[ Equations in integers ]

Solve the equation $x^{2}-5 y^{2}=1$ in integers.

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Solve the equation in non-negative integers, and then arbitrarily place the signs. From now on, only such solutions are considered.

Note that $(x, y)$ is a solution if and only if $(x+y \sqrt{5})(x-y \sqrt{5})=1$. It follows that if $(x, y)$ and $(u, v)$ are solutions and

$s+t \sqrt{5}=(x+y \sqrt{5})(u+v \sqrt{5})$, then $(s, t)$ is also a solution. Indeed, $s-t \sqrt{5}=(x-y \sqrt{5})(u-v \sqrt{5})$, so $(s+t \sqrt{5})(s-t \sqrt{5})=1$.

Two particular solutions are $(1,0)$ and $(9,4)$. From this, we obtain a series of solutions $\left(x_{n}, y_{n}\right)$, where $x_{n}+y_{n} \sqrt{5}=(9+4 \sqrt{5})^{n}$. We will prove that there are no other solutions.

Note that if $(x, y)$ and $(u, v)$ are solutions and $x<u$, then $y<v$.

Let $(u, v)$ be a solution different from those already found. In this case, $(9+4 \sqrt{5})^{n}<u+v \sqrt{5}<(9+4 \sqrt{5})^{n+1}$ for some $n$. Then $(s, t)$, where

$s+t \sqrt{5}=(u+v \sqrt{5})(9-4 \sqrt{5})^{n}$ is also a solution, and $1<s+t \sqrt{5}<9+4 \sqrt{5}$. Therefore, $1<s<9$. However, a direct check shows that there are no solutions satisfying this condition. Contradiction.

## Answer

$\pm\left(x_{n}, \pm y_{n}\right)$, where $x_{n}+y_{n} \sqrt{5}=(9+4 \sqrt{5})^{n}, n=0,1,2,3, \ldots$