Math 390 Project Skills

Continued Fractions

Group 2C

Abstract

Within this report we will show the basic properties of continued fractions and how to convert numbers into

continued fractions. We will also introduce the notion of infinite continued fractions and see if and where they

converge. Finally we will see how continued fractions are related to other well known areas of mathematics such

as quadratic irrationals and Pell’s equation.

1 Introduction

So what are continued fractions? Even though they have been around for centruries, the foundations for continued

fractions were only laid in the late 17th century. To put it simply, they are just another way of representing a

real number, rational or irrational, in terms of a sequence of integers. Arising naturally, we shall see the intriguing

secrets they possess coming from trying to solve the famous Pell equation.

The name continued fraction first came about in 1653. Now used to describe the representation of the reals, English

mathematician John Wallis in his Arithmetica infinitorum talked about the idea of continued fractions. Since then

they have grown predominantly in the field of number theory as an alternative way of representing numbers. We

begin with the basics of what actually are continued fractions, and some fundamental features. Our results shall

then take us to Diophantine equations, in particular Pell’s equation, and see what interesting features we can find

along the way.

As you already know, there are many ways of representing numbers; for example decimals, series and fractions.

Each of the systems have unique qualities making them interesting to examine. Take the fractions for example,

then we notice that15

11= 1.36363636 . . .

But we could also express this fraction as 1 411 . We could take this one step further and get

14

11= 1 +

4

11= 1 +

1114

= 1 +1

2 + 34

;

since we can express 114 as 2 + 3

4 . Now we begin to see a glimpse of a new number representation, continued

fractions. What would happen if we continue in this way? And what results would we find? We shall explore these

questions and more in this report.

1

2 Continued fractions

Definition 2.0.1. Continued fractions are all expressions of the form

x = a0 +b0

a1 +b1

a2 + . . .

,

where ai and bi can either be real or complex numbers, for all i, and a0 ≥ 0.

For now we will only address finite continued fractions but will introduce the notion of infinite continued fractions

later.

Definition 2.0.2. A simple continued fraction is the specific case where bi = 1 for all i;

x = a0 +1

a1 +1

a2 + . . .

,

where ai are integers, for all i and a0 ≥ 0. [3]

We can define a finite continued fraction [a0; a1, a2, . . . , an] as follows [11],

[a0; a1, a2, . . . , an] = a0 +1

a1 +1

a2 +1

. . .+1

an

.

Let’s take another look at our first example of the fraction 1511 . We saw that 15

11 = 1 +1

2 + 34

. We proceed in a

similar fashion as before

15

11= 1 +

1

2 +3

4

= 1 +1

2 +143

= 1 +1

2 +1

1 + 13

= [1; 2, 1, 3]

using the notation defined above.

Note, if

x = a0 +b0

a1 +b1

a2 + . . .

,

then a0 is bxc, where bxc is the floor function.

Also observe that none of the ai = 0, with the exception of a0 if x < 1.

2

We can go further with the fraction 1511 by saying:

[1; 2, 1, 3] = 1 +1

2 +1

1 + 13

= 1 +1

2 +1

1 +1

2 + 11

= [1; 2, 1, 2, 1].

Taking a closer look at the last fraction of any finite continued fraction [4] we can see that

1

an=

1

(an − 1) + 1=

1

(an − 1) +1

1

;

so we can end on a 1, i.e [. . . , an] = [. . . , an − 1, 1].

This means the representation of a number as a finite continued fraction need not be unique. We shall not prove

this here but the finite continued fraction form is unique up to the last two terms and there are only two possible

ways of expressing a finite continued fraction: [a0; a1, a2, . . . , an] or [a0; a1, a2, . . . , an − 1, 1].

As we have witnessed, transforming a rational number into the continued fraction form is not difficult. It simply

boils down to breaking up a fraction into an integer part and another fraction, and that’s all there is to it.

3 Euclidian Algorithm

An easier and simpler method lies with the famous mathematician Euclid at around 300 BC, in his book Euclid’s

Elements. The algorithm he described is one of the oldest currently in common use. Principally, it is used to find

the highest common factor of two numbers. But here we shall see another side of the process.

Theorem 3.0.3. (Division with remainder) Given a ∈ Z and d ∈ N, there exist unique q, r ∈ Z such that a = qd+r

and 0 ≤ r < d.

