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 length $$s$$.

Let $$\frac {p_n} {q_n}$$ be a convergent of $$\sqrt n$$.

Then:
 * $$p_{rs}^2 - n p_{rs}^2 = \left({-1}\right)^{rs}$$ 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 $\frac p q$ 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:
 * $$\left[{a \left \langle{b_1, b_2, \ldots, b_{m-1}, b_m, b_{m-1}, \ldots, b_2, b_1, 2 a}\right \rangle}\right]$$

or
 * $$\left[{a \left \langle{b_1, b_2, \ldots, b_{m-1}, b_m, b_m, b_{m-1}, \ldots, b_2, b_1, 2 a}\right \rangle}\right]$$.

For each $$r \ge 1$$ we can write $$\sqrt n$$ as the (non-simple) finite continued fraction:
 * $$\sqrt n \left[{a \left \langle{b_1, b_2, \ldots, b_2, b_1, 2 a, b_1, b_2, \ldots, b_2, b_1, x}\right \rangle}\right]$$

which has a total of $$rs + 1$$ 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:
 * $$\frac {p_{rs - 1}} {q_{rs - 1}}, \quad \frac {p_{rs}} {q_{rs}}, \quad \frac {x p_{rs} + p_{rs - 1}} {x q_{rs} + q_{rs - 1}}$$

The last one of these equals $$\sqrt n$$ itself. So:
 * $$\sqrt n \left({x q_{rs} + q_{rs - 1}}\right) = \left({x p_{rs} + p_{rs - 1}}\right)$$.

Substituting $$a + \sqrt n$$ for $$x$$, we get:
 * $$\sqrt n \left({\left({a + \sqrt n}\right) q_{rs} + q_{rs - 1}}\right) = \left({\left({a + \sqrt n}\right) p_{rs} + p_{rs - 1}}\right)$$.

This simplifies to:
 * $$\sqrt n \left({a q_{rs} + q_{rs - 1} - p_{rs}}\right) = a + p_{rs} + p_{rs - 1} - n q_{rs}$$.

The RHS of this is an integer while the LHS 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 $$p_{rs}$$, $$(2)$$ by $$q_{rs}$$ and then subtracting:
 * $$p_{rs}^2 - n q_{rs}^2 = p_{rs} q_{rs - 1} = p_{rs - 1} q_{rs}$$

By Properties of Convergents of Continued Fractions, the RHS of this is $$\left({-1}\right)^{rs}$$.

When the cycle length $$s$$ of the continued fraction of $$\sqrt n$$ is even, we have $$\left({-1}\right)^{rs} = 1$$.

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

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