Exercise 7.8.4 (Solutions of $x^2 -dy^2 = 1$, with $k \mid y$)

Proof. Since $d$ is not a perfect square, so is $d^{\prime }=d{k}^2$. By Theorem 7.25 applied to $d^{\prime }$, there are infinitely many integers solutions $(u,v)$ of $u^2-d^{\prime }{v}^2=1$, that is $u^2-d{k}^2{v}^2=1$. Put $x=u,y=\mathit{kv}$. Then $(x,y)$ is a solution of $x^2-d{y}^2=1$, with $k\mid y$. Since $(u,v)\mapsto (u,\mathit{kv})$ is one-to-one, there are infinitely many solutions in integers of $x^2-d{y}^2=1$, with $k\mid y$. □