Pell’s equation

Pell’s equation

Pell’s equation refers generally to the equation

$$x^2-A{y}^2=1$$

where $A$ is a positive integer not a square. This equation is now known to have an infinite number of solutions in positive integers $x$ and $y$.

Fermat, in 1657, proposed to English mathematicians that they find a method of solving this equation in general, for integral $x$ and $y$. Fermat apparently already possessed a method, for doing so. According to [1], the Greeks and Indians had detailed knowledge of this problem; the solutions

${577}^2-2\cdot {408}^2=1$

${1151}^2-92\cdot {120}^2=1$

were both known by Indian mathematicians B.C. In fact, the solution given above for $A=92$ is the smallest solution for this $A$, and was given by Brahmagupta (b. 598 A.D.) together with a method for its derivation. The cyclic method, described in detail below, was given by Bháscara Achárya (b. 1114 A.D.).

The English, notably Wallis, developed a different but provably almost identical method, presumably in response to Fermat’s challenge. Both methods produce, in time, all integral solutions to the equation.

As can be seen from the entry on Pell’s equation and simple continued fractions, solutions of Pell’s equation can also be found by developing the continued fraction for $\sqrt{A}$.

Here is a table of the minimum value of $y$ for some values of $A$; a complete table for can be found in [1]:

A	y
2	2
22	42
37	12
61	226153980
109	15140424455100
136	3
143	1

According to [1], Pell had nothing to do with the development of this problem or its solution; its attribution to Pell is due to an error on the part of Euler.

The Cyclic Method

The cyclic method relies on successive approximation and on a hope that the approximation process will converge. Of course, it can be proven that the process does converge and that it results eventually in all solutions of the Pell equation. The method relies on the well-known identity

$$(x^2-A{y}^2)(w^2-A{z}^2)={(xw+Ayz)}^2-A{(xz+yw)}^2$$ (1)

Given an equation

$$c^2-A{d}^2=k$$

this identity can be used to produce another solution.

So, for example, let’s take the case $A=22$. Start by choosing $x$ so that $x^2-22\cdot 1^2$ is as small in absolute value as possible, i.e. $x=5$ and $5^2-22\cdot 1^2=3$. If we multiply that by the equation $r^2-22\cdot 1^2=s$, we get, using (1),

$${(5r+22)}^2-22\cdot {(r+5)}^2=3s$$

Remember that we are trying to find values so that the right-hand is $1$. However, simply choosing a value of $r$ that makes $s$ as small as possible will not achieve any progress, since we will then have $r=5$ and $s=3$, so we will get a right-hand side of $9$. Instead, we will choose $r$ so that $r+5$ is divisible by $3$; then $5r+22$ must be divisible by $3$ as well, so $s$ is divisible by $3$ and we may divide the resulting equation through by $9$. To make $r+5\equiv 0(mod3)$ we need $r\equiv 1(mod3)$. We want to make $s$ as small as possible in magnitude subject to this condition, so we choose $r=4$ and then $s=-6$. The resulting equation, after dividing through by $9$, is

$${14}^2-22\cdot 3^2=-2$$

We are not done, so let’s repeat the operation. Multiply again by $r^2-22\cdot 1^2=s$ to get

$${(14r+66)}^2-22\cdot {(3r+14)}^2=-2s$$

and we must choose $r$ such that $3r+14\equiv 0(mod2)$ (i.e. $r\equiv 0(mod2)$) to make $s$ as small as possible in magnitude. This happens for $r=4,s=-6$; substituting and dividing through by $2^2=4$ we get

$${61}^2-22\cdot {13}^2=3$$

Repeating, we get

$${(61r+22\cdot 13)}^2-22\cdot {(13r+61)}^2=3s$$

so $13r+61\equiv 0(mod3)$, or $r\equiv 3(mod3)$. Minimizing $s$ leads to $r=5$, $s=3$, whence, dividing the result by $3^2=9$ we get

$${197}^2-22\cdot {42}^2=1$$

and we are done. The sequence of equations was

$$1^2-22\cdot 0^2=1$$

$$5^2-22\cdot 1^2=3$$

$${14}^2-22\cdot 3^2=-2$$

$${61}^2-22\cdot {13}^2=3$$

$${197}^2-22\cdot {42}^2=1$$

The fact that the sequence of differences is palindromic is not an accident; it can be proven that this is always the case.

The English Method

The English method uses the same basic algorithm as the cyclic method, except that when choosing $r$, it the largest $r$ such that , which . Thus if two successive equations in this method are

$$x^2-A{y}^2=k$$

$$X^2-A{Y}^2=K$$

it is clear that

$$K=\frac{r^2-A}{k}$$

and thus that $K$ and $k$ have different signs. So in the English method, the values of $k$ alternate in sign.

For the case $A=22$, above, the sequence of equations is

$$1^2-22\cdot 0^2=1$$

$$4^2-22\cdot 1^2=-6$$

$$5^2-22\cdot 1^2=3$$

$${14}^2-22\cdot 3^2=-2$$

$${61}^2-22\cdot {13}^2=3$$

$${136}^2-22\cdot {29}^2=-6$$

$${197}^2-22\cdot {42}^2=1$$

In general, the English method results in the same solutions as the cyclic method, but it gets there more slowly - in the English method, an additional entry is inserted in between any two entries that have the same sign.

Title	Pell’s equation
Canonical name	PellsEquation
Date of creation	2013-03-22 16:28:07
Last modified on	2013-03-22 16:28:07
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	8
Author	rm50 (10146)
Entry type	Topic
Classification	msc 11D09