A talk given at School of Mathematics and Systems ScienceChinese Academy of Sciences (Beijing, Jan. 29, 2018)and Wuhan Univ. (June 15, 2018)and St. Petersburg Dept. of Steklov Math. Instituteof Russian Academy of Sci. (July 20, 2018)and Huaiyin Normal College (Sept. 28, 2018)and South China Univ. of Technology (Dec. 7, 2018)

Problems and Results on Sums of Squares

Zhi-Wei Sun

Nanjing UniversityNanjing 210093, P. R. China

zwsun@nju.edu.cn

http://math.nju.edu.cn/∼zwsun

Dec. 7, 2018

Abstract

Due to the efforts of Fermat, Euler, Legendre and Gauss, it isknown what natural numbers can be written as the sum of twosquares or three squares. Lagrange’s four-square theorem proved in1770 states that each natural number can be expressed as the sumof four squares. In the talk we will first review classical results onsums of two or three or four squares. Then we turn to thespeaker’s recent discoveries which refine Lagrange’s four-squaretheorem or the Gauss-Legendre theorem on sums of three squares.In particular we will introduce our results refining Lagrange’sfour-square theorem as well as some recent conjectures of thespeaker one of which states that any integer greater than one canbe written as the sum of two squares, a power of 3 and a power of5.

2 / 63

Part I. Sums of Two, Three, Four Squares

3 / 63

An assertion of Fermat

Squares: 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, . . .

For n ∈ N = {0, 1, 2, . . .}, clearly

(2n)2 ≡ 0 (mod 4) and (2n + 1)2 = 4n(n + 1) + 1 ≡ 1 (mod 8).

If x , y ∈ Z, then x2 + y2 is congruent to 0 + 0 or 0 + 1 or 1 + 1modulo 4, and hence x2 + y2 6≡ 3 mod 4. Clearly, 2 = 12 + 12.

Theorem (claimed by Fermat in 1640 and proved by Euler in1752). Each prime p ≡ 1 (mod 4) can be written as the sum oftwo squares.

If p = x2 + y2 with x , y ∈ Z, then (xy∗)2 ≡ −(yy∗)2 ≡ −1(mod p) where y∗ is the unique integer in {1, . . . , p − 1} withyy∗ ≡ 1 (mod p). Let q = p−1

2 !. Then

q2 ≡(p−1)/2∏

r=1

r(p − r) = (p − 1)! ≡ −1 (mod p)

with the help of Wilson’s theorem.4 / 63

Proof of Fermat’s assertion

Two of the (b√pc+ 1)2 > p numbers

x + qy (x , y = 0, . . . , b√pc),

say x1 + qy1 and x2 + qy2 (with x1, y1, x2, y2 ∈ {0, . . . , b√

pc}, andx1 6= x2 or y1 6= y2), are congruent modulo p. Thus

(x1 − x2)2 ≡ (q(y2 − y1))2 ≡ −(y1 − y2)2 (mod p).

Let x = x1 − x2 and y = y1 − y2. Then x2 + y2 ≡ 0 (mod p) and

0 < x2 + y2 6 b√pc2 + b√pc2 < 2p.

So we must have x2 + y2 = p.

5 / 63

Jacobsthal’s Theorem

It can be shown that any prime p ≡ 1 (mod 4) can be writtenuniquely in the form x2 + y2 with x , y ∈ N and x 6 y .

Recall that for an odd prime p and an integer a, the Legendresymbol ( a

p ) is defined as follows:

(a

p

)=

0 if p | a,

1 if p - a and x2 ≡ a (mod p) for some x ∈ Z,−1 if p - a and x2 6≡ a (mod p) for all x ∈ Z.

Jacobsthal’s Theorem (1907). Let p be any prime congruent to1 mod 4. Let s, t ∈ Z with ( s

p ) = 1 and ( tp ) = −1. Then

p =

( (p−1)/2∑x=0

(x(x2 + s)

p

))2

+

( (p−1)/2∑x=0

(x(x2 + t)

p

))2

.

6 / 63

Sums of Two Squares

The set {x2 + y2 : x , y ∈ Z} is closed under multiplication. Infact,

(a2 + b2)(c2 + d2) = (ac ± bd)2 + (ad ∓ bc)2.

Thus those n ∈ N with ordp(n) even for any prime p ≡ 3 (mod 4)can be written as the sum of two squares. In 1749 Euler showedthat other numbers cannot be written as the sum of two squares.

Theorem (Jacobi, 1834). Let n ∈ Z+ = {1, 2, 3, . . .} and define

r2(n) := |{(x , y) ∈ Z2 : x2 + y2 = n}|.Then we have

r2(n) = 4∑2-d |n

(−1)(d−1)/2.

Note that∑∞

n=0 r2(n)qn = ϕ(q)2, where

ϕ(q) :=+∞∑

x=−∞qx2 .

7 / 63

Four-square theorem

Four-Square Theorem. Each n ∈ N = {0, 1, 2, . . .} can bewritten as the sum of four squares.

Examples. 3 = 12 + 12 + 12 + 02 and 7 = 22 + 12 + 12 + 12.

A. Diophantus (AD 299-215, or AD 285-201) was aware of thistheorem as indicated by examples given in his book Arithmetica.

In 1621 Bachet translated Diophantus’ book into Latin and statedthe theorem in the notes of his translation.

In 1748 L. Euler found the four-square identity

(x21 + x2

2 + x23 + x2

4 )(y21 + y2

2 + y23 + y2

4 )

=(x1y1 + x2y2 + x3y3 + x4y4)2 + (x1y2 − x2y1 − x3y4 + x4y3)2

+ (x1y3 − x3y1 + x2y4 − x4y2)2 + (x1y4 − x4y1 − x2y3 + x3y2)2.

and hence reduced the theorem to the case with n prime.

Euler also proved for each odd prime p there is a positive integerm < p such that pm = x2 + y2 + 12 + 02 for some x , y ∈ Z.

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Lagrange first proved the Four-Square Theorem

On the basis of Euler’s work, in 1770 J. L. Lagrange firstcompleted the proof of the four-square theorem. The celebratedtheorem is now well known as Lagrange’s four-square theorem.

A crucial step in Lagrange’s proof is to show that for any oddprime p if pm = x2 + y2 + z2 + w2 for some m ∈ {2, . . . , p − 1}and x , y , z ,w ∈ Z then there is a positive integer m0 < m suchthat pm0 can be written as the sum of four squares.

It is known that only the following numbers have a uniquerepresentation as the sum of four unordered squares:

1, 3, 5, 7, 11, 15, 23

and22k+1m (k = 0, 1, 2, . . . and m = 1, 3, 7).

For example, 4k × 14 = (2k3)2 + (2k+1)2 + (2k)2 + 02.

9 / 63

The representation function r4(n)

Jacobi used his triple-product formula

∞∏n=1

(1−q2n)(1+q2n−1z)(1+q2n−1z−1) =+∞∑

n=−∞znqn2 (|q| < 1, z 6= 0)

to study the fourth power of ϕ(q) =∑∞

n=−∞ qn2 , and this led himto deduce that

r4(n) = 8∑

d |n & 4-d

d for all n ∈ Z+,

where

r4(n) := |{(w , x , y , z) ∈ Z4 : w2 + x2 + y2 + z2 = n}|.

