was true: does S2 imply S1? This was answered in the negative by Delorme who gave one more condition, and Choudhry who gave a parametric example that solved S2 but not S1. Theorem 2: (Delorme) The system S2 implies S1 if it is also the case that, ad(a2-d2) + be(b2-e2) + cf(c2-f2) = 0 S. Brudno This is a simple example that solves both S1 and S2 ,

ak + bk + ck = dk + (b+c)k + (b-c)k, k = 2,6 Note that for k = 2, one must solve a2 = b2+c2+d2. The complete soln for k = 2,6 is then, {c,d} = {ax/(a2+9b2), 3bx/(a2+9b2)} where {a,b,x} satisfies the elliptic curve, (a2-b2)(a2+9b2) = x2

Labelled as xi, yi, the terms also satisfy x1y1(x1

2-y12) + x2y2(x2

2-y22)

+ x3y3(x32-y3

2) = 0. One small soln, among infinitely many, is {a,b} = {5/4, 1} which yields, 65k + 52k + 15k = 36k + 67k + 37k Piezas This is another simple example that solves both systems, (ad+b)k + (c-2d)k + (de+f)k = (de-f)k + (c+2d)k + (ad-b)k, k = 2,6 where {a,b,c,d,e,f} = {-4x+11y, y, 2(x-3y)(x-2y)+y2, x-y, 2x-3y, 2x-5y}, and {x,y} satisfies x2-6y2 = 1. It is quite interesting that this particular Pell equation appears in the context of 6th powers. However, since the eqn is homogeneous, one can just as well solve it in the rationals. This is also discussed several sections below. A. Bremner, Piezas

It turns out the complete soln of S1, S2 can be given in terms of a quartic polynomial that is to be made a square, hence can be treated as an elliptic curve. R. Guy in the same book mentions that Bremner gave a method that can find all parametric solns. I don’t have access to this paper yet but after some experimentation found a procedure. It suffices to use S2 and one of the eqns of S1. Using the general form,

(p+q)k + (r+s)k + (t+u)k = (t-u)k + (r-s)k + (p-q)k

for k = 2,6 and one of the others, say c2+cf-f2 = -(a2-ad-d2) with variables changed appropriately, the complete soln is then,

{q,u} = {-rs(p+2t)/w, rs(2p-t)/w}, where w = p2+t2 and {p,r,s,t} satisfy, ((p2-11pt-t2)s2 + w2)r2 = (p2+pt-t2+s2)w2

(Perhaps not surprisingly, the form F5:= x2+xy-y2 appears again.) Thus, the problem is to find {p,t,s} such that the expression, ((p2-11pt-t2)s2 + (p2+t2)2)(p2+pt-t2+s2) = y2

which is only a quartic in the variable s is a square, though some {p,t,s} are trivial. So, given an initial soln, this can be treated as an elliptic curve to generate more. For ex, let, {p,t,s} = {1, n, n-1} This is trivial with respect to the original sextic eqn S2 but, using the same {p,t}, we can find more rational points s on the curve with the next one (using Fermat’s method) non-trivial of deg-18. There are also small non-trivial solns,

{p,t,s} = {-n-1, (n-1)(n-2), 5+n2}{p,t,s} = {(-3+n)(2+n), 3+3n+2n2, 3+6n+n2} either one of which yields scaled versions of the 4th-deg Brudno-Delorme identity given below. Again this can be used to generate more rational points with the first one yielding a rational 18-deg. (Using the same {p,t}, there is another rational point s that is also only a quadratic, s = 3-2n+n2 for the first and s = 9+4n+n2 for the second, but these are trivial.) Other solns are, {p,t,s} = {2(1+n)(-1-n+n2)(-1+n+n2), (1+n)(1-n+n2)(1-5n+n2), (-1+n)(1+2n+7n2+2n3+n4)}{p,t,s} = {2(3+3n+2n2)(3+2n+2n2), (3+3n+2n2)(3-2n+4n2), (1+n)(9+4n2)} The first, after a small adjustment, yields Delorme’s 5th-deg identity given below, while the second yields a different 5th deg. Delorme in his paper gave identities of deg n for all 4 ≤ n ≤ 11, except n = 6,10. Q: Can anyone give an identity with polynomials of deg n = 6? Note: It can be shown that the system ak+bk+ck = dk+ek+fk, for k = 2,6 with, a2+nad-d2 = -(b2-nbe-e2)b2+nbe-e2 = -(c2-ncf-f2)c2+ncf-f2 = -(a2-nad-d2) has non-trivial solns only for n = ±1. Proof: Using the general form and the method described above, and elimination of appropriate terms, the final resultant eqn has either trivial factors, or a quadratic factor solvable over √(-1), or if (n+1)(n-1) = 0. Since the first two can be

disregarded then only the last applies, proving the assertion. S. Brudno, J. Delorme The 4th-deg Brudno-Delorme identity also has the side condition 3a+b+c = d+e+3f, call this S3. Let, ak+bk+ck = dk+ek+fk, for k = 2,6 where, {a,b,c} = {-n4-n3-5n2+8n+8, (n3+7n-2)(n+2), 3(3n2+2n+4)}{d,e,f} = {(n2-n+3)(n+2)2, -4n3-5n2-8n+8, -n4+n2+14n+4} (modified to reduced the size of the coefficients), then these satisfy S1, S2, S3. The smallest non-trivial example is n = -3 which, after removing common factors, yields {-23, 10, 15} = {3, 19, -22}, the example given earlier. Interestingly, it can be shown that this identity depends on the equation x2-6y2 = z2. This author found that some of the relations are enough to derive a version using: one eqn from S2 (at k = 2), two from S1, and S3. Let, (ad+b)k + (c-2d)k + (de+f)k = (de-f)k + (c+2d)k + (ad-b)k

and using this substitution also on the side conditions, one can derive the variables as, {a,b,c,d,e,f} = {-4x+11y, y, 2(x-3y)(x-2y)+y2, x-y, 2x-3y, 2x-5y}, where x2-6y2 = 1. One can easily solve for {x,y} in the integers as a Pell equation or, since the eqn is homogeneous, in the rationals. Q: Is it possible to solve S2, S3 without solving S1? J. Delorme

Delorme gave identities of deg n for all 4 ≤ n ≤ 11 , except n = 6,10, that solves S1, S2, one of which is the 5th deg,

{x1, x2, x3} = {3n5+8n4+9n3-4n2-9n-2, -2n5-n4+12n3+13n2+4n-1, -

n5-9n4-13n3-7n2-7n-3}{y1, y2, y3} = {2n5+9n4+4n3-9n2-8n-3, -3n5-7n4-7n3-13n2-9n-1, -

n5+4n4+13n3+12n2-n-2} Q: Any linear relations between the xi and yi? Update (7/14/09): Since the xi, yi are polynomials in n, it turns out the problem is equivalent to finding six unknown rational constants pi such that p1x1+p2x2+p3x3+p4y1+p5y2+p6y3 = m for some constant m, preferably m = 0, for all n. Expanding and collecting powers of n, set the coefficients, which are polynomials in the pi, equal to zero. For this particular identity, we can't find pi such that m = 0, but instead find m = 40 given by the linear relation, 5x1-12x2-7x3-12y1+5y2+7y3 = 40 (End note) Update (6/24/09): A. Choudhry The fact that solving the systems S1, S2 is reducible to an elliptic curve implies that one soln can lead to another. Choudhry ("On Equal Sums of Sixth Powers", 1994) has given an explicit construction for this. Given the system, xk+yk+zk = uk+ vk+ wk, k = 2,6 with, x2+xu-u2 = w2+wz-z2

y2+yv-v2 = u2+ux-x2

z2+zw-w2 = v2+vy-y2

Let {x,y,z,u,v,w} = {x1, x2, x3, y1, y2, y3} be a soln, then a new one is given by, {x,y,z,u,v,w} = {ap+x1q, bp+x2q, cp+x3q, dp+y1q, ep+y2q, y3q}, where {p,q} and {a,b,c,d,e} are, {p,q} = {-a(2x1-y1)-b(2x2+y2)+d(x1+2y1)-e(x2-2y2), a2+b2-d2-

e2-ad+be} {a,b,c,d,e} = {(r1

2+2r1r2)s2, (r32-2r3r4)s1, s1s2, -(r1

2+r22)s2,

(r32+r4

2)s1} {s1, s2} = {r2

2+r2r1-r12, r3

2+r3r4-r42}, and {r1, r2, r3, r4} = {x1-

y3, x3-y1, x2-y3, x3-y2}. Example: The initial soln {x1, x2, x3, y1, y2, y3} = {3, 22, -19, 23, 15, 10} gives {x,y,z,u,v,w} = {4513, -104, 4693, -2273, -5099, 3352} after removing the common factor 3*392002. (End note.) A. Choudhry There is a 3-parameter soln for k = 1,2,6 which entails solving an elliptic curve. This is a class that solves S2, but not S1. x1

k+x2k+x3

k = y1k+y2

{x1, x2, x3} = { 2(α+β)m+(α-β+t)n, -2αm+(α+β+t)n, -2βm-(α+β-t)n}{y1, y2, y3} = {-2(α+β)m+(α-β+t)n, 2αm+(α+β+t)n, 2βm-(α+β-t)n}

for k = 1,2 is true for all variables, though is not a complete soln. We can do the substitution {α+β, α-β, t} = {a,b,c} to get the slightly more symmetric, {x1, x2, x3} = { 2am+(b+c)n, -(a+b)m+(a+c)n, -(a-b)m-(a-c)n}{y1, y2, y3} = {-2am+(b+c)n, (a+b)m+(a+c)n, (a-b)m-(a-c)n} When expanded for k = 6 this has the condition, (Poly1)m2+(Poly2)n2 = 0 where, Poly1:= b(11a2+b2)+5(3a2+b2)c; Poly2:= (a2+b2)(b+5c)+10(bc2+c3) linear and cubic in c, respectively. The problem then is: Given constants a,b, find c such that, y2 = -(Poly1)(Poly2) avoiding the cases c = -b/3 and y = Poly1 = 0 since these give trivial results. For a non-trivial ex., let {α,β} = {1, 7}, hence {a,b} = {8,-6} giving, y2 = -6(-74+19c)(-60+50c-6c2+c3) one soln being c = 38/23 with this elliptic curve having an infinite number of rational points. (Choudhry gave more than twenty {a,b} with another as {11,-9}.) Piezas While Choudhry did not give the complete soln of k = 1,2,6, it can be

shown that the general soln entails making a certain quartic polynomial into a square (just like for the case k = 2,6 plus S2). We can show this in two ways. First method: Using a small variant of the general form, (p+q)k + (r+s)k + (t+u)k = (-t+u)k + (-r+s)k + (-p+q)k

and letting r = np for some rational constant n, one can completely solve k = 1,2 with, {t, u} = {-p(n+1), (q+ns)/(n+1)} With these values, k = 6 becomes a polynomial of form (Poly1)p2+(Poly2) = 0, where Poly1 is linear and Poly2 is cubic in {q,s}, so their product is a quartic and the problem is reduced to finding, y2 = -(Poly1)(Poly2) a situation similar to Choudhry’s, though the tricky part is finding appropriate rational n which may be the ratio of relatively large integers. Second method: We simply use the old form L1 again. Expanding for k = 2, the non-trivial condition is b+c = 0, hence we modify it to, (a+bp+q)k + (b-bp+q)k + (-b+ap+q)k = (a-bp+q)k + (b+ap+q)k + (-b+bp+q)k

which solves k = 1,2. Expanding for k = 6 results in a quadratic in b of form (Poly1)b2+(Poly2) = 0, where Poly1 is linear and Poly2 is cubic in {p,q}. The objective then is to make its discriminant D a square,

y2 = -(Poly1)(Poly2) similar to the first method, but now the expressions are simple enough to write down, Poly1:= a(1+p)(1+p2) + 5(1-p+p2)qPoly2:= a3(1+p)(1+p2) + 5a2(1+p+p2)q + 10a(1+p)q2 + 10q3

Thus any solution to k = 1,2,6 must satisfy D = y2. For example, using a result of Choudhry’s, 43k + (-372)k + 371k = 307k + (-405)k + 140k

for k = 1,2,6, we equate its first four terms with that of the modified L1 to get, {a,b,p,q} = {291, -388, 33/97, -116} Using the values for a,p gives the elliptic curve in q, y2 = -(818844+7297q)(7369596+123291q+780q2+2q3) where a square numerical factor has been removed. A rational point, of course, is Choudhry's q = -116 (with another one as q = -130 though this gives a trivial result). From these, more points q can then be computed though typically they are rational numbers with many digits. Note: Incidentally, if we are to add more constraints to this system, then there are no non-trivial rational solns to k = 1,2,6 plus any of the eqns of S1, for ex. c2+cf-f2 = -(a2-ad-d2). These four eqns are enough to linearly derive a soln. Using the general form, (p+q)k + (r+s)k + (t+u)k = (t-u)k + (r-s)k + (p-q)k

then, {q,r,u} = {(p+2t)s/v, -(p2+t2)/v, -(2p-t)s/v}, where v = p-3t and {p,s,t} satisfies p3-2p2t+pt2+8t3-(p-8t)s2 = 0. One must then solve the elliptic curve, (p3-2p2t+pt2+8t3)(p-8t) = y2

and as far as I checked this has only trivial solns. However, the curve, (p3-2p2t+pt2+8t3)(p-8t) = -y2

does have non-trivial ones, like {p,t,y} = {24, 13, 1280} and many more thus giving an imaginary value to s. So this system is solvable in terms of Gaussian rationals.

Piezas We can also consider the more general system, na+b+c = nd+e+f (eq.1)ak+bk+ck = dk+ek+fk (eq.2) with k = 2,6 for any n. To solve this, we can use solns to n = 1 and exploit the fact that sign changes to the terms ai do not affect eq.2. The form F1 that gave the complete soln for the case k = 2,4 is also useful here, (p+qu)k + (r+su)k + (t+u)k = (p-qu)k + (r-su)k + (t-u)k

hence {a,b,c,d,e,f} = {p+qu, r+su, t+u, p-qu, r-su, t-u} and we will

solve the system, a+b-c = d+e-fna+b+c = nd+e+fa2+b2+c2 = d2+e2+f2

giving {t,s,q} = {-pq-rs, -1-nq, 2/(1-n)}. Using these, F1 at k = 6 reduces to the form, (Poly1)u2+(Poly2) = 0 where Poly1 is linear in {p,r} and Poly2 is cubic given by, Poly1:= (-3+n)(5-2n+n2)p + 4n(1+n2)rPoly2:= (-3+n)(5-2n+n2)p3 + 2(5+11n-5n2+n3)p2r + (-5-7n-15n2+3n3)pr2 + 4n(1+n2)r3 Thus, again, one should solve, y2 = – (Poly1)(Poly2) a situation similar to the previous section. Trivial solns should be avoided such as y = Poly1 = 0 or r = -p(n-3)/(2n) with the latter, using Fermat’s method to compute a second rational point, just yielding the former. (Note that the constraint excludes n = 3, though this in fact is solvable using another method.) There are other non-trivial points though. For ex, let n = 12, then, y2 = -75(75p+464r)(225p3+458p2r+587pr2+1392r3) and with {p,r} = {-104, 17} this gives, (-127)k + 496k + 393k = 23k + (-479)k + (-432)k

From this initial soln, subsequent ones can then be found proving that the system, a+b-c = d+e-f12a+b+c = 12d+e+fak+bk+ck = dk+ek+fk

for k = 2,6 has an infinite number of solns. Later, it will be shown that this approach can be extended to solve the system, ma+b+c+d = me+f+g+hak+bk+ck+dk = ek+fk+gk+hk

for any m for k = 2,4,6. Note 1: For what small integer n does this method provide non-trivial solns? After some experimentation, I found n = 12, 15, 21, 30, 33, 135 for n < 150 though there could be skips since I was focusing only on n = 3m and assuming relatively small initial solns. Other than n=1, is there an n ≠ 3m with an infinite number of solns? Note 2: Excluding the case n=1 of which the complete soln is known, are there other general methods for other n? For example, the one discussed in this section cannot deal with n=3, though there is a polynomial identity for this. Note 3: Given a parametric soln to k = 1,2,6, it seems one can modify this to solve na+b+c = nd+e+f. For ex, using Choudhry’s soln to x1

+x2k+x3

k = y1k+y2

k+y3k, with {xi, yi} as,

{x1, x2, x3} = { 2(α+β)m+(α-β+t), -2αm+(α+β+t), -2βm-(α+β-t)}{y1, y2, y3} = {-2(α+β)m+(α-β+t), 2αm+(α+β+t), 2βm-(α+β-t)}

discussed previously which is already true for k = 1,2, a linear combination of terms involving only the constants α,β should be found. Three possibilities are, nx1+x2-x3 = ny1+y2-y3, x1+nx2-x3 = y1+ny2-y3, x1-x2+nx3 = y1-y2+ny3, where n = (α-β)/(α+β), n = (α+2β)/α, and n = (α+2β)/β, respectively. If either α,β = 1, then the last two yield integer n. For α = 1, Choudhry found β = {7, 10, 16, 67} so using the second formula this gives n = {15, 21, 33, 135}, all n = 3m and consistent with the odd n found by this author. Note 4: If there is an initial soln for a certain n, is there always an infinite number for this n? Note 5: A small variation of the method and which mostly yields the same results is to use different terms of F1 which are to be multiplied by n and have their signs changed. Again let {a,b,c,d,e,f} = {p+qu, r+su, t+u, p-qu, r-su, t-u} but to solve the system, -a+b+c = -d+e+fa+b+nc = d+e+nfak+bk+ck = dk+ek+fk, k = 2 giving, {t,s,q} = {-pq-rs, -n-q, (1-n)/2} and F1 at k = 6 again reduces to the form, (Poly11)u2+(Poly12) = 0 though now defined diffently,

Poly11:= (-3+n)(5-2n+n2)p + (3+n)(5+2n+n2)rPoly12:= (-3+n)(5-2n+n2)p3 + (25-7n-5n2+3n3)p2r + (-25-7n+5n2+3n3)pr2 + (3+n)(5+2n+n2)r3 As before, solve y2 = – (Poly11)(Poly12) with trivial soln {p,r} = {-(n+3), n-3}. For ex, for comparison, again let n = 12, y2 = – 75(75p+173r)(225p3+881p2r+1159pr2+519r3) with non-trivial soln {p,r} = {-39, 17} which has the same r and yields the same terms as in the first curve. To compare the two methods, for n = 12, 15, 21, 30, 33, 135, non-trivial points {p,r} are, First method: {-104, 17}, {-183, 34}, {-53, 17}, {-24, 13}, {-137, 39}, {-229, 106}Second method: {-39, 17}, {-65, 34}, {-24, 17}, {?, ?}, {-50, 39}, {-111, 106} Notice that the value for r is the same. If the two methods give the same results, why is there apparently no small non-trivial soln for n = 30 using the latter one? Furthermore, Poly2 and Poly12 as cubic polynomials have different but related discriminants Di being, D1 = (5-2n+n2)(5+2n+n2)

D2 = (5-2n+n2)(1+n2) respectively. Do the two methods then for the same n always give the same results? Q: Any other forms for equal sums of sixth powers with six terms? 6.3 Seven terms

It is unknown if the unbalanced multi-grade eqn, x1

k+x2k+x3

k = x4k+x5

k+x6k+x7

k, for k = 2,4,6 has non-trivial solns. Q: Anyone can provide one, preferably parametric? (Update, 2/22/10): In response to a post in sci.math, Giovanni Resta made a search and found that there are no solns with terms < 845. James Waldby would later extend the search radius to < 1280 with still no soln. However, there are a few that are good only for k = 2,6, namely, [73, 58, 41] = [70, 65, 32, 15][85, 74, 61] = [87, 71, 56, 26][218, 167, 29] = [224, 107, 102, 65][351, 265, 221] = [336, 309, 169, 73][493, 335, 65] = [497, 297, 155, 16] Some exhibit interesting side relations between its terms, such as the smallest one with x1-x2 = x7, x1-x3 = x6, though with appropriate substitutions, they are still not enough to reduce it to a parametric form. (End update.) 6.4 Eight terms Some general identities analogous to the quadratic-quartic system Q1 are: Theorem (Piezas): Let ak+bk+ck+dk = ek+fk+gk+hk, for k = 2,4,6.

Define {x,y,z} = {a2+b2+c2+d2, a4+b4+c4+d4, a6+b6+c6+d6}. Then, a8+b8+c8+d8-e8-f8-g8-h8 = -4(a2-e2)(b2-e2)(c2-e2)(d2-e2) = -4(a2b2c2d2-e2f2g2h2) 4(a10+b10+c10+d10-e10-f10-g10-h10) = 5(a8+b8+c8+d8-e8-f8-g8-h8)(x) 8(a12+b12+c12+d12-e12-f12-g12-h12) = 6(a8+b8+c8+d8-e8-f8-g8-h8)(x2+y) 24(a14+b14+c14+d14-e14-f14-g14-h14) = 7(a8+b8+c8+d8-e8-f8-g8-h8)(x3+3xy+2z) The first one is true regardless of whether e is replaced by any of {f,g,h}. The next two can be combined together to an 8-12-10 Identity. Define n as, a4+b4+c4+d4 = n(a2+b2+c2+d2)2 then, 25(a8+b8+c8+d8-e8-f8-g8-h8)(a12+b12+c12+d12-e12-f12-g12-h12) = 12(n+1)(a10+b10+c10+d10-e10-f10-g10-h10)2 This is analogous to Ramanujan’s 6-10-8 Identity though since Ramanujan was able to use a linear relationship between the terms ai, namely a+b = ±c, to define n as the constant ½, his result was then much neater. However, work by Euler shows that, a4+b4+c4+d4 = n(a2+b2+c2+d2)2 does have a parametric soln for n = ½, given in the section on Fourth

Powers, though it is much more complicated than for the case d = 0. Chernick (and Escott, independently) There is the aesthetically-pleasing identity, (-u+7w)k + (u-2v+w)k + (-3u-w)k + (3u+2v+w)k =(u+7w)k + (-u+2v+w)k + (3u-w)k + (-3u-2v+w)k

for k = 1,2,4,6, if u2+uv+v2 = 7w2

with signs changed by this author so it would be valid for k = 1. This was derived using the equally simple quartic identity, uk + vk + (u+v)k = wk + (2w)k + (3w)k for k = 2,4 with the same condition. Since the eqn is homogeneous, one can do the substitution {u,v,w} = {y+z, y-z, x} and the condition becomes, 7x2-3y2 = z2

where the binary quadratic form has square-free discriminant D = 21. This is the only known quadratic identity (in contrast to the system k = 1,3,5,7 which has three known D). So how did Chernick and Escott find this? One technique is to keep applying the transformation theorems discussed in Chap II to gradually increase the range of the exponents and, if the chosen starting identity has special properties, the same terms may appear on both sides and cancel out, keeping down the number of terms. For example, using the quartic identity and Theorem 5 with T = 0, this becomes, [u, v, u+v, -u, -v, -u-v] = [w, 2w, 3w, -w, -2w, -3w], k = 1,2,..5

If the Tarry-Escott theorem is then used, generally this doubles the number of terms. But for this particular instance, using T = w, the number of terms on either side of the equation instead of doubling, increases only up to eight but the range is now k = 1,2,3,4,5,6. A second application of the same theorem, with T = u, does not increase the number of terms at all and produces a symmetric ideal soln for k = 1,2,..7. Since this has appropriate pairs with a common sum, using Theorem 5 in reverse it is easy to recover a (k.4.4) for k = 2,4,6 and which is the identity given at the start of this section. Using a similar method, Tarry earlier found the first ideal soln (symmetric) for k =1,2,..7. Starting with [1, 9] = [4, 6] which is only for k = 1, he used five successive applications of the Tarry-Escott theorem with T = 2, 1, 7, 8, 13, respectively, to get, [1, 5, 10, 16, 27, 28, 38, 39] = [2, 3, 13, 14, 25, 31, 36, 40], k = 1,2,…6 Note that this is a symmetric soln with the common sum 41, implying this is dependent on an identity valid for k = 1,3,5. Then using a last transformation with T = 11 found the ideal soln, [1, 5, 10, 24, 28, 42, 47, 51] = [2, 3, 12, 21, 31, 40, 49, 50], k = 1,2,…7 It turns out this can be generalized with solns depending on the nice eqn x2+7y2 = 8z2. Piezas To simplify matters we will start with the sixth deg. Rearranging terms, this is [5, 16, 27, 38, 28, 39, 1, 10] = [3, 14, 25, 36, 2, 13, 40, 31], k = 1,2,…6

which makes it easy to see that the first six terms (with a skip after the fourth) of each side involve an arithmetic progression, differing by 11. This is then an example of a “special” property. To recall, by Frolov’s theorem we can always set a1 = 0 so, to generalize Tarry's result, we are to solve the system, [0, a, 2a, 3a, a+c, 2a+c, x1, x2] = [b, a+b, 2a+b, 3a+b, a+b-c, 2a+b-c, y1, y2], k = 1,2,…6 with unknowns {a,b,c,x1,x2,y1,y2}. Using the Tarry-Escott theorem, with T = a the properties of the system ensure there is no increase in the number of terms and this becomes, [0, a+c, x1, x2, 4a+b, 3a+b-c, a+y1, a+y2] = [b, a+b-c, y1, y2, 4a, 3a+c, a+x1, a+x2] now for k = 1,2,…7. But since we wish this to be symmetric, pairs must have a common sum, 0 + (4a+b) = (a+c) + (3a+b-c) = (x1) + (a+y1) = (x2) + (a+y2)b + (4a) = (a+b-c) + (3a+c) = (y1) + (a+x1) = (y2) + (a+x2) where the first two pairs have the sum 4a+b while the last two imply x1+y1 = x2+y2. (In Tarry’s example, this is 1+40 = 10+31.) This can be satisfied by letting {x1,x2,y1,y2} = {t+u, t+v, t-u, t-v} and, if these are to have the same common sum as the others, then it must be the case that, (t+u) + (a+(t-u)) = 4a+b so t = (3a+b)/2. We have reduced our unknowns to just five:

{a,b,c,u,v}. (In fact, it can be assumed c = 1 without loss of generality.) Using Theorem 5 in reverse, if we subtract (4a+b)/2 from the first four terms on each side, namely {0, a+c, x1, x2} and {b, a+b-c, y1, y2}, this gives us, (4a-b)k + (2a-b+2c)k + (a+2u)k + (a+2v)k = (4a+b)k + (2a+b-2c)k + (a-2u)k + (a-2v)k

for k = 2,4,6. To simplify, let {a,b,c} = {2p, -2q, -q+r} and rearranging terms, we get, (p+u)k + (p+v)k + (2p+r)k + (4p+q)k = (p-u)k + (p-v)k + (2p-r)k + (4p-q)k, (eq.1) Using this simpler system, we can then solve for the unknowns. This is given by, {p,q,r,u,v} = {2x2+x+1, x2+6x+1, -8(x2+2x), 6x2+4x-2+y, 6x2+4x-2-y} where x,y must satisfy, y2 = 15x4+104x3+223x2+54x+4 with integral solns x = -3,-2,0,1,5 which are trivial though a non-trivial one is x = -1/7. From these one can then compute more rational points. This was derived by expanding eq.1 at k = 2,4,6 and eliminating {u,v} to get the first condition 8p2-8q2-5qr-r2 = 0. Letting {p,q,r} = {z, 2y, x-5y}, this becomes, x2+7y2 = 8z2

with initial soln {x,y,z}= {1,1,1} and is easily solved parametrically.