Theorem 3.0.4. (Euclidean Algorithm) Let a and b be integers, b 6= 0 and |a| > |b|, with {r1, . . . , rn−1} ∈ N and

{q1, . . . , qn} ∈ Z. Then the following chain of divisions yields the highest common factor of a and b,

a = bq1 + r1, 0 ≤ r1 < b

b = r1q2 + r2, 0 ≤ r2 < r1

r1 = r2q3 + r3, 0 ≤ r3 < r2...

......

rn−3 = rn−2qn−1 + rn−1, 0 ≤ rn−1 < rn−2

rn−2 = rn−1qn + 0.

Suppose we have a rational number x say, and we wish it to be written as a continued fraction. By the definition

of being rational, we can write x = ab , where a and b are integers. Now we apply the algorithm described above.

From the first line we see thata

b= q1 +

r1b,

3

by dividing through by b. We can rearrange this to get

a

b= q1 +

1br1

.

In the same way

b

r1= q2 +

r2r1

⇒ b

r1= q2 +

1r1r2

.

Generally, we haveab = q1 + r1

b ,

br1

= q2 + r2r1,

r1r2

= q3 + r3r2,

......

rn−3

rn−2= qn−1 + rn−1

rn−2,

rn−2

rn−1= qn + 0.

Hence we have the continued fraction of a rational number which is formed from successive divisions. Note that

the terms of our continued fraction are exactly the quotients from the division.

x =a

b= q0 +

1

q1 +1

q2 +1

. . .+1

qn

= [q0; q1, q2, . . . , qn].

Take the fraction 317 . By the Euclidean algorithm we have

317 = 4 + 3

7 ,

73 = 2 + 1

3 ,

31 = 3 + 0

1 .

And so we have that 317 = [4; 2, 3].

Since the Euclidean algorithm is a finite process, it must eventually terminate when ri = 0. Therefore all rational

numbers can be expressed as finite continued fractions. Furthermore, by the uniqueness property of the Euclidean

algorithm, all of the quotients, and hence the continued fraction expression is unique up to the last two terms.

We have formalised the notion of generating continued fractions. Now we shall discuss the idea of a limit for

continued fractions in the cases of both rationals and irrationals.

4

4 Convergence

The idea of convergence appears in a diverse range of topics within mathematics. It is fundamental in order for

us to build a better understanding of these unique type of number. We shall now address the notion of infinite

continued fractions [5]. This may raise questions of convergence and whether infinite continued fractions have a

meaning [12]. We summarise our concerns in the following theorems, the proofs of which are quite involved, and

so we do not give them here.

Definition 4.0.5. The value of the infinite continued fraction [a0; a1, a2, . . .] is

limk→∞

ck .

Theorem 4.0.6. An infinite continued fraction converges and defines a real number. There is a one-to-one

correspondence between

• all continued fractions [a0; a1, a2, . . .] with an integer a0 and positive integers ak for k > 0 (and the last term

an > 1 in the case of finite continued fractions); and

• real numbers.

It is important to state this theorem. It covers all areas of uncertainty regarding the fundamentals of continued

fractions. Not only does it clarify the notion of the continued fraction, but also addresses underlying uncertainties

regarding uniqueness and convergence of all continued fractions, both finite and infinite. To secure our foundations

upon this subject we shall state here, without proof, the following results.

Theorem 4.0.7. The continued fraction expansion of a real number is finite if and only if the real number is

rational.

Proof. This is a result of the Euclidean algorithm using the division with remainder theorem.

Theorem 4.0.8. The continued fraction expansion of a real number is infinite if and only if the real number is

irrational.

Now we have a sure footing on the topic of continued fractions we can move on to some more interesting properties.

Definition 4.0.9. We define the kth convergent of a continued fraction [a0; a1, a2, . . .] as the value of [a0, a1, a2, . . . , ak]

which is denoted ck.

For example take the number π, then π = [3; 7, 15, 1, 292, 1, 1, 1 . . .]. Then π can be roughly estimated as π ≈[3; 7, 15, 1, 292], [8]. Therefore [3; 7, 15, 1, 292] is the 5th convergent of π.