This is related to modular forms of weight two. Let τ be a complexnumber with positive real part and set θ(τ) = ϕ(e2πiτ ). Then

θ

(τ

4τ + 1

)=√

4τ + 1 θ(τ) and hence θ4(

τ

4τ + 1

)= (4τ+1)2θ4(τ).

10 / 63

Sums of three squaresSuppose that n = x2 + y2 + z2 with x , y , z ∈ Z. If 4 | n, thenx , y , z are all even and

n

4=(x

2

)2+(y

2

)2+(z

2

)2.

If n ≡ 3 mod 4, then x , y , z are odd and hence n ≡ 3 (mod 8).

Gauss-Legendre Theorem. n ∈ N can be written as the sum ofthree squares if and only if n is not of the form 4k(8l + 7) withk , l ∈ N.

Let n ∈ Z+ and let h(−n) denote the class number of the fieldQ(√−n). For

R3(n) := |{(x , y , z) ∈ Z3 : x2+y2+z2 = n and gcd(x , y , z) = 1}|,Gauss proved that

R3(n) =

12h(−n) if n > 3 and n ≡ 1, 2 (mod 4),

24h(−n) if n > 3 and n ≡ 3 (mod 8),

0 if 4 | n or n ≡ 7 (mod 8).11 / 63

Part II. New Problems for Sums of Four Squares

12 / 63

Universal sums of four mixed powers

If any n ∈ N can be written as f (x1, . . . , xn) with x1, . . . , xn in N(or Z), then we say that f is universal over N (or Z).

Theorem (Sun [J. Number Theory 175(2017)]) For any a ∈ {1, 4}and k ∈ {4, 5, 6}, awk + x2 + y2 + z2 is universal over N.

Theorem (Z.-W. Sun [Nanjing Univ. J. Math. Biquarterly34(2017)]) Let a, b, c , d ∈ Z+ with a 6 b 6 c 6 d , and leth, i , j , k ∈ {2, 3, . . .} with at most one of h, i , j , k equal to two.Suppose that h 6 i if a = b, i 6 j if b = c , and j 6 k if c = d . Iff (w , x , y , z) = awh + bx i + cy j + dzk is universal over N, thenf (w , x , y , z) must be among the following 9 polynomials

w2 + x3 + y4 + 2z3, w2 + x3 + y4 + 2z4, w2 + x3 + 2y3 + 3z3,

w2 + x3 + 2y3 + 3z4, w2 + x3 + 2y3 + 4z3, w2 + x3 + 2y3 + 5z3,

w2 + x3 + 2y3 + 6z3, w2 + x3 + 2y3 + 6z4, w3 + x4 + 2y2 + 4z3.

Conjecture (Sun [Nanjing Univ. J. Math. Biquarterly 34(2017)])All the 9 polynomials are universal over N.

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Discoveries on April 8, 2016Motivated by my conjecture that any n ∈ N can be written as

x31 + x3

2 + 2x33 + 2x3

4 + 3x35 (x1, x2, x3, x4, x5 ∈ N)

(which is stronger than the result g(3) = 9 for Waring’s problem),on April 8, 2016 I considered to write n ∈ N as

∑5i=1 aix

2i (xi ∈ N)

with certain restrictions on x1, . . . , x5.

Conjecture (Z.-W. Sun) Let n > 1 be an integer.

(i) n can be written as

x21+x2

2+x23+2x2

4+2x25 = x2

1+x22+x2

3+(x4 + x5)2+(x4−x5)2 (xi ∈ N)

with x1 + x2 + x3 + x4 + x5 prime.

(ii) We can write n as

x21 + 2x2

2 + 3x23 + 4x2

4 + 5x25 (x1, x2, x3, x4, x5 ∈ N)

with x1 + x2 + x3 + x4 a square.

Remark. Squares are sparser than prime numbers.14 / 63

1-3-5-Conjecture (1350 US dollars for the first solution)

1-3-5-Conjecture (Z.-W. Sun, April 9, 2016): Any n ∈ N can bewritten as x2 + y2 + z2 + w2 with x , y , z ,w ∈ N such thatx + 3y + 5z is a square.

Examples.

7 =12 + 12 + 12 + 22 with 1 + 3× 1 + 5× 1 = 32,

8 =02 + 22 + 22 + 02 with 0 + 3× 2 + 5× 2 = 42,

31 =52 + 22 + 12 + 12 with 5 + 3× 2 + 5× 1 = 42,

43 =12 + 52 + 42 + 12 with 1 + 3× 5 + 5× 4 = 62.

The conjecture has been verified by Qing-Hu Hou for all n 6 1010.

We guess that, if a, b, c are positive integers with gcd(a, b, c)squarefree such that any n ∈ N can be written asx2 + y2 + z2 + w2 (x , y , z ,w ∈ N) with ax + by + cz a square,then we must have {a, b, c} = {1, 3, 5}.

15 / 63

Graph for the number of such representations of n

16 / 63

无 解

数字几时有，

把酒问青天。

一二三四五，

自然藏玄机。

四个平方和，

遍历自然数。

奇妙一三五，

更上一层楼。

苍天捉弄人，

数论妙无穷。

吾辈虽努力，

难解一三五!

时势唤英雄，

攻关需豪杰。

人间若无解，

天神会证否？

17 / 63

Refinements of the 1-3-5 conjecture

Conjecture (Z.-W. Sun, Dec. 12, 2016): For any positive integern, there is a prime p such that there are exactly n ways to write pas x2 + y2 + z2 + w2 (x , y , z ,w ∈ N) with x + 3y + 5z a square.

Remark. I have verified this for n = 1, . . . , 500.

Conjecture (Z.-W. Sun, Feb. 17, 2017): Each n ∈ N can bewritten as x2 + y2 + z2 + w2 (x , y , z ,w ∈ N) such thatx + 3y + 5z and (at least) one of y , z ,w are squares.

Conjecture (Z.-W. Sun, March 12, 2018): Each n ∈ Z+ can bewritten as x2 + y2 + z2 + w2 (x , y , z ,w ∈ N) such that x + 3y + 5zis a positive square and 2x (or 3x) or y or z is a square.

Examples:

8 =02 + 22 + 22 + 02 with 0 + 3× 2 + 5× 2 = 42 and 0 = 02;

188 =72 + 92 + 32 + 72 with 7 + 3 · 9 + 5 · 3 = 72 and 9 = 32;

248 =102 + 22 + 02 + 122 with 10 + 3 · 2 + 5 · 0 = 42 and 0 = 02;

808 =122 + 142 + 182 + 122 with 12 + 3 · 14 + 5 · 18 = 122 & 3 · 12 = 62.18 / 63

1-2-3-Conjecture (Companion of 1-3-5-Conjecture)

1-2-3-Conjecture (Z.-W. Sun, July 24, 2016). Any n ∈ N can bewritten as x2 + y2 + z2 + 2w2 with x , y , z ,w ∈ N such thatx + 2y + 3z is a square.

