# Pell's equation

(Redirected from Pellian equation)
Pell's equation for n = 2 and six of its integer solutions

Pell's equation (also called the Pell–Fermat equation) is any Diophantine equation of the form

${\displaystyle x^{2}-ny^{2}=1\,}$

where n is a given positive nonsquare integer and integer solutions are sought for x and y. In Cartesian coordinates, the equation has the form of a hyperbola; solutions occur wherever the curve passes through a point whose x and y coordinates are both integers, such as the trivial solution with x = 1 and y = 0. Joseph Louis Lagrange proved that, as long as n is not a perfect square, Pell's equation has infinitely many distinct integer solutions. These solutions may be used to accurately approximate the square root of n by rational numbers of the form x/y.

This equation was first studied extensively in India, starting with Brahmagupta, who developed the chakravala method to solve Pell's equation and other quadratic indeterminate equations in his Brahma Sphuta Siddhanta in 628, about a thousand years before Pell's time. His Brahma Sphuta Siddhanta was translated into Arabic in 773 and was subsequently translated into Latin in 1126. Bhaskara II in the 12th century and Narayana Pandit in the 14th century both found general solutions to Pell's equation and other quadratic indeterminate equations. Solutions to specific examples of the Pell equation, such as the Pell numbers arising from the equation with n = 2, had been known for much longer, since the time of Pythagoras in Greece and to a similar date in India. The name of Pell's equation arose from Leonhard Euler's mistakenly attributing Lord Brouncker's solution of the equation to John Pell.[1]

## History

As early as 400 BC in India and Greece, mathematicians studied the numbers arising from the n = 2 case of Pell's equation,

${\displaystyle x^{2}-2y^{2}=1}$

and from the closely related equation

${\displaystyle x^{2}-2y^{2}=-1}$

because of the connection of these equations to the square root of two.[2] Indeed, if x and y are positive integers satisfying this equation, then x/y is an approximation of √2. The numbers x and y appearing in these approximations, called side and diameter numbers, were known to the Pythagoreans, and Proclus observed that in the opposite direction these numbers obeyed one of these two equations.[2] Similarly, Baudhayana discovered that x = 17, y = 12 and x = 577, y = 408 are two solutions to the Pell equation, and that 17/12 and 577/408 are very close approximations to the square root of two.

Later, Archimedes approximated the square root of 3 by the rational number 1351/780. Although he did not explain his methods, this approximation may be obtained in the same way, as a solution to Pell's equation.[2] Archimedes' cattle problem involves solving a Pellian equation. It is now generally accepted that this problem is due to Archimides. [3][4]

Around AD 250, Diophantus considered the equation

${\displaystyle a^{2}x^{2}+c=y^{2},}$

where a and c are fixed numbers and x and y are the variables to be solved for. This equation is different in form from Pell's equation but equivalent to it. Diophantus solved the equation for (a, c) equal to (1, 1), (1, −1), (1, 12), and (3, 9). Al-Karaji, a 10th-century Persian mathematician, worked on similar problems to Diophantus.

In Indian mathematics, Brahmagupta discovered that

${\displaystyle (x_{1}^{2}-Ny_{1}^{2})(x_{2}^{2}-Ny_{2}^{2})=(x_{1}x_{2}+Ny_{1}y_{2})^{2}-N(x_{1}y_{2}+x_{2}y_{1})^{2}=(x_{1}x_{2}-Ny_{1}y_{2})^{2}-N(x_{1}y_{2}-x_{2}y_{1})^{2}}$

(see Brahmagupta's identity). Using this, he was able to "compose" triples ${\displaystyle (x_{1},y_{1},k_{1})}$ and ${\displaystyle (x_{2},y_{2},k_{2})}$ that were solutions of ${\displaystyle x^{2}-Ny^{2}=k}$, to generate the new triples

