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

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

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:
 * $\displaystyle \frac {p_{rs - 1}} {q_{rs - 1}}, \quad \frac {p_{rs}} {q_{rs}}, \quad \frac {x p_{rs} + p_{rs - 1}} {x q_{rs} + q_{rs - 1}}$

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.

However, the attribution (a mistake by Euler - even Homer nods) is incorrect. It was in fact William Brouncker who was the first European to solve it.

The equation had been extensively investigated by Brahmagupta and subsequent mathematicians of the Indian tradition as far back as 628 C.E.: subsequently Bhāskara II in the 12th century and Narayana Pandit in the 14th found solutions.