Examples:

14 =12 + 12 + 22 + 2× 22 with 1 + 2× 1 + 3× 2 = 32,

30 =32 + 22 + 32 + 2× 22 with 3 + 2× 2 + 3× 3 = 42,

33 =12 + 02 + 02 + 2× 42 with 1 + 2× 0 + 3× 0 = 12,

84 =42 + 62 + 02 + 2× 42 with 4 + 2× 6 + 3× 0 = 42,

169 =102 + 62 + 12 + 2× 42 with 10 + 2× 6 + 3× 1 = 52,

225 =102 + 62 + 92 + 2× 22 with 10 + 2× 6 + 3× 9 = 72.

Another Conjecture (Sun, 2017). Any n ∈ N can be written asx2 + y2 + z2 + 2w2 with x , y , z ,w ∈ Z and x + 2y + 3z = 1.

Remark. This was proved by Hai-Liang Wu and Z.-W. Sun in 2017for sufficient large integers n.

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Graph for the number of such representations of n

20 / 63

Diagonal ternary quadratic forms

For a, b, c ∈ Z+ = {1, 2, 3, . . .}, we define

E (a, b, c) := {n ∈ N : n 6= ax2 + by2 + cz2 for any x , y , z ∈ N}.

It is known that E (a, b, c) is an infinite set.

Gauss-Legendre Theorem. E (1, 1, 1) = {4k(8l + 7) : k , l ∈ N}.

There are totally 102 diagonal ternary quadratic formsax2 + by2 + cz2 with a, b, c ∈ Z+ and gcd(a, b, c) = 1 for whichthe structure of E (a, b, c) is known explicitly. For example,

E (1, 1, 2) ={4k(16l + 14) : k , l ∈ N},E (1, 1, 5) ={4k(8l + 3) : k , l ∈ N},E (1, 2, 3) ={4k(16l + 10) : k , l ∈ N},E (1, 2, 6) ={4k(8l + 5) : k , l ∈ N}.

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Sums of a fourth power and three squares

Theorem (Z.-W. Sun, March 27, 2016). Each n ∈ N can bewritten as w4 + x2 + y2 + z2 with w , x , y , z ∈ N.

Proof. For n = 0, 1, 2, . . . , 15, the result can be verified directly.Now let n > 16 be an integer and assume that the result holds forsmaller values of n.

Case 1. 16 | n.By the induction hypothesis, we can write

n

16= x4 + y2 + z2 + w2 with x , y , z ,w ∈ N.

It follows that n = (2x)4 + (4y)2 + (4z)2 + (4w)2.

Case 2. n = 4kq with k ∈ {0, 1} and q ≡ 7 (mod 8).In this case, n − 1 6∈ E (1, 1, 1), and hence n = 14 + y2 + z2 + w2

for some y , z ,w ∈ N.

Case 3. 16 - n and n 6= 4k(8l + 7) for any k ∈ {0, 1} and l ∈ N.In this case, n 6∈ E (1, 1, 1) and hence there are y , z ,w ∈ N suchthat n = 04 + y2 + z2 + w2.

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Suitable polynomials

Definition (Z.-W. Sun, 2016). A polynomial P(x , y , z ,w) withinteger coefficients is called suitable if any n ∈ N can be written asx2 + y2 + z2 + w2 with x , y , z ,w ∈ N such that P(x , y , z ,w) is asquare.

We have seen that both x and 2x are suitable polynomials. The1-3-5-Conjecture says that x + 3y + 5z is suitable.

We conjecture that there only finitely many a, b, c, d ∈ Z withgcd(a, b, c , d) squarefree such that ax + by + cz + dw is suitable,and we have found all such quadruples (a, b, c , d).

23 / 63

x − y and 2x − 2y are suitable

Let a ∈ {1, 2}. We claim that any n ∈ N can be written asx2 + y2 + z2 + w2 with x , y , z ,w ∈ N such that a(x − y) is asquare, and want to prove this by induction.

For every n = 0, 1, . . . , 15, we can verify the claim directly.

Now we fix an integer n > 16 and assume that the claim holds forsmaller values of n.

Case 1. 16 | n.In this case, by the induction hypothesis, there are x , y , z ,w ∈ Nwith a(x − y) a square such that n/16 = x2 + y2 + z2 + w2, andhence n = (4x)2 + (4y)2 + (4z)2 + (4w)2 with a(4x − 4y) asquare.

Case 2. 16 - n and n 6∈ E (1, 1, 2).In this case, there are x , y , z ,w ∈ N with x = y andn = x2 + y2 + z2 + w2, thus a(x − y) = 02 is a square.

24 / 63

x − y and 2x − 2y are suitable

Case 3. 16 - n and n ∈ E (1, 1, 2) = {4k(16l + 14) : k , l ∈ N}.

In this case, n = 4k(16l + 14) for some k ∈ {0, 1} and l ∈ N. Notethat n/2− (2/a)2 6∈ E (1, 1, 1). So, n/2− (2/a)2 = t2 + u2 + v2

for some t, u, v ∈ N with t > u > v . As n/2− (2/a)2 > 8− 4 > 3,we have t > 2 > 2/a. Thus

n =2

((2

a

)2

+ t2

)+ 2(u2 + v2)

=

(t +

2

a

)2

+

(t − 2

a

)2

+ (u + v)2 + (u − v)2

with

a

((t +

2

a

)−(

t − 2

a

))= 22.

This proves that x − y and 2x − 2y are both suitable.

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Suitable polynomials of the form ax ± by

Conjecture (Z.-W. Sun, April 14,2016) Let a, b ∈ Z+ withgcd(a, b) squarefree.

(i) The polynomial ax + by is suitable if and only if{a, b} = {1, 2}, {1, 3}, {1, 24}.

(ii) The polynomial ax − by is suitable if and only if (a, b) isamong the ordered pairs

(1, 1), (2, 1), (2, 2), (4, 3), (6, 2).

Remark. In 2016, I proved that any n ∈ N can be written asx2 + y2 + z2 + w2 with x , y , z ,w ∈ Z such that x + 2y is a square(or a cube). In a joint paper with my student Yu-Chen Sun [ActaArith. 183(2018)], we managed to show that x + 2y is indeedsuitable.

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Write n = x2 + y 2 + z2 + w 2 with x + 3y a square

In 1916 Ramanujan conjectured that

(1) the only positive even numbers not of the form x2 + y2 + 10z2

are those 4k(16l + 6) (k , l ∈ N)

and

(2) sufficiently large odd numbers are of the form x2 + y2 + 10z2.

In 1927 L. E. Dickson [Bull. AMS] proved (1). In 1990 W. Dukeand R. Schulze-Pillot [Invent. Math.] confirmed (2). In 1997 K.Ono and K. Soundararajan [Invent. Math.] proved that under theGRH (Generalized Riemann Hypothesis) any odd number greaterthan 2719 has the form x2 + y2 + 10z2.