5

So to get better approximations of our number x say, we take parts of the continued fraction of larger sizes until

we are close enough to x. Going back to our example of π then [3; 7, 15, 1, 292, 1, 1, 1] is a better approximation

than [3; 7, 15, 1, 292]. Hence with each successive convergent we get closer to the number we want to find.

Figure 1: kth convergence of a continued fraction.

From the above figure, we can also see that ck gets smaller for odd values of k and bigger for even values. We may

say c0 < c2 < c4 < . . . < c5 < c3 < c1.

We can express ck as a fraction pkqk

. This means we now have two sequences of real numbers (pk) and (qk), which

together can be used to find the partial sum of an irrational expression.

For example, take√

2. Then we see that√

2 = [1; 2, 2, 2, 2, . . .]. So the convergents are

k 1 2 3 4 5 6 7 8 9

ck32

75

1712

4129

9970

239169

577408

1393985

33632378

pk 3 7 17 41 99 239 577 1393 3363

qk 2 5 12 29 70 169 408 985 2378

We see both the sequences (pk) and (qk) are divergent for our example of an irrational number. However the ratio

of pkqk

converges to√

2. There are more intriguing features of the sequences (pk) and (qk), left for the interested

reader to find.

We have now addressed the key issue of uniqueness and convergence of our sequences of numbers. Although we

have only briefly talked about this topic, all continued fractions do converge to real numbers, and later we will see

that the above expansion of√

2 is in fact true.

6

5 Quadratic Irrationals

Upon initial observation the topic of quadratic irrationals may seem an usual appearance in this report. We shall

soon see how continued fractions can be used to solve monic quadratic equations.

Consider the quadratic equation

x2 − ax− b = 0 ; a, b ∈ Z .

By rearranging and dividing through by x, we can rewrite this as

x = a+b

x.

Now, if we reconstitute in for x on the right hand side, we have an infinite continued fraction

x = a+b

a+b

a+ . . .

.

This means that all monic quadratics can be solved using continued fractions. This is assuming that the solution

is in fact real. If it is complex, then it cannot be represented by a continued fraction. This is because, for our

purposes, a continued fraction can only represent a real number.

Definition 5.0.10. If the function f(x) = ax2 + bx + c = 0 can be solved by the irrational number α, then α is

known as a quadratic irrational; where a, b, c ∈ Z, a 6= 0.

From this definition, it follows from a general quadratic equation that

α =−b±

√d

2a,

where d = b2 − 4ac. Because α is irrational, d cannot be a perfect square and since α is real d > 0.

Definition 5.0.11. Suppose we have the quadratic irrational α = −b+√d

2a and its conjugate α′ = −b−√d

2a . Then α

is said to be reduced if α > 1 and −1α′ > 1.

Rearranging the second inequality from this definition we have the following two conditions for α to a reduced

quadratic irrational:

• α > 1

• −1 < α′ < 0

Consider the quadratic f(x) = 16x2−16x−3. Suppose we have f(α) = 0 for some α. Solving the quadratics yields

α = 2±√7

4 . Notice that 2+√7

4 > 1 and −1 < 2−√7

4 < 0, and so we see α is a reduced quadratic irrational. However,

if we look at the quadratic g(x) = x2 + x − 1. Then g has roots −1±√5

2 which doesn’t satisfy either inequality to

be a reduced quadratic irrational.

7

Definition 5.0.12. A continued fraction is periodic if the sequence ak, ak+1, . . . , ak+n eventually repeats. i.e

[a0, a1, . . . , ak−1, ak, . . . , ak+n, ak, . . . , ak+n, ak, . . . , ak+n, . . .],

This is often abbreviated as

[a0, a1, ..., ak−1, ak, ..., ak+n]

Definition 5.0.13. A continued fraction is purely periodic if a0 = a0+n, a1 = a1+n, a2 = a2+n · · · ak− = ak+n,

i.e

[a0, ..., an]

For example we have the continued fraction of√

2 = [1; 2] which is periodic and the continued fraction of the golden

ratio, [1; 1, 1, 1, . . .] = [1], is purely periodic.

Theorem 5.0.14. An irrational number α is a quadratic irrational if and only if the continued fraction form of α

is periodic.

The proof of this theorem involves recursive relations for the sequences pk and qk and a little algebraic manipulation.