${\displaystyle (x_{1}x_{2}+Ny_{1}y_{2},x_{1}y_{2}+x_{2}y_{1},k_{1}k_{2})}$ and ${\displaystyle (x_{1}x_{2}-Ny_{1}y_{2},x_{1}y_{2}-x_{2}y_{1},k_{1}k_{2}).}$

Not only did this give a way to generate infinitely many solutions to ${\displaystyle x^{2}-Ny^{2}=1}$ starting with one solution, but also, by dividing such a composition by ${\displaystyle k_{1}k_{2}}$, integer or "nearly integer" solutions could often be obtained. For instance, for ${\displaystyle N=92}$, Brahmagupta composed the triple (10, 1, 8) (since ${\displaystyle 10^{2}-92(1^{2})=8}$) with itself to get the new triple (192, 20, 64). Dividing throughout by 64 ('8' for ${\displaystyle x}$ and ${\displaystyle y}$) gave the triple (24, 5/2, 1), which when composed with itself gave the desired integer solution (1151, 120, 1). Brahmagupta solved many Pell equations with this method; in particular he showed how to obtain solutions starting from an integer solution of ${\displaystyle x^{2}-Ny^{2}=k}$ for k = ±1, ±2, or ±4.[5]

The first general method for solving the Pell equation (for all N) was given by Bhaskara II in 1150, extending the methods of Brahmagupta. Called the chakravala (cyclic) method, it starts by choosing two relatively prime integers ${\displaystyle a}$ and ${\displaystyle b}$, then composing the triple ${\displaystyle (a,b,k)}$ (that is, one which satisfies ${\displaystyle a^{2}-Nb^{2}=k}$) with the trivial triple ${\displaystyle (m,1,m^{2}-N)}$ to get the triple ${\displaystyle (am+Nb,a+bm,k(m^{2}-N))}$, which can be scaled down to

${\displaystyle \left({\frac {am+Nb}{k}}\,,\,{\frac {a+bm}{k}}\,,\,{\frac {m^{2}-N}{k}}\right).}$

When ${\displaystyle m}$ is chosen so that ${\displaystyle (a+bm)/k}$ is an integer, so are the other two numbers in the triple. Among such ${\displaystyle m}$, the method chooses one that minimizes ${\displaystyle (m^{2}-N)/k}$, and repeats the process. This method always terminates with a solution (proved by Lagrange in 1768). Bhaskara used it to give the solution x = 1766319049, y = 226153980 to the notorious N = 61 case.[5]

Several European mathematicians rediscovered how to solve Pell's equation in the 17th century, apparently unaware that it had been solved almost five hundred years earlier in India. Fermat found how to solve the equation and in a 1657 letter issued it as a challenge to English mathematicians. In a letter to Digby, Bernard Frénicle de Bessy said that Fermat found the smallest solution for N up to 150, and challenged John Wallis to solve the cases N = 151 or 313. Both Wallis and Lord Brouncker gave solutions to these problems, though Wallis suggests in a letter that the solution was due to Brouncker.

Pell's connection with the equation is that he revised Thomas Branker's translation (Rahn 1668) of Johann Rahn's 1659 book "Teutsche Algebra" into English, with a discussion of Brouncker's solution of the equation. Euler mistakenly thought that this solution was due to Pell, as a result of which he named the equation after Pell.

The general theory of Pell's equation, based on continued fractions and algebraic manipulations with numbers of the form ${\displaystyle P+Q{\sqrt {a}},}$ was developed by Lagrange in 1766–1769.[6]

## Solutions

### Fundamental solution via continued fractions

Let ${\displaystyle {\tfrac {h_{i}}{k_{i}}}}$ denote the sequence of convergents to the regular continued fraction for ${\displaystyle {\sqrt {n}}}$. This sequence is unique. Then the pair (x1,y1) solving Pell's equation and minimizing x satisfies x1 = hi and y1 = ki for some i. This pair is called the fundamental solution. Thus, the fundamental solution may be found by performing the continued fraction expansion and testing each successive convergent until a solution to Pell's equation is found.