With the help of the Ono-Soundararajan result, the speaker hasproved the following result.

Theorem (Z.-W. Sun, 2016) Under the GRH, any n ∈ N can bewritten as n = x2 + y2 + z2 + w2 (x , y , z ,w ∈ Z) with x + 3y asquare.

27 / 63

Suitable ax − by − cz or ax + by − czConjecture (Z.-W. Sun, April 14, 2016): (i) Let a, b, c ∈ Z+ withb 6 c and gcd(a, b, c) squarefree. Then ax − by − cz is suitable ifand only if (a, b, c) is among the five triples

(1, 1, 1), (2, 1, 1), (2, 1, 2), (3, 1, 2), (4, 1, 2).

(ii) Let a, b, c ∈ Z+ with a 6 b and gcd(a, b, c) squarefree. Thenax + by − cz is suitable if and only if (a, b, c) is among thefollowing 52 triples

(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (1, 2, 3), (1, 3, 1),

(1, 3, 3), (1, 4, 4), (1, 5, 1), (1, 6, 6), (1, 8, 6), (1, 12, 4), (1, 16, 1),

(1, 17, 1), (1, 18, 1), (2, 2, 2), (2, 2, 4), (2, 3, 2), (2, 3, 3), (2, 4, 1),

(2, 4, 2), (2, 6, 1), (2, 6, 2), (2, 6, 6), (2, 7, 4), (2, 7, 7), (2, 8, 2),

(2, 9, 2), (2, 32, 2), (3, 3, 3), (3, 4, 2), (3, 4, 3), (3, 8, 3), (4, 5, 4),

(4, 8, 3), (4, 9, 4), (4, 14, 14), (5, 8, 5), (6, 8, 6), (6, 10, 8), (7, 9, 7),

(7, 18, 7), (7, 18, 12), (8, 9, 8), (8, 14, 14), (8, 18, 8), (14, 32, 14),

(16, 18, 16), (30, 32, 30), (31, 32, 31), (48, 49, 48), (48, 121, 48).28 / 63

Linear restrictions involving cubes

Conjecture (Z.-W. Sun, 2016). For each c = 1, 2, 4, any n ∈ Ncan be written as w2 + x2 + y2 + z2 with w , x , y , z ∈ N and y 6 zsuch that 2x + y − z = ct3 for some t ∈ N.

Examples.

13 =22 + 02 + 32 + 02 with 2× 2 + 0− 3 = 13,

2976 =202 + 162 + 482 + 42 with 2× 20 + 16− 48 = 23.

Conjecture (Z.-W. Sun, March 17, 2018). Any n ∈ Z+ can bewritten as x2 + y2 + z2 + w2 with x , y , z ∈ N and w ∈ Z+ suchthat x + 3y + 9z = 2km3 for some k ,m ∈ N.

Examples.

79 =22 + 72 + 12 + 52 with 2 + 3× 7 + 9× 1 = 2233,

496 =42 + 82 + 42 + 202 with 4 + 3× 8 + 9× 4 = 43.

29 / 63

n = x2 + y 2 + z2 + w 2 with x + y + z a square (or a cube)

Theorem (Z.-W. Sun, April-May, 2016) Let c ∈ {1, 2} andm ∈ {2, 3}. Then any n ∈ N can be written as x2 + y2 + z2 + w2

with x , y , z ,w ∈ Z such that x + y + cz = tm for some t ∈ Z.

Proof for the Case c = 1. For n = 0, . . . , 4m − 1 we can easilyverify the desired result directly.

Now let n ∈ N with n > 4m. Assume that any r ∈ {0, . . . , n − 1}can be written as x2 + y2 + z2 + w2 with x , y , z ,w ∈ Z such thatx + y + z ∈ {tm : t ∈ Z}. If 4m | n, then there are x , y , z ,w ∈ Zwith x2 + y2 + z2 + w2 = n/4m such that x + y + z = tm for somet ∈ Z, and hence

n = (2mx)2 + (2my)2 + (2mz)2 + (2mw)2

with 2mx + 2my + (2mz) = 2m(x + y + z) = (2t)m. Below wesuppose that 4m - n.

30 / 63

Continued the proofIt suffices to show that there are x , y , z ∈ Z and δ ∈ {0, 1, 2m}such that

n = x2 +(y +z)2 +(z−y)2 +(δ−2z)2 = x2 +2y2 +6z2−4δz +δ2.

(Note that (y + z) + (z − y) + (δ − 2z) = δ ∈ {tm : t ∈ Z}.)Suppose that this fails for δ = 0. As

E (1, 2, 6) = {4k(8l + 5) : k , l ∈ N},

n = 4k(8l + 5) for some k, l ∈ N with k < m. Clearly,

3n − 1 =

{3(8l + 5)− 1 = 2(12l + 7) if k = 0,

3× 4(8l + 5)− 1 = 8(12l + 7) + 3 if k = 1.

Thus, if k ∈ {0, 1}, then 3n − 1 does not belong to

E (2, 3, 6) = {3q + 1 : q ∈ N} ∪ {4k(8l + 7) : k , l ∈ N},

31 / 63

Continue the proof

hence for some x , y , z ∈ Z we have

3n− 1 = 3x2 + 6y2 + 2(3z − 1)2 = 3(x2 + 2y2 + 2(3z2 − 2z)) + 2

and thus

n = x2 + 2y2 + 6z2− 4z + 1 = x2 + (y + z)2 + (z − y)2 + (1− 2z)2

as desired.

When k = 2 and m = 3, we have

3n − 64 = 3× 16(8l + 5)− 64 = 42(8(3l + 1) + 3) 6∈ E (2, 3, 6),

and hence there are x , y , z ∈ Z such that

3n−43 = 3x2+6y2+2(3z−8)2 = 3(x2+2y2+2(3z2−16z))+2×43

and thusn = x2+2y2+6z2−32z +64 = x2+(y +z)2+(z−y)2+(23−2z)2

as desired.32 / 63

Suitable ax + by + cz − dw or ax + by − cz − dwConjecture (Z.-W. Sun, April 14, 2016): Let a, b, c , d ∈ Z+ witha 6 b 6 c and gcd(a, b, c , d) squarefree. Then ax + by + cz − dwis suitable if and only if (a, b, c , d) is among the 12 quadruples

(1, 1, 2, 1), (1, 2, 3, 1), (1, 2, 3, 3), (1, 2, 4, 2),

(1, 2, 4, 4), (1, 2, 5, 5), (1, 2, 6, 2), (1, 2, 8, 1),

(2, 2, 4, 4), (2, 4, 6, 4), (2, 4, 6, 6), (2, 4, 8, 2).

Conjecture (Z.-W. Sun, April 14, 2016): Let a, b, c , d ∈ Z+ witha 6 b and c 6 d , and gcd(a, b, c , d) squarefree. Thenax + by − cz − dw is suitable if and only if (a, b, c , d) is amongthe 9 quadruples

(1, 2, 1, 1), (1, 2, 1, 2), (1, 3, 1, 2), (1, 4, 1, 3),

(2, 4, 1, 2), (2, 4, 2, 4), (8, 16, 7, 8), (9, 11, 2, 9), (9, 16, 2, 7).