By recovering {u,v} there is a second quadratic condition to fulfill hence the elliptic curve. As was alluded to earlier, this symmetric sixth degree implied a fifth degree identity. Thus, in summary, if, (3p+q)k + (p+q)k + (-p+r)k + (p+r)k = (3p-q)k + (p-q)k + (-u)k + (-v)k, for k = 1,3,5, then, (p-u)k + (p-v)k + (2p-r)k + (-4p+q)k = (-p-u)k + (-p-v)k + (-2p-r)k + (4p+q)k, for k = 1,2,4,6. Note that the second system satisfies x1-y1 = x2-y2 = (x3-y3)/2 = -(x4-y4)/4 and a minor change in signs made it valid for k = 1 as well. From this one can find an ideal seventh degree soln such as the beautifully concise, qk + (2p+r)k + (3p+u)k + (3p+v)k + (5p-u)k + (5p-v)k + (6p-r)k + (8p-q)k =(-q)k + (2p-r)k + (3p-u)k + (3p-v)k + (5p+u)k + (5p+v)k + (6p+r)k + (8p+q)k

for k = 1,2,…7 where {p,q,r,u,v} are as defined above. (Update, 8/4/09): Alternatively, the system can be expressed in a more symmetric form, Theorem: If (a-b-c)k + (a-b+c)k + (b+3c)k + (b+c)k = (a+b-d)k + (a+b+d)k + (-b+3c)k + (-b+c)k, for k = 1,3,5, then, (a+b+c+d)k + (a+b+c-d)k + (-a+b+2c)k + (b-4c)k = (a+b-c+d)k + (a+b-c-d)k + (-a+b-2c)k + (b+4c)k, for k = 1,2,4,6,

where {a,b,c,d} satisfy a2+3ab+4b2 = 8c2 and a2-5ab+36b2 = 8d2. Note how in the second system, the LHS difers from the RHS only in the sign of the variable c. This identity is easily proven by solving for

{c,d} in terms of {a,b} and subsituting into the two systems, though the ratio a/b = {-4, -2, -1, 1, 4} should be avoided as they are trivial, with a small non-trivial soln as {a,b} = {14,1}. (End note.) Note: Eq.1 can be generalized to the form, (an+p)k + (bn+q)k + (n+r)k + (n+s)k = (an-p)k + (bn-q)k + (n-r)k + (n-s)k for some constants a,b. Another soln can be found when {a,b} = {1,7}. Deriving the other variables we get {r,s} = {-p-4q, -3q} where p,q,n satisfy p2+4pq+7q2 = 28n2. Let {p,q,n} = {2x-y, y, z} and this becomes, x2+xy+y2 = 7z2

and an inspection reveals this is just the Chernick-Escott identity in disguise. I am not aware of another a,b such that the form has a non-trivial soln though there may be more. The next example also involves a quartic elliptic curve.

(Update, 1/25/10): Piezas In the previous section, it was shown how a k = 1,3,5 can lead to a k = 2,4,6. This can be brought higher. Given a (k.4.4) for k = 2,4,6, if its terms {xi , yi} are such that x1-x2 = y1-y2 = y2-y3 , then it can lead to a (k.5.5) for k = 1,3,5,7. This is exemplified by the infinite family, (3a+2b+c)k + (3a+2b-c)k + (-2a-b+d)k + (-2a-b-d)k = (3a+b+2c)k + (3a+b)k + (3a+b-2c)k + (-7a-b)k, for k = 1,2,4,6 which leads to the higher system, [3a+b+3c, 3a+2b-2c, -7a-b+c, -2a-b-c-d, -2a-b-c+d]k =[3a+b-3c, 3a+2b+2c, -7a-b-c, -2a-b+c-d, -2a-b+c+d]k, for k = 1,2,3,5,7 where, for both, 10a2-b2 = c2, and 55a2-6b2 = d2, with trivial ratios a/b = {1, 1/3, -5/13}, and small non-trivial soln {a,b} = {5, 13}. It is tempting to speculate that there may be higher-order version of this for (k.5.5), k = 2,4,6,8, with the form, [pa+qb+c, pa+qb-c, ra+sb+c, ra+sb-c, ta+wb]k =[ua+vb+2c, ua+vb, ua+vb-2c, xa+yb+d, xa+yb-d]k (System 3) where m1a2+m2b2 = c2, and m3a2+m4b2 = d2, for constants {p,q,r,s,t,u,v,w,x,y} and {m1, m2, m3, m4}. System 3 has x1-x2 = x3-x4 = y1-y2 = y2-y3 and would automatically lead to a (k.6.6) for k = 1,3,5,7,9. While there is numerical example with the right constraints, it is unknown if it belongs to a family with an infinite number of rational solns. (End update.) Crussol Given the system of eqns for indefinite k, x1

k + x2k + x3

k + x4k = y1

k + y2k + y3

k + y4k

Chernick used certain expressions to solve the case k = 1,2,3 while Lander focused on k = 1,3,5, (which Choudhry showed could be extended to k = 1,3,7). But prior to this, Crussol took the case k = 1,2,4,6 and used the version, (a+x)k + (a-x)k + (b+y)k + (b-y)k = (a+z)k + (a-z)k + (b+t)k + (b-t)k (eq.0) which in one sense is a simpler form and gave a partial soln. (Later we will give the complete one.) Crussol established that for k = 2,4,6, one should satisfy the beautifully simple conditions, x2+y2 = z2+t2

5(a2+b2) = x2+y2

3(a2-b2) = y2-z2

The terms of eq.0, expressed as {xi, yi}, obviously is identically true for the side conditions given by Lander, namely x1+x2 = y1+y2, x3+x4 = y3+y4 which makes it already valid for k = 1. However, if also true for k = 2,4,6, then it satisfies the following additional relations, x1x2+x3x4 = y1y2+y3y4 x1

2+3x1x2+x22+x3

2+3x3x4+x42 = 0

2+3y1y2+y22+y3

2+3y3y4+y42 = 0

To solve the first condition, Crussol used {x,y,z,t} = {de+cf, ce-df, de-cf, ce+df} which reduced the other two to, 5(a2+b2) = (c2+d2)(e2+f2) (eq.1)3(a2-b2) = (c2-d2)(e2-f2) (eq.2) Note that eq.1 is of the form m(a2+b2) = (c2+d2)(e2+f2) for some constant m and can be solved, though not completely (see section on Sums of Two Squares), as, {a,b,c,d} = {fu-ev, eu+fv, 2u+v, -u+2v} for free variables {e,f,u,v}. Using these expressions on (eq.2) results in a quadratic in e (or f) and solving for this gives the last two unknowns, {e,f} = {3u2+4uv-3v2, 3uv±w} where {u,v,w} satisfy the quartic elliptic curve, (3u2+4uv-3v2)2 + (3uv)2 = w2

two small solns of which are {u,v} = {5,3} or {9,16}, etc. Note 1: For comparison, when dealing with odd powers, one can move terms around and it makes no difference to use the related system, (a+x)k + (a-x)k + (a+z)k + (a-z)k = (b+y)k + (b-y)k + (b+t)k + (b-t)k

where now it has x1+x2 = x3+x4, y1+y2 = y3+y4. But for even powers, this only has trivial solns if simultaneous true for k = 2,4,6. Note 2: The basic form of eq.0 can also be used for just k = 1,2,6. For example, Shuwen found, 31k + 126k + 62k + 107k = 38k + 119k + 51k + 118k, which has 31+126 = 38+119, 62+107 = 51+118 and apparently is a particular instance of a parametric identity though I haven’t been able to derive it yet. (Anyone can provide it?) Whether Crussol’s form can be extended to eighth powers for k = 1,8 or 1,2,8 still remains to be seen. (The only known numerical soln to (8.4.4) found by Kuosa in 2006 is not of this form.) Note 3: Using a different approach, the complete soln to k = 1,2,4,6 with x1+x2 = y1+y2, x3+x4 = y3+y4 will be discussed in the subsequent sections. (Update, 1/18/10): Piezas Crussol’s soln can be given in a more aesthetic manner as two quadratics to be made squares. Let, [c+be+af, c-be-af, d+ae-bf, d-ae+bf]k =[c+be-af, c-be+af, d+ae+bf, d-ae-bf]k

for k = 1,2,4,6, where, ma2+nb2 = c2 (eq.1)na2+mb2 = d2 (eq.2) with {m,n} = {(4e2-f2)/15, (4f2-e2)/15} for some constants {e,f}. These simultaneous equations, eq.1 and eq.2, define an elliptic curve. For ex, let {e,f} = {13, 1} and we get, (a+13b+c)k + (-a-13b+c)k + (13a-b+d)k + (-13a+b+d)k =(-a+13b+c)k + (a-13b+c)k + (13a+b+d)k + (-13a-b+d)k

where 45a2-11b2 = c2, and -11a2+45b2 = d2. These conditions are the same as the ones for the only known [k.6.6] identity for k = 1,2,4,6,8,10 discussed in Tenth Powers. More generally, given eq.1 and eq.2, which lead to an elliptic curve E(m,n), it is quite easy to find a k = 1,2,4,6 which uses this curve since the constants {e,f} can be derived from {m,n} as, {4m+n, m+4n} = {e2, f2} hence, if the two expressions themselves are squares, then it can be used for a k = 1,2,4,6. (End update.) (Update, 2/1/10) This update covers a family that includes 6th, 8th, and 10th power multi-grades. Whether it can be extended for 12th powers with a minimal number of terms remains to be seen. Piezas (6th powers) Given the general form, [xy+ax+3by-c, xy-ax-3by-c, xy-3bx+ay+c, xy+3bx-ay+c]k =[xy+ay+3bx-c, xy-ay-3bx-c, xy-3by+ax+c, xy+3by-ax+c]k

Notice how {x,y} just swap places or alternatively, “a” in one side is replaced by “3b” in the other side. With terms expressed as {pi, qi}, this form identically is true for the following constraints, p1+p2 = q1+q2,p3+p4 = q3+q4,

p1k+p2

k+p3k+p4

k = q1k+q2

k+q3k+q4

k, k = 1,2 We can make this valid for k = 1,2,4,6. For k = 4, let c = ab. For k = 6 and some constants {a,b}, then {x,y} must satisfy the simple eqn, 10a2b2-(a2+9b2)(x2+y2)+10x2y2 = 0 This is an elliptic “curve” in disguise. Let {x,y} = {u, v/(10u2-a2-9b2)} and one is to solve, (a2+9b2-10u2)(10a2b2-(a2+9b2)u2) = v2

which is a quartic polynomial in u to be made a square. Trivial ratios are a/b = {0,1,3,9}. It can be proven that for any rational {a,b} that do not have a trivial ratio, one can always find an infinite number of rational {x,y}. Proof: A polynomial soln can be given starting from the trivial point u = 3b. This yields the non-trivial, u = 9b(a2+15b2)/(5a2-117b2) from which one can then successively compute an infinite sequence of rational polynomials. Chris Smyth (8th powers) This is a variation of the Letac-Sinha 8th power multi-grade. For k = 1,2,4,6,8, [xy+ax+ay-d, -(xy-ax-ay-d), xy-bx+by+d, -(xy+bx-by+d), cx+cy]k =[xy+bx+by-d, -(xy-bx-by-d), xy-ax+ay+d, -(xy+ax-ay+d), cx-cy]k

where {a,b,c,d} = {1, 3, 4, 11} and {x,y} satisfies the eqn, x2y2-13(x2+y2)+112 = 0 One can note its affinity with the 6th power multi-grade and see also in the general form how “a” in one side has just been replaced with “b” in the other side. With terms expressed as {pi,qi}, this obeys the constraints, p1-p2 = q1-q2p3-p4 = q3-q4 If we set that, p1

k+p2k+p3

k+p4k+p5

k = q1k+q2

k+q3k+q4

k+q5k, k = 1,2

then one must have 2ak-2bk+ck = 0, for k = 1,2, and the consequence is that, p1

k+p2k+p5

k = q1k+q2

p3k+p4

k = q3k+q4

for k = 1,2 which, naturally enough, is the same set of constraints obeyed by the Letac-Sinha identity. Expanding the system for k = 4,6,8, it will be seen that the only non-trivial soln in the rationals is {a,b,c,d} = {m, 3m, 4m, 11m2} where one can set m = 1 without loss of generality and {x,y} satisfies the eqn, x2y2-13(x2+y2)+112 = 0 Let {x,y} = {u, v/(u2-13)} and this is the elliptic curve, (u2-13)(13u2-112) = v2

Smyth’s form for (k.5.5) can be generalized by adding two terms on each side to create a (k.7.7). A possible addition, given in italics, may be, [xy+ax+ay-c, -(xy-ax-ay-c), xy-bx+by+c, -(xy+bx-by+c), dx+dy, ex+fy, fx-ey]k =[xy+by+bx-c, -(xy-by-bx-c), xy-ay+ax+c, -(xy+ay-ax+c), dx-dy, ey+fx, ex-fy]k which is the same pair successfully added to the (k.4.4). For non-zero {e,f}, whether this has a non-trivial soln up to k = 8, 10, or even 12 is unknown. Choudhry, Wroblewski (10th powers) The 6th power (k.4.4) identity can be generalized to the form (k.6.6) by adding the italized pair of terms, [-(axy+bx+cy-d), axy-bx-cy-d, axy-cx+by+d, -(axy+cx-by+d), ex+fy, fx-ey]k =[-(axy+by+cx-d), axy-by-cx-d, axy-cy+bx+d, -(axy+cy-bx+d), ey+fx, ex-fy]k where again, {x,y} merely swap places. (Since the eqn is homogeneous, one may assume the leading coefficient of the terms as a = 1 without loss of generality.) This obeys the same basic constraints (up to sign changes), p1-p2 = q1-q2,p3-p4 = q3-q4,

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

If we assume, p1+p2+p5 = q1+q2+q5, p3+p4+p6 = q3+q4+q6, then it must be the case that 2(b-c) = e-f (eq.1). Choudhry and Wroblewski found for k = 2,4,6,8,10, the equivalent non-trivial solns, {a,b,c,d,e,f} = {1, 1, 2, 14, 3, 5}, where 2x2y2-17(x2+y2)+392 = 0{a,b,c,d,e,f} = {2, 1, 2, 7, 3, 5}, where 8x2y2-17(x2+y2)+98 = 0 Disregarding eq.1, two other distinct non-trivial solns found (by Piezas), though only up to k = 2,4,6,8, are, {a,b,c,d,e,f} = {5, 3, 7, 3, 2, 10}, where 125x2y2-53(x2+y2)+45 = 0{a,b,c,d,e,f} = {5, 9, 11, 11, 10, 12}, where 125x2y2-221(x2+y2)+605 = 0 though this author is not certain if these are the same 8th power multi-grades found by Wroblewski. (End update.)

Piezas To find an infinite family of polynomial solns to k = 1,2,4,6, we will use the form F2, (a+bh)k + (c+dh)k + (e+fh)k + (g+h)k = (a-bh)k + (c-dh)k + (e-fh)k + (g-h)k

the big brother of F1 (which was for six terms with k = 2,4 and k = 1,2,6). To make F2 valid for k = 1,2, set g = -(ab+cd+ef) and f = -(1+b+d). Expanding for k = 4,6, we get, (Poly11)h2 + (Poly12) = 0 (eq.1)(Poly21)h4 + (Poly22)h2 + (Poly23) = 0 (eq.2) respectively, where the Polyi are expressions in the variables vi = {a,b,c,d,e}. There are two ways to solve these two: first, to make Poly11 = Poly12 = 0, which will also cause eq.2 to factor into two quadratics, or second, to find vi so eq.2 will factor into the same quadratic as eq.1. To find a common root between eq.1 and 2, it is a simple matter to eliminate the variable h using Mathematica’s Resultant[ ] function. (And since h is conveniently in even powers, it can be set h = √v to shorten the calculation.) The resultant is a factorable 6th deg in e and, surprisingly, this has three non-trivial linear factors given by, (a+ab-c+cd-be-de) (-a+ab+c+cd-be-de) (a+ab+c+cd-2e-be-de) = 0 plus an irreducible cubic factor. (One can get more solns if a linear root of the cubic can be found such as for special cases like d = -b when the above three become trivial but the cubic now has a single non-trivial linear factor.) In the general case though, solving for one appropriate variable of either of the three (b,d,e, respectively) and substituting into Poly11 = 0, one can solve for a second variable. These two can then set Poly11 = Poly12 = 0, and cause eq.2 to factor: two linear and one quadratic in h, with only the latter being non-trivial. Sparing the reader the rest of the algebra, given F2, (a+bh)k + (c+dh)k + (e+fh)k + (g+h)k = (a-bh)k + (c-dh)k + (e-fh)k + (g-h)k

where g = -(ab+cd+ef) and f = -(1+b+d), then for k = 1,2,4,6, if, A. Linear factors: First factor: b = (1+d)(a+2e)/(a-e), c = (2+d)(a+e)/(1-d), where {a,d,e} satisfy (Poly1)(-1+d)2h2-(Poly2)(a-e)2 = 0 Poly1:= a2d-(9+20d+9d2)ae+de2; Poly2:= 9a2d-(1-20d+d2)ae+9de2

For arbitrary {a,d,e,h}, this already solves k = 1,2,4. For k = 6, the quadratic in h must be solved hence the problem is reduced to making its discriminant D a square, or D = (Poly1)(Poly2) = y2. (Note that the Polyi are curiously palindromic.) Since D is a quartic polynomial with a square leading and constant term, this is easily made a square. For example, using Fermat’s method, one soln is, {a,e} = {9d(41+80d+41d2), -(5+4d)(4+5d)(20+59d+20d2)} though presumably smaller polynomials may exist. Using this initial rational point on the curve D = y2, subsequent ones can then be found. Second factor: a = (2+b)(c+e)/(1-b), d = (1+b)(c+2e)/(c-e), where {b,c,e} satisfy (Poly3)(-1+b)2h2-(Poly4)(c-e)2 = 0 Poly3:= bc2-(9+20b+9b2)ce+be2; Poly4:= 9bc2-(1-20b+b2)ce+9be2

so to solve (Poly3)(Poly4) = y2. (After an exchange of variables this is essentially the same as in the first.) Third factor: b = -(a+2c)(a+c-2e)/(a2-c2+ae-ce), d = (2a+c)(a+c-2e)/(a2-c2+ae-ce), where {a,c,e} satisfy (Poly5)h2-(Poly6)(a-c)2(a+c+e)2 = 0 Poly5:= (a+c)2(2a+c)(a+2c)+(a+c)(a2-38ac+c2)e-(a2-38ac+c2)e2; Poly6:= (2a+c)(a+2c)+(a+c)e-e2

so to solve (Poly5)(Poly6) = y2. This again has a square constant term so is easily made a square. B. Alternative method Still more solns can be found by solving Poly11 = Poly12 = 0 alternatively. Instead of eliminating h between eq.1 and 2, we eliminate the variable e between Poly11 and 12. This has seven non-trivial factors, with three essentially the same as the ones already discussed. The other four are the simpler, (2+b+d)(1+b)(1+d)(b+d) = 0 The first is a special case of Poly11 = Poly12 = 0, giving a simple soln to F2, (a+bh)k + (c+dh)k + (e+fh)k + (g+h)k = (a-bh)k + (c-dh)k + (e-fh)k + (g-h)k

still with g = -(ab+cd+ef) and f = -(1+b+d), where, {c,d,e,h} = {-ab(1-b)/w, -2-b, a(1-b)(3+2b)/w, ay/w}, with w = (2+b)(3+b) and b2+2b+21 = y2. This gives terms as quartic polynomials. (Alternatively, one can set d = 1 to get similar results.) The next three factors are variations on the same theme, namely a system that also satisfies x1+x2 = y1+y2 and x3+x4 = y3+y4 which imply an F2 with d = -b; f = -1. Solving Poly11 = Poly12 = 0 gives a soln but is trivial. However, by using the previous method of eliminating h between eq.1 and eq.2, the resultant gives three factors (squared this time), the first two trivial but not the third. This, in fact, is the linear root of the irreducible cubic which reduces for the special case d = -b. Using this root, eq.2 factors into two quadratics, one identical to eq.1 so the problem is reduced to just solving this. Thus, the complete soln to, x1

k+x2k+x3

k+x4k = y1

k+y2k+y3

for k = 1,2,4,6 where x1+x2 = y1+y2 and x3+x4 = y3+y4 is given by F2, (a+bh)k + (c+dh)k + (e+fh)k + (g+h)k = (a-bh)k + (c-dh)k + (e-fh)k + (g-h)k

with {d, f} = {-b, -1} where {a,g,h} = {v1(bc+e)e/w, v2(bc+e)c/w, y/(2w)}, and w = v1be+v2c, and {b,c,e} satisfying the quartic, (2v2c2-15bce+2v1e2)2 - v1v2c2e2 = y2