This has been left to the interested reader.

The result means that all real solutions to quadratic equations can be expressed as continued fractions with the

periodic property. It can be deduced that all irrational numbers of the form√d must also be periodic continued

fractions, since√d = b+ 2aα. This brings us onto the next section, square roots.

6 Square roots

So far we have discussed the convergence of rational and irrational numbers. We shall now move on to looking

at square roots of positive integers. For example, let’s begin with a square root√

2. To write this as a continued

fraction, we first notice that 1 <√

2 < 2, so we start with rewriting√

2 as

√2 = 1 +

1

x

for some x > 1. So to find the continued fraction of√

2 we need to find x. Rearranging the equation, we have that

x =1√

2− 1=√

2 + 1.

This means that

x =√

2 + 1 = 1 +1

x+ 1 = 2 +

1

x.

By substituting 2 + 1x wherever we see x, we now have our continued fraction for x:

x = 2 +1

x= 2 +

1

2 +1

x

= . . .

Hence the continued fraction of√

2 is [1; 2].

8

Now we shall take a look at the continued fraction of some other integers d.

d Periodic Continued Fraction of√d Continued Fraction of

√d

2 [1; 2] [1; 2, 2, 2, 2, 2, 2, 2, . . .]

8 [2; 1, 4] [2; 1, 4, 1, 4, 1, 4, . . .]

19 [4; 2, 1, 3, 1, 2, 8] [4; 2, 1, 3, 1, 2, 8, 2, 1, 3, 1, 2, 8, . . .]

23 [4; 1, 3, 1, 8] [4; 1, 3, 1, 8, 1, 3, 1, 8, . . .]

45 [6; 1, 2, 2, 2, 1, 12] [6; 1, 2, 2, 2, 1, 12, 1, 2, 2, 2, 1, 12, . . .]

53 [7; 3, 1, 1, 3, 14] [7; 3, 1, 1, 3, 14, 3, 1, 1, 3, 14, . . .]

It is evident that square roots of integers have a periodic continued fraction representation. However, we often

write square roots of numbers in decimal form. In decimal form, there is no sequence visible in the nth decimal

place of the square root. Instead, the continued fraction form gives an interesting repeating pattern.

Moreover, you may notice that the final term of the period is double the first term. We shall see this is always the

case by our next two theorems.

Theorem 6.0.15. Suppose we have a positive non- square integer d. Then the continued fraction of√d is periodic

with√d = [a0; a1, a2, ..., an−1, 2a0] .

Proof. Take b√dc = a0. It follows that

√d + a0 > 1 and −1 < −

√d + a0 < 0. This means a0 +

√d is a

reduced quadratic irrational. By theorem 5.0.14, that the continued fraction form of√d + a0 is purely periodic,

giving us√d + a0 = [2a0; a1, a2, ..., an−1, an]. Equivalently,

√d + a0 = [2a0; a1, a2, ..., an−1, an, 2a0], which implies

√d = [a0; a1, a2, ..., an−1, 2a0].

Theorem 6.0.16. Apart for the term 2a1, the periodic part of the continued fraction of√d is symmetrical.

We shall omit the proof to the second theorem here. The proof to this theorem can be found in notes by Alexandra

Ioana Gliga [2] on Continued Fractions of the Square Root of Prime Numbers. The proof is relatively straight

forward using ideas from this section and the previous one on the periodicity of continued fractions.

9

7 Pell’s equation

As early as 400 BC in India and Greece, mathematicians studied Pell’s equation. This is because of the link between

Pell’s equation and the irrational number√

2. Despite Pell’s equation being named after the English mathematician

John Pell, the credit should have been given to Viscount William Brouncker. The mistaken identity was made by

Leonard Euler [9]. Unfortunately for Brouncker, the name Pell’s equation stuck.

We begin with first defining the equation. Pell’s equation is a particular form of a diophantine equation [7] [10].

These are equations of two or more variables with integer solutions.

Definition 7.0.17. Suppose d is a positive integer. Then for some x, y ∈ Z,

x2 − dy2 = ±1

is known as Pell’s equation.

We call the equation x2 − dy2 = −1 the negative Pell equation [14].

Let us first concentrate on the positive Pell equation; i.e. x2 − dy2 = 1.