Conjecture (Z.-W. Sun, April 2016) For any a, b, c , d ∈ Z+ thereare infinitely many positive integers not of the formx2 + y2 + z2 + w2 (x , y , z ,w ∈ N) with ax + by + cz + dw asquare. 33 / 63

A general theorem joint with Yu-Chen Sun

Theorem (Yu-Chen Sun and Z.-W. Sun, 2016) Let a, b, c, d ∈ Zwith a, b, c , d not all zero. Let λ ∈ {1, 2} and m ∈ {2, 3} Thenany n ∈ N can be written as x2 + y2 + z2 + w2 withx , y , z ,w ∈ Z/(a2 + b2 + c2 + d2) such thatax + by + cz + dw = λrm for some r ∈ N.

Proof. Let n ∈ N. By a result of Z.-W. Sun, we can write(a2 + b2 + c2 + d2)n as (λrm)2 + t2 + u2 + v2 with r , t, u, v ∈ N.Set s = λrm, and define x , y , z ,w by

x = as−bt−cu−dva2+b2+c2+d2 ,

y = bs+at+du−cva2+b2+c2+d2 ,

z = cs−dt+au+bva2+b2+c2+d2 ,

w = ds+ct−bu+ava2+b2+c2+d2 .

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Proof of the general theorem

Then ax + by + cz + dw = s,

ay − bx + cw − dz = t,

az − bw − cx + dy = u,

aw + bz − cy − dx = v .

With the help of Euler’s four-square identity,

x2 + y2 + z2 + w2 =s2 + t2 + u2 + v2

a2 + b2 + c2 + d2= n

andax + by + cz + dw = s = λrm.

This concludes the proof.

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Joint work with Yu-Chen Sun

Theorem (Y.-C. Sun and Z.-W. Sun [Acta Arith. 183(2018)]) (i)Let m ∈ Z+. Then any n ∈ N can be written as x2 + y2 + z2 + w2

(x , y , z ,w ∈ Z) with x + y + z + w an m-th power if and only ifm 6 3.

(ii) Let λ ∈ {1, 2} and m ∈ {2, 3}. Then any n ∈ N can be writtenas x2 + y2 + z2 + w2 (x , y , z ,w ∈ Z) with x + y + z + 2w = λrm

(or x + y + 2z + 3w = λrm) for some r ∈ N.

(iii) Let λ ∈ {1, 2} and m ∈ {2, 3}. Then any n ∈ N can bewritten as x2 + y2 + z2 + w2 (x , y , z ,w ∈ Z) with x + 2y + 3z (orx + y + 3z , or x + 2y + 2z) in the set {λrm : r ∈ N}.

(iv) (Progress on the 1-3-5-Conjecture) Let λ ∈ {1, 2}, m ∈ {2, 3}and n ∈ N. Then we can write n as x2 + y2 + z2 + w2 withx , y , 5z , 5w ∈ Z such that x + 3y + 5z ∈ {λrm : r ∈ N}. Also,any n ∈ N can be written as x2 + y2 + z2 + w2 withx , y , z ,w ∈ Z/7 such that x + 3y + 5z ∈ {λrm : r ∈ N}.

36 / 63

A Lemma

The proof of the Theorem needs several lemmas and some previousresults of Z.-W. Sun. Here is one of them.

Lemma. Define x = s−t−u−2v

7 ,

y = s+t+2u−v7 ,

z = s−2t+u+v7 ,

w = 2s+t−u+v7 .

Then

x2 + y2 + z2 + w2 =s2 + t2 + u2 + v2

7.

Also,

x + y + z + 2w =s,

w + 2x + 3z =s − t,

x + 3y + 5w =2s + t.

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Joint work with Hai-Liang Wu

Besides the 1-3-5 conjecture, I also conjectured that any n ∈ N canbe written as x2 + y2 + z2 + w2 with x , y , z ,w ∈ N such that

|x + 3y − 5z | ∈ {4k : k ∈ N}.

In 2017, Hai-Liang Wu and the speaker used the theory of ternaryquadratic forms and modular forms to obtain the followingprogress on the 1-3-5 conjecture.

Theorem (H.-L. Wu and Z.-W. Sun, 2017). Any sufficiently largeinteger n not divisible by 16 can be written as x2 + y2 + z2 + w2

with x , y , z ,w ∈ Z such that x + 3y + 5z ∈ {1, 4}.

Let

A =⋃k∈N{4k + 1, 4k + 2, 8k + 4} and B =

⋃k∈N{4k + 3, 16k + 8}.

Then A ∪ B = {n ∈ N : 16 - n}.38 / 63

Wu and Sun’s work

Suppose that n ∈ A is large enough. Then we are able to showthat 70n− 2 = 5r20 + 7r21 + 70w2 for some r0, r1,w ∈ Z. As r20 ≡ 1(mod 7) and r21 ≡ 22 (mod 5), without loss of generality we mayassume that r0 ≡ 1 (mod 7) and r1 ≡ 2 (mod 5). Apparently,r0 ≡ r1 (mod 2).

If r0 ≡ 1 (mod 14), then r0 = 14u + 1 and r1 = 10v − 3 for someu, v ∈ Z, hence 70n − 2 = 5(14u + 1)2 + 7(10v − 3)2 + 70w2 andthus n = x2 + y2 + z2 + w2, where x = 1 + u − 3v , y = 3u + vand z = −2u. Note that x + 3y + 5z = 1.

If r0 ≡ 8 (mod 14), then r0 = 14u + 8 and r1 = 10v + 2 for someu, v ∈ Z, hence 70n − 2 = 5(14u + 8)2 + 7(10v + 2)2 + 70w2 andthus n = x2 + y2 + z2 + w2, where x = u − 3v , y = 2 + 3u + vand z = −1− 2u. Obviously, x + 3y + 5z = 1.

Similarly, we can show that if n ∈ B is large enough then there arex , y , z ,w ∈ Z such that n = x2 + y2 + z2 + w2 andx + 3y + 5z = 4.

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Suitable polynomials of the form ax2 + by 2 + cz2

Conjecture (Z.-W. Sun, April 9, 2016): (i) Any natural numbercan be written as x2 + y2 + z2 + w2 with x , y , z ,w ∈ N and x > ysuch that ax2 + by2 + cz2 is a square, provided that the triple(a, b, c) is among

(1, 8, 16), (4, 21, 24), (5, 40, 4), (9, 63, 7), (16, 80, 25),

(16, 81, 48), (20, 85, 16), (36, 45, 40), (40, 72, 9).

(ii) ax2 + by2 + cz2 is suitable if (a, b, c) is among the triples

(1, 3, 12), (1, 3, 18), (1, 3, 21), (1, 3, 60), (1, 5, 15),

(1, 8, 24), (1, 12, 15), (1, 24, 56), (3, 4, 9), (3, 9, 13),

(4, 5, 12), (4, 5, 60), (4, 9, 60), (4, 12, 21), (4, 12, 45), (5, 36, 40).