where {v1, v2} = {b2-4, 1-4b2}. Since again this has a square leading and constant term, this is easily made a square with one small soln {c,e} = {8b, 1-4b2} giving, {a,g,h} = {(b2-4)(1+4b2)/u, 8b(1+4b2)/u, (4-13b2+4b4)/u}, with u = b(b2+4) and yielding terms as 5th degree polynomials in b. From this initial polynomial soln, one can then generate an infinite more. (In contrast to Crussol's approach which can only give infinite numerical solns.) If we are to consider the more general system, nx1+x2+x3+x4 = ny1+y2+y3+y4

x1k+x2

k+x3k+x4

k = y1k+y2

k+y3k+y4

for k = 2,4,6, using our previous results for k = 1,2,4,6 it can be proven this has a soln for any n. We simply use the same method of changing the signs of appropriate terms since this does not affect the validity of the even powers. Thus, the system valid at k = 1, (a+bh) + (c+dh) + (e+fh) + (g+h) = (a-bh) + (c-dh) + (e-fh) + (g-h) where f = -1-b-d, is also good for, (-a-bh) + n(c+dh) + (-e-fh) + (g+h) = (-a+bh) + n(c-dh) + (-e+fh) + (g-h) if d is specialized as d = -2/(n+1). For example, an explicit soln to (Poly1)(Poly2) = y2 for k = 1,2,4,6 (using the first factor) was given earlier as, {a,e} = {9d(41+80d+41d2), -(5+4d)(4+5d)(20+59d+20d2)} for any d. This can be used for k = 2,4,6 with (-x1)+nx2+(-x3)+x4 = (-y1)+ny2+(-y3)+y4, by letting d = -2/(n+1), hence the system can be solved for any n as claimed. (If n = -1, division by zero can be avoided by simply changing the sign of x2,y2 to change the sign of n.) Some concise examples for special n, derived slightly differently, are, (a+bh)k + (c+dh)k + (e+fh)k + (g+h)k = (a-bh)k + (c-dh)k + (e-fh)k + (g-h)k

where a = 1, f = -1-bn-d, g = -(b+cd+ef), valid for k = 2,4,6 if, 1) {b,c,d,e,h} = {-2/(1+n), -2(-2+n)(3+n)/w, 2/(1+n), (1+n)(3+n)/w, (1+n)y/w}, where w = 2n(-5+n) and y2 = 81-2n+n2. 2) {b,c,d,e,h} = {-2/(1+n), -(3+n)(-1+3n)/w, -1, -2(3+n)/w, (1+n)y/w}, where w = 2n(1+3n) and y2 = 21+38n+21n2. 3) {b,c,d,e,h} = {-2/(1+n), -(3+n)(-2+3n)/w, -n/(1+n), -(3+n)/w, (1+n)y/w}, where w = 2n(-1+3n) and y2 = -15+14n+33n2. which also are true for both, n(a+bh) + (c+dh) + (e+fh) + (g+h) = n(a-bh) + (c-dh) + (e-fh) + (g-h) (a+bh) + (-c-dh) + (-e-fh) + (g+h) = (a-bh) + (-c+dh) + (-e+fh) + (g-h) Note 1: In contrast, as was seen, an infinite number of primitive solns to the system, nx1+x2+x3 = ny1+y2+y3

x1k+x2

k+x3k = y1

k+y2k+y3

for k = 2,6 and integer n,xi,yi can be proven only for some n which, other than the exceptional case n = 1, for the moment are all n = 3m. Whether there is an infinite number for some n ≠ 3m remains to be seen. Note 2: If the terms are signed, all known solns to S6 are valid for k = 1 as well. (In fact, also for S8 and S10.) Is there signed S6 that is not for k=1? A simple computer search may be able to settle this. Note 3: Lastly, as was mentioned, the system k = 1,2,4,6 with side conditions x1+x2 = y1+y2; x3+x4 = y3+y4 also has solns for just k = 1,2,6. The complete soln to this would be to ignore eq.1 and just solve eq.2 which, to recall, is a quartic with only even powers. To factor this into two quadratics entails the sufficient (but not necessary)1 condition of making its discriminant D a square. As a polynomial in the variable a, D is a quartic with a square leading term so it is quite easy to do this. A simpler case is when b = -2 since the constant term of D also becomes a square. Set c = 1 without loss of generality to get, D: = (3-4e+2e2)2 + (-39+88e-64e2+16e3)a + (73-88e+28e2)a2 + 3(-13+8e)a3 + 9a4

Three small non-trivial solns to D = y2 are, a1 = 8(2e-e2)/(-17+16e), a2 = -(21-22e+8e2)/(-19+8e), a3 =

(13-18e+8e2)/(-13+16e), giving three families solving the specialized form of F2 for just k = 1,2,6, (a-2h)k + (1+2h)k + (e-h)k + (g+h)k = (a+2h)k + (1-2h)k + (e+h)k + (g-h)k

where g = -2+2a+e and either, 1) a = 8(2e-e2)/(-17+16e), h = y1/(-17+16e), and y1

2 = (17-23e

+8e2)(17+13e+8e2) 2) a = -(21-22e+8e2)/(-19+8e), h = y2/(-19+8e), and y2

-53+82e-67e2+32e3

3) a = (13-18e+8e2)/(-13+8e), h = y3/(3(-13+8e)), and y3

(2873-7878e+8415e2-4096e3+768e4) Footnote 1: While it is a reasonable assumption that P: = x4+bx2+c = 0 can be completely factored into two quadratics by finding b,c such that the expression D = b2-4c is a square, a second approach is to find b,c such that it factors as P = (x2+mx+n)(x2-mx+n) = 0 in which case it generally does not have a D that is a square.

k + x2k + x3

k + x4k = y1

k + y2k + y3

k + y4k

5(a2+b2) = x2+y2

3(a2-b2) = y2-z2

2+3x1x2+x22+x3

2+3x3x4+x42 = 0

2+3y1y2+y22+y3

2+3y3y4+y42 = 0

p1k+p2

k+p3k+p4

k = q1k+q2

k+q3k+q4

k+p2k+p3

k+p4k+p5

k = q1k+q2

k+q3k+q4

k+q5k, k = 1,2

k+p2k+p5

k = q1k+q2

p3k+p4

k = q3k+q4

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

k+x2k+x3

k+x4k = y1

k+y2k+y3

x1k+x2

k+x3k+x4

k = y1k+y2

k+y3k+y4

x1k+x2

k+x3k = y1

k+y2k+y3

2 = (17-23e

+8e2)(17+13e+8e2) 2) a = -(21-22e+8e2)/(-19+8e), h = y2/(-19+8e), and y2

-53+82e-67e2+32e3

3) a = (13-18e+8e2)/(-13+8e), h = y3/(3(-13+8e)), and y3

k + x2k + x3

k + x4k = y1

k + y2k + y3

k + y4k

5(a2+b2) = x2+y2

3(a2-b2) = y2-z2

2+3x1x2+x22+x3

2+3x3x4+x42 = 0

2+3y1y2+y22+y3

2+3y3y4+y42 = 0

p1k+p2

k+p3k+p4

k = q1k+q2

k+q3k+q4

k+p2k+p3

k+p4k+p5

k = q1k+q2

k+q3k+q4

k+q5k, k = 1,2

k+p2k+p5

k = q1k+q2

p3k+p4

k = q3k+q4

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

k+x2k+x3

k+x4k = y1

k+y2k+y3

x1k+x2

k+x3k+x4

k = y1k+y2

k+y3k+y4

x1k+x2

k+x3k = y1

k+y2k+y3

2 = (17-23e

+8e2)(17+13e+8e2) 2) a = -(21-22e+8e2)/(-19+8e), h = y2/(-19+8e), and y2

-53+82e-67e2+32e3

3) a = (13-18e+8e2)/(-13+8e), h = y3/(3(-13+8e)), and y3

k + x2k + x3

k + x4k = y1

k + y2k + y3

k + y4k

5(a2+b2) = x2+y2

3(a2-b2) = y2-z2

2+3x1x2+x22+x3

2+3x3x4+x42 = 0

2+3y1y2+y22+y3

2+3y3y4+y42 = 0

p1k+p2

k+p3k+p4

k = q1k+q2

k+q3k+q4

k+p2k+p3

k+p4k+p5

k = q1k+q2

k+q3k+q4

k+q5k, k = 1,2

k+p2k+p5

k = q1k+q2

p3k+p4

k = q3k+q4

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

k+x2k+x3

k+x4k = y1

k+y2k+y3

x1k+x2

k+x3k+x4

k = y1k+y2

k+y3k+y4

x1k+x2

k+x3k = y1

k+y2k+y3

2 = (17-23e

+8e2)(17+13e+8e2) 2) a = -(21-22e+8e2)/(-19+8e), h = y2/(-19+8e), and y2

-53+82e-67e2+32e3

3) a = (13-18e+8e2)/(-13+8e), h = y3/(3(-13+8e)), and y3

k + x2k + x3

k + x4k = y1

k + y2k + y3

k + y4k

5(a2+b2) = x2+y2

3(a2-b2) = y2-z2

2+3x1x2+x22+x3

2+3x3x4+x42 = 0

2+3y1y2+y22+y3

2+3y3y4+y42 = 0

p1k+p2

k+p3k+p4

k = q1k+q2

k+q3k+q4

k+p2k+p3

k+p4k+p5

k = q1k+q2

k+q3k+q4

k+q5k, k = 1,2

k+p2k+p5

k = q1k+q2

p3k+p4

k = q3k+q4

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

k+x2k+x3

k+x4k = y1

k+y2k+y3

x1k+x2

k+x3k+x4

k = y1k+y2

k+y3k+y4

x1k+x2

k+x3k = y1

k+y2k+y3

2 = (17-23e

+8e2)(17+13e+8e2) 2) a = -(21-22e+8e2)/(-19+8e), h = y2/(-19+8e), and y2

-53+82e-67e2+32e3

3) a = (13-18e+8e2)/(-13+8e), h = y3/(3(-13+8e)), and y3

k + x2k + x3

k + x4k = y1

k + y2k + y3

k + y4k

5(a2+b2) = x2+y2

3(a2-b2) = y2-z2

2+3x1x2+x22+x3

2+3x3x4+x42 = 0

2+3y1y2+y22+y3

2+3y3y4+y42 = 0

p1k+p2

k+p3k+p4

k = q1k+q2

k+q3k+q4

k+p2k+p3

k+p4k+p5

k = q1k+q2

k+q3k+q4

k+q5k, k = 1,2

k+p2k+p5

k = q1k+q2

p3k+p4

k = q3k+q4

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

k+x2k+x3

k+x4k = y1

k+y2k+y3

x1k+x2

k+x3k+x4

k = y1k+y2

k+y3k+y4

x1k+x2

k+x3k = y1

k+y2k+y3

2 = (17-23e

+8e2)(17+13e+8e2) 2) a = -(21-22e+8e2)/(-19+8e), h = y2/(-19+8e), and y2

-53+82e-67e2+32e3

3) a = (13-18e+8e2)/(-13+8e), h = y3/(3(-13+8e)), and y3

k + x2k + x3

k + x4k = y1

k + y2k + y3

k + y4k

5(a2+b2) = x2+y2

3(a2-b2) = y2-z2

2+3x1x2+x22+x3

2+3x3x4+x42 = 0

2+3y1y2+y22+y3

2+3y3y4+y42 = 0

p1k+p2

k+p3k+p4

k = q1k+q2

k+q3k+q4

k+p2k+p3

k+p4k+p5

k = q1k+q2

k+q3k+q4

k+q5k, k = 1,2

k+p2k+p5

k = q1k+q2

p3k+p4

k = q3k+q4

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

k+x2k+x3

k+x4k = y1

k+y2k+y3

x1k+x2

k+x3k+x4

k = y1k+y2

k+y3k+y4

x1k+x2

k+x3k = y1

k+y2k+y3

2 = (17-23e

+8e2)(17+13e+8e2) 2) a = -(21-22e+8e2)/(-19+8e), h = y2/(-19+8e), and y2

-53+82e-67e2+32e3

3) a = (13-18e+8e2)/(-13+8e), h = y3/(3(-13+8e)), and y3

k + x2k + x3

k + x4k = y1

k + y2k + y3

k + y4k

5(a2+b2) = x2+y2

3(a2-b2) = y2-z2

2+3x1x2+x22+x3

2+3x3x4+x42 = 0

2+3y1y2+y22+y3

2+3y3y4+y42 = 0

p1k+p2

k+p3k+p4

k = q1k+q2

k+q3k+q4

k+p2k+p3

k+p4k+p5

k = q1k+q2

k+q3k+q4

k+q5k, k = 1,2

k+p2k+p5

k = q1k+q2

p3k+p4

k = q3k+q4

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

k+x2k+x3

k+x4k = y1

k+y2k+y3

x1k+x2

k+x3k+x4

k = y1k+y2

k+y3k+y4

x1k+x2

k+x3k = y1

k+y2k+y3

2 = (17-23e

+8e2)(17+13e+8e2) 2) a = -(21-22e+8e2)/(-19+8e), h = y2/(-19+8e), and y2

-53+82e-67e2+32e3

3) a = (13-18e+8e2)/(-13+8e), h = y3/(3(-13+8e)), and y3

k + x2k + x3

k + x4k = y1

k + y2k + y3

k + y4k

5(a2+b2) = x2+y2

3(a2-b2) = y2-z2

2+3x1x2+x22+x3

2+3x3x4+x42 = 0

2+3y1y2+y22+y3

2+3y3y4+y42 = 0

p1k+p2

k+p3k+p4

k = q1k+q2

k+q3k+q4

k+p2k+p3

k+p4k+p5

k = q1k+q2

k+q3k+q4

k+q5k, k = 1,2

k+p2k+p5

k = q1k+q2

p3k+p4

k = q3k+q4

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

k+x2k+x3

k+x4k = y1

k+y2k+y3

x1k+x2

k+x3k+x4

k = y1k+y2

k+y3k+y4

x1k+x2

k+x3k = y1

k+y2k+y3

2 = (17-23e

+8e2)(17+13e+8e2) 2) a = -(21-22e+8e2)/(-19+8e), h = y2/(-19+8e), and y2

-53+82e-67e2+32e3

3) a = (13-18e+8e2)/(-13+8e), h = y3/(3(-13+8e)), and y3

k + x2k + x3

k + x4k = y1

k + y2k + y3

k + y4k

5(a2+b2) = x2+y2

3(a2-b2) = y2-z2

2+3x1x2+x22+x3

2+3x3x4+x42 = 0

2+3y1y2+y22+y3

2+3y3y4+y42 = 0

p1k+p2

k+p3k+p4

k = q1k+q2

k+q3k+q4

k+p2k+p3

k+p4k+p5

k = q1k+q2

k+q3k+q4

k+q5k, k = 1,2

k+p2k+p5

k = q1k+q2

p3k+p4

k = q3k+q4

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

k+x2k+x3

k+x4k = y1

k+y2k+y3

x1k+x2

k+x3k+x4

k = y1k+y2

k+y3k+y4

x1k+x2

k+x3k = y1

k+y2k+y3

2 = (17-23e

+8e2)(17+13e+8e2) 2) a = -(21-22e+8e2)/(-19+8e), h = y2/(-19+8e), and y2

-53+82e-67e2+32e3

3) a = (13-18e+8e2)/(-13+8e), h = y3/(3(-13+8e)), and y3

k + x2k + x3

k + x4k = y1

k + y2k + y3

k + y4k

5(a2+b2) = x2+y2

3(a2-b2) = y2-z2

2+3x1x2+x22+x3

2+3x3x4+x42 = 0

2+3y1y2+y22+y3

2+3y3y4+y42 = 0

p1k+p2

k+p3k+p4

k = q1k+q2

k+q3k+q4

k+p2k+p3

k+p4k+p5

k = q1k+q2

k+q3k+q4

k+q5k, k = 1,2

k+p2k+p5

k = q1k+q2

p3k+p4

k = q3k+q4

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

k+x2k+x3

k+x4k = y1

k+y2k+y3

x1k+x2

k+x3k+x4

k = y1k+y2

k+y3k+y4

x1k+x2

k+x3k = y1

k+y2k+y3

2 = (17-23e

+8e2)(17+13e+8e2) 2) a = -(21-22e+8e2)/(-19+8e), h = y2/(-19+8e), and y2

-53+82e-67e2+32e3

3) a = (13-18e+8e2)/(-13+8e), h = y3/(3(-13+8e)), and y3

k + x2k + x3

k + x4k = y1

k + y2k + y3

k + y4k

5(a2+b2) = x2+y2

3(a2-b2) = y2-z2

2+3x1x2+x22+x3

2+3x3x4+x42 = 0

2+3y1y2+y22+y3

2+3y3y4+y42 = 0

p1k+p2

k+p3k+p4

k = q1k+q2

k+q3k+q4

k+p2k+p3

k+p4k+p5

k = q1k+q2

k+q3k+q4

k+q5k, k = 1,2

k+p2k+p5

k = q1k+q2

p3k+p4

k = q3k+q4

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

k+x2k+x3

k+x4k = y1

k+y2k+y3

x1k+x2

k+x3k+x4

k = y1k+y2

k+y3k+y4

x1k+x2

k+x3k = y1

k+y2k+y3

2 = (17-23e

+8e2)(17+13e+8e2) 2) a = -(21-22e+8e2)/(-19+8e), h = y2/(-19+8e), and y2

-53+82e-67e2+32e3

3) a = (13-18e+8e2)/(-13+8e), h = y3/(3(-13+8e)), and y3

k + x2k + x3

k + x4k = y1

k + y2k + y3

k + y4k

5(a2+b2) = x2+y2

3(a2-b2) = y2-z2

2+3x1x2+x22+x3

2+3x3x4+x42 = 0

2+3y1y2+y22+y3

2+3y3y4+y42 = 0

p1k+p2

k+p3k+p4

k = q1k+q2

k+q3k+q4

k+p2k+p3

k+p4k+p5

k = q1k+q2

k+q3k+q4

k+q5k, k = 1,2

k+p2k+p5

k = q1k+q2

p3k+p4

k = q3k+q4

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

k+x2k+x3

k+x4k = y1

k+y2k+y3

x1k+x2

k+x3k+x4

k = y1k+y2

k+y3k+y4

x1k+x2

k+x3k = y1

k+y2k+y3

2 = (17-23e

+8e2)(17+13e+8e2) 2) a = -(21-22e+8e2)/(-19+8e), h = y2/(-19+8e), and y2

-53+82e-67e2+32e3

3) a = (13-18e+8e2)/(-13+8e), h = y3/(3(-13+8e)), and y3

k + x2k + x3

k + x4k = y1

k + y2k + y3

k + y4k

5(a2+b2) = x2+y2

3(a2-b2) = y2-z2

2+3x1x2+x22+x3

2+3x3x4+x42 = 0

2+3y1y2+y22+y3

2+3y3y4+y42 = 0

p1k+p2

k+p3k+p4

k = q1k+q2

k+q3k+q4

k+p2k+p3

k+p4k+p5

k = q1k+q2

k+q3k+q4

k+q5k, k = 1,2

k+p2k+p5

k = q1k+q2

p3k+p4

k = q3k+q4

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

k+x2k+x3

k+x4k = y1

k+y2k+y3

x1k+x2

k+x3k+x4

k = y1k+y2

k+y3k+y4

x1k+x2

k+x3k = y1

k+y2k+y3

2 = (17-23e

+8e2)(17+13e+8e2) 2) a = -(21-22e+8e2)/(-19+8e), h = y2/(-19+8e), and y2

-53+82e-67e2+32e3

3) a = (13-18e+8e2)/(-13+8e), h = y3/(3(-13+8e)), and y3

k + x2k + x3

k + x4k = y1

k + y2k + y3

k + y4k

5(a2+b2) = x2+y2

3(a2-b2) = y2-z2

2+3x1x2+x22+x3

2+3x3x4+x42 = 0

2+3y1y2+y22+y3

2+3y3y4+y42 = 0

p1k+p2

k+p3k+p4

k = q1k+q2

k+q3k+q4

k+p2k+p3

k+p4k+p5

k = q1k+q2

k+q3k+q4

k+q5k, k = 1,2

k+p2k+p5

k = q1k+q2

p3k+p4

k = q3k+q4

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

k+x2k+x3

k+x4k = y1

k+y2k+y3

x1k+x2

k+x3k+x4

k = y1k+y2

k+y3k+y4

x1k+x2

k+x3k = y1

k+y2k+y3

2 = (17-23e

+8e2)(17+13e+8e2) 2) a = -(21-22e+8e2)/(-19+8e), h = y2/(-19+8e), and y2

-53+82e-67e2+32e3

3) a = (13-18e+8e2)/(-13+8e), h = y3/(3(-13+8e)), and y3

k + x2k + x3

k + x4k = y1

k + y2k + y3

k + y4k

5(a2+b2) = x2+y2

3(a2-b2) = y2-z2

2+3x1x2+x22+x3

2+3x3x4+x42 = 0

2+3y1y2+y22+y3

2+3y3y4+y42 = 0

p1k+p2

k+p3k+p4

k = q1k+q2

k+q3k+q4

k+p2k+p3

k+p4k+p5

k = q1k+q2

k+q3k+q4

k+q5k, k = 1,2

k+p2k+p5

k = q1k+q2

p3k+p4

k = q3k+q4

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

k+x2k+x3

k+x4k = y1

k+y2k+y3

x1k+x2

k+x3k+x4

k = y1k+y2

k+y3k+y4

x1k+x2

k+x3k = y1

k+y2k+y3

2 = (17-23e

+8e2)(17+13e+8e2) 2) a = -(21-22e+8e2)/(-19+8e), h = y2/(-19+8e), and y2

-53+82e-67e2+32e3

3) a = (13-18e+8e2)/(-13+8e), h = y3/(3(-13+8e)), and y3

k + x2k + x3

k + x4k = y1

k + y2k + y3

k + y4k

5(a2+b2) = x2+y2

3(a2-b2) = y2-z2

2+3x1x2+x22+x3

2+3x3x4+x42 = 0

2+3y1y2+y22+y3

2+3y3y4+y42 = 0

p1k+p2

k+p3k+p4

k = q1k+q2

k+q3k+q4

k+p2k+p3

k+p4k+p5

k = q1k+q2

k+q3k+q4

k+q5k, k = 1,2

k+p2k+p5

k = q1k+q2

p3k+p4

k = q3k+q4

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

k+x2k+x3

k+x4k = y1

k+y2k+y3

x1k+x2

k+x3k+x4

k = y1k+y2

k+y3k+y4

x1k+x2

k+x3k = y1

k+y2k+y3

2 = (17-23e

+8e2)(17+13e+8e2) 2) a = -(21-22e+8e2)/(-19+8e), h = y2/(-19+8e), and y2

-53+82e-67e2+32e3

3) a = (13-18e+8e2)/(-13+8e), h = y3/(3(-13+8e)), and y3

k + x2k + x3

k + x4k = y1

k + y2k + y3

k + y4k

5(a2+b2) = x2+y2

3(a2-b2) = y2-z2

2+3x1x2+x22+x3

2+3x3x4+x42 = 0

2+3y1y2+y22+y3

2+3y3y4+y42 = 0

p1k+p2

k+p3k+p4

k = q1k+q2

k+q3k+q4

k+p2k+p3

k+p4k+p5

k = q1k+q2

k+q3k+q4

k+q5k, k = 1,2

k+p2k+p5

k = q1k+q2

p3k+p4

k = q3k+q4

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

k+x2k+x3

k+x4k = y1

k+y2k+y3

x1k+x2

k+x3k+x4

k = y1k+y2

k+y3k+y4

x1k+x2

k+x3k = y1

k+y2k+y3

2 = (17-23e

+8e2)(17+13e+8e2) 2) a = -(21-22e+8e2)/(-19+8e), h = y2/(-19+8e), and y2

-53+82e-67e2+32e3

3) a = (13-18e+8e2)/(-13+8e), h = y3/(3(-13+8e)), and y3

PART 10. Sum / Sums of Seventh Powers 7.1 Eight terms For k = 7, the first soln to, x1

k + x2k + x3

k + x4k = y1

k + y2k + y3

k + y4k (eq.1)

was found in 1996 by Ekl as [10, 14, 123, 149] = [15, 90, 129, 146]. In 1999, Nuutti Kuosa found [184, 443, 556, 698] = [230, 353, 625, 673] for k = 1,3,7 though it seems he didn't realize it was multigrade. (As it turns out, neither does it belong to the family given by Choudhry below.) Update (6/26/09): Special forms: a) x1

7 + x27 + x3

7 + x47 + x5

7 + x67 + x7

7 = x87

Only three solns are known to the case of 7 positive 7th powers equal to a 7th power, the first one found by Mark Dodrill in 1999: 1277 + 2587 + 2667 + 4137 + 4307 + 4397 + 5257 = 5687

7 + x27 + x3

7 + x47 + x5

7 + x67 + x7

7 = 1 Just like three 3rd powers and five 5th powers, when signed, can sum to 1, so can seven 7th powers, two of which found by Kuosa, [1, 1146, 1348, 2816] = [130, 1031, 1951, 2787][1, 1140, 1823, 3189] = [485, 621, 1859, 3188] (End note) c) x1

7 + x27 + x3

7 + x47 + x5

7 + x67 + x7

7 = 0 Still unknown. (Though since five 5th powers can sum to both zero and 1, it may be a general property of odd n nth powers for n > 3.) A.J. Choudhry Choudhry found a multi-variable cubic polynomial such that when it has a rational root, it yields a k = 1,3,7. While no polynomial soln is yet known, it can be shown that if xi and yi are complex numbers with integer components, then there are families of polynomial solns (to be given later). Given, x1

k + x2k + x3

k + x4k = y1

k + y2k + y3

k + y4k, (eq.1)

{x1, x2, x3, x4} = {a+b+c, a-b-c, -a-b+c, -a+b-c}{y1, y2, y3, y4} = {d+e+f, d-e-f, -d-e+f, -d+e-f} a form which gives x1+x2+x3+x4 = y1+y2+y3+y4 = 0. Note that the xi

(or yi) also satisfies, 5(x1

2+x22+x3

2+x42)(x1

3+x23+x3

3+x43) = 6(x1

5+x25+x3

5+x45)

Theorem 1: Eq. 1 is true for k = 1,3,7 if abc = def and, 2(a4+b4+c4) - 5(a2+b2+c2)2 = 2(d4+e4+f4) - 5(d2+e2+f2)2 This is the same form used by Chernick for ideal solns of k = 1,2,3 and Lander for fifth powers k = 1,3,5, though as one can see the condition is a little more complicated. For the special case d+e = f, the system simplifies to satisfying the two eqns, abc = de(d+e)2(a4+b4+c4) - 5(a2+b2+c2)2 = 16(d2+de+e2)2 though whether this has non-trivial solns in the rationals is not known. If there is, then one of the terms xi or yi vanishes so would be the first counter-example to Euler’s conjecture for seventh powers. However, if this constraint is not given, after a clever substitution which reduced the problem to finding a rational root of a cubic, Choudhry found several solns to Theorem 1, one is {a,b,c,d,e,f} = {324, 5439, 893, 5217, 1539, 196} which gives, [3328, 2111, -3004, -2435] = [3476, 1741, -3280, -1937] which is for k = 1,3,7 as well as satisfying x1+x2+x3+x4 =

y1+y2+y3+y4 = 0, and 5(x12+x2

2+x32+x4

2)(x13+x2

3+x33+x4

3) = 6

(x15+x2

5+x35+x4

5). Note 1: Kuosa's 1999 soln for k = 1,3,7 does not belong to this family as there is no transposition of four appropriate xi such that their sum is equal to zero.