Through observation, we can almost immediately see that if (x, y) is a solution, then so too are (−x,−y) and

(±x,∓y). So if we have one solution then we must have at least 3 other solutions, unless either x or y is equal to

zero.

We can separate the positive Pell equation into two cases. One where d is a perfect square, and one where it is not.

Suppose first that d is a perfect square. Then d = m2, for some integer m. Hence we have

x2 − dy2 = x2 −m2y2 = (x+my)(x−my) = 1.

Which, in turn, implies that either

x+my = x−my = 1

or

x+my = x−my = −1

Solving these equations simultaneously yields x = ±1 and y = 0. It is worth noting that this result is independent

of our initial choice of d. This solution is known as the trivial solution and always solves the equation.

Now we turn to the possibility that d is non-square. We begin with looking the graph of Pell’s equation with

differing values of d to get a visual interpretation of Pell’s equation.

From the graph in Figure 2 we see that increasing the value of d causes the graph to stretch parallel to the x axis.

This is of course expected as changing the value of d is a simple transformation. We also notice that the graph

is symmetrical in both the x and y axes, which should have been clear with the presence of the squares in Pell’s

equation.

10

Figure 2: This is graph shows Pell’s equations with different values of d. The red plot is d = 1 , giving x2−y2 = 1 .

The green plot is d = 2 , so x2 − 2y2 = 1 . The final blue plot is d = 16 , a square number, giving x2 − 16y2 = 1 .

The dots on curves are integer solutions of that curve.

Close to the origin, we see that all of our graphs curve into the points (±1, 0), the trivial solutions. However, if

we just focus on the higher values of x and y, we see that the curves are almost straight lines. This suggests the

presence of asymptotes. Our asymptote looks to be of linear form, i.e. y = ax + b for some real a and b. Setting

this equal to to the equation of the positive Pell equation we find that y = ± 1√d

√x2 − 1. Because

√x2 − 1→ x as

x→∞, the equations of the asymptotes are y = ± 1√dx.

Now we shall take a look at the integer solutions to the Pell equation x2 − dy2 = 1. As we can see from the graph

above, we have that two of the curves have integer solutions, which are marked on our graph. From the equation

x2− 2y2 = 1, it need not be true that each curve can have a maximum of one integer solution. There could indeed

be more integer solutions to the equations with higher x and y values. In the case that d = 1, we have the curve

x2 − y2 = 1. No integer solutions are marked on the graph, but could there be any integer solutions with higher x

and y values? It turns out that the equation does not have an integer solution other than the trivial one. This is

because the difference between two square integers is at least 3, i.e. 22 − 12. This difference increases with higher

x ad y values.

We have looked at x2 − dy2 = 1, now we consider the negative Pell’s equation x2 − dy2 = −1.

We see a similar story in Figure 3, on the next page, as to our first graph Figure 2. With larger values of d we

again have a stretch in the x axis. We have the same symmetry properties.

However there are some noticeable differences. One being in Figure 3, the graph doesn’t touch the horizontal axis.

This is because when we have y = 0, it follows that x2 = −1. Since no such real number exists, the graph could

not possibly touch the x axis.

Furthermore, we do not have a trivial solution to our Pell equation x2 − dy2 = −1. Rather than the graph curving

to the points (±1, 0), they do not approach one particular point. Examining this further, we have when x = 0 that

dy2 = 1, which yields the results y = ± 1√d. So rather than approaching one point independent to d, it approaches

some variable point depending on the initial choice of d. We see as we increase the value of d then this point

approaches the origin from both sides. But we do know that this point lies somewhere within the interval [-1,1],

11

Figure 3: This graph shows Pell’s equations with the same values of d as the graph above Figure 2, however this

time using x2 − dy2 = −1 . Again, the red plot is d = 1 . The green plot is d = 2 . The final blue plot is when

d = 16 , a square number.

since d > 1.

Again looking at the higher values of x and y, we find that the shape of the graph is almost linear. In a similar

way as before we find that we have the equation of the asymptotes are y = ± 1√dx. This is identical to the positive

Pell equation. This may be an indication of a link between the positive and negative Pell equations. This brings

us to our next theorems.