(iii) If a, b, c are positive integers with ax2 + by2 + cz2 suitable,then a, b, c cannot be pairwise coprime.

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Suitable polynomials related to Pythagorean triples

Conjecture (Z.-W. Sun, April 12, 2016). Any n ∈ Z+ can bewritten as w2 + x2 + y2 + z2 with w ∈ Z+ and x , y , z ∈ N suchthat (10w + 5x)2 + (12y + 36z)2 is a square.

Remark: In 2017, Yu-Chen Sun and Z.-W. Sun proved that anyn ∈ N can be written as w2 + x2 + y2 + z2 with w , x , y , z integerssuch that (10w + 5x)2 + (12y + 36z)2 is a square.

Conjecture (Z.-W. Sun, May 15, 2016). (i) Any positive integer ncan be written as x2 + y2 + z2 + w2 with x , y , z ,w ∈ N and y > zsuch that (x + y)2 + (4z)2 is a square.

(ii) Any integer n > 5 can be written as x2 + y2 + z2 + w2 withx , y , z ,w ∈ N such that 8x + 12y and 15z are the two legs of aright triangle with positive integer sides.

Theorem (Z.-W. Sun, May 16, 2016). Any n ∈ Z+ can be writtenas x2 + y2 + z2 + w2 with x , y , z ,w ∈ N and y > 0 such thatx + 4y + 4z and 9x + 3y + 3z are the two legs of a right trianglewith positive integer sides.

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A conjecture involving mixed terms

Conjecture (Z.-W. Sun, 2016) (i) Any n ∈ N can be written asx2 + y2 + z2 + w2 with x , y , z ,w ∈ N such that xy + 2zw orxy − 2zw is a square.

(ii) Any n ∈ Z+ can be written as w2 + x2 + y2 + z2 with w ∈ Z+

and x , y , z ∈ N such that w2 + 4xy + 8yz + 32zx is a square.

(iii) Let a, b, c ∈ N with 1 6 a 6 c and gcd(a, b, c) squarefree.Then awx + bxy + cyz is suitable if and only if

(a, b, c) = (1, 2, 2), (2, 1, 4), (2, 8, 4).

(iv) Let a, b, c ∈ Z+ with gcd(a, b, c) squareferee. Then thepolynomial axy + byz + czx is suitable if and only if {a, b, c} isamong the sets

{1, 2, 3}, {1, 3, 8}, {1, 8, 13}, {2, 4, 45},{4, 5, 7}, {4, 7, 23}, {5, 8, 9}, {11, 16, 31}.

(v) 36x2y + 12y2z + z2x , w2x2 + 3x2y2 + 2y2z2 andw2x2 + 5x2y2 + 80y2z2 + 20z2w2 are suitable.

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Suitable polynomials of the form ax4 + by 3z

The following conjecture sounds very mysterious!

Conjecture (Z.-W. Sun, 2016) Let a and b be nonzero integerswith gcd(a, b) squarefree. Then the polynomial ax4 + by3z issuitable if and only if (a, b) is among the ordered pairs

(1, 1), (1, 15), (1, 20), (1, 36), (1, 60), (1, 1680) and (9, 260).

Examples:9983 = 632 + 542 + 172 + 532

with 634 + 543 × 17 = 42932, and

20055 = 472 + 62 + 772 + 1092

with 474 + 1680× 63 × 77 = 57292.

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Other suitable polynomials

Conjecture (Z.-W. Sun, 2016) (i) Any n ∈ Z+ can be written asx2 + y2 + z2 + w2 (w ∈ Z+ and x , y , z ∈ N) with x3 + 4yz(y − z)(or x3 + 8yz(2y − z)) a square.

(ii) The polynomials w(x + 2y + 3z), w(x2 + 8y2 − z2) andx2 + 3y2 + 5z2 − 8w2 are suitable.

(iii) Any n ∈ Z+ can be written as x2 + y2 + z2 + w2 withx , y , z ,w ∈ N and z < w such that 4x2 + 5y2 + 20zw is a square.

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Restrictions involving powers of two

Theorem (Z.-W. Sun, arXiv:1701.05868) (i) Any n ∈ Z+ can bewritten as x2 + y2 + z2 + w2 with x , y , z ,w ∈ N and|x + y − z | ∈ {4k : k ∈ N} (or |x − 2y | ∈ {4k : k ∈ N}).

(ii) For each c = 1, 2, any n ∈ Z+ can be written asx2 + y2 + z2 + w2 with x , y , z ,w ∈ Z andx + y + 2z ∈ {c4k : k ∈ N}.

(iii) Any n ∈ Z+ can be written as x2 + y2 + z2 + w2 withx , y , z ,w ∈ Z and x + 2y + 2z ∈ {4k : k ∈ N}.

(iv) Any integer n > 1 can be written as x2 + y2 + z2 + w2 withx , y , z ,w ∈ Z and x + 2y + 2z ∈ {3× 2k : k ∈ N}.

Conjecture (Z.-W. Sun, 2016). Any n ∈ Z+ can be written asx2 + y2 + z2 + w2 with x , y , z ,w ∈ N such that x + 2y − 2z is apower of four (including 40 = 1).

Remark. Qing-Hu Hou has verified this for n up to 109.

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The 24-conjecture

24-Conjecture (Z.-W. Sun, Feb. 4, 2017). Each n ∈ N can bewritten as x2 + y2 + z2 + w2 with x , y , z ,w ∈ N such that both xand x + 24y are squares.

Remark. Qing-Hu Hou has verified this for n 6 1010. I would liketo offer 2400 US dollars as the prize for the first proof.

12 =12 + 12 + 12 + 32 with 1 = 12 and 1 + 24× 1 = 52,

23 =12 + 22 + 32 + 32 with 1 = 12 and 1 + 24× 2 = 72,

24 =42 + 02 + 22 + 22 with 4 = 22 and 4 + 24× 0 = 22,

47 =12 + 12 + 32 + 62 with 1 = 12 and 1 + 24× 1 = 52,

71 =12 + 52 + 32 + 62 with 1 = 12 and 1 + 24× 5 = 112,

168 =42 + 42 + 62 + 102 with 4 = 22 and 4 + 24× 4 = 102,

344 =42 + 02 + 22 + 182 with 4 = 22 and 4 + 24× 0 = 22,

632 =02 + 62 + 142 + 202 with 0 = 02 and 0 + 24× 6 = 122,

1724 =252 + 12 + 32 + 332 with 25 = 52 and 25 + 24× 1 = 72.46 / 63

A similar conjecture

Conjecture (Z.-W. Sun, Feb. 4, 2017). (i) Each n ∈ N can bewritten as x2 + y2 + z2 + w2 with x , y , z ,w ∈ N such that both xand 49x + 48(y − z) are squares.

(ii) Each n ∈ N can be written as x2 + y2 + z2 + w2 withx , y , z ,w ∈ N such that both x and 121x + 48(y − z) are squares.