As Lenstra (2002) describes, the time for finding the fundamental solution using the continued fraction method, with the aid of the Schönhage–Strassen algorithm for fast integer multiplication, is within a logarithmic factor of the solution size, the number of digits in the pair (x1,y1). However, this is not a polynomial time algorithm because the number of digits in the solution may be as large as √n, far larger than a polynomial in the number of digits in the input value n (Lenstra 2002).

### Additional solutions from the fundamental solution

Once the fundamental solution is found, all remaining solutions may be calculated algebraically from

${\displaystyle x_{k}+y_{k}{\sqrt {n}}=(x_{1}+y_{1}{\sqrt {n}})^{k},}$

expanding the right side, equating coefficients of ${\displaystyle {\sqrt {n}}}$ on both sides, and equating the other terms on both sides. This yields the recurrence relations

${\displaystyle \displaystyle x_{k+1}=x_{1}x_{k}+ny_{1}y_{k},}$
${\displaystyle \displaystyle y_{k+1}=x_{1}y_{k}+y_{1}x_{k}.}$

### Concise representation and faster algorithms

Although writing out the fundamental solution (x1, y1) as a pair of binary numbers may require a large number of bits, it may in many cases be represented more compactly in the form

${\displaystyle x_{1}+y_{1}{\sqrt {n}}=\prod _{i=1}^{t}(a_{i}+b_{i}{\sqrt {n}})^{c_{i}}}$

using much smaller integers ai, bi, and ci.

For instance, Archimedes' cattle problem is equivalent to the Pell equation ${\displaystyle x^{2}-410286423278424y^{2}=1}$, the fundamental solution of which has 206545 digits if written out explicitly. However, the solution is also equal to

${\displaystyle x_{1}+y_{1}{\sqrt {n}}=u^{2329},}$

where

${\displaystyle u=x'_{1}+y'_{1}{\sqrt {4729494}}}$

and ${\displaystyle x'_{1}}$ and ${\displaystyle y'_{1}}$ only have 45 and 41 decimal digits, respectively. (Alternatively, one may write even more concisely

${\displaystyle u=(300426607914281713365{\sqrt {609}}+84129507677858393258{\sqrt {7766}})^{2}.}$ [7])

Methods related to the quadratic sieve approach for integer factorization may be used to collect relations between prime numbers in the number field generated by √n, and to combine these relations to find a product representation of this type. The resulting algorithm for solving Pell's equation is more efficient than the continued fraction method, though it still takes more than polynomial time. Under the assumption of the generalized Riemann hypothesis, it can be shown to take time

${\displaystyle \exp O({\sqrt {\log N\log \log N}}),}$

where N = log n is the input size, similarly to the quadratic sieve (Lenstra 2002).

### Quantum algorithms

Hallgren (2007) showed that a quantum computer can find a product representation, as described above, for the solution to Pell's equation in polynomial time. Hallgren's algorithm, which can be interpreted as an algorithm for finding the group of units of a real quadratic number field, was extended to more general fields by Schmidt & Völlmer (2005).

## Example

As an example, consider the instance of Pell's equation for n = 7; that is,

${\displaystyle x^{2}-7y^{2}=1}$

The sequence of convergents for the square root of seven are

h / k (Convergent) h2 −7k2 (Pell-type approximation)
2 / 1 −3
3 / 1 +2
5 / 2 −3
8 / 3 +1

Therefore, the fundamental solution is formed by the pair (8, 3). Applying the recurrence formula to this solution generates the infinite sequence of solutions

(1, 0); (8, 3); (127, 48); (2024, 765); (32257, 12192); (514088, 194307); (8193151, 3096720); (130576328, 49353213); ... (sequence A001081 (x) and A001080 (y) in OEIS)

The smallest solution can be very large. For example, the smallest solution to ${\displaystyle x^{2}-313y^{2}=1}$ is (32188120829134849, 1819380158564160), and this is the equation which Frenicle challenged Wallis to solve.[8] Values of n such that the smallest solution of ${\displaystyle x^{2}-ny^{2}=1}$ is greater than the smallest solution for any smaller value of n are

1, 2, 5, 10, 13, 29, 46, 53, 61, 109, 181, 277, 397, 409, 421, 541, 661, 1021, 1069, 1381, 1549, 1621, 2389, 3061, 3469, 4621, 4789, 4909, 5581, 6301, 6829, 8269, 8941, 9949, ... (sequence A033316 in the OEIS)

(For these records, see (x), and (y)).

