Exercise 7.8.2 (If $x^2 - dy^2 = N$ has one solution, then it has infinitely many)

Proof. Here $d$ is a positive integer which is not a perfect square.

Suppose that $x^2-d{y}^2=N$ has an integer solution $(x,y)\in \mathbb{Z}\times \mathbb{Z}$ (then $N\ne 0$). Since $|x{\text{|}}^2-d|y{\text{|}}^2=N$, we may assume that $x\ge 0$ and $y\ge 0$ (but $(x,y)\ne (0,0)$).

By Theorem 7.25, there exist positive integers $x_1$ and $y_1$ such that $x_1^2-y_1^{2}d=1$ and $x_1>1$. We define the integers $x_n$ and $y_n$ by $x_n+y_n\sqrt{d}={(x_1+y_1\sqrt{d})}^n$. Then $x_n>0$, $y_n>0$ and $x_n^2-y_n^{2}d=1$. Therefore

So

is a solution of $x^2-d{y}^2=N$ for every $n\ge 1$.

We must prove that there are infinitely many such solutions.

Since $x_{n+1}+y_{n+1}\sqrt{d}=(x_n+y_n\sqrt{d})(x_1+y_1\sqrt{d})$, we obtain

Here $y_1>0,x_n>0,y_n>0,d>0$ and $x_1>1$, thus for every $n>0$,

so the sequences ${(x_n)}_{n\in {\mathbb{N}}^{\text{∗}}}$ and ${(y_n)}_{n\in {\mathbb{N}}^{\text{∗}}}$ are strictly increasing.

Let $n,m$ be two positive integers such that $n>m$, and assume for the sake of contradiction that $(X_n,Y_n)=(X_m,Y_m)$. By (1),

therefore

Moreover $x^2-d{y}^2=N$, so $(x,y)\ne (0,0)$. Hence

that is

If $y_n-y_m\ne 0$, then the integer $d$ is the square of a rational number, hence $d$ is a perfect square, in contradiction with the hypothesis. Hence $y_n-y_m=0$ and $x_n-x_m=d{(y_n-y_m)}^2=0$, so $(x_n,y_n)=(x_m,y_m).$

But the sequences ${(x_n)}_{n\in {\mathbb{N}}^{\text{∗}}}$ and ${(y_n)}_{n\in {\mathbb{N}}^{\text{∗}}}$ are strictly increasing thus $x_n>x_m$ and $y_n>y_m$. This contradiction shows that

Therefore the solutions $(X_n,Y_n)$ of $x^2-d{y}^2=N$ are distinct.

If $x^2-d{y}^2=N$ has one solution, then it has infinitely many. □