Note 2: Theorem 1 can also be satisfied by ak+bk+ck = dk+ek+fk, for k = 2,4 where abc = def. But, using resultants, it is easily proven that this restricted case has only trivial solns. Piezas Express eq.1 in the form F2, (a+bh)k + (c+dh)k + (e+fh)k + (g+h)k = (a-bh)k + (c-dh)k + (e-fh)k + (g-h)k

To satisfy the side condition x1+x2 = y1+y2; x3+x4 = y3+y4 which is essentially the same as Choudhry’s, let {b,f} = {-d, -1}. This makes it true for k = 1. Expanding for k = 3, we find the eqn, d = -(e2-g2)/(a2-c2) Substituting this into k = 7 we get a quartic in h with only even exponents, call this H4, with coefficients in {a,c,e,g} as, H4: = 3(v1

4-v34)h4 + 5(v1

2v4-v2v32)h2v1

2 + (a2c2-e2g2-v22+v4

= 0 where {v1, v2, v3, v4} = {a2-c2, a2+c2, e2-g2, e2+g2}. From Choudhry's result, we know H4 can have non-trivial rational roots, one of which is h = 98 when {a,c,e,g} = {3166, 2273, 3378, 1839}. To find a polynomial identity, as was pointed out in the previous section, there are two ways to factor quartics with only even exponents, one of which is to make its discriminant a square. One soln is a+c+e+g = 0. Let {a,c,e,g} = {p+q+r, p-q-r, -p+q-r, -p-q+r} for

symmetry, and H4 has the non-trivial quadratic factor, 3h2(q2+r2) + (5p2+3q2+3r2)(q+r)2 = 0 Unfortunately, this can only be solved by complex values. If we let h = hi where i is the imaginary unit, then we are to find, 3(5p2+3q2+3r2)(q2+r2) = y2

As it is only a quadratic in p with a square constant term, this is easily done. In summary, for the moment, only particular values (extrapolated from Choudhry's cubic) are known such that H4 has a non-trivial rational h. If an appropriate general relation can be found between {a,c,e,g} such that H4 has a quadratic factor with real roots, then it may be possible to find an infinite family of polynomials for k = 1,3,7, just as there is one for k = 1,5. Note: For k = 1,3,5,7 with ten terms, a similar situation arose, a parametrization dependent on a multi-variable quartic polynomial with only even exponents. However, this author was able to find two relations between the variables such that the quartic factored. See discussion in the subsequent sections. 7.2 Nine terms If x1 = 0, there are only two known primitive solns to, x1

k+x2k+x3

k+x4k+x5

k = y1k+y2

k+y3k+y4

k+y5k, for k = 1,3,5,7

These lead to ideal solns of deg 8. Interestingly, terms can be transposed such that x1+x2+x3+x4+x5 = y1+y2+y3+y4+y5 = 0. (There is a situation similar for k = 1,3,5 wherein the only known solns with x1

= 0 can be transposed in the same manner. It is interesting to consider what the first k = 1,3,5,7,9 with x1 = 0 would look like.) Letac in 1942 found the two, [0, 34, 58, 82, 98] = [13, 16, 69, 75, 99][0, 63, 119, 161, 169] = [8, 50, 132, 148, 174] (How did he find it prior to computer searching?) (Update, 8/15/09): Christian Bau did a search for terms < 200 and James Waldby extended it to < 450 with no new primitive solns being found.) It seems unknown if there is a parametric soln. But there are two points to be noted. First, as pointed out by Bailey et al, they can be transposed so an equal number of terms on either side sum to zero. For the first, it is, [0, -13, -16, -69, 98] = [75, 99, -34, -58, -82] The second one has more "structure" as it can be re-arranged in four distinct ways, [0, 63, 119, -8, -174] = [-161, -169, 148, 50, 132][0, 63, 119, -50, -132] = [-161, -169, 148, 8, 174] since 8+174 = 50+132, and, [63, -50, -174, 161, 0] = [-119, 132, 148, -169, 8][-119, 132, 148, -161, 0] = [63, -50, -174, 169, -8] since 169-8 = 161-0. As the left hand side involves only four terms, this (and only this side) also obeys, 5(x1

2+x22+x3

2+x42)(x1

3+x23+x3

3+x43) = 6(x1

5+x25+x3

5+x45)

a general identity first mentioned in k = 1,3,5. Second point, for the first soln, re-arranged as,

[-99, -13, 0, 34, 98] = [-82, -58, 16, 69, 75] Shuwen found this is good for one even exponent, k = 2. Later we shall see that both these phenomena appear in the only (and presumably smallest) soln for k = 1,3,5,7,9. Piezas Conjecture: "Define Fk:= x1

k+x2k+x3

k+x4k-(y1

k+y2k+y3

k+y4k+y5

k) where x1+x2+x3+x4 = y1+y2+y3+y4+y5 = 0. If Fk = 0 for k = 1,3,5,7,

then F22 = -2F4."

This works for the two known solns. I don’t know whether this is true in general for systems of this form or is just a peculiarity of Letac’s method. 7.3 Ten terms The next step is S7 and S8, x1

k+x2k+x3

k+x4k+x5

k = y1k+y2

k+y3k+y4

for k = 1,3,5,7, or k = 2,4,6,8. It can be proven by computer search that the smallest solution to the first system, at least in distinct integers, is given by, [3, 19, 37, 51, 53] = [9, 11, 43, 45, 55] This turns to have already been known (pre-computer era) by Sinha as belonging to an infinite family. However, there are just a few methods known to parametrically solve S7 (at least previously) and just one for

S8, which is deplorable. With the advent of computer algebra systems, the Internet, and hopefully more people working on the field of equal sums of like powers (ESLP), perhaps this can be rectified eventually. Ironically, compared to S6, it seems solns to S7 are rather plentiful since while only one binary quadratic form identity (Chernick’s) is known for the former, several are known for the latter. (Then again, it could be merely a lack of knowledge about S6.) For example, one soln found by this author is the beautifully simple, Sagan's Identity, 1 + 5k + (3+2y)k + (3-2y)k + (-3+3y)k + (-3-3y)k = (-2+x)k + (-2-x)k + (5-y)k + (5+y)k

for k = 1,3,5,7 if x2-10y2 = 9. Note that this also satisfies x1+x2 = x3+x4 = -(x5+x6) as well as some other obvious ones. While the complete soln uses rational x,y, if the condition is treated as a Pell eqn, then there are integral ones giving an infinite number of solns where two terms {1, 5} are constants. Update, 11/7/09: Last year, I dedicated an algebraic identity I found to the astronomer and science writer Carl Sagan (1934-1996). I read his "Broca's Brain" and "The Dragons of Eden" in my late teens, and read and eventually saw "Contact". Since I already dedicated one article I wrote on Degen's Eight-Square Identity to the novelist Katherine Neville, author of the amazing book "The Eight", I thought it was only fitting I name one of the identities I found after Sagan. After all, it has billions and billions of solutions. (You need to be a Sagan fan to understand the previous remark.) Today is the first Carl Sagan Day, and I decided to revive the article I wrote about Sagan's Identity that perished when Geocities was closed by Yahoo last Oct. 26. (End update.) Another is,

(3x+2y)k + (-3x+7y)k + (3x-2)k + (3+2x)k + (-5x-3y)k =(-3x+2y)k + (3x+7y)k + (-3x-2)k + (3-2x)k + (5x-3y)k

for k = 1,2,3,5,7 if x2+6y2 = 1 which has the distinction of being valid for k = 2 as well. Finally, a few others have the condition x2-70y2 = 1. All known parametrizations for k = 1,3,5,7 in terms of binary quadratic forms involve the unsigned discriminants D = 6, 10, 70. In short, the set D only employs the small prime factors p = 2,3,5,7. Whether this is coincidence I do not know. (Update, 1/25/10): Piezas Previously, it was seen how a k = 1,3,5 can lead to a k = 2,4,6. This can be brought higher. Given a (k.4.4) for k = 2,4,6, if its terms {xi , yi} are such that x1-x2 = y1-y2 = y2-y3 , then it can lead to a (k.5.5) for k = 1,3,5,7. This is exemplified by the infinite family, (3a+2b+c)k + (3a+2b-c)k + (-2a-b+d)k + (-2a-b-d)k = (3a+b+2c)k + (3a+b)k + (3a+b-2c)k + (-7a-b)k, for k = 1,2,4,6 which leads to the higher system, [3a+b+3c, 3a+2b-2c, -7a-b+c, -2a-b-c-d, -2a-b-c+d]k =[3a+b-3c, 3a+2b+2c, -7a-b-c, -2a-b+c-d, -2a-b+c+d]k, for k = 1,2,3,5,7 where, for both, 10a2-b2 = c2, and 55a2-6b2 = d2, with trivial ratios a/b = {1, 1/3, 5/13}. It is tempting to speculate that there may be higher-order version of this for (k.5.5), for k = 2,4,6,8. (End update.) It can be shown that given the system k = 1,3,5,7,

x1k+x2

k+x3k+x4

k+x5k = y1

k+y2k+y3

k+y4k+y5

if it has either of the constraints,

1) x1-x2 = x3-x4 = y1-y2 2) x1+x2 = x3+x4 = y1+y2

(and equivalent forms after transposition such as the one above), then an infinite sequence of polynomial solns can be found by finding non-trivial rational points on a quartic polynomial that is to be made a square. I. Constraint: x1-x2 = x3-x4 = y1-y2 (where x1+x2+x3+x4+x5 = y1+y2+y3+y4+y5) Here is a beautiful theorem by T. Sinha. (See update in next page, bottom section.) If, (a+3c)k + (b+3c)k + (a+b-2c)k = (c+d)k + (c+e)k + (-2c+d+e)k, for k = 2,4, then, ak + (a+2c)k + bk + (b+2c)k + (-c+d+e)k = (a+b-c)k + (a+b+c)k + dk + ek + (3c)k, for k = 1,3,5,7, excluding the trivial case c = 0. Note that this also satisfies, x1-x2 = x3-x4 = y1-y2 = -2c, 2(x1+x3) = y1+y2 x5 = y3+y4-(1/3)y5

a set of conditions which is enough to give the complete soln. An example is, {a,b,c,d,e} = {43, 9, 1, 19, 37} yielding, [43, 45, 9, 11, 55] = [51, 53, 19, 37, 3] which is the smallest solution mentioned previously. Sinha’s problem then is to solve, p1

k+p2k+p3

k = q1k+q2

for k = 2,4 where p1+p2-p3 = 2(q1+q2-q3) ≠ 0. He gave a single polynomial soln to this in terms of binary quadratic forms but this author showed there are four such solns, though there could be more. 1. (5x+2y+z)k + (5x+2y-z)k + (-14x-4y)k = (14x+3y)k + (4x-y)k + (6x-2y)k

where x2-50xy-5y2 = z2, D = 70. (Sinha’s) 2. (-3x+2y+z)k + (-3x+2y-z)k + (2x-4y)k = (8x-y)k + (2x+3y)k + (14x-2y)k

where 121x2-10xy-5y2 = z2, D = 70. 3. (6x+3y)k + (4x+9y)k + (2x-12y)k = (-x+3y+3z)k + (-x+3y-3z)k + (-6x-6y)k

where x2+10y2 = z2, D = -10. (This one, after minor changes, gives the example at the start of this section.)

4. (2x+3y)k + (-4x+y)k + (-10x-4y)k = (3x+y+z)k + (3x+y-z)k + (2x-2y)k

where 49x2+40xy+10y2 = z2, D = -10. where D is the discriminant of the binary quadratic form. Furthermore, treated as rational points on a certain elliptic curve, it can be showed that an infinite sequence can be derived from these thus providing polynomial solns of increasing degree. (See discussion on Fourth Powers.) The example below is based on Sinha’s soln [1], (2p+q+r)k + (-4p-q+r)k + (5p+q)k + (9p-q)k + (11p+3q)k =(-2p-q+r)k + (4p+q+r)k + (p-2q)k + (9p+3q)k + (11p+2q)k

for k = 1,3,5,7, if p2-50pq-5q2 = r2. Note: To compare, for eighth powers, other than the Letac-Sinha identity, there are only two solns known found by computer search with all terms xi < 520. One obeys the same constraints as Sinha’s identity, x1-x2 = x3-x4 = y1-y2, (and an additional x4-x2 = y4-y2). This is, [366, 103, 452, 189, -515] = [508, 245, -18, 331, -471], for k = 1,2,4,6,8 but whether this has a parametric soln is unknown. II. Constraint: x1+x2 = x3+x4 = y1+y2 (where x1+x2+x3+x4+x5 = y1+y2+y3+y4+y5) An example is one by Gloden,

[9, 45, 11, 43, 55] = [3, 51, 19, 37, 53] based on an identity given in the next section and simple transposition of terms, [-51, 9, 11, 55, -53] = [3, -45, -43, 19, 37] gives the equivalent rule, x1-y1 = -(x2-y2) = -(x3-y3) useful for the case k = 1,2,3,5,7, like this numerical example by Shuwen, {-71, 143, -17, -163, 121} = {103, -31, 157, -47, -169} One can use this to find a polynomial identity. This eqn has a lot of structure. Expressed in terms of xi and yi, then, x1-y1 = -(x2-y2) = x3-y3

x32+x4

2 = y32+y4

the first of which, after negating, is the same as Gloden’s. Using these constraints, we find the identity given earlier, (3x+2y)k + (-3x+7y)k + (3x-2)k + (3+2x)k + (-5x-3y)k =(-3x+2y)k + (3x+7y)k + (-3x-2)k + (3-2x)k + (5x-3y)k

for k = 1,2,3,5,7, if x2+6y2 = 1. Shuwen's was just the case {x,y} = {29/35, 8/35}, though some rational values should be avoided as they are trivial. However, one can find a more general result,

Theorem 1: (Piezas) "The system x1k+x2

k+x3k+x4

k+x5k = y1

+y3k+y4

k+y5k, for k = 1,2,3,5,7, with,

x1-y1 = -(x3-y3) = -(x4-y4) has a complete soln in terms of making two quadratic polynomials into squares." This is given by, (a+bj)k + (c+dj)k + (e+fj)k + (g+hj)k + (i+j)k = (a-bj)k + (c-dj)k + (e-fj)k

+ (g-hj)k + (i-j)k where, {a, i, d, f, h} = {1, -(b+cd+ef+gh), b-1, -b, -b} e = ((b-2c+bc)+u) j/v g = ((b-2c+bc)-u) j/v j = v/(3+2b) and {u,v} are expressed in the free variables {b,c} as, (-9-9b+27b2+21b3+2b4) - 2(-1+2b)(6-9b-6b2+b3)c + (32-64b-b2+33b3+2b4)c2 = nu2

(-9-15b+2b2) + 2(-1+2b)(-3+5b)c + (23-23b+18b2)c2 = nv2

where n = (1+b)(-1+2b). Hence, two quadratic polynomials in the variable c must be made squares. One polynomial soln is given by, c = 2(2b2+18b-9)/(14b2-59b+27) This, and a special case, will be explained more below. Also, since the theorem involves an even power, then we can give this relevant

conjecture, Conjecture: "Define Fk:= x1

k+x2k+x3

k+x4k+x5

k – (y1k+y2

+y4k+y5

k). If Fk = 0 for k = 1,2,3,5,7, then (11F9)

(x12+x2

2+x32+x4

2+x52) = 9F11."

The conjecture is the generalization of the proven theorem for k = 1,2,3,5 and, using the soln given above, is easily shown to be true for this special case. However, I do not know how to prove this in general. Proof (for theorem): Using F3, the more general version of F2, (a+bj)k + (c+dj)k + (e+fj)k + (g+hj)k + (i+j)k = (a-bj)k + (c-dj)k + (e-fj)k

+ (g-hj)k + (i-j)k where set a = 1 without loss of generality. For k = 1,2, let, i = -(ab+cd+ef+gh), d = -(1+b+f+h) To satisfy the two constraints, let, (a+bj) - (a-bj) = -(e+fj) + (e-fj) (a+bj) - (a-bj) = -(g+hj) + (g-hj) so f = -b, h = -b. Expanding F3 for k = 3,5,7, we get, (Poly01)j2 + (Poly02) = 0 (eq.1)(Poly11)j4 + (Poly12)j2 + (Poly13) = 0 (eq.2)(Poly21)j6 + (Poly22)j4 + (Poly23)j2 + (Poly24) = 0 (eq.3) Eliminate j between eq.1 and 2, then between eq.1 and 3 (set j = √v for

ease of computation) to get two eqs, call these the auxiliary resultants, (Poly01)g + (Poly02) = 0 (eq.4)(Poly11)g3 + (Poly12)g2 + (Poly13)g + Poly(14) = 0 (eq.5) where the first fortunately is only linear in the variable g. (It can get as high as sextic in the general case, with the second resultant even higher.) Eliminate g between the two and we get the final resultant which has a non-trivial factor only a quadratic in e, and some trivial linear ones. This has discriminant D that is only a quadratic in c so is low enough to be handled by standard algebraic methods. One can then finally express {e,g,j} in terms of free variables {b,c}, the explicit expressions given earlier. Note that {e,g} are the two roots of the same quadratic. The problem then is reduced to making two quadratic polynomials in c as squares, with the first as the discriminant D of the final resultant. Two special cases were found by this author: one is when b = 3/5 which makes D a square and yields the identity dependent on x2+6y2 = 1 given previously; the second is based on a numerical result by Gloden, [39, 65, 1, -57, -53] = [-33, 11, 73, 15, -71] re-arranged by Shuwen to make it valid for k = 1,2,3,5,7. This author noticed it had another side condition x3+x4 = y4+y5 so using this constraint found a polynomial for c given at the start as, c = 2(2b2+18b-9)/(14b2-59b+27) which makes both quadratics as squares and should yield terms as quartic polynomials, with Gloden’s soln as b = 4. In fact, there is an infinite family of polynomial c though what is desirable are those of small degree. In general, it’s long been known that given an initial soln

to a quadratic polynomial that is to be made a square, P1(x) = y2, then the complete soln can be given as a binary quadratic form, x = p1m2+p2mn+p3n2, where we can suppress n for conciseness. If there is a second quadratic also in x which is also to be made a square, P2(x)

= z2, by substituting this value for x into it, one gets a quartic expression of form, q1m4+q2m3+q3m2+q4m+q5 = z2

Using an initial polynomial soln m, one can then compute an infinite sequence for the system k = 1,2,3,5,7. (In our case, one initial value c, which incidentally makes both quadratics as squares, though trivial, is c = b/(1-b). Appropriate application of what was discussed can then generate an infinite sequence of polynomial solns.) Note: This approach can also be used for k = 1,2,4,6,8 with the same constraints. Unfortunately, neither of the auxiliary resultants is linear in some variable but instead are of quite high degree. However, using a simple algebraic trick on the variables {e,g} discussed in the next section, these can be reduced and the final resultant turns out to be a quartic in one variable with odd exponents so, in general, needs the cube roots of unity. I do not know if this is non-trivially solvable. Even more generally, the odd system k = 1,3,5,7, (a+bj)k + (c+dj)k + (e-bj)k + (g-bj)k + (i+j)k = (a-bj)k + (c-dj)k + (e+bj)k

+ (g+bj)k + (i-j)k where a=1 for convenience and which has x1-y1 = -(x3-y3) = -(x4-y4), other than one exceptional case b(c±e) – c = ±i, then this has a complete soln where d = b-1, j is a root of a quadratic, and {e,g} are two roots of a single quartic, both only in even exponents with coefficients in the remaining variables {b,c,i}. In other words, the radical soln of this system does not employ the cube roots of unity. (Of

course, for certain variables this radical soln becomes rational.) Using the same approach as the one above, we find the final resultant of this system factors as the four cases of b(c±e) – c = ±i and an irreducible quartic factor, call this Q4, with only even exponents. Two of the roots of Q4 turn out to define {e,g}, though unfortunately is too tedious to write down here. To find j, use the eqn above at k = 3 to get, b-c2+bc2-b(e2+g2)+i2+(b-b2)j2 = 0 and using the appropriate two roots of Q4 as {e,g}, it transforms to, -c2+i2-b(-3+c2+2i2) + 3b(1+b)(-1+2b)j2 = 0 so we have all the unknowns {e,g,j} in terms of {b,c,i}. If the exceptional case b(c±e) – c = ±i (or equivalently, b(c±g) – c = ±i) is used to define i, two of the variables {e,g,j} are roots of two different quartics while the third variable is a root of a quadratic. This system can have a non-trivial radical soln but I don’t know if it can be rational. However, if an appropriate relationship can be found between {b,c,i}, such as, (3b-3c+5bc) + (3+2b)i = 0 then Q4 can factor into two quadratics as in the explicit identity for k = 1,2,3,5,7 given earlier where {e,g} are the two roots of one quadratic factor. To find a second relationship, using one soln by Gloden, [-51, -19, 9, 43, 55] = [3, 53, -45, -11, 37] in addition to the afore-mentioned constraints, this also has x1+x3 = -(y2+y4) and we can have a second theorem, Theorem 2: (Piezas) "The odd system x1

k+x2k+x3

k+x4k+x5

k = y1k

+y2k+y3

k+y4k+y5

k, for k = 1,3,5,7 where, x1-y1 = -(x3-y3) = -(x4-y4), as well as x1+x3 = -(y2+y4), other than one exceptional case, has a complete soln in terms of making two quadratic polynomials into squares." (See update below.) Proof: Given F3 again, (a+bj)k + (c+dj)k + (e+fj)k + (g+hj)k + (i+j)k = (a-bj)k + (c-dj)k + (e-fj)k