## The smallest solution of Pell equations

The following is a list of the smallest solution to ${\displaystyle x^{2}-ny^{2}=1}$ with n ≤ 128. For square n, there is no solution except (1, 0). (sequence A002350 (x) and A002349 (y) in OEIS, or (x) and (y) (for nonsquare n))

n x y n x y n x y n x y
1 - - 33 23 4 65 129 16 97 62809633 6377352
2 3 2 34 35 6 66 65 8 98 99 10
3 2 1 35 6 1 67 48842 5967 99 10 1
4 - - 36 - - 68 33 4 100 - -
5 9 4 37 73 12 69 7775 936 101 201 20
6 5 2 38 37 6 70 251 30 102 101 10
7 8 3 39 25 4 71 3480 413 103 227528 22419
8 3 1 40 19 3 72 17 2 104 51 5
9 - - 41 2049 320 73 2281249 267000 105 41 4
10 19 6 42 13 2 74 3699 430 106 32080051 3115890
11 10 3 43 3482 531 75 26 3 107 962 93
12 7 2 44 199 30 76 57799 6630 108 1351 130
13 649 180 45 161 24 77 351 40 109 158070671986249 15140424455100
14 15 4 46 24335 3588 78 53 6 110 21 2
15 4 1 47 48 7 79 80 9 111 295 28
16 - - 48 7 1 80 9 1 112 127 12
17 33 8 49 - - 81 - - 113 1204353 113296
18 17 4 50 99 14 82 163 18 114 1025 96
19 170 39 51 50 7 83 82 9 115 1126 105
20 9 2 52 649 90 84 55 6 116 9801 910
21 55 12 53 66249 9100 85 285769 30996 117 649 60
22 197 42 54 485 66 86 10405 1122 118 306917 28254
23 24 5 55 89 12 87 28 3 119 120 11
24 5 1 56 15 2 88 197 21 120 11 1
25 - - 57 151 20 89 500001 53000 121 - -
26 51 10 58 19603 2574 90 19 2 122 243 22
27 26 5 59 530 69 91 1574 165 123 122 11
28 127 24 60 31 4 92 1151 120 124 4620799 414960
29 9801 1820 61 1766319049 226153980 93 12151 1260 125 930249 83204
30 11 2 62 63 8 94 2143295 221064 126 449 40
31 1520 273 63 8 1 95 39 4 127 4730624 419775
32 17 3 64 - - 96 49 5 128 577 51

## Connections

Pell's equation has connections to several other important subjects in mathematics.

### Algebraic number theory

Pell's equation is closely related to the theory of algebraic numbers, as the formula

${\displaystyle x^{2}-ny^{2}=(x+y{\sqrt {n}})(x-y{\sqrt {n}})}$

is the norm for the ring ${\displaystyle \mathbb {Z} [{\sqrt {n}}]}$ and for the closely related quadratic field ${\displaystyle \mathbb {Q} ({\sqrt {n}})}$. Thus, a pair of integers ${\displaystyle (x,y)}$ solves Pell's equation if and only if ${\displaystyle x+y{\sqrt {n}}}$ is a unit with norm 1 in ${\displaystyle \mathbb {Z} [{\sqrt {n}}]}$. Dirichlet's unit theorem, that all units of ${\displaystyle \mathbb {Z} [{\sqrt {n}}]}$ can be expressed as powers of a single fundamental unit (and multiplication by a sign), is an algebraic restatement of the fact that all solutions to the Pell equation can be generated from the fundamental solution. The fundamental unit can in general be found by solving a Pell-like equation but it does not always correspond directly to the fundamental solution of Pell's equation itself, because the fundamental unit may have norm −1 rather than 1 and its coefficients may be half integers rather than integers.

