Pell's Equation

Definition
The Diophantine equation:
 * $x^2 - n y^2 = 1$

is known as Pell's equation.

Solution
Let the continued fraction of $\sqrt n$ have a cycle whose length is $s$:


 * $\sqrt n = \sqbrk {a_1 \sequence {a_2, a_3, \ldots, a_{s + 1} } }$

Let $a_n = \dfrac {p_n} {q_n}$ be a convergent of $\sqrt n$.

Then:
 * ${p_{r s} }^2 - n {q_{r s} }^2 = \paren {-1}^{r s}$ for $r = 1, 2, 3, \ldots$

and all solutions of:
 * $x^2 - n y^2 = \pm 1$

are given in this way.

Proof
First note that if $x = p, y = q$ is a positive solution of $x^2 - n y^2 = 1$ then $\dfrac p q$ is a convergent of $\sqrt n$.

The continued fraction of $\sqrt n$ is periodic from Continued Fraction Expansion of Irrational Square Root and of the form:
 * $\sqbrk {a \sequence {b_1, b_2, \ldots, b_{m - 1}, b_m, b_{m - 1}, \ldots, b_2, b_1, 2 a} }$

or
 * $\sqbrk {a \sequence {b_1, b_2, \ldots, b_{m - 1}, b_m, b_m, b_{m - 1}, \ldots, b_2, b_1, 2 a} }$

For each $r \ge 1$ we can write $\sqrt n$ as the (non-simple) finite continued fraction:
 * $\sqrt n = \sqbrk {a \sequence {b_1, b_2, \ldots, b_2, b_1, 2 a, b_1, b_2, \ldots, b_2, b_1, x} }$

which has a total of $r s + 1$ partial quotients.

The last element $x$ is of course not an integer.

What we do have, though, is:

The final three convergents in the above FCF are:
 * $\dfrac {p_{r s - 1} } {q_{r s - 1} }, \quad \dfrac {p_{r s} } {q_{r s} }, \quad \dfrac {x p_{r s} + p_{r s - 1} } {x q_{r s} + q_{r s - 1} }$

The last one of these equals $\sqrt n$ itself.

So:
 * $\sqrt n \paren {x q_{r s} + q_{r s - 1} } = \paren {x p_{r s} + p_{r s - 1} }$

Substituting $a + \sqrt n$ for $x$, we get:
 * $\sqrt n \paren {\paren {a + \sqrt n} q_{r s} + q_{r s - 1} } = \paren {\paren {a + \sqrt n} p_{r s} + p_{r s - 1} }$

This simplifies to:
 * $\sqrt n \paren {a q_{r s} + q_{r s - 1} - p_{r s} } = a p_{r s} + p_{r s - 1} - n q_{r s}$

The of this is an integer while the  is $\sqrt n$ times an integer.

Since $\sqrt n$ is irrational, the only way that can happen is if both sides equal zero.

This gives us:

Multiplying $(1)$ by $p_{r s}$, $(2)$ by $q_{r s}$ and then subtracting:
 * $p_{r s}^2 - n q_{r s}^2 = p_{r s} q_{r s - 1} - p_{r s - 1} q_{r s}$

By Difference between Adjacent Convergents of Simple Continued Fraction, the of this is $\paren {-1}^{r s}$.

When the cycle length $s$ of the continued fraction of $\sqrt n$ is even, we have $\paren {-1}^{r s} = 1$.

Hence $x = p_{r s}, y = q_{r s}$ is a solution to Pell's Equation for each $r \ge 1$.

When $s$ is odd, though:
 * $x = p_{r s}, y = q_{r s}$ is a solution of $x^2 - n y^2 = -1$ when $r$ is odd
 * $x = p_{r s}, y = q_{r s}$ is a solution of $x^2 - n y^2 = 1$ when $r$ is even.

Also known as
A particular instance of Pell's equation can also be seen referred to as a Pellian equation.