+ (g-hj)k + (i-j)k The case to be avoided is the same b(c±e) – c = ±i. Set a = 1 and {d,f,h} = {b-1, -b, -b} to satisfy the usual conditions and k = 1. The new constraint then entails making a+c+e+g+j = 0. To satisfy this and k = 3,5,7, let, e = ((1+b+bc)+u) i/vg = ((1+b+bc)-u) i/vi = -v/(3+2b)j = -(1+3c)/(3+2b) where {u,v} are, (8+24b+15b2+2b3) – 2(-1+2b)(-6-2b+b2)c + (-36+12b+23b2+2b3)c2 = (2b-1)u2

(24b+39b2+18b3) + 18b(1+b)(-1+2b)c + (1+b)(-9-39b+50b2)c2 = (2b-1)v2

so the problem is reduced to making these two quadratic polynomials in c as squares. By completely solving one, say the second, using a binary quadratic form (an initial soln for both is c = -1) then solving the

other one entails making a quartic polynomial (with a square leading term) a square, just like for Theorem 1. By using Fermat’s method an infinite sequence of polynomial solns can then be easily found. This also has a special case at b = -1/6 where the two become, (4/81)(-119+406c+1009c2) = u2

(4/81)(-9+5c)2 = v2

so one merely has to solve the first quadratic condition. Explicitly, the identity is, (17+3p)k + (7+37p)k + (-6-2p-q)k + (-6-2p+q)k + (-15-13p)k =(15-3p)k + (-7-5p)k + (-4+4p-q)k + (-4+4p+q)k + (-3+23p)k for k = 1,3,5,7, if -119+406p+1009p2 = q2 (which has discriminant D = 70). Note 1: As one can see, in Theorem 2 the relationship between {b,c,i} is quadratic while that of {b,c,j} is linear. (In Theorem 1, this is reversed.) Note 2: Sometimes it is a matter of choosing the right form to solve a particular system of eqns. For ex, suppose one chose the more obvious form, (a+b)k + (c+d)k + (e+f)k + (g+h)k + (i+j)k = (a-b)k + (c-d)k + (e-f)k + (g-h)k + (i-j)k To satisfy the three constraints and k = 1, set f = -b, h = -b, g = -(a+b+c-d+e), j = b-d, where a = 1 for convenience. Expanding for k = 3,5,7, by eliminating {i,c} one gets a final resultant that is a quadratic in e with a quartic discriminant in {b,d}. Recovering i, this is also a quadratic with a different quartic discriminant in {b,d}. Thus, one ends up with two quartic polynomials to be made as squares which is hard

to solve when in fact using the first approach, the polynomials can be reduced to just two quadratics. It makes one wonder whether in other situations where one ends up with two quartic polynomials, if there is a right approach to simplify the problem. Note 3: Authors who have written on the seventh degree are Sinha, Letac, Gloden, Moessner, etc:A. Letac, Gazeta Matematica, 48 (1942), p. 68-69.A. Gloden, Mehrgradige Gleichungen, Groningen, 1944; Two theorems on multi-degree equalities, Amer. Math. Monthly, 55 (1948), p. 86-88.A. Moessner, On equal sums of powers, Math. Student, 15 (1947), p. 83-88.I’ll add the information in these sources once I can access them (or if someone is nice enough to send me a pdf copy), though I presume Gloden’s and Moessner’s are just special cases of the computer-assisted general theorems in this section. On the other hand, Letac’s has one term xi = 0 with different constraints so he must have used another method. (Update, 7/26/09): I just realized that Sinha's multi-grade system has an extra side-condition such that it is essentially equivalent to Theorem 2. With a small linear change of variables, it is quite easy to find the complete soln. Let, (a+4e)k + (b+4e)k + (a+b)k = (c+2e)k + (d+2e)k + (c+d)k, for k = 2,4 then, (-a-b-e)k + (-c-e)k + (b+e)k + (a+e)k + (c+d+e)k = (3e)k + (a+b+3e)k + (-a-3e)k + (-b-3e)k + (d+e)k, for k = 1,3,5,7 (eq.1) Re-arranged this way, note that eq.1 also satisfies, x1-y1 = -(x3-y3) = -(x4-y4), as well as x1+x3 = -(y2+y4),

just like in Theorem 2. The complete soln for this simple version is also much simpler given by, let r = 2a-d, then, b = (-(a+4e)r+u)/(2r)c = (-(d+2e)r+v)/(2r) where {u,v} are in the free variables {a,d,e} as, 2a3+3a2d-4d3 + 8(2a-d)(a+d)e + 32(a+d)e2 = (1/r)u2

8a3-6ad2-d3 + 4(2a-d)(4a+d)e + 4(26a-d)e2 = (1/r)v2

hence, naturally enough, involves making two quadratics as squares, though, ad(a+d)(a-d)(a+2e)(a-2e)(d+4e)(d-4e) = 0 and a few other relations are trivial. A non-trivial example is, {a,b,c,d,e} = {2x-9y, -3y, -3(x+y-z), -3(x+y+z), x+3y}{u,v} = {3(2x-y)(7x-15y+3z), 40x2-84xy-135y2+54yz} where {x,y,z} satisfies, x2+10y2 = z2

One can wonder if Theorem 1 has a simpler version as well. It can also be shown that there is no permutation of the {xi, yi} of Sinha's system such that it is valid for k = 2 for all {a,d,e}, except for special cases, so it is not equivalent to Theorem 1. (End update.) Note: By a suitable change of variables, two cubic polynomials to be made squares may be reduced to just two quadratics, and hence amenable to being treated as an elliptic curve. For example, given, 2b3+3b2c-4c3-2(7b2+8bc-8c2)x+32(b+c)x2-96x3 = u2

8b3-6b2c-c3-2(26b2-20bc-c2)x+8(19b-8c)x2-288x3 = v2

Let {b,c} = {a+mx, d+nx}, and we get, P3x3+P2x2+P1x+P0 = u2

Q3x3+Q2x2+Q1x+Q0 = v2

where the Pi and Qi are polynomials in {a,d,m,n}. The objective is to eliminate the cubic term by setting P3 = Q3 = 0. Since these are only in the {m,n}, one can get their resultant and, for these pair, it turns out the suitable values are {m,n} = {4,2}. (Of course, not all cubic pairs will fortuitously have rational {m,n}, but it will be worth checking.)

PART 11. Sum / Sums of Eighth Powers 8.1 Eight terms8.2 Nine terms8.3 Ten terms8.4 Twelve terms8.5 Fourteen terms8.6 Sixteen terms8.7 Eighteen terms 8.1 Eight terms The first soln to, x1

k+x2k+x3

k+x4k = y1

k+y2k+y3

k+y4k, k = 8

was found by Nuutti Kuosa in 2006, 8618 + 19538 + 20128 + 31138 = 11288 + 25578 + 27678 + 28238

though it is not known if this the smallest possible as this is the only soln so far. It is also not known if k = 1,8 or k = 1,2,8 is possible. Based on known solns to k = 1,3,5, k = 1,2,6, and k = 1,3,7, this author is assuming a k = 1,8 will be found that satisfies the side conditions: x1+x2 = y1+y2, and x3+x4 = y3+y4. 8.2 Nine terms The first and only soln so far to 8 eighth powers equal to an eighth power, x1

8+x28+x3

8+x48+x5

8+x68+x7

8+x88 = z8

was found by Scott Chase in 2000, 908 + 2238 + 4788 + 5248 + 7488 + 10888 + 11908 + 13248 = 14098

Whether there are seven 8th powers equal to an 8th power is unknown, though they are conjectured to be possible (the Lander, Parkin, and Selfridge Conjecture in Eulernet). No more k positive kth powers equal to a kth power is known for k > 8 (with the case k = 6 unknown as well), though J. Wroblewski has come close for k = 9,10. 8.3 Ten terms Other than particular instances of the Letac-Sinha identity, only two solns are known for multi-grade k = 2,4,6,8 (found by Borwein, Lisonek, and Percival in 2000) namely,

[71, -131, -180, 307, 308] = [301, 99, 100, 188, -313] (eq.1)[366, 103, 452, 189, -515] = [508, 245, -18, 331, -471] (eq.2) which, as signed terms, are also valid for k = 1 hence all known solns are for k = 1,2,4,6,8. (Note: After all, there are 32 possible sums of ± x1 ± x2 ± x3 ± x4 ± x5 and an equal number for the other side, and the chances might be that two of the sums will be the same. In general, it raises the question whether systems k = 2,4,…2n as n increases also tend to be valid for k = 1 if the terms are signed.) Eq.1 has two side conditions: x1-x2 = y1-y2 and x4-x5 = y2-y3. Eq. 2 is more structured, satisfying five: x1-x2 = x3-x4 = y1-y2 = y2-y3, and x1-x4 = y1-y4, and x2-x4 = y2-y4. It is easily shown that, together with the five eqns k = 1,2,4,6,8, (eq. 2) is the only non-trivial soln of this system of 10 eqns in 10 unknowns. A. Letac, T. Sinha (a+c)k + (a-c)k + (3b+d)k + (3b-d)k + (4a)k = (3a+c)k + (3a-c)k + (b+d)k + (b-d)k + (4b)k

for k = 1,2,4,6,8, where a2+12b2 = c2 (eq.1), and 12a2+b2 = d2 (eq.2). The ratio a/b = {1/2, 2} must be avoided as it yields trivial solns, with the smallest non-trivial one as {a,b} = {218, 11869}. Labeled as, x1

k+x2k+x3

k+x4k+x5

k = y1k+y2

k+y3k+y4

this also satisfies x1-x2 = y1-y2, and x3-x4 = y3-y4, and interestingly, for k = 1,2, x1

k+x2k+x5

k = y1k+y2

k; x3k+x4

k = y3k+y4

k+y5k,

for any {a,b,c,d}. (In fact, it can be shown that the complete soln to the system with all these constraints is given by this identity.) The first pair of constraints implies that the system k = 2,4,6,8 may have the “side condition”, x1 ± x2 = y1 ± y2; x3 ± x4 = y3 ± y4 just like for sixth powers k = 2,4,6. To solve the Letac-Sinha identity, since the complete soln to eq.1 is given by {a,b} = {m2-12n2, 2mn}, if we substitute this into eq.2 we get, 4(3m4-71m2+432) = d2

where it was set n=1 without loss of generality. Trivial rational solns are given by m = {2, 3, 4, 6}. However, we can use these to generate subsequent ones that are non-trivial, such as m = 109, and so on. (Update, 2/1/10): Chris Smyth This is a variation of the Letac-Sinha 8th power [k.5.5] multi-grade. For k = 1,2,4,6,8, [xy+ax+ay-d, -(xy-ax-ay-d), xy-bx+by+d, -(xy+bx-by+d), cx+cy]k =[xy+bx+by-d, -(xy-bx-by-d), xy-ax+ay+d, -(xy+ax-ay+d), cx-cy]k

where {a,b,c,d} = {1, 3, 4, 11} and {x,y} satisfies the eqn x2y2-13(x2+y2)+112 = 0. One can see in the general form how “a” in one side has just been replaced with “b” in the other side. With terms expressed as {pi,qi}, this obeys,

p1-p2 = q1-q2p3-p4 = q3-q4 If we set that, p1

k+p2k+p3

k+p4k+p5

k = q1k+q2

k+q3k+q4

k+q5k, k = 1,2

k+p2k+p5

k = q1k+q2

p3k+p4

k = q3k+q4

Smyth’s form for (k.5.5) can be generalized by adding two terms on each side to create a (k.7.7). A possible addition, given in italics, may be, [xy+ax+ay-c, -(xy-ax-ay-c), xy-bx+by+c, -(xy+bx-by+c), dx+dy, ex+fy, fx-ey]k =[xy+by+bx-c, -(xy-by-bx-c), xy-ay+ax+c, -(xy+ay-ax+c), dx-dy, ey

+fx, ex-fy]k which is the same pair successfully added to the (k.4.4). For non-zero {e,f}, whether this has a non-trivial soln up to k = 8, 10, or even 12 is unknown. See also Sixth Powers for the context of this identity. (End update.) (Update, 1/13/10): Wroblewski Using a special case of a general theorem, Wroblewski used a k = 2,4,6,8 to derive a k = 1,3,5,7,9! “If [x1, x2, x3, x4, x5]k = [y1, y2, y3, y4, y5]k, for k = 2,4,6,8, where x1-x2 = x3-x4 = y1-y2 = y2-y3 = N, then, [a+x5, -a+y4, -a+y3, a+x1, a+x3, -a+y5]k = [-a+x5, a+y4, a+y1, -a+x2,

-a+x4, a+y5]k, for k = 1,3,5,7,9 where a = N/2.” It can be seen that the second eqn (or eq. 2) found by Borwein et al, [366, 103, 452, 189, -515] = [508, 245, -18, 331, -471] satisfies this and yields, [-767, 399, -299, 995, 1167, -1205] = [-1293, 925, 1279, -57, 115, -679] which is the second k = 1,3,5,7,9 since the first one was found in 2000 by Shuwen via computer search. Since then, Wroblewski has found a

few others also by computer search. See also Ninth Powers. (End update.) (Update, 1/27/10): Piezas It can be shown that the system, [x1, x2, x3, x4, x5]k = [y1, y2, y3, y4, y5]k, for k = 2,4,6,8, with the constraint x1-x2 = x3-x4 = y1-y2 = y2-y3 can be completely solved in terms of nested square roots. The soln is, (1-e)k + (-1-e)k + (1-f)k + (-1-f)k + ak = (2-c)k + (-c)k + (-2-c)k + bk + dk a = 4(u+v),b = 4(u-v)2c2 = -3+t+31uv-3w2d2 = 53+t+71uv-11w{e2, f2} = {p-q, p+q} and, p = (1/2)(13+t+9uv-5w)(1/2)q2 = 20+4t+301uv+35u2v2-5(4+11uv)w where w2 = -15+2t+22uv-135u2v2, and t = 32(u2+v2), for two free variables {u,v}. Thus {e,f} are two roots of an octic polynomial. This then leads to the 9th deg multigrade, (1+a)k + (-1+b)k + (-3+c)k + (-1+d)k + (2+e)k + (2+f)k =(-1+a)k + (1+b)k + (3+c)k + (1+d)k + (-2+e)k + (-2+f)k, for k = 1,3,5,7,9

Six variables S = {c,d,e,f,q,w} must be squares, and ±w is equally valid, though only one sign may yield a rational value. One non-trivial soln is {u,v} = {46/263, 423/526} which gives the original equation that inspired the form. Even with six expressions to be made squares, due to the symmetric nature of the system, there is in fact an infinite number of {u,v} such that, with the sign of ±w chosen appropriately, all S are squares, one of which is the family, v = 4(u-1)/(12u-7) but unfortunately just yields trivial terms. (To find two others, one can expand at k = 11.) However, it may be possible there are other non-trivial solns. If we add one more constraint to the 8th deg system, namely x2-x4 = y2-y4, we lose one free variable but the problem simplifies to four quadratics to be made squares. This is given by, [b-c-d, b+c-d, -b-c-d, -b+c-d, f]k = [a+b-2c, a+b, a+b+2c, a-b, e]k, for k = 2,4,6,8 where {c,d,e,f} are, 32c2 = 5a2-10ab+8b2

32d2 = 29a2+6ab+72b2

4e2 = 9a2-6ab+36b2

4f2 = 13a2+2ab+4b2

Since the eqns are homogeneous, we can set b = 1 without loss of generality and "a" is just the only free variable. This system has also been investigated by Wroblewski, see below. There are some trivial {a,b} like the ratio a/b = {2, 6, etc). A non-trivial soln is {a,b} = {-288, 43} which makes all {c,d,e,f} as squares and yields the original multi-grade. It is easy to make {c,d} as squares using an elliptic curve, though apparently it is only by chance that {e,f} are

squares as well. But this can provide a rational family to, x1

k + x2k + x3

k + x4k + x5

k/2 = y1k + y2

k + y3k + y4

k + y5k/2, for k =

2,4,6,8 Considering how this result has a much simpler form, it may be possible that, with the right constraint, the more general case can be simplified still. But if we add yet one more condition and require that k = 1 hold as well (for signed terms), then it can be shown that the original eqn is the only non-trivial soln to this system of 9 eqns (4+5) in 9 unknowns (since one variable can be set equal to 1 without loss of generality). (End update.) Piezas Theorem: If ak+bk+ck+dk+ek = fk+gk+hk+ik+jk, for k = 2,4,6,8, define n as, a4+b4+c4+d4+e4 = n(a2+b2+c2+d2+e2)2, then, 72(a10+b10+c10+d10+e10-f10-g10-h10-i10-j10)(a14+b14+c14+d14+e14-f14-g14-h14-i14-j14) = 35(n+1)(a12+b12+c12+d12+e12-f12-g12-h12-i12-j12)2 which is a part of a family that starts with Ramanujan’s 6-10-8 Identity and depends on a system that is valid for k = 2,4,…2n. Note 1: It can be shown that the system, (pa+c)k + (pa-c)k + (qb+d)k + (qb-d)k +(ra)k = (qa+c)k + (qa-c)k + (pb+d)k + (pb-d)k +(rb)k, for k = 1,2,4,6,8, where c = √(ma2+nb2) and d = √(na2+mb2) is non-trivial only for {p,q,r} = {1, 3, 4}, and {m,n} = {1, 12}. Note 2: Just like for k = 1,3,5,7, the general k = 1,2,4,6,8 can be completely parametricized by the form,

(a+bj)k + (c+dj)k + (e+fj)k + (g+hj)k + (i+j)k = (a-bj)k + (c-dj)k + (e-fj)k

+ (g-hj)k + (i-j)k where for simplicity set a = 1 without loss of generality. For the system which also has x1-x2 = y1-y2, x3-x4 = y3-y4 it can be shown this entails the necessary but not sufficient condition of finding a non-trivial rational root of a 10th deg resultant. Let d = b, h = f. To satisfy k = 1,2, one can then use the variables i,f. Expanding for k = 4,6,8 gives the eqns, (Poly01)j2 + (Poly02) = 0 (eq.1)(Poly11)j4 + (Poly12)j2 + (Poly13) = 0 (eq.2)(Poly21)j6 + (Poly22)j4 + (Poly23)j2 + (Poly24) = 0 (eq.3) Eliminate j between eqs.1,2, then between eq.1,3, (as before, let j = √x for ease of computation) to get two auxiliary resultants, eq.4a and eq.5a, which are 6th and 9th deg in the variable g, respectively. There are still 4 variables, b,c,e,g, and it is at this point that the system for k = 1,2,4,6 with a similar “side condition” has a resultant with a single non-trivial linear factor and some trivial ones. For k = 1,2,4,6,8, one more variable has to be eliminated but since eq.4a and 5a are of high degree, these are horrendous to resolve directly. A trick to reduce the degree (which works for this and similar constraints) is to let {e,g} = {u+v, u-v} to get, (Poly31)v4 + (Poly32)v2 + (Poly33) = 0 (eq.4b)(Poly41)v6 + (Poly42)v4 + (Poly43)v2 + (Poly44) = 0 (eq.5b) where again let v = √y to shorten the calculation. Eliminate y and the final resultant turns out to have a non-trivial 10th-deg factor in c, with trivial linear ones, showing a qualitative difference from the previous system which had a non-trivial linear factor. One must then find appropriate {b,u} such that this decic has a non-trivial rational root and

the Letac-Sinha identity guarantees there are infinitely many. Whether there is another family such that this decic has rational factors is not known. In summary, the system, x1

k+x2k+x3

k+x4k+x5

k = y1k+y2

k+y3k+y4

as S7 for k = 1,3,5,7 and S8 for k = 2,4,6,8 can have either of the side-conditions,

1) x1-x2 = x3-x4 = y1-y22) x1+x2 = x3+x4 = y1+y2 (or equivalent forms)3) x1-x2 = y1-y2; x3±x4 = y3±y4

with S7 having solns for all three and S8 only for (1), (3). Whether all three can apply to both remains to be seen. Furthermore, for powers greater than the sixth, it seems more constraints are needed to decrease the number of variables so that: a) a parametric soln is possible; or b) the final resultant is not of so high degree. For example, the Letac-Sinha identity which is the only one known for eighth powers already satisfies four or five side conditions. This implies that for the 12-term nonic and decic systems, and analogous higher ones, a lot of algebraic ingenuity must be employed to find a parametrization. If there is any. Like solving univariate eqns in the radicals, there might come a point when there are simply too many variables than can be handled by the constraints. (See update in Tenth Powers.) (Update, 1/12/10): Piezas, Wroblewski Starting with the known soln for k = 2,4,6,8, [-508, -245, 331, 18, 471] = [-366, 189, -103, 452, 515]

Wroblewski suggested the general form, [x, x+a+b, x+b+d, x+2a+2b, p]k = [x+b, x+d, x+a+2b, x+a+b+d, q]k

where, for this case, {a,b,d,p,q,x} = {121, 142, 697, 471, 515, -508}. This system can be resolved into a final eqn that is only a quartic, though the expressions are messy. The first author translated the general form into a more symmetric version and found the complete soln in terms of four quadratics to be made squares. Let, [a+b, -a+b, a+b+2c, a+b-2c, e]k = [b-c-d, b+c-d, b-c+d, b+c+d, f]k (eq.1) for k = 2,4,6,8, then with {a,b}as free variables, {c,d,e,f} are given as, 32c2 = 5a2-10ab+8b2

32d2 = 29a2+6ab+72b2

4e2 = 9a2-6ab+36b2

4f2 = 13a2+2ab+4b2

with some {a,b} as trivial like the ratio a/b = {2, 6, etc). A non-trivial soln is {a,b} = {-288, 43} which makes all {c,d,e,f} as squares and yields the multi-grade which inspired this form. It is easy to make {c,d} as squares using an elliptic curve, though apparently it is only by chance that {e,f} are squares as well. But this can provide a rational family to the system, x1

k + x2k + x3

k + x4k + x5

k/2 = y1k + y2

k + y3k + y4

k + y5k/2, for k =

2,4,6,8 (End update.) 8.4 Twelve terms

(Update, 1/17/10): I. x1

k+x2k + … + x6

k = x1k+x2

k + … + x6k , for k

= 2,4,8 Wroblewski completely solved the system, [ap+d, ap-d, bq+c, bq-c, ar, bs]k = [bp+c, bp-c, aq+d, aq-d, as, br]k, for k = 2,4,8 with the palindromic conditions ma2+nb2 = c2 and na2+mb2 = d2 for some constants {p,q,r,s; m,n} as an elliptic curve imbedded in another elliptic curve. The soln is, {p,q,r,s; m,n} = {u+v, u-v, u-2v, u+2v; w, v2} where {u,v,w} satisfy the elliptic curve, 3u4+20u2v2+16v4 = w2 (eq.1) The first point is {u,v,w} = {2,1,12} which yields {p,q,r,s; m,n} = {3,1,0,4 ; 12, 1} and solving the simultaneous quadratics 12a2+b2 = c2 and a2+12b2 = d2 requires another elliptic curve. Since r = 0, this in fact is the known (8.5.5) identity discussed in Ten Terms, hence is valid for k = 2,4,6,8. The second point is {u,v,w} = {6,1,68} which gives {7,5,4,8; 68, 1} so, [7a+d, 7a-d, 5b+c, 5b-c, 4a, 8b]k = [7b+c, 7b-c, 5a+d, 5a-d, 8a, 4b]k

for k = 2,4,8 where 68a2+b2 = c2 and a2+68b2 = d2. And so on for all rational points of eq.1. (End update.) (Update, 1/12/10): II. x1

k+x2k + … + x6

k = x1k+x2

k + … + x6k , for

k = 2,4,6,8 Wroblewski gave three new beautiful identities: 1. If 64a2-11b2 = 5c2 and -11a2+64b2 = 5d2, then (2a+5b+c)k + (2a+5b-c)k + (5a-2b+d)k + (5a-2b-d)k + (4a+6b)k + (6a-4b)k =(2a-5b+c)k + (2a-5b-c)k + (5a+2b+d)k + (5a+2b-d)k + (4a-6b)k + (6a+4b)k

for k = 2,4,6,8. Trivial ratios are a/b = {1, 2}. These two quadratic conditions define an elliptic curve, hence the system has an infinite number of rational solns. 2. If 248a2-27b2 = 5c2 and -27a2+248b2 = 5d2, then (a+10b+c)k + (a+10b-c)k + (10a-b+d)k + (10a-b-d)k + (a+11b)k + (11a-b)k =(a-10b+c)k + (a-10b-c)k + (10a+b+d)k + (10a+b-d)k + (a-11b)k + (11a+b)k

for k = 2,4,6,8. Trivial ratios are a/b = {1, 3}, while a non-trivial soln is {a,b} = {9533, 3439} from which an infinite more can be computed. These belong to the general form, [pa+qb+c, pa+qb-c, qa-pb+d, qa-pb-d, ra+sb, sa-rb]k =[pa-qb+c, pa-qb-c, qa+pb+d, qa+pb-d, ra-sb, sa+rb]k

where ma2+nb2 = c2 and na2+mb2 = d2 (two quadratics with palindromic coefficients) which also appeared in Tenth Powers by this author. Notice how the variable b is just negated in the RHS. Thus, there are now three known non-trivial solns,