### Chebyshev polynomials

Demeyer (2007) mentions a connection between Pell's equation and the Chebyshev polynomials: If Ti (x) and Ui (x) are the Chebyshev polynomials of the first and second kind, respectively, then these polynomials satisfy a form of Pell's equation in any polynomial ring R[x], with n = x2 − 1:

${\displaystyle T_{i}^{2}-(x^{2}-1)U_{i-1}^{2}=1.\,}$

Thus, these polynomials can be generated by the standard technique for Pell equations of taking powers of a fundamental solution:

${\displaystyle T_{i}+U_{i-1}{\sqrt {x^{2}-1}}=(x+{\sqrt {x^{2}-1}})^{i}.\,}$

It may further be observed that, if (xi,yi) are the solutions to any integer Pell equation, then xi = Ti (x1) and yi = y1Ui − 1(x1) (Barbeau, chapter 3).

### Continued fractions

A general development of solutions of Pell's equation ${\displaystyle x^{2}-ny^{2}=1}$ in terms of continued fractions of ${\displaystyle {\sqrt {n}}}$ can be presented, as the solutions x and y are approximates to the square root of n and thus are a special case of continued fraction approximations for quadratic irrationals.

The relationship to the continued fractions implies that the solutions to Pell's equation form a semigroup subset of the modular group. Thus, for example, if p and q satisfy Pell's equation, then

${\displaystyle {\begin{pmatrix}p&q\\nq&p\end{pmatrix}}}$

is a matrix of unit determinant. Products of such matrices take exactly the same form, and thus all such products yield solutions to Pell's equation. This can be understood in part to arise from the fact that successive convergents of a continued fraction share the same property: If pk−1/qk−1 and pk/qk are two successive convergents of a continued fraction, then the matrix

${\displaystyle {\begin{pmatrix}p_{k-1}&p_{k}\\q_{k-1}&q_{k}\end{pmatrix}}}$

has determinant (−1)k.

Størmer's theorem applies Pell equations to find pairs of consecutive smooth numbers. As part of this theory, Størmer also investigated divisibility relations among solutions to Pell's equation; in particular, he showed that each solution other than the fundamental solution has a prime factor that does not divide n.

As Lenstra (2002) describes, Pell's equation can also be used to solve Archimedes' cattle problem.

## The negative Pell equation

The negative Pell equation is given by

${\displaystyle x^{2}-ny^{2}=-1}$

It has also been extensively studied; it can be solved by the same method of continued fractions and will have solutions if and only if the period of the continued fraction has odd length. However it is not known which roots have odd period lengths and therefore not known when the negative Pell equation is solvable. A necessary (but not sufficient) condition for solvability is that n is not divisible by 4 or by a prime of form 4k + 3.[9] Thus, for example, x2 − 3ny2 = −1 is never solvable, but x2 − 5ny2 = −1 may be.

The first few numbers n for which x2 − ny2 = −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, 106, 109, 113, 122, 125, 130, 137, 145, 149, 157, 170, 173, 181, 185, 193, 197, 202, 218, 226, 229, 233, 241, 250, ... (sequence A031396 in the OEIS).

Cremona & Odoni (1989) demonstrate that the proportion of square-free n divisible by k primes of the form 4m + 1 for which the negative Pell equation is solvable is at least 40%. If the negative Pell equation does have a solution for a particular n, its fundamental solution leads to the fundamental one for the positive case by squaring both sides of the defining equation:

${\displaystyle (x^{2}-ny^{2})^{2}=(-1)^{2}}$

implies

${\displaystyle (x^{2}+ny^{2})^{2}-n(2xy)^{2}=1.}$

## Transformations

The related equation

${\displaystyle u^{2}-dv^{2}=\pm 2}$ (*)

can be used to find solutions to the positive Pell equation for certain d. Legendre proved that all primes of form d = 4m + 3 solve one case of (*), with the form 8m + 3 solving the negative, and 8m + 7 for the positive. Their fundamental solution then leads to the one for x2dy2 = 1. This can be shown by squaring both sides of (*)