Theorem 7.0.18. Suppose we have a solution (x, y) = (x0, y0) to the Pell equation x2−dy2 = 1, then all solutions

(x, y) to the positive Pell equation are

x+ y√d = (x0 + y0

√d)n.

We also have a similar result for the negative Pell equation.

Theorem 7.0.19. Suppose we have a solution (x, y) = (x0, y0) to the negative Pell equation x2 − dy2 = −1, then

all solutions (x, y) to the positive Pell equation are

x+ y√d = (x0 + y0

√d)n,

for all odd integers n ≥ 1

The proof to this theorem is similar to the one above. We shall leave these to the interested reader. Our next

theorem provides us with the link between continued fractions and Pell’s equation.

Theorem 7.0.20. Suppose x and y are integer solutions to Pell’s equation x2 − dy2 = 1. Then xy is convergent to

√d.

Proof. If x and y are solutions to Pell’s equation, we have x2−dy2 = 1, which implies (x+dy)(x−dy) = 1. Notice

12

we have that

x− y√d =

1

x+ y√d> 0 ,

so it follows x > y√d. Now consider

|√d− x

y| = x− y

√d

y=

1

y(x− y√d)

<1

y(y√d− y

√d)

=1

2y2√d<

1

2y2.

Hence xy is convergent to

√d.

Now we come to the issue of the existence of a solution.

Theorem 7.0.21. Suppose we have a non-square number d, then the equation x2 − dy2 = 1 has a non trivial

solution.

This means that Pell’s equations always have positive integer solutions regardless of our choice of d. We shall omit

the proof for length reasons. This was first proved by Lagrange in 1768. 100 years earlier Fermat claimed to have

a proof and challenged other mathematicians in Europe to prove it [1].

Let us now look if a similar result holds for the negative Pell equation x2 − dy2 = −1. Unfortunately, a full and

comprehensive set of solutions to the negative Pell equation is not yet known. However, some interesting results

have surfaced upon our pursuit of the final missing piece. Most notably x2− dy2 = −1 is solvable if and only if the

period length of the continued fraction for√d is odd. We shall come to this result later. Other results include:

• If d = 3p for some integer p then Pell’s equation doesn’t have a solution.

• In 1785, Legendre proved that if d is a prime and is congruent to 1 mod 4 then the negative Pell equation is

solvable.

• The negative Pell equation x2 − dy2 = 1 is solvable if and only if there exist a primitive Pythagorean triple

(A,B,C such that A2+B2 = C2) with A and B co prime; so by the Euclidean algorithm there exists a, b ∈ Z,

d = a2 + b2 and |aA− bB| = 1. [6]

• Numbers d for which x2 − dy2 = −1 is solvable are :

1, 2, 5, 10, 13, 17, 26, 29, 37, 41, 50, 53, 58, 61, 65, 73, 74, 82, 85, 89, 97, 101 . . . [13]

To illustrate the point of not all negative Pell equations having solutions, let’s look at an example. Take the negative

Pell equation x2 − 3y2 = −1. We shall show this has no solution. To do this we use an argument by contradiction

using modular arithmetic. So suppose an integer solution exists for some x and y satisfying x2 − 3y2 = −1. Then

we can reduce both sides by mod 3 to get x2 = −1 mod 3. This congruence has no solution: the only squares mod

3 are 0 and 1. Thus x2 − 3y2 = −1 has no solutions.

Let’s take a look at the continued fraction of a non-square number, for example√

153. We can see that

√153 = [12; 2, 1, 2, 2, 2, 1, 2, 24]

The first few convergents are

k 1 2 3 4 5 6 7 8 9 10 11 12

ck = pkqk

12 252

373

998

23519

56946

80465

2177176

530524289

1082818754

16133313043

43094734840

13

We see that pkqk

converges to√d, in this case

√153 . Therefore pk

qk≈√d for large enough k. Rearranging this to

get p2k − dq2k, we would expect this to be fairly close to 0 for large enough k.

k pk qk p2k − dq2k1 12 1 −9

2 25 2 13

3 37 3 −8

4 99 8 9

5 235 19 −8

6 569 46 13

7 804 65 −9

8 2177 176 1

9 53052 4289 −9

10 108281 8754 13

11 161333 13043 −8

12 430947 34840 9

However, as the table above shows, this need not be the case. We see that the final column isn’t close nor converging

to zero. In fact, we do see a repeating pattern occurring through the increasing values of k.