{p,q,r,s} = {1, 3, 2, 8}; {m,n}= {45, -11}; for k = 2,4,6,8,10{p,q,r,s} = {2, 5, 4, 6}; {m,n}= {64/5, -11/5}{p,q,r,s} = {1, 10, 1, 11}; {m,n}= {248/5, -27/5} and whether there are more is unknown, though Wroblewski has done a numerical search for {p,q,r,s} within a bound. (By expanding the system, one can linearly express {m,n} in terms of {p,q,r,s}, so the latter are the true unknowns.) 3. If -15a2+96b2 = 9c2 and -96a2+825b2 = 9d2, then (a+8b+c)k + (a+8b-c)k + (2a-b+d)k + (2a-b-d)k + (3a-2b+c)k + (3a-2b-c)k =(a-8b+c)k + (a-8b-c)k + (2a+b+d)k + (2a+b-d)k + (3a+2b+c)k + (3a+2b-c)k

for k = 2,4,6,8. Trivial ratios are a/b = {1, 2, 3, 5/2}. This generalizes the form above into, [pa+qb+c, pa+qb-c, ra+sb+d, ra+sb-d, ta+ub+c, ta+ub-c]k =[pa-qb+c, pa-qb-c, ra-sb+d, ra-sb-d, ta-ub+c, ta-ub-c]k

where p1a2+p2b2 = c2, and p3a2+p4b2 = d2. Again, the variable b has simply been negated in the RHS. Whether there are other non-trivial solns of this form, or if it can be extended up to k = 10, is unknown. 8.5 Fourteen terms Birck-Sinha Theorem

Theorem: If ak+bk+ck = dk+ek+fk, k = 2,4, where a+b ≠ c; d+e ≠ f, then, (a+b+c)k + (-a+b+c)k + (a-b+c)k + (a+b-c)k + (2d)k + (2e)k + (2f)k + 0 =0 + (2a)k + (2b)k + (2c)k + (-d+e+f)k + (d-e+f)k + (d+e-f)k + (d+e+f)k

for k = 1,2,4,6,8. Note that for any {a,b,c}, then, 0 + (2a)k + (2b)k + (2c)k = (a+b+c)k + (-a+b+c)k + (a-b+c)k + (a+b-c)k, for k = 1,2 and likewise for the {d,e,f}. The zero term is added to show that, if expressed in terms of {xi, yi}, then sums of appropriate terms from opposite sides are equal, namely, x1+y1 = x2+y2 = x3+y3 = x4+y4 = a+b+cx5+y5 = x6+y6 = x7+y7 = x8+y8 = d+e+f which also indicates that it is just a special case of a more general theorem given in the next section. Proof: (Sinha) Define {s2, s4}:= {a2+b2+c2, a4+b4+c4}, and Fk:= (a

+b+c)k + (-a+b+c)k + (a-b+c)k + (a+b-c)k - (2a)k - (2b)k - (2c)k and we find that: F1 = 0, F2 = 0, F4 = 12(s2

2-2s4), F6 = 60s2

(s22-2s4), F8 = 28(7s2

2+2s4)(s22-2s4)

Given analogous functions t1, t2 in the variables {d,e,f}, one can

evaluate Fk(d,e,f) similarly. Since it is given that ak+bk+ck = dk+ek+fk for k = 2,4, then Fk(a,b,c) = Fk(d,e,f) for k = 1,2,4,6,8. (End proof.) Also, using F4 and F6, we get a Ramanujan-type identity, 5[a2+b2+c2][(a+b+c)4 + (-a+b+c)4 + (a-b+c)4 + (a+b-c)4 - (2a)4 - (2b)4 - (2c)4] =[(a+b+c)6 + (-a+b+c)6 + (a-b+c)6 + (a+b-c)6 - (2a)6 - (2b)6 - (2c)6] for arbitrary {a,b,c}. Using a similar analysis, Hirschhorn showed Ramanujan may have used this method to find his 6-10-8 Identity and, in the process, Hirschhorn found an analogous 3-7-10 Identity. We can also use the Birck-Sinha theorem to find identities with fewer terms. To recall, we have, (a+b+c)k + (-a+b+c)k + (a-b+c)k + (a+b-c)k + (2d)k + (2e)k + (2f)k + 0 =0 + (2a)k + (2b)k + (2c)k + (-d+e+f)k + (d-e+f)k + (d+e-f)k + (d+e+f)k

If we equate a pair of terms (in color) on one side to the other side, (a+b-c) = (2c)(2f) = (d+e-f) we get, in addition to the requirement ak+bk+ck = dk+ek+fk, for k = {2,4}, the auxiliary conditions that, a+b = 3cd+e = 3f Fortunately, such a system is solvable by, {a,b,c} = {3p+u, 3p-u, 2p}{d,e,f} = {3q+v, 3q-v, 2q}

(3p+u)k + (3p-u)k + (2p)k = (3q+v)k + (3q-v)k + (2q)k, k = {2,4} where, p2+12q2 = u2

12p2+q2 = v2

and the Birck-Sinha theorem becomes the known (k.5.5) Birck-Sinha Identity discussed previously, (4p)k + (p-u)k + (p+u)k + (2p)k + (3q+v)k + (3q-v)k + (2q)k + 0 =0 + (3p+u)k + (3p-u)k + (2p)k + (q-v)k + (q+v)k + (2q)k + (4q)k

for k = {1,2,4,6,8} since two terms (as well as the zero term) on either side cancel out. One cannot use the theorem to find another (k.5.5) identity (as trying to cancel other combinations of terms yield trivial results). Wroblewski instead looked for more parametric solutions to, x1

k + x2k + x3

k = y1k + y2

k + y3k, for k = 2,4

where at least x1+x2 = 3x3 since this is enough to yield a (k.6.6) identity. Equivalently, let, {x1, x2, x3} = {3p+u, 3p-u, 2p} and the condition is, y1

2 + y22 + y3

2 = 2(11p2+u2)

y14 + y2

4 + y34 = 2(89p4+54p2u2+u4)

though any relation that yields xi = yi must be avoided as it is trivial.

There are only seven infinite families known so far. The first is by Birck & Sinha, the second is by Piezas, while the last five are by Wroblewski: 1. y1+y2 = 3y3: (3p+u)k + (3p-u)k + (2p)k = (3q+v)k + (3q-v)k + (2q)k

p2+12q2 = u2

12p2+q2 = v2

Ex. {p,q} = {218, 11869} 2. x1+x2 = y1+y2: (3p+u)k + (3p-u)k + (2p)k = (3p+v)k + (3p-v)k + (2q)k

26p2-3q2 = -u2 24p2-q2 = -v2

Ex. {p,q} = {143, 749} 3. x1-x2 = y1-y2: (3p+u)k + (3p-u)k + (2p)k = (u+v)k + (-u+v)k + (2q)k

8p2-3q2 = -3u2 11p2-2q2 = v2

Ex. {p,q} = {153, 343} Note: This yields the nice (k.6.6):

(4p)k + (p-u)k + (p+u)k + (u+v)k + (-u+v)k + (2q)k =(3p+u)k + (3p-u)k + (q-u)k + (q+u)k + (-q+v)k + (q+v)k, for k = {1,2,4,6,8} 4. 3(x1-x2) = 2(y1-y2): (3p+u)k + (3p-u)k + (2p)k = (3u+v)k + (-3u+v)k + (2q)k

8p2-3q2 = 5u2 3p2+q2 = v2

Ex. {p,q} = {323, 397} OR, p2-q2 = -9u2 104p2-23q2 = 9v2

Ex. {p,q} = {208, 233} 5. x1-x2 = 3y1-2y2: (3p+u)k + (3p-u)k + (2p)k = (2u-2v)k + (2u-3v)k + (2q)k

33u2-134uv+133v2 = 153p2

-192u2+112uv+937v2 = 612q2

Ex. {u,v} = {7,3} 6. 2x1-x2 = 3(y1-y2): (3p+u)k + (3p-u)k + (2p)k = (1/2)k ((p+u+v)k + (-p-u+v)k + (2q)k)

19p2+22pu+3u2 = 3q2 91p2-50pu+3u2 = 3v2 Ex. {p,u} = {9, -8}, or {p,u} = {45, 76} 7. 4x1+x2 = 3(y1-y2): (3p+u)k + (3p-u)k + (2p)k = (1/2)k ((5p+u+v)k + (-5p-u+v)k + (2q)k) 67p2+38pu+3u2 = 3q2 77p2+106pu-3u2 = -3v2 Ex. {p,u} = {9, -8} The two quadratic conditions of each family define an elliptic curve and, from an initial rational point, one can find an infinite more. Q. Any others? For the latest data on higher powers with a restricted number of terms, see Wroblewski's "A Collection of Numerical Solutions of Multi-grade Equations". (End update.) (Update, 1/17/10): Wroblewski The Birck-Sinha Theorem can be expressed in the form, (p+q)k + (-p+q)k + (-q+v)k + (q+v)k + (-r+u)k + (r+u)k + (2s)k + 0 =(-p+v)k + (p+v)k + (2q)k + 0 + (r+s)k + (-r+s)k + (-s+u)k + (s+u)k

This still has the expected common sums, x1+y1 = x2+y2 = x3+y3 = x4+y4 = q+v

x5+y5 = x6+y6 = x7+y7 = x8+y8 = s+u By letting, {p,q,r,s} = {-a+b, c, -d+e, f}; {u,v} = {d+e, a+b} one recovers the Birck-Sinha Theorem. Wroblewski gave the special case such that the above is valid for k = 1,2,4,6,8 as, ps = qrp2+3q2+s2 = u2

q2+r2+3s2 = v2

though there are certain {p,q,r,s} such that the system is trivial. Proof: Simply solve for {u,v}, express s in terms of {p,q,r}, and substitute into the system for k = 1,2,4,6,8. Extending to k = 10 will give the trivial values. <End proof> Assuming r = pt, s = qt, it reduces to two conditions, p2+3q2+(qt)2 = u2

q2+(pt)2+3(qt)2 = v2

which are two quadratic polynomials to be made squares and can easily be solved as an elliptic curve. An example is {p,q,t} = {9, 8, 2}. Alternatively, by assuming p = 3q, r = 3s, this equates two terms of one side to the other side, {-p+q, -r+s} = {-2q, -2s}. The conditions become, 12q2+s2 = u2

q2+12s2 = v2

Since two terms from each side cancel out, this reduces from a (k.7.7) to the (k.5.5) Birck-Sinha Identity discussed in a previous section.

(End update.) (Update, 2/2/10): Piezas More generally, given the same form above, (p+q)k + (-p+q)k + (-q+v)k + (q+v)k + (-r+u)k + (r+u)k + (2s)k + 0 =(-p+v)k + (p+v)k + (2q)k + 0 + (r+s)k + (-r+s)k + (-s+u)k + (s+u)k

for k = 1,2,4,6,8, then {p,q,r,s,u,v} must satisfy two simultaneous eqns, p2+2q2+v2 = r2+2s2+u2 (eq.1)(p2-q2)(q2-v2) = (r2-s2)(s2-u2) (eq.2) though avoiding trivial cases. Proof: Simply solve for {u,v} in terms of {p,q,r,s}, and substitute into the system. Also, this can be shown as the Birck-Sinha theorem in disguise by letting {p,q,r,s} = {-a+b, c, -d+e, f}; {u,v} = {d+e, a+b}, and (eq.1) and (eq.2) are satisfied if, ak+bk+ck = dk+ek+fk, for k = 2,4 For the special case u = v, then eq.1 becomes, p2+2q2 = r2+2s2

the complete soln of which reduces eq.2 to the easy problem of making a quadratic polynomial into a square. (End update.) 8.6 Sixteen terms Piezas

Theorem 1. If ak+bk+ck+dk = ek+fk+gk+hk, for k = 2,4 and abcd = efgh, where a+b+c+d ≠ 0 and e+f+g+h ≠ 0, (call the entire system V1) then,

(-a+b+c+d)k + (a-b+c+d)k + (a+b-c+d)k + (a+b+c-d)k + (2e)k + (2f)k + (2g)k + (2h)k =(2a)k + (2b)k + (2c)k + (2d)k + (-e+f+g+h)k + (e-f+g+h)k + (e+f-g+h)k + (e+f+g-h)k for k = 1,2,4,6,8. (Call this derived system V2.) This has the common sums, x1+y1 = x2+y2 = x3+y3 = x4+y4 = a+b+c+dx5+y5 = x6+y6 = x7+y7 = x8+y8 = e+f+g+h Thus if a+b+c+d = e+f+g+h = 0, then xi = -yi and the system becomes trivial so this case should be avoided. However, one can set d = h = 0, and this (k.8.8) reduces to the (k.7.7) Birck-Sinha Theorem. Proof: Following Sinha’s approach, define, {s1, s2, s4} = {abcd, a2+b2+c2+d2, a4+b4+c4+d4} and Fk = (a+b+c-

d)k + (a+b-c+d)k + (a-b+c+d)k + (-a+b+c+d)k - (2a)k - (2b)k - (2c)k - (2d)k

and we can evaluate Fk for certain k in terms of the si. Set s0 =

-8s1+s22-2s4, then,

F1 = F2 = 0, F4 = 12s0, F6 = 60s0s2, F8 = 28s0

(-24s1+7s22+2s4),

For the variables e,f,g,h, assume analogous functions t1, t2, t4, t6. Obviously Fk(e,f,g,h) at the same k will be evaluated identically. Since

it is given that ak+bk+ck+dk = ek+fk+gk+hk, for k = 2,4, and abcd = efgh, this implies Fk(a,b,c,d) = Fk(e,f,g,h) for k = 1,2,4,6,8, and we complete the proof. Also, using F4 and F6, we can get the more general identity, 5[a2+b2+c2+d2]*[(a+b+c-d)4 + (a+b-c+d)4 + (a-b+c+d)4 + (-a+b+c+d)4 - (2a)4 - (2b)4 - (2c)4 - (2d)4] =[(a+b+c-d)6 + (a+b-c+d)6 + (a-b+c+d)6 + (-a+b+c+d)6 - (2a)6 - (2b)6 - (2c)6 - (2d)6] for arbitrary {a,b,c,d}. Note: The theorem can be extended to 10th powers if, ak+bk+ck+dk = ek+fk+gk+hk, for k = 2,4,6 and abcd = efgh, but this has only trivial solns. Proof: Define Fk = ak

+bk+ck+dk-(ek+fk+gk+hk). If Fk = 0 for k = 2,4, then, F8 = -4(abcd+efgh)(abcd-efgh) + (4/3)(F6)(a2+b2+c2+d2) Thus, if both abcd-efgh = 0 and F6 = 0, this implies F8 = 0 as well, in contradiction to Bastien’s Theorem. So, if Fk = 0 for k = 2,4,6, then abcd ≠ efgh. Or if abcd = efgh, then Fk = 0 only for k = 2,4. <End proof.>

(Update, 2/15/10): Piezas, Wroblewski Using a numerical search with constraints, Wroblewski found hundreds of solns to a multi-grade octic chain of form, x1

k+x2k+x3

k+x4k-(x5

k+x6k+x7

k+x8k) = y1

k+y2k+y3

k+y4k-(y5

+y7k+y8

k) = z1k+z2

k+z3k+z4

k-(z5k+z6

k+z7k+z8

k) valid for k = 2,4,6,8. An example is, [221, 93, 73, 9]k-[219, 117, 31, 17]k = [198, 116, 96, 32]k-[192, 144, 58, 44]k = [189, 125, 105, 23]k-[175, 161, 75, 27]k

Note that it is also true x1

k+x2k+x3

k+x4k = x5

k+x6k+x7

k+x8k, for k =

2,4 and x1x2x3x4 = x5x6x7x8, relationships which also hold for the yi and zi. It turns out the general case could be explained in the context of the theorem given by the author in section 8.6 above. Moving half of the terms of one side to the other yields, (2a)k + (2b)k + (2c)k + (2d)k - ((2e)k + (2f)k + (2g)k + (2h)k) =(a+b+c-d)k + (a+b-c+d)k + (a-b+c+d)k + (-a+b+c+d)k - ((e+f+g-h)k + (e+f-g+h)k + (e-f+g+h)k + (-e+f+g+h)k) and since either sign of {±d, ±h} can be used and still satisfy abcd = efgh, then the RHS has two distinct octuplets for a given {a,b,c,d,e,f,g,h}. Thus, given, p1

k+p2k+p3

k+p4k = q1

k+q2k+q3

k+q4k, k = 2,4

where p1p2p3p4 = q1q2q3q4, then this automatically leads to an octic chain of length 3. Wroblewski’s solns satisfies the constraints,

p1p2 = q3q4p3p4 = q1q2

p12+p2

2 = q12+q2

p32+p4

2 = q32+q4

p14+p2

4+p34+p4

4 = q14+q2

4+q34+q4

The first author found the complete soln to this system as an elliptic “curve”, namely, {p1, p2, p3, p4} = {ar+bc, ac-br, ds+ef, df-es}{q1, q2, q3, q4} = {ar-bc, ac+br, ds-ef, df+es} r = (abc2+def2)(d2-e2)(f/y), s = (abc2+def2)(a2-b2)(c/y) and {a,b,c,d,e,f} must satisfy, (abc2+def2)(c2de)(a2-b2)2 + (abc2+def2)(abf2)(d2-e2)2 = y2

which is only a quartic polynomial in {c,f} to be made a square. Given one soln, then an infinite more can be found. Piezas Another set of conditions is, p1p4 = q1q4p2p3 = q2q3p1+p2 = q1+q2

p1k+p2

k+p3k+p4

k = q1k+q2

k+q3k+q4

k, k = 2,4 The soln can be given as,

{p1, p2, p3, p4} = {a(p+q), b(r-s), c(r+s), d(p-q)}{q1, q2, q3, q4} = {a(p-q), b(r+s), c(r-s), d(p+q)} {p,q,r,s} = {a(b2-c2), be, (a2-d2)b, ae} and where {a,b,c,d,e} satisfy, a2(3b2-c2)+e2 = (b2+c2)d2

This can be easily solved parametrically as quadratic forms as discussed in Form 17, Fourth Powers . (End update) (Update, 1/25/10): Wroblewski Given, (a2)k + (b2)k + (a2+b2)k = (c2)k + (d2)k + (c2+d2)k, for k = 2,4 (eq.1) or equivalently, a4+a2b2+b4 = c4+c2d2+d4

then we get the [k.8.8] identity, [ap+cq, -ap+cq, bp+dq, -bp+dq, dp+aq, dp-aq, cp+bq, cp-bq]k = [ap+dq, -ap+dq, bp+cq, -bp+cq, cp+aq, cp-aq, dp+bq, dp-bq]k, for k = 1,2,4,8 (Notice how {c,d} of the LHS is replaced by {d,c} in the RHS.) for arbitrary {p,q} and some constants {a,b,c,d}, hence giving an identity in terms of linear forms. An example is {a,b,c,d} = {1, 26, 17, 22}, though Choudhry has given a 7th deg identity for,

a4+ma2b2+b4 = c4+mc2d2+d4

for arbitrary m. See Form 3 here. To find a [8.7.7], since {p,q} are arbitrary, one can equate appropriate xp = yq so a pair of terms cancel out. Note: Eq. 1, or more generally its unsquared version, is at the core of the Ramanujan 6-10-8 Identity. (End update.) (Update, 1/17/10):8.5 Eighteen terms Lander (28k+4 + 1)8 = (28k+4-1)8 + (27k+4)8 + (2k+1)8 + 7(25k+3)8 + 7(23k+2)8 Note: This has a counterpart for 4th powers. I misplaced my notes but, if I remember correctly, 7 is replaced by 3(?), and the powers of 2 are changed appropriately. If someone can come up with something, pls submit it. (End update) (Update, 1/18/10): Wroblewski suggested the family which starts with Pythagorean triples, (2xy)2 + (x2-y2)2 = (x2+y2)2 2(4x3y)4 + 2(2xy3)4 + (4x4-y4)4 = (4x4+y4)4 (16x7y)8 + (2xy7)8 + 7(8x5y3)8 + 7(4x3y5)8 + (16x8-y8)8 = (16x8+y8)8

where Lander’s 8th powers identity was just the special case {x,y} = {2k, 1}. If we continue and try to decompose (256x16+1)16 - (256x16-1)16 as a sum of 16th powers, we find the coefficients have

prime factors other than 2 and/or 2k-1, hence such neat decompositions of nth = 2k powers stop at n = 8. (End update)

PART 12. Ninth Powers 9.1 Nine terms9.2 Ten terms9.3 Twelve terms9.4 Fourteen terms9.5 Sixteen terms 9.1 Nine Terms It is conjectured as a special case of the Lander-Parkin-Selfridge Conjecture that, for odd k > 3, the eqn, x1

k + x2k + …+ xk

k = 0 has non-trivial solns in positive and negative integers. Three are known for k = 5, but none have yet been found for k = 7,9, or higher. 9.2 Ten Terms Results for 9th powers and higher with a minimal number of terms came only in the last decade. For an equal sum of five 9th powers, x1

k+x2k+x3

k+x4k+x5

k = y1k+y2

k+y3k+y4

k+y5k (eq.1)

the first for k = 9 was found by Ekl in 1997 and is given by,

{192, 91, 101, 26, 30} = {12, 180, 17, 175, 116} Surprisingly this also satisfies a lot of side conditions, namely x1 = x2+x3 = y1+y2 = y3+y4. The first for k = 1,3,9 was found by Wroblewski in 2004, {51, 253, 412, 600, 624} = {100, 187, 429, 603, 621} which also satisfies x1+x2+x3 = y1+y2+y3; x4+x5 = y4+y5. While an analogous k = 1,3,5 is trivial, there is a non-trivial k = 1,3,7 and now a k = 1,3,9, so it is tempting to speculate there is a k = 1,3,n for all odd n > 5. There are now 22 known solns to ten 9th powers equal to zero, though Wroblewski's is the only one that is multi-grade, or even valid for k = 1. See his tables here. a) Whether there are nine 9th powers equal to zero remains unknown.b) There is yet no solution to x1+x2+x3+x4+x5 = y1+y2+y3+y4+y5 = 0, unlike for 5th and 7th powers where analogous systems are known.c) In general, whether there are k positive kth powers equal to a kth power for k > 8 is also unknown, though Wroblewski has come close with ten 9th powers and twelve 10th powers: 429 + 999 + 1799 + 4759 + 5429 + 5749 + 6259 + 6689 + 8229 + 8519

= 9179

6210 +11510 +17210 +24510 +29510 +53310 +68910 +92710 +101110

+123410 +160310 +168410 = 177210

(Update, 2/18/10): On a hunch, this author re-checked Wroblewski’s soln for (k.5.5) for k = 1,3,9 to see if there was a partition such that it would be good for k = 2 as well. It turns out there was. Thus, [51, 253, 412, -621, 600]k = [187, 100, 429, 603, -624]k, for k = 1,2,3,9

with terms obeying the constraints, 1. x1+x2+x3 = y1+y2+y32. x4+x5 = y4+y53. (x1-y1)(x2-y2) = -(x3-y3)(x4-y4)4. (x1-y1)(x2-y2) = (x3-y3)(x5-y5) with the last two suggested by Wroblewski, though [4] is just a consequence of [2] and [3]. Why this “numerical curiosity” obeys such symmetric constraints is unknown. (Note that the difference xi-yi for all five pairs is divisible by 17.) (End update.) 9.3 Twelve Terms (Update, 1/11/10): For twelve terms, or an equal sum of six 9th powers, x1

k+x2k+x3

k+x4k+x5

k+x6k = y1

k+y2k+y3

k+y4k+y5

the multi-grade system k = 1,3,5,7,9 used to have just one known soln, but Wroblewski increased the total to five in 2009. (If the special case x1 = 0 is found, this will lead to an ideal soln of deg 10 of the Prouhet-Tarry-Escott problem.) Interestingly, all but one have a re-arrangement such that it is good for k = 2 as well, [-269, -173, -7, 29, 311, 313] = [-247, -193, -59, 91, 289, 323][57, 679, 1293, -115, -925, -1279] = [299, 767, 1205, -399, -995, -1167][407, 163, 341, -37, -371, -119] = [221, 311, 403, -23, -181, -347] [365, 1115, 1325, -305, -731, -1037] = [13, 23, 1319, 1177, -689, -1111][43, 161, 217, 335, 391, 463] = [85, 91, 283, 287, 403, 461] (only for

k = 1,3,5,7,9) The first was found by Shuwen in 2000 after two months of computer time! Wroblewski derived the second one from a special k = 2,4,6,8 eqn, and found the next three via a numerical search. (End update) Two things can be noted. First, analogous to the situation for even systems k = 2,4,…2n which, if terms are signed, can be made valid for k = 1, it seems odd systems k = 1,3,…2n+1 can be made true for k = 2 since one can “move” terms around and there might be a balanced partition, x1

k+x2k+…+xm

k = y1k+y2

k+…+ymk

also true for k = 2. The consequence of letting a system valid for an even power is that one can no longer transpose them arbitrarily and it may "behave" more predictably. Recall that we extended a result for fifth powers and gave a conjecture for the seventh as, "Let Fk:= x1

k+x2k+x3

k+x4k+x5

k – (y1k+y2

k+y3k+y4

k+y5k). If Fk = 0

for k = 1,2,3,5,7, then (11F9)(x12+x2

2+x32+x4

2+x52) = 9F11."