${\displaystyle (u^{2}-dv^{2})^{2}=(\pm 2)^{2}}$

to get

${\displaystyle (u^{2}+dv^{2})^{2}-d(2uv)^{2}=4.}$

Since ${\displaystyle dv^{2}=u^{2}\mp 2}$ from (*), it follows that

${\displaystyle (u^{2}\mp 1)^{2}-d(uv)^{2}=1,}$

and so the fundamental solutions to (*) are smaller than those of the associated negative Pell equation. For example, u2-3v2 = -2 is {u,v} = {1, 1}, so x2 − 3y2 = 1 has {x,y} = {2, 1}. On the other hand, u2 − 7v2 = 2 is {u,v} = {3,1}, so x2 − 7y2 = 1 has {x,y} = {8,3}.

Another related equation,

${\displaystyle u^{2}-dv^{2}=\pm 4}$

can also be used to find solutions to Pell equations for certain d, this time for the positive and negative case. For the following transformations,[10] if fundamental {u,v} are both odd, then it leads to fundamental {x,y}.

1. If u2 − dv2 = −4, and {x,y} = {(u2 + 3)u/2, (u2 + 1)v/2}, then x2 − dy2 = −1.

Ex. Let d = 13, then {u,v} = {3, 1} and {x,y} = {18, 5}.

2. If u2 − dv2 = 4, and {x,y} = {(u2 − 3)u/2, (u2 − 1)v/2}, then x2 − dy2 = 1.

Ex. Let d = 13, then {u,v} = {11, 3} and {x,y} = {649, 180}.

3. If u2 − dv2 = −4, and {x,y} = {(u4 + 4u2 + 1)(u2 + 2)/2, (u2 + 3)(u2 + 1)uv/2}, then x2 − dy2 = 1.

Ex. Let d = 61, then {u,v} = {39, 5} and {x,y} = {1766319049, 226153980}.

Especially for the last transformation, it can be seen how solutions to {u,v} are much smaller than {x,y}, since the latter are sextic and quintic polynomials in terms of u.

## Notes

1. ^ Lettre IX. Euler à Goldbach, dated 10 August 1750 in: P. H. Fuss, ed., Correspondance Mathématique et Physique de Quelques Célèbres Géomètres du XVIIIeme Siècle … (Mathematical and physical correspondence of some famous geometers of the 18th century), vol. 1, (St. Petersburg, Russia: 1843), pp. 35-39 ; see especially page 37. From page 37: "Pro hujusmodi quaestionibus solvendis excogitavit D. Pell Anglus peculiarem methodum in Wallisii operibus expositam." (For solving such questions, the Englishman Dr. Pell devised a singular method [which is] shown in Wallis' works.)
2. ^ a b c Knorr, Wilbur R. (1976), "Archimedes and the measurement of the circle: a new interpretation", Archive for History of Exact Sciences, 15 (2): 115–140, MR 0497462, doi:10.1007/bf00348496.
3. ^ Fraser, P.M. (1972). Ptolemaic Alexandria. Oxford University Press.
4. ^ Weil, A. (1972). Number Theory, an Approach Through History. Birkhäuser.
5. ^ a b John Stillwell (2002), Mathematics and its history (2nd ed.), Springer, pp. 72–76, ISBN 978-0-387-95336-6
6. ^ "Solution d'un Problème d'Arithmétique", in J.-A. Serret (Ed.), Oeuvres de Lagrange, vol. 1, pp. 671–731, 1867.
7. ^
8. ^ Prime Curios!: 313
9. ^ This is because the Pell equation implies that −1 is a quadratic residue modulo n.
10. ^ A Collection of Algebraic Identities: Pell Equations.