From the table, we observe that the number one appears in the final column. This is when k = 8, which coincides

with the length of the period of the continued fraction of√

153. This reflects that 2177− 153· 176 = 1. Therefore

we can say the convergent 2177176 to

√153 is a solution (2177, 176) to the Pell equation x2 − 153y2 = 1. Furthermore

we see that

[12; 2, 1, 2, 2, 2, 1, 2] =2177

176.

This is further evidence towards the link between the convergences of the continued fraction of√d and solutions

to the Pell equation x2 − dy2 = 1.

We see that with d = 153 the solution to the Pell equation was the continued fraction of√

153 with only one period

and removing the last entry of the period. This leads us to believe that a similar method could be used to find the

solution to Pell’s equation for an arbitrary value of d.

√d Continued Fraction of

√d Length of Period(k) [a0; a1, . . . , ak−1] = pk−1

qk−1p2k−1 − dq2k−1√

153 [12; 2, 1, 2, 2, 2, 1, 2, 24] 8 [12; 2, 1, 2, 2, 2, 1, 2] = 2177176 1

√73 [8; 1, 1, 5, 5, 1, 1, 16] 7 [8; 1, 1, 5, 5, 1, 1] = 1068

125 −1√

220 [14; 1, 4, 1, 28] 4 [14; 1, 4, 1] = 896 1

√111 [10; 1, 1, 6, 1, 1, 20] 6 [10; 1, 1, 6, 1, 1] = 295

28 1√

137 [11; 1, 2, 2, 1, 1, 2, 2, 1, 22] 9 [11; 1, 2, 2, 1, 1, 2, 2, 1] = 1744149 −1

From the table, it appears that there is a link between the convergences of a continued fraction of√d and solutions

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to the Pell equation. However, some of the results are solutions to Pell’s equation equalling -1 rather than 1. This

is due to the length of the period of the continued fraction of√d. If the length is odd, then we have solutions to

x2 − dy2 = −1, and if it is even we have solutions to x2 − dy2 = 1. Our results can be summarised by the next

theorem.

Definition 7.0.22. The solution to x2 − dy2 = ±1 in positive integers for which x+ y√d is smallest is called the

fundamental solution of this equation.

Theorem 7.0.23. Let the continued fraction of√d = [a0; a1, a2, ..., ak1, ak] and let pk−1

qk−1= [a0; a1, a2, . . . , ak−1].

Then the smallest solution in positive integers to Pell’s equation x2 − dy2 = 1 is given by

(x1, y1) =

(pk−1, qk−1) if k is even

(p2k−1 + q2k−1d, 2pk−1qk−1) if k is odd

We conclude our exploration of the world of continued fractions by solving the seemingly innocent Pell equation

x2 − 29y2 = 1.

With some calculation, we see that the continued fraction of√

29 = [5; 2, 1, 1, 2, 10]. Through observation we see

that the length of the period k is 5, which is odd. Thus by the above theorem we have that the fundamental

solution is

(x1, y1) = (p24 + q24d, 2p4q4) = (702 + 29 ∗ 132, 2 ∗ 70 ∗ 13) = (9801, 1820).

Substituting this into our Pell equation, we see that this is a solution. Now, if we wanted to find all other solutions,

by theorem 7.0.18 we simply raise the fundamental solution to a power. So our solution set is (xn, yn) such that

xn is the integer part of (9801 + 1820√

29)n and yn the coefficient of√

29.

The next smallest solution is (192119201, 35675640).

8 Conclusion

In this project, we have seen a new way of representing real numbers in terms of a sequence of positive integers.

In continued fraction form, rationals are finite continued fractions, irrationals are infinite, and square roots are

periodic. Square roots in particular possess some interesting properties that are not seen in decimal form, such

as their symmetrical periodic nature. This brings us to Pell’s equation, within this chapter we have seen several

results. Most notably that the continued fraction of√d solves the positive Pell equation. Also we have seen that

if one solution does exist, other than the trivial one, we have infinitely many solutions. Otherwise we simply have

the trivial case. With regards to the negative Pell equation, we know that if the length of the continued fraction

is odd, then we have a solution, but there is no simple solution if the length is even.

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