Its analogue for ninth powers is, "Let Fk:= x1

k+x2k+x3

k+x4k+x5

k+x6k – (y1

k+y2k+y3

k+y4k+y5

+y6k). If Fk = 0 for k = 1,2,3,5,7,9, then (13F11)

(x12+x2

2+x32+x4

2+x52+x6

2) = 11F13. It is readily verified that the first four solns satisfies this. The pattern is quite obvious and can be extrapolated for systems with powers 11th, 13th, and so on, though a proof is desired that indeed it is generally true. Second, there may be a partition such that the sum of an equal

number of terms on either side is zero, or x1+x2+…+xm = y1+y2+…+ym = 0. It turns out that for all five solns, it can be done in at least one way. (Update, 12/21/09): Wroblewski observed that the conjecture can be generalized as (13F11)(x1

2+x22+...+x6

2+y12+y2

2+...+y62)/2 = 11F13,

and similarly for other powers, even if Fk is not zero for k = 2. Of course, if it is for k = 2, then it reduces to the form given above. (End update.) (Update, 1/13/10): Wroblewski found the second k = 1,3,5,7,9 using the following special case of a general theorem, “If [x1, x2, x3, x4, x5]k = [y1, y2, y3, y4, y5]k, for k = 2,4,6,8, where x1-x2 = x3-x4 = y1-y2 = y2-y3 = N, then, [a+x5, -a+y4, -a+y3, a+x1, a+x3, -a+y5]k = [-a+x5, a+y4, a+y1, -a+x2,

-a+x4, a+y5]k, for k = 1,3,5,7,9 where a = N/2.” Only one known k = 2,4,6,8 has terms with these conditions (see Eighth Powers for details) and yields, [-767, 399, -299, 995, 1167, -1205] = [-1293, 925, 1279, -57, 115, -679] Piezas If the generic derived 9th degree system is arranged in the above manner, and with terms labeled as {pi, qi}, the form will

satisfy the ff five symmetric constraints, p1+p2 = q1+q2p1+p6 = q1+q6p4-p5 = q4-q5p1+p3+p5 = q1+q3+q5p2+p4+p6 = q2+q4+q6 Together with k = 3,5,7,9 (since k = 1 is implied), this is a system of 9 eqns in 11 unknowns (since, by scaling, one term can be set equal to unity without loss of generality.) This can be completely solved as, (1+a)k + (-1+b)k + (-3+c)k + (2+f)k + (2+e)k + (-1+d)k =(-1+a)k + (1+b)k + (3+c)k + (-2+f)k + (-2+e)k + (1+d)k

for k = 1,3,5,7,9 and two free variables {u,v} where, a = 4(u+v),b = 4(u-v)2c2 = -3+t+31uv-3w2d2 = 53+t+71uv-11w{e2, f2} = {p-q, p+q} and p = (1/2)(13+t+9uv-5w)(1/2)q2 = 20+4t+301uv+35u2v2-5(4+11uv)w where w2 = -15+2t+22uv-135u2v2, and t = 32(u2+v2). Thus, six variables S = {c,d,e,f,q,w} must be made squares, with {e,f} being two roots of an octic, and ±w is equally valid, though only one

sign may yield a rational value. One non-trivial soln is {u,v} = {46/263, 423/526} which gives the original equation that inspired the form. Even with six expressions to be made squares, due to the symmetric nature of the system, there is in fact an infinite number of {u,v} such that, with the sign of ±w chosen appropriately, all S are squares, one of which is the family, v = 4(u-1)/(12u-7) but unfortunately just yields trivial terms. (To find two others, one can expand at k = 11.) However, it may be possible there are other non-trivial solns to this system. (End update.) (Update, 7/25/09): Recent work by A. Bremner and J. Delorme ("On Equal Sums of Ninth Powers", Math. of Comp, July 2009) have shown that the multi-grade system, x1

k+x2k+x3

k+x4k+x5

k+x6k = y1

k+y2k+y3

k+y4k+y5

k+y6k (eq.1)

for k = 1,2,3,9 has an infinite number of distinct and non-trivial solns. They solved this by assuming, {x1, x2, x3, x4, x5, x6} = {u1+w, u2+w, u3+w, v1-w, v2-w, v3-w}{y1, y2, y3, y4, y5, y6} = {u1-w, u2-w, u3-w, v1+w, v2+w, v3+w} Note that this satisfies the particular set of constraints, x1-y1 = x2-y2 = x3-y3 = -(x4-y4) = -(x5-y5) = -(x6-y6) = 2w and, x1-y4 = -x4+y1; x2-y5 = -x5+y2; x3-y6 = -x6+y3. Eq.1 is identically true for k = 1 while k = 2,3 are also if, u1

n+u2n+u3

n = v1n+v2

n+v3n, for n = 1,2 (eq.2)

After finding {ui, vi} such that this is satisfied, one can then find a relationship between them and the free variable w so that it is valid for k = 9 as well. One can use two approaches: Method 1. Bremner-Delorme Define, {u1, u2, u3} = {a-b+c, c+bt, a+c+at}{v1, v2, v3} = {a-b+c+at, a+c+bt, c} These make eq.1 valid for k = 1,2,3. In the paper, after a lot of clever math, Bremner and Delorme gave the explicit xi, yi. This author started with those and derived the {ui, vi} given above. Expanding for k = 9, one gets a quintic (and some trivial linear factors) in {a,b,c}, but a quartic in w. One can factor this polynomial into a quadratic and cubic using the relation, w = 2a-b+3c+(a+b)t The quadratic can be disregarded as it yields only complex solns. The cubic factor, rather complicated to explicitly write here, is homogeneous in {a,b,c} and can be treated as an elliptic surface such that from an initial soln, one can compute an infinite more. One set among many given by Bremner and Delorme is, {a,b,c,t} = {-1, 8 , -5, 3} and with w = -4, this yields, [-18, 15, -13, -13, 22, -1] = [-10, 23, -5, -21, 14, -9], for k = 1,2,3,9. Note: For convenience, this author changed the paper's variables from {q1, q2, q4} to {-a, b, c}. However, values {a,b,c,t} such that w = 0 are trivial.

Method 2. Piezas An alternative way is that k = 9 becomes a quartic that factors into two quadratics. Eq.2 can be completely solved by, {u1, u2, u3} = {m-pr+q, -r-mp+q, pr+q+r}{v1, v2, v3} = {m+pr+q, r-mp+q, -pr+q-r} Note that the r in the ui are just negated in the vi. (This complete parametrization was also used by this author for the multi-grade system [k,3,3] for k = 1,2,6.) Expanding eq.1 for k = 9 results in a quartic in r with only even exponents plus the trivial linear factors, (1+p)(m+r)(m-r)prw = 0 (eq.3) To factor the quartic, use the relationship, m(p-1)-3q+w = 0 Solve for q and substitute when k = 9. One quadratic factor in r (call this Q1) can be disregarded since, as before, it only yields complex solns. The other one (call this Q2), for rational solns, its discriminant D which is only a quartic polynomial must be made a square. For convenience, set w = n/2 and we find this as the rather simple form, D: = f2m4-2efm3n-3e2m2n2-7fmn3-7en4 = y2

where {e, f} = {(7/2)(p2+p+1), 2(p-1)(p+2)(2p+1)} Treating D as an elliptic "surface", given an initial soln, one can then find an infinity, as in the previous method. (In fact, the relationship given above was tediously derived from that method.) Thus, any rational value of m,n,p such that y is rational but does not involve the trivial factors of eq.3, then non-trivially solves eq.1 for k = 1,2,3,9. An

example is {p,n} = {-1/4, -2} which, after letting {m, r} = {a, b} for aesthetic reasons, gives {xi, yi} as, xi = {-16+7a+3b, -16-2a-12b, -16-5a+9b; 8+7a-3b, 8-2a+12b, 8-5a-9b}yi = {8+7a+3b, 8-2a-12b, 8-5a+9b; -16+7a-3b, -16-2a+12b, -16-5a-9b} (Note that the b's in the xi are just negated in the yi.) Expanding for k = 9, one must solve the quadratic, -128-39a2+5a3-9(13+5a)b2 = 0 The discriminant D of this (after removing a square numerical factor) is, D: = (13+5a)(-128-39a2+5a3) = y2

which, if to be made a square, is an elliptic curve. Two small solns are a = {-5, -7/2}. From these initial values, one can then compute more. These give, up to permutation, the same xi, yi in the first method. Using the original variables, finding w from n, and deriving r from Q2, we have all the unknowns as {m,p,q,r,w} = {-5, -1/4, 7/4, 4, -1} and get an initial soln to eq.1 as [-13, -18, 15, -13, 22, -1] = [-5, -10, 23, -21, 14, -9], for k = 1,2,3,9. Using the other rational points on this elliptic curve will then yield an infinite number of solns to eq.1. Note 1: Of course, other choices of {p,n} will give different elliptic curves.Note 2: Whether in the first or second method, one can extend the range of exponents up to k = 1,2,3,5,9, but all solns, unfortunately, are now complex. To cover k = 7 as well did not yield any non-trivial linear relationship.Note 3: This author is of the opinion that with a different set of side

conditions, it will be possible to find an identity, whether as an elliptic curve or as polynomials, for the multi-grade system k = 1,3,5,7,9, just like its higher counterpart k = 2,4,6,8,10. (End update.) 9.4 Fourteen Terms (No identities are yet known.) 9.5 Sixteen Terms (Update, 2/15/10): Wroblewski, Piezas A multi-grade [k.4.4] for k = 2,4 can be used to find a nonic [k.8.8] for k = 1,2,3,4,5,9. Given six variables {p,q,s,t,y,z}, System 1:[10p+2q-s+t, -10p-2q-s+t, 2p-10q+y-z, -2p+10q+y-z]k =[10p-2q-s+t, -10p+2q-s+t, 2p+10q+y-z, -2p-10q+y-z]k, for k = 2,4 System 2:[5p+q+s, -5p-q+s, 5p+q+t, -5p-q+t, p-5q+y, -p+5q+y, p-5q+z, -p+5q+z]k =[5p-q+s, -5p+q+s, 5p-q+t, -5p+q+t , p+5q+y, -p-5q+y, p+5q+z, -p-5q+z]k, for k = 1,2,3,4,5 is identically true using the substitutions, {p,q,s,t,y,z} = {u/2, v/2, a+b, a-b, a+c, a-c}{b,u,c,v} = {eg-2fh, eh+fg, eg+2fh, eh-fg} for arbitrary {a,e,f,g,h}. It can be shown that [2] is derived from [1] using Theorem 5 of the Prouhet-Tarry-Escott problem. A) One can make the first system valid for k = 6 and the second for k =

6,7 as well if, (10g2-13h2)e2-(13g2-40h2)f2 = 0 for arbitrary a. This is reducible to an elliptic curve, with one soln as {e,f,g,h} = {171, 97, 7, 47}, and an infinite more can then be computed. B) However, if System 2 is to be true only for k = 1,2,3,4,5,9, then one must solve a quadratic in a with a sextic discriminant that must be made a square. The only known soln found by Wroblewski is {a,e,f,g,h} = {104, 7, 3/2, -18, 17}, though others may exist. Whether with the right constraints this can be reduced to a quartic polynomial to be made a square, hence an elliptic curve, is unknown. (End update) (Update, 2/16/10): As observed by Bremner and Delorme, the smallest [k.6.6] for k = 9, found by Lander et al through computer search back in 1967, is in fact good for k = 1,2,3,9, [18, 23, 13, -10, -15, -5]k = [21, 22, 9, -13, -14, -1]k

Notice also how appropriate pairs all have the same sum, 18-10 = 23-15 = 13-5 = 21-13 = 22-14 = 9-1 = 8 This “numerical curiosity” has an explanation, which lies in the simple eqn, 142 + 192 + 92 = 172 + 182 + 52

and is just a special instance of Theorem 5 of the Prouhet-Tarry-Escott Problem. Define, Fn = an+bn+cn+dn-(en+fn+gn+hn) and the [k.8.8],

[x+a, x+b, x+c, x+d, x-a, x-b, x-c, x-d]k = [x+e, x+f, x+g, x+h, x-e, x-f, x-g, x-h]k

1) If F2 = 0, and a variable x such that 42F4x4+28F6x2+3F8 = 0, then the system is for k = 1,2,3,9. 2) If F2 = F4 = 0 and 28F6x2+3F8 = 0, then it is for k = 1,2,3,4,5,9. Notice how appropriate pairs of terms all have the same sum 2x. Also, it reduces to a [k.6.6] if a pair from opposite sides, say d = h = 0. Lander’s soln was simply the case, {a,b,c,d} = {14, 19, 9, 0}{e,f,g,h} = {17, 18, 5, 0} and x = 4. Bremner and Delorme found a relationship between the terms such that the quartic in x factored into two quadratics. Making its discriminant a square involved an elliptic curve, so there are an infinite number of solns to [1]. This is discussed more in Section 9.3 above. For non-zero {d,h}, no example is yet known for [1], but for [2], Wroblewski directly found a nonic [k.8.8] for k = 1,2,3,4,5,9, which yields, {a,b,c,d} = {240, 63, 197, 122}{e,f,g,h} = {167, 10, 168, 243} and x = 52. Since, these variables can be reduced in size with appropriate transformations given in the previous update, it is possible there are others. (End update.)

PART 13. Tenth Powers 10.1 Ten terms10.2 Twelve Terms10.3 Fourteen Terms10.4 Sixteen Terms10.5 Twenty Terms 10.1 Ten Terms No non-trivial solution in the rationals is yet known to equal sums of five 10th powers, x1

10+x210+x3

10+x410+x5

10 = y110+y2

10+y310+y4

10+y510

10.2 Twelve Terms The eqn, x1

k+x2k+x3

k+x4k+x5

k+x6k = y1

k+y2k+y3

k+y4k+y5

k+x6k (eq.1)

for k = 10 has relatively small solns. The first one was found by Randy Ekl in 1997, [5, 23, 34, 34, 85, 92] = [16, 25, 28, 32, 71, 95] There are many more and whether these belong to a parametric family is unknown. However, a family has been found for the even multi-grade system k = 2,4,6,8,10 (pls see update below). The numerical example was first found in 1999, {22, 61, 86, 127, 140, 151} = {35, 47, 94, 121, 146, 148} by Kuosa and Meyrignac as k = 10, and Shuwen (who noticed it was not just for k = 10 but was multi-grade) which automatically leads to an ideal soln of deg 11. To see if this is also for k = 1 if the terms are signed, there are now 64 possible sums of ± a ± b ± c ± d ± e ± f (and an equal number for the terms on the other side), and Shuwen found that sums of one side are equal to the other in seven ways, a+b+c-d+e+f = -g+h-i+j+m+n = 333a+b-c-d+e+f = g+h-i-j+m+n = 161a+b+c-d-e+f = g+h+i-j+m-n = 53a+b-c+d-e+f = -g-h+i+j-m+n = 135a-b+c+d-e+f = -g-h+i-j+m+n = 185a-b+c-d+e-f = g-h+i+j-m-n = -91a-b-c+d-e+f = g-h-i+j+m-n = 13 not counting seven other combinations which simply negate the terms. Whether this is: (1) a fluke, or (2) a property of the family which this example belongs to, or (3) solns to even systems with signed terms tend to be valid for k = 1 the more terms there are, is not known. The second combination (arranged differently) also satisfies x1+x2 = y1+y2; x3+x4+x5+x6 = y3+y4+y5+y6 though again it is unknown if two or all three, x1 ± x2 = y1 ± y2x3 ± x4 = y3 ± y4x5 ± x6 = y5 ± y6 can be satisfied by another soln to this decic system. Note: Kuosa and Meyrignac were looking at 10.6.6 and among the 14 solns they found, one turned out to be for k = 2,4,6,8,10. If this was just a statistical fluke, one might expect some of the others to be also valid for, say, k = 2 or 4. This author checked and none is good for any other even exponent. So there is the situation of a small soln that is suddenly multi-grade for six exponents, counting k = 1. Together with its other properties, it seems there might be more to it than meets the eye. Update (June 6, 2009): Choudhry, Wroblewski It has been brought to my attention (by J. Wroblewski) that the case of an equal sum of six tenth powers, a1

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

2,4,6,8,10 has already been cracked (in 2008) and shown to have an infinity of solns by reducing it, like the other systems, to a particular elliptic curve, though the Kuusa-Meyrignac-Shuwen soln does not belong to this family. (See their paper "Ideal Solutions of the Tarry-Escott Problem of Degree Eleven with Applications to Sums of Thirteenth Powers", Hardy-Ramanujan Journal, Vol. 31, 2008.) Choudhry and Wroblewski defined ai and bi as, a1 = 2xy+x+2y-7; b1 = 2xy+2x+y-7a2 = 2xy-x-2y-7; b2 = 2xy-2x-y-7a3 = 2xy-2x+y+7; b3 = 2xy-x+2y+7a4 = 2xy+2x-y+7; b4 = 2xy+x-2y+7a5 = 3x+5y; b5 = 5x+3ya6 = 5x-3y; b6 = 3x-5y where a variable z has been set z = 1 without loss of generality. Substituting these values into the system, the higher eqns hold if the ff condition is satisfied, 8x2y2-17x2-17y2+98 = 0 Do the simple change of variables {x, y} = {u, v/(8u2-17)} and the condition reduces to, (-17+8u2)(-98+17u2) = v2

or a quartic polynomial in u that is to be made a square. As Choudhry and Wroblewski proved, this is an elliptic curve. One trivial soln is u = 1 but from this initial point, we can compute an infinite number of non-trivial ones such as u = -457/353, etc. However, by looking at the solns ai and bi, one can see certain relationships and symmetries between them. Using those relationships, this author will show that it can be simplified to a form analogous to the Letac-Sinha identity for k = 2,4,6,8. (Update, 2/1/10): Choudhry, Wroblewski The 6th power (k.4.4) identity can be generalized to the form (k.6.6) by adding the italized pair of terms, [-(axy+bx+cy-d), axy-bx-cy-d, axy-cx+by+d, -(axy+cx-by+d), ex+fy, fx-ey]k =[-(axy+by+cx-d), axy-by-cx-d, axy-cy+bx+d, -(axy+cy-bx+d), ey+fx, ex-fy]k where again, {x,y} merely swap places. (Since the eqn is homogeneous, one may assume the leading coefficient of the terms as a = 1 without loss of generality.) This obeys the same basic constraints (up to sign changes), p1-p2 = q1-q2,p3-p4 = q3-q4,

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

If we assume, p1+p2+p5 = q1+q2+q5, p3+p4+p6 = q3+q4+q6, then it must be the case that 2(b-c) = e-f (eq.1). Choudhry and Wroblewski found for k = 2,4,6,8,10, the equivalent non-trivial solns, {a,b,c,d,e,f} = {1, 1, 2, 14, 3, 5}, where 2x2y2-17(x2+y2)+392 = 0{a,b,c,d,e,f} = {2, 1, 2, 7, 3, 5}, where 8x2y2-17(x2+y2)+98 = 0 Disregarding eq.1, two other distinct non-trivial solns found (by Piezas), though only up to k = 2,4,6,8, are, {a,b,c,d,e,f} = {5, 3, 7, 3, 2, 10}, where 125x2y2-53(x2+y2)+45 = 0{a,b,c,d,e,f} = {5, 9, 11, 11, 10, 12}, where 125x2y2-221(x2+y2)+605 = 0 though this author is not certain if these are the same 8th power multi-grades found by Wroblewski. See also Sixth Powers for the context of this identity. (End update.) Piezas (a+3b+c)k+(-a-3b+c)k+(3a-b+d)k+(-3a+b+d)k+(2a+8b)k+(-8a+2b)k =(-a+3b+c)k+(a-3b+c)k+(3a+b+d)k+(-3a-b+d)k+(-2a+8b)k+(-8a-2b)k

for k = 2,4,6,8,10, where 45a2-11b2 = c2 and -11a2+45b2 = d2. where the ratio a/b = {2, 1/2, 3/2, 2/3} must be avoided as it yields trivial solns. Labelling terms as ai and bi, then a small tweak in the sign of {a1, a4, b1, b4} can make it valid for k = 1 as well. Note that, just like in the Letac-Sinha identity, the two quadratic polynomials to be made squares have palindromic or reversible coefficients. One can easily verify this soln by solving {c,d} in radicals for arbitrary {a,b} and substituting them into the system. Of course, if they are to be rational, then appropriate {a,b} must be chosen, with the smallest non-trivial one being {a,b} = {186, 331} giving, [886, -293, 1180, 953, 1510, -413] = [700, -107, 1511, 622, 1138, -1075] after removing a small common factor. By solving the first condition as a quadratic form {a, b} = {u2+11v2, 2u2+2uv-22v2} using this on the other will result in a quartic polynomial that is to be made a square, with some {u,v} as trivial, but a non-trivial is {u,v} = {12, -5}. Treating this as an elliptic curve, from this initial point, an infinite number of rational solns can then be computed. This identity was found using the relationships between the ai and bi and, whether in the Choudhry-Wroblewski version or by this author, they satisfy seven side conditions: a1+a2 = b1+b2a3+a4 = b3+b4a1-a2-a5 = b1-b2-b5a1-a2-a3+a4-a5-a6 = b1-b2-b3+b4-b5-b6a1+a2+a3-a4+a5+a6 = b1+b2-b3+b4+b5-b6

a1k+a2

k+a3k+a4

k = b1k+b2

k+b3k+b4

k, for k = 1,2 and together with the five eqns, a1

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

2,4,6,8,10 we have 12 (or 11) eqns in 12 unknowns, call this augmented system as S10+. By solving this system, it can be shown that the identity given by this author is the only non-trivial soln to S10+. (There may be just 11 eqns as one may be a consequence of the others.) Note 1: By negating {a1, a4, b1, b4}, the identity in fact is also valid for k = 1. Also, by eliminating or changing one of the side conditions, it may be possible to come up with a different identity from this one. What would be an appropriate change, I do not yet know.Note 2: I guess my postscript at the final section need to be modified. (However, the case k = 1,3,5,7,9 still needs to be cracked.) (Update, 1/12/10): The identity has the form, [pa+qb+c, pa+qb-c, qa-pb+d, qa-pb-d, ra+sb, sa-rb]k =[pa-qb+c, pa-qb-c, qa+pb+d, qa+pb-d, ra-sb, sa+rb]k

where ma2+nb2 = c2 and na2+mb2 = d2. Notice how the variable b is just negated in the RHS. Wroblewski did a search and found two new solns, though only for k = 2,4,6,8: {p,q,r,s; m,n} = {1, 3, 2, 8; 45, -11}; for k = 2,4,6,8,10{p,q,r,s; m,n} = {2, 5, 4, 6; 64/5, -11/5}{p,q,r,s; m,n} = {1, 10, 1, 11; 248/5, -27/5} See also Eighth Powers. (End update.) (Update, 1/11/10) 10.3 Fourteen Terms The smallest soln to, x1

k+x2k+x3

k+x4k+x5

k+x6k+x7

k = y1k+y2

k+y3k+y4

k+y5k+y6

+y7k (eq.1)

for k = 10 is, [1, 8, 15, 26, 26, 33, 38]10 = [22, 23, 24, 29, 32, 35, 36]10

given by Ekl (“New results in equal sums of like powers”, 1998). However, as far back in 1939, Moessner already managed to solve eq.1 (pre-computer searching!). The second(?) smallest soln, and the smallest in distinct integers, turns out to be multigrade, [32, -1, -61, -55, -31, -13, -28, 68]k = [44, -13, -49, -67, 20, -64, 23, 17]k

for k = 1,2,4,6,8,10, and belongs to an infinite family. Note that its terms obey the relations, -32+44 = -1+13 = 61-49 = -55+67 = 1231+20 = -13+64 = 28+23 = 68-17 = 51