(iii) Each n ∈ N can be written as x2 + y2 + z2 + w2 withx , y , z ,w ∈ N such that both x and −7x − 8y + 8z + 16w aresquares.

Remark. Qing-Hu Hou has verified parts (i)-(ii) and part (iii) for nup to 109 and 108.

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Other conjectures involving joint restrictions

Conjecture (Z.-W. Sun, Feb. 2017). (i) Each n ∈ N can bewritten as x2 + y2 + z2 + w2 with x , y , z ,w ∈ N and x ≡ y(mod 2) such that both x and x2 + 62xy + y2 are squares.

(ii) Any n ∈ Z+ can be written as x4 + y2 + z2 + w2 withx , y , z ∈ Z and w ∈ Z+ such that 8y2 − 8yz + 9z2 is a square.

Conjecture (Z.-W. Sun, March 2, 2017). (i) Any n ∈ N can bewritten as x2 + y2 + z2 + w2 with x ,w ∈ N and y , z ∈ Z such thatboth x + 2y and z + 2w are squares.

(ii) Each n ∈ N can be written as x2 + y2 + z2 + w2 withx , y , z ∈ Z and w ∈ N such that both x + 3y and z + 3w aresquares.

(iii) Any n ∈ Z+ can be written as x2 + y2 + z2 + w2 withx , y , z ∈ Z and w ∈ Z+ such that both 2x + y and 2x + z aresquares.

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Other conjectures involving joint restrictions

Conjecture (Z.-W. Sun, March 13-4 2018). (i) Each n ∈ Z+ canbe written as x2 + y2 + z2 + w2 with x , y , z ,w ∈ N and w ∈ Z+

such that both (3x)2 + (4y)2 + (12z)2 is a square and one ofz , 2z , 3z is a square.

(ii) Any n ∈ Z+ can be written as x4 + y2 + z2 + w2 withx , y , z ∈ N and w ∈ Z+ such that (12x)2 + (15y)2 + (20z)2 is asquare and one of x , y , z is a square.

(iii) Any n ∈ N can be written as x2 + y2 + z2 + w2 withx , y , z ,w ∈ N such that (12x)2 + (21y)2 + (28z)2 is a square andone of x , 2y , z is a square.

Examples:

671 = 182 + 172 + 32 + 72 with (3 · 18)2 + (4 · 17)2 + (12 · 3)2 = 942;

39 = 52 + 22 + 12 + 32 with (12 · 5)2 + (15 · 2)2 + (20 · 1)2 = 702;

and 1244 = 142 + 22 + 122 + 302 with

(12 · 14)2 + (21 · 2)2 + (28 · 12)2 = 3782.49 / 63

Conjectures involving cubic diophantine equations

Conjecture (Z.-W. Sun, March 2017). (i) Any n ∈ N can bewritten as x2 + y2 + z2 + w2 with x , y , z ,w ∈ N and y ≡ z(mod 2) such that 72x3 + (y − z)3 is a square.

(ii) Let a, b ∈ Z+ with gcd(a, b) squarefree. Then, each n ∈ N canbe written as x2 + y2 + z2 + w2 with x , y , z ,w ∈ N such thatax3 + b(y − z)3 is a square, if and only if (a, b) is among theordered pairs

(1, 1), (1, 9), (2, 18), (8, 1), (9, 5), (9, 8),

(9, 40), (16, 2), (18, 16), (25, 16), (72, 1).

(iii) Let a and b > a be positive integers with gcd(a, b) squarefree.Then, every n ∈ N can be written as x2 + y2 + z2 + w2

(x , y , z ,w ∈ Z) with ax3 + by3 a square, if and only if (a, b) isamong the ordered pairs

(1, 2), (1, 8), (2, 16), (4, 23), (4, 31), (5, 9),

(8, 9), (8, 225), (9, 47), (25, 88), (50, 54).50 / 63

Restricted sums of four squares involving primes

The following conjecture implies the twin prime conjecture.

Conjecture (Z.-W. Sun, August 23, 2017). Any positive oddinteger can be written as x2 + y2 + z2 + w2 with x , y , z ,w ∈ Zsuch that p = x2 + 3y2 + 5z2 + 7w2 and p − 2 are twin prime.

Example.39 = 12 + 32 + 52 + 22

with 12 + 3 · 32 + 5 · 52 + 7 · 22 = 181 and 181− 2 = 179 twinprime. Also,

123 = 72 + 32 + 72 + 42

with 72 + 3 · 32 + 5 · 72 + 7 · 42 = 433 and 433− 2 = 431 twinprime.

Conjecture (Z.-W. Sun, August 28, 2017). Any integer n > 1 notdivisible by 4 can be written as x2 + y2 + z2 + w2 (x , y , z ,w ∈ N)such that p = x + 2y + 5z , p − 2, p + 4 and p + 10 are all prime.

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Restricted sums of four squares involving primes

The following conjecture implies the twin prime conjecture, and Ihave verified it for all odd numbers below 4× 107.

Conjecture (Z.-W. Sun, August 23, 2017). Any positive oddinteger can be written as x2 + y2 + z2 + w2 with x , y , z ,w ∈ Zsuch that p = x2 + 3y2 + 5z2 + 7w2 and p − 2 are twin prime.

Example.39 = 12 + 32 + 52 + 22

with 12 + 3 · 32 + 5 · 52 + 7 · 22 = 181 and 181− 2 = 179 twinprime. Also,

123 = 72 + 32 + 72 + 42

with 72 + 3 · 32 + 5 · 72 + 7 · 42 = 433 and 433− 2 = 431 twinprime.

Conjecture (Z.-W. Sun, August 28, 2017). Any integer n > 1 notdivisible by 4 can be written as x2 + y2 + z2 + w2 (x , y , z ,w ∈ N)such that p = x + 2y + 5z , p − 2, p − 4 and p + 10 are all prime.

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Restricted sums of four squares involving primes

Conjecture (Z.-W. Sun, August 19, 2017). Any positive oddinteger can be written as x2 + y2 + z2 + 4w2 with x , y , z ,w ∈ Nsuch that 2x + 2y + 2z + 1 is prime.

Example. 143 = 12 + 52 + 92 + 4 · 32 with 21 + 25 + 29 + 1 = 547prime.

Conjecture (Z.-W. Sun, August 20, 2017). Any odd integer n > 1can be written as x2 + y2 + z2 + w2 (x , y , z ,w ∈ N) such that2x+y + 2z+w + 1 is prime.

Example. 197 = 62 + 62 + 22 + 112 with26+6 + 22+11 + 1 = 12289 prime. And

2× 6998538 + 1 = 1222 + 2202 + 2082 + 37272

with 2122+220 + 2208+3727 + 1 = 2342 + 23935 + 1 a prime of 1185decimal digits.

I have verified both conjectures for positive odd integers not morethan 2× 107.

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Part III. Restricted Sums of Three Squares

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On the representation n = x2 + y 2 + z(3z − 1)/2

Those numbers z(3z − 1)/2 (z ∈ Z) are called generalizedpentagonal numbers.

