# Pell's Equation/Examples/13

## Theorem

$x^2 - 13 y^2 = 1$

has the smallest positive integral solution:

$x = 649$
$y = 180$

## Proof

$\sqrt {13} = \left[{3, \left \langle{1, 1, 1, 1, 6}\right \rangle}\right]$

The cycle is of length is $5$.

By the solution of Pell's Equation, the only solutions of $x^2 - 13 y^2 = 1$ are:

${p_{5 r} }^2 - 13 {q_{5 r} }^2 = \left({-1}\right)^{5 r}$

for $r = 1, 2, 3, \ldots$

When $r = 1$ this gives:

${p_5}^2 - 13 {q_5}^2 = -1$

which is not the solution required.

When $r = 2$ this gives:

${p_{10} }^2 - 13 {q_{10} }^2 = 1$
$p_{10} = 649$
$q_{10} = 180$

although on that page the numbering goes from $p_0$ to $p_9$, and $q_0$ to $q_9$.

$\blacksquare$