Wroblewski, Piezas Let a2+b2 = c2 and a2+52b2 = d2, a pair of conditions which is a concordant form. Then a (k,7,7) for k = 1,2,4,6,8,10 is, (8b)k + (5a-4b)k + (-a-2d)k + (a-2d)k + (-5a-4b)k + (a-4b)k + (-12b+4c)k + (12b+4c)k =(4a+8b)k + (a-4b)k + (3a-2d)k + (-3a-2d)k + (-4a+8b)k + (-16b)k + (a+4c)k + (-a+4c)k

With terms as {xi, yi}, the identity obeys the constraints, x1-y1 = -(x2-y2) = x3-y3 = -(x4-y4) = -4ax5-y5 = -(x6-y6) = x7-y7 = -(x8-y8) = -(a+12b)

This was derived from a (k.8.8) discussed in the next section but since two terms (in blue) cancel out, then it reduces to a (k.7.7). However, ratios such as a/b = {4, 12, 12/5, 4/3} should be avoided as the system becomes trivial, with the smallest non-trivial soln as the smallest Pythagorean triple {a,b,c} = {3, 4, 5}, and d = 29. This yields,

[32, -1, -61, -55, -31, -13, -28, 68]k = [44, -13, -49, -67, 20, -64, 23, 17]k This was cited by Ekl (1998) as the smallest soln in distinct integers, but it was not mentioned in the paper that it was also multigrade. It is quite interesting that the smallest multigrade (k.7.7) for k = 2,4,6,8,10 belongs to a family, as it is not known if the smallest multigrade (k.6.6) belongs to a family or not. The two quadratic polynomials generally define an elliptic curve, and there is an infinite number of solns, with other small {a,b} as {612, 35} and {783, 56}. A pair of algebraic forms {a2+b2, a2+Nb2} that is to made squares is called a concordant form. (However, found throughout this book, is the more general case of {pa2+qb2, ra2+sb2} to be made squares.) This identity is a special case of a more general one involving 16 terms given below. 10.4 Sixteen Terms Piezas, Wroblewski The latter author suggested the system S1, [-a-c+x, a+c+x, -a+b+c-y, a-b-c-y, -a-b-c-t, a+b+c-t, -b+c+z, b-c+z]k =[a-c+x, -a+c+x, a+b+c-y, -a-b-c-y, -a+b-c-t, a-b+c-t, b+c+z, -b-c+z]k which, for some {a,b,c,x,y,z,t}, is valid for k = 1,2,4,6,8,10. This has a lot of symmetries and its terms {xi, yi} obeys the constraints, x1-y1 = -(x2-y2) = x3-y3 = -(x4-y4) = -2ax5-y5 = -(x6-y6) = x7-y7 = -(x8-y8) = -2bx1-x2+y1-y2 = -x3+x4+y5-y6 = -x7+x8-y7+y8 = -4cx5-x6 = -y3+y4 = -2(a+b+c) An example by Wroblewski is, [-31, 193, 75, -100, -179, 164, 72, 51]k = [53, 109, 159, -184, -60, 45, 191, -68]k

which is for k = 1,2,3,4,6,8,10. The former author found the complete radical soln using Mathematica. This entails making four quartic polynomials as squares. Set a = 1 without loss of generality and define {x,y,z,t} = {√p, √q, √r, √s}. Then {p,q,r,s} can be rationally expressed in terms of the free variables {b,c}. As the actual expressions are unwieldy, first define, p-s = (-2b+9b2+8c+2bc-8c2)/3q-s = 8(1-b)(1-c)(-b+c)c/(3b+3c+3bc)q-r = (-9+2b-2c-8bc+8c2)/3 and it is easy to solve for {p,q,r}. Substituting these into the system and factoring at either k = 6,8,10, one will find a common linear factor for the last unknown s. Expanding for k = 12 will yield the trivial forms of {b,c}. A non-trivial one which makes {p,q,r,s} squares is {b,c} = {17/12, 5/3} which gives the numerical example above. However, Wroblewski found that if b+c = 1/2, such as, {a,b,c} = {1, (u+12v)/(4u), (u-12v)/(4u)} then {p,s} are squares, {q,r} become just quadratic polynomials, and two terms {x6, y2} are identical, so it reduces to the simpler one in section 10.3. (End update) (Update, 1/17/10) 10.5 Twenty Terms Wroblewski added four more terms to each side of the (k.6.6) to get the (k.10.10), [pa+qb+c, pa+qb-c, qa-pb+d, qa-pb-d, ra+sb, sa-rb, ta+ub, ua-tb, va+wb, wa-vb]k =[pa-qb+c, pa-qb-c, qa+pb+d, qa+pb-d, ra-sb, sa+rb, ta-ub, ua+tb, va-wb, wa+vb]k for k = 2,4,6,8,10, where ma2+nb2 = c2 and na2+mb2 = d2. Again, the variable b is just negated in the RHS. Wroblewski found three non-trivial solns, {p,q,r,s,t,u,v,w; m,n} = {4, 4, 3, 11, 5, 7, 7, -3; 45, -11}{p,q,r,s,t,u,v,w; m,n} = {4, 4, 9, 17, 15, 13, 13, 7; 189, 85}{p,q,r,s,t,u,v,w; m,n} = {3, 3, 11, 17, 16, 14, 14, 10; 220, 136} though certain ratios a/b must be avoided and which can determined by expanding for k = 12. Interestingly, the first (k.10.10) soln uses the same {m,n} as the one for (k.6.6)! As these two quadratic conditions define an elliptic curve, it is unknown why certain curves are "favored", appearing again and again (it also appears in a (k.4.4) for k = 1,2,4,6). It is also remains to be known whether: a) there are others, b) if there is w = v = 0 so reduces to 16 terms, or c) if it can be non-trivially valid for k = 12. (End update) (Update, 3/1/10): Piezas This author noticed that, other than the obvious relations p = q, u = v, the above also obey, r = -q+vs = q+vmn = rstw Combined with the conditions k = 2,4,6,8,10, these were enough to find a formula for these variables. Thus, {p,q,r,s; t,u,v,w} = {y, y, -y+z, y+z; x+y, z, z, x-y}{m,n} = {x2+yz+y2, x2-yz+y2} where {x,y,z} satisfies the simple quadratic condition x2+3y2 = z2. Wroblewski’s were the cases, {x,y} = {1, 4}; {11, 4}; {13, 3} However, there is an infinite number of them given by {x,y,z} = {e2-3f2, 2ef, e2+3f2}. So a fourth one using {x,y} = {11, 5} gives, {p,q,r,s; t,u,v,w; m,n} = {5, 5, 9, 19; 16, 14, 14, 6; 216, 76} with one soln to 216a2+76b2 = c2 and 76a2+216b2 = d2 as, {a,b} = {495753715, 352750681} though presumably smaller solns may exist. From this initial rational point, one can then compute an infinite more. And so on for an infinite number of other formulas using different {m,n} and other elliptic curves, though expanding this family at k = 12 or 14 generates only trivial solns. (End update.)

10+x210+x3

10+x410+x5

10 = y110+y2

10+y310+y4

10+y510

k+x2k+x3

k+x4k+x5

k+x6k = y1

k+y2k+y3

k+y4k+y5

k+x6k (eq.1)

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

a1k+a2

k+a3k+a4

k = b1k+b2

k+b3k+b4

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

k+x2k+x3

k+x4k+x5

k+x6k+x7

k = y1k+y2

k+y3k+y4

k+y5k+y6

+y7k (eq.1)

for k = 10 is, [1, 8, 15, 26, 26, 33, 38]10 = [22, 23, 24, 29, 32, 35, 36]10

10+x210+x3

10+x410+x5

10 = y110+y2

10+y310+y4

10+y510

k+x2k+x3

k+x4k+x5

k+x6k = y1

k+y2k+y3

k+y4k+y5

k+x6k (eq.1)

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

a1k+a2

k+a3k+a4

k = b1k+b2

k+b3k+b4

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

k+x2k+x3

k+x4k+x5

k+x6k+x7

k = y1k+y2

k+y3k+y4

k+y5k+y6

+y7k (eq.1)

for k = 10 is, [1, 8, 15, 26, 26, 33, 38]10 = [22, 23, 24, 29, 32, 35, 36]10

10+x210+x3

10+x410+x5

10 = y110+y2

10+y310+y4

10+y510

k+x2k+x3

k+x4k+x5

k+x6k = y1

k+y2k+y3

k+y4k+y5

k+x6k (eq.1)

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

a1k+a2

k+a3k+a4

k = b1k+b2

k+b3k+b4

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

k+x2k+x3

k+x4k+x5

k+x6k+x7

k = y1k+y2

k+y3k+y4

k+y5k+y6

+y7k (eq.1)

for k = 10 is, [1, 8, 15, 26, 26, 33, 38]10 = [22, 23, 24, 29, 32, 35, 36]10

10+x210+x3

10+x410+x5

10 = y110+y2

10+y310+y4

10+y510

k+x2k+x3

k+x4k+x5

k+x6k = y1

k+y2k+y3

k+y4k+y5

k+x6k (eq.1)

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

a1k+a2

k+a3k+a4

k = b1k+b2

k+b3k+b4

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

k+x2k+x3

k+x4k+x5

k+x6k+x7

k = y1k+y2

k+y3k+y4

k+y5k+y6

+y7k (eq.1)

for k = 10 is, [1, 8, 15, 26, 26, 33, 38]10 = [22, 23, 24, 29, 32, 35, 36]10

10+x210+x3

10+x410+x5

10 = y110+y2

10+y310+y4

10+y510

k+x2k+x3

k+x4k+x5

k+x6k = y1

k+y2k+y3

k+y4k+y5

k+x6k (eq.1)

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

a1k+a2

k+a3k+a4

k = b1k+b2

k+b3k+b4

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

k+x2k+x3

k+x4k+x5

k+x6k+x7

k = y1k+y2

k+y3k+y4

k+y5k+y6

+y7k (eq.1)

for k = 10 is, [1, 8, 15, 26, 26, 33, 38]10 = [22, 23, 24, 29, 32, 35, 36]10

10+x210+x3

10+x410+x5

10 = y110+y2

10+y310+y4

10+y510

k+x2k+x3

k+x4k+x5

k+x6k = y1

k+y2k+y3

k+y4k+y5

k+x6k (eq.1)

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

a1k+a2

k+a3k+a4

k = b1k+b2

k+b3k+b4

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

k+x2k+x3

k+x4k+x5

k+x6k+x7

k = y1k+y2

k+y3k+y4

k+y5k+y6

+y7k (eq.1)

for k = 10 is, [1, 8, 15, 26, 26, 33, 38]10 = [22, 23, 24, 29, 32, 35, 36]10

10+x210+x3

10+x410+x5

10 = y110+y2

10+y310+y4

10+y510

k+x2k+x3

k+x4k+x5

k+x6k = y1

k+y2k+y3

k+y4k+y5

k+x6k (eq.1)

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

a1k+a2

k+a3k+a4

k = b1k+b2

k+b3k+b4

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

k+x2k+x3

k+x4k+x5

k+x6k+x7

k = y1k+y2

k+y3k+y4

k+y5k+y6

+y7k (eq.1)

for k = 10 is, [1, 8, 15, 26, 26, 33, 38]10 = [22, 23, 24, 29, 32, 35, 36]10

10+x210+x3

10+x410+x5

10 = y110+y2

10+y310+y4

10+y510

k+x2k+x3

k+x4k+x5

k+x6k = y1

k+y2k+y3

k+y4k+y5

k+x6k (eq.1)

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

a1k+a2

k+a3k+a4

k = b1k+b2

k+b3k+b4

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

k+x2k+x3

k+x4k+x5

k+x6k+x7

k = y1k+y2

k+y3k+y4

k+y5k+y6

+y7k (eq.1)

for k = 10 is, [1, 8, 15, 26, 26, 33, 38]10 = [22, 23, 24, 29, 32, 35, 36]10

10+x210+x3

10+x410+x5

10 = y110+y2

10+y310+y4

10+y510

k+x2k+x3

k+x4k+x5

k+x6k = y1

k+y2k+y3

k+y4k+y5

k+x6k (eq.1)

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

a1k+a2

k+a3k+a4

k = b1k+b2

k+b3k+b4

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

k+x2k+x3

k+x4k+x5

k+x6k+x7

k = y1k+y2

k+y3k+y4

k+y5k+y6

+y7k (eq.1)

for k = 10 is, [1, 8, 15, 26, 26, 33, 38]10 = [22, 23, 24, 29, 32, 35, 36]10

10+x210+x3

10+x410+x5

10 = y110+y2

10+y310+y4

10+y510

k+x2k+x3

k+x4k+x5

k+x6k = y1

k+y2k+y3

k+y4k+y5

k+x6k (eq.1)

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

a1k+a2

k+a3k+a4

k = b1k+b2

k+b3k+b4

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

k+x2k+x3

k+x4k+x5

k+x6k+x7

k = y1k+y2

k+y3k+y4

k+y5k+y6

+y7k (eq.1)

for k = 10 is, [1, 8, 15, 26, 26, 33, 38]10 = [22, 23, 24, 29, 32, 35, 36]10

10+x210+x3

10+x410+x5

10 = y110+y2

10+y310+y4

10+y510

k+x2k+x3

k+x4k+x5

k+x6k = y1

k+y2k+y3

k+y4k+y5

k+x6k (eq.1)

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

a1k+a2

k+a3k+a4

k = b1k+b2

k+b3k+b4

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

k+x2k+x3

k+x4k+x5

k+x6k+x7

k = y1k+y2

k+y3k+y4

k+y5k+y6

+y7k (eq.1)

for k = 10 is, [1, 8, 15, 26, 26, 33, 38]10 = [22, 23, 24, 29, 32, 35, 36]10

10+x210+x3

10+x410+x5

10 = y110+y2

10+y310+y4

10+y510

k+x2k+x3

k+x4k+x5

k+x6k = y1

k+y2k+y3

k+y4k+y5

k+x6k (eq.1)

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

a1k+a2

k+a3k+a4

k = b1k+b2

k+b3k+b4

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

k+x2k+x3

k+x4k+x5

k+x6k+x7

k = y1k+y2

k+y3k+y4

k+y5k+y6

+y7k (eq.1)

for k = 10 is, [1, 8, 15, 26, 26, 33, 38]10 = [22, 23, 24, 29, 32, 35, 36]10

10+x210+x3

10+x410+x5

10 = y110+y2

10+y310+y4

10+y510

k+x2k+x3

k+x4k+x5

k+x6k = y1

k+y2k+y3

k+y4k+y5

k+x6k (eq.1)

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

(-p1)k+p2k+p3

k+(-p4)k = (-q1)k +q2k+q3

k+(-q4)k, k = 1,2

p52+p6

2 = q52+q6

a1k+a2

k+a3k+a4

k = b1k+b2

k+b3k+b4

k+a2k+a3

k+a4k+a5

k+a6k = b1

k+b2k+b3

k+b4k+b5

k+b6k, for k =

k+x2k+x3

k+x4k+x5

k+x6k+x7

k = y1k+y2

k+y3k+y4

k+y5k+y6

+y7k (eq.1)

for k = 10 is, [1, 8, 15, 26, 26, 33, 38]10 = [22, 23, 24, 29, 32, 35, 36]10

PART 31. Eleventh and Higher Powers (Update, 1/11/10): I. Eleventh Powers 11.1 Eleven and Twelve terms (No soln is yet known.) 11.2 Fourteen Terms It is conjectured, as a special balanced case of the Lander-Parkin-Selfridge (LPS) Conjecture, that the equal sums of like powers, x1

k+x2k+...+xn

k = y1k+y2

k+...+ynk

for odd k where n = (k+3)/2 always has non-trivial solns. There are parametric identities for all odd k < 11 but, for k = 11, only a single numerical example where all terms are positive is known. Interestingly, at least for the first few k, "first" solns have relatively small terms and valid also for k = 1, excepting the one for k = 11, k = 1,3: [1, 5, 5] = [2, 3, 6]k = 1,5; [5, 6, 6, 8] = [4, 7, 7, 7]; (Subba Rao, 1934)

k = 1,7; [2, 12, 15, 17, 18] = [8, 8, 13, 16, 19]; (LPS, 1967)k = 1,9; [1, 13, 13, 14, 18, 23] = [5, 9, 10, 15, 21, 22]; (LPS, 1967)k = 11; [46, 52, 115, 119, 249, 566, 614] = [127, 152, 175, 212, 441, 487, 629]; (Nuutti Kuosa, 2004) As k = 11 was not found by an exhaustive search (?), it may be possible there is a smaller soln valid for both k = {1, 11}, just like the rest in this list. In fact, the one for k = 9 belongs to a family found by Bremner and Delorme (2009) that is for k = 1,3,9. If indeed first solns are relatively small, it may be feasible to find the one for k = 13. 11.3 Twenty Terms (Update, 1/18/10): Wroblewski Using a variant of Theorem 5 discussed in Equal Sums of Like Powers, he used the 10th power [k,6,6] identity to get an 11th power [k,10,10] using, NewLeftTerms = {OldLeftTerms+c, OldRightTerms-c}NewRightTerms = {OldLeftTerms-c, OldRightTerms+c} Normally, this doubles the number of terms but since four terms cancel out, one gets, [a+3b+2c, -a+3b-2c, 2a-8b-c, -2a-8b+c, -8a-2b-c, -8a+2b+c, 3a+b-c-d, 3a-b+c-d, 3a+b-c+d, 3a-b+c+d]k =[a+3b-2c, -a+3b+2c, 2a-8b+c, -2a-8b-c, -8a-2b+c, -8a+2b-c, 3a+b+c-d, 3a-b-c-d, 3a+b+c+d, 3a-b-c+d]k

for k = 1,2,3,5,7,9,11 and same conditions 45a2-11b2 = c2, 45b2-11a2 = d2. Naturally, c has just been negated in the RHS. The process can be continued, with c replaced by d in the transformation, to find a [k,16,16] for k = 1,2,3,4,6,8,10,12, since eight terms cancel. (End update.)

II. Twelfth Powers 12.1 Twelve Terms (No solution is yet known.) 12.2 Fourteen Terms Similar to odd k, it is conjectured that the equal sums of like powers, x1

k+x2k+...+xn

k = y1k+y2

k+...+ynk

for even k where n = (k+2)/2 always has non-trivial solns. Identities are known for all even k < 12, but there is only a single numerical example for k = 12. "First" solns have relatively small terms: k = 2: [1, 7] = [5, 5]k = 4: [2, 4, 7] = [3, 6, 6]k = 6: [2, 2, 9, 9] = [3, 5, 6, 10]; (Subba Rao, 1934)k = 8: [1, 10, 11, 20, 43] = [5, 28, 32, 35, 41]; (LPS, 1967)k = 10: [5, 23, 34, 34, 85, 92] = [16, 25, 28, 32, 71, 95]; (Randy Ekl, 1997)k = 12: [3, 37, 42, 48, 88, 89, 95] = [30, 54, 73, 73, 74, 77, 99]; (Greg Childers, 2000) No example is known for k = 14 as a (14,8,8) though, if the pattern continues, the first soln might have terms just in the lower end of 3 digits. (End update.) III. Thirteenth Powers (Coming)

IV. Higher Powers Update (8/23/09): B. D. Bhargava ([email protected]) gave interesting multi-grade identities higher than tenth powers. For more details, see section "Maths -- A Unique Equation" of his website: http://bhargavabd.hpage.com. Given any value x, then, A. For k = {1,2,3,...9}: 4(x+2)k + 2(x+5)k + 2(x+6)k + 4(x+8)k + 9(x+11)k + (x+13)k - 7(x+1)k - 3(x+4)k - 4(x+7)k - (x+9)k - 6(x+10)k - 5(x+12)k =-4(x-2)k - 2(x-5)k - 2(x-6)k - 4(x-8)k - 9(x-11)k - (x-13)k + 7(x-1)k + 3(x-4)k + 4(x-7)k + (x-9)k + 6(x-10)k + 5(x-12)k - 8xk

B. For k = {1,2,3,...10}: (x-14)k + 8(x-12)k + 10(x-9)k + 6(x-6)k + 7(x-3)k + 4(x-1)k + 7(x+2)k + 2(x+4)k + 5(x+5)k + 2(x+7)k + 3(x+8)k + 10(x+10)k + (x+11)k + 5(x+13)k =(x+14)k + 8(x+12)k + 10(x+9)k + 6(x+6)k + 7(x+3)k + 4(x+1)k + 7(x-2)k + 2(x-4)k + 5(x-5)k + 2(x-7)k + 3(x-8)k + 10(x-10)k + (x-11)k + 5(x-13)k

C. For k = {1,2,3,...11}: 15(x+2)k + 5(x+5)k + 6(x+6)k + 13(x+8)k + 29(x+11)k + 7(x+13)k - 26(x+1)k - (x+3)k - 8(x+4)k - 14(x+7)k - 20(x+10)k - 20(x+12)k - (x+14)k =-15(x-2)k - 5(x-5)k - 6(x-6)k - 13(x-8)k - 29(x-11)k - 7(x-13)k + 26

(x-1)k + (x-3)k + 8(x-4)k + 14(x-7)k + 20(x-10)k + 20(x-12)k + (x-14)k - 30xk

D. For k = {1,2,3,... 12}: 7(x-14)k + 22(x-12)k + 33(x-9)k + (x-7)k + 19(x-6)k + 23(x-3)k + 15(x-1)k + 25(x+2)k + 6(x+4)k + 14(x+5)k + 8(x+7)k + 14(x+8)k + 29(x+10)k + 19(x+13)k + (x+15)k =7(x+14)k + 22(x+12)k + 33(x+9)k + (x+7)k + 19(x+6)k + 23(x+3)k + 15(x+1)k + 25(x-2)k + 6(x-4)k + 14(x-5)k + 8(x-7)k + 14(x-8)k + 29(x-10)k + 19(x-13)k + (x-15)k

and so on. One can see the symmetry and how the systems differ when it ends either in an odd or even power. As pointed out by Bhargava, it is possible to construct an unlimited number of such equations up to any degree k. (End update)

a collection of algebraic identities

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