In a paper published in 2015, I noted that any n ∈ N can bewritten as x2 + y2 + z(3z − 1)/2 with x , y , z ∈ Z. Surprising, thiscan be further refined.

Conjecture (Z.-W. Sun, March 3, 2017). Any n ∈ N can bewritten as x2 + y2 + z(3z − 1)/2 with x , y , z ∈ Z such that x + 2yis a square.

Example.

803 = (−17)2+132+(−15)(3(−15)− 1)

2with (−17)+2×13 = 32.

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Refining the Gauss-Legendre theorem

Gauss-Legendre Theorem. A nonnegative integer n can beexpressed as the sum of three squares if and only if it is not of theform 4k(8l + 7) with k , l ∈ N.

Conjecture (Z.-W. Sun, March 4, 2017). (i) Any n ∈ Z+ withord2(n) odd can be written as x2 + y2 + z2 with x , y , z ∈ Z suchthat x + 3y + 5z is a square.

(ii) Let n ∈ N \ {63}. Then 4n + 1 can be written as x2 + y2 + z2

with x , y , z ∈ Z such that x + 3y + 5z is a square.

(iii) Any n ∈ N not of the form 4k(8l + 7) with k, l ∈ N can bewritten as x2 + y2 + z2 with x , y , z ∈ Z such that x + 2y + 3z is asquare or twice a square.

Example.

1430 = (−13)2+(−6)2+352 with (−13)+3×(−6)+5×35 = 122.

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Restricted representations for positive odd numbers

It is known that any positive odd numbers can be written asx2 + y2 + 2z2 with x , y , z ∈ Z. Also, we can replace x2 + y2 + 2z2

by x2 + 2y2 + 3z2.

Conjecture (Z.-W. Sun, March 5, 2017). (i) Any positive oddinteger can be written as x2 + y2 + 2z2 with x , y , z ∈ Z such that2x + y + z is a square or a power of two.

(ii) Any positive odd integer can be written as x2 + 2y2 + 3z2 withx , y , z ∈ Z such that x + y + z is a square or twice a square.

Examples:

2×1143 + 1 = (−22)2 + 2×302 + 3×12 with (−22) + 30 + 1 = 32,

and2× 6408 + 1 = (−22)2 + 2× 752 + 3× 192

with(−22) + 75 + 19 = 2× 62.

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Part IV. Sums of Two Squares and Two Other Terms

58 / 63

Write n > 5 as x2 + y 2 + 2z + 5× 2w

Conjecture (Z.-W. Sun, Feb. 22, 2018). For any n ∈ Z+ we canwrite n2 as x2 + y2 + z2 + w2 with x , y , z ,w ∈ N andx + 2y + 3z ∈ {4k : k ∈ N}.

Conjecture (Z.-W. Sun, April 22, 2018). Any integer n > 3 canbe written as x2 + 2y2 + 3× 2z + 4w with x , y , z ,w ∈ N.

Remark. I have verified this for n up to 1010. The first value ofn > 1 with n2 not of the form x2 + 2y2 + 3× 2z is 5884015571.

R. Crocker [Colloq. Math. 2008]: For each integer a > 2, thereare infinitely many positive integers none of which is the sum oftwo squares and at most two powers of a.

Conjecture (Z.-W. Sun, April 28, 2018). Any integer n > 5 canbe written as x2 + y2 + 2z + 5× 2w with x , y , z ,w ∈ N.

Remark. I have verified this for n up to 5× 109.

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Write n = a2 + b2 + 3c + 5d

Conjecture (Z.-W. Sun, April 28, 2018). Any integer n > 1 canbe written as a2 + b2 + 3c + 5d with a, b, c , d ∈ N = {0, 1, 2, . . .}.Remark. I have verified this for n up to 2× 1010, and I’d like tooffer 3500 US dollars as the prize for the first proof of thisconjecture. I also conjecture that 5d in the conjecture can bereplaced by 2d .

Example.

2 = 02 +02 +30 +50, 5 = 02 +12 +31 +50, 25 = 12 +42 +31 +51.

Conjecture (Z.-W. Sun, April 26, 2018). Any integer n > 1 canbe written as the sum of two squares and two central binomialcoefficients.

Remark. I have verified this for n up to 1010.

Example.

2435 = 322 + 332 +

(2× 4

4

)+

(2× 5

5

).

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Sums of two triangular numbers and two powers of 5

Recall that those Tn = n(n + 1)/2 with n ∈ N are called triangularnumbers. As claimed by Fermat and proved by Gauss, each n ∈ Nis the sum of three triangular numbers.

Conjecture (Z.-W. Sun, April 23, 2018). Any integer n > 1 canbe written as Ta + Tb + 5c + 5d with a, b, c , d ∈ N.

Remark. I have verified this for n up to 1010.

Conjecture (Z.-W. Sun, April 23, 2018). Any integer n > 1 can bewritten as the sum of p5(a) + p5(b) + 3c + 3d with a, b, c , d ∈ N,where p5(k) denotes the pentagonal number k(3k − 1)/2.

Remark. I have verified this for n up to 7× 106.

Example.

285 = p5(1) + p5(11) + 33 + 34, 13372 = p5(17) + p5(65) + 34 + 38.

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Two new conjectures involving primes

Conjecture (Z.-W. Sun, May 2018) Any even number n > 2 canbe written as the sum of a prime, a power of 2 and a power of 5.Furthermore, we can write every integer n > 7 asp + 2k + (1 + (n mod 2))× 5m, where p is an odd prime, andk ,m ∈ N with 2k + (1 + (n mod 2))× 5m squarefree.

Remark. I have verified this for n up to 2× 1010.

Recall that the Fibonacci numbers F0,F1, . . . and the Lucasnumbers L0, L1, . . . are respectively given by

F0 = 0, F1 = 1, Fn+1 = Fn + Fn−1 (n = 1, 2, 3, . . .)

and

L0 = 2, L1 = 1, Ln+1 = Ln + Ln−1 (n = 1, 2, 3, . . .).

Conjecture (Z.-W. Sun, June 2018) Any integer n > 3 can bewritten as p + FkLm, where p is an odd prime, and k and m arepositive integers.

Remark. I have verified this for n up to 5× 109.62 / 63

References

For the main sources of my above conjectures and related results,you may look at the following papers:

1. Zhi-Wei Sun, Refining Lagrange’s four-square theorem, J.Number Theory 175(2017), 167–190. arXiv:1604.06723

2. Yu-Chen Sun and Zhi-Wei Sun, Some variants of Lagrange’sfour squares theorem, Acta Arith. 183(2018), no.4, 339-356. Seealso http://arxiv.org/abs/1605.03074.

3. Zhi-Wei Sun, Restricted sums of four squares,arXiv:1701.05868 , http://arxiv.org/abs/1701.05868.

4. Hai-Liang Wu and Zhi-Wei Sun, On the 1-3-5 conjecture andrelated topics, arXiv:1710.08763, available from the websitehttp://arxiv.org/abs/1710.08763.

Thank you!63 / 63