Exercise 1.5.5

Proof. Note that $(x_0,y_0)=(1,1)$ is a solution of $x^2-2y^2=-1$, and $(3,2)$ is a solution of $x^2-2y^2=1$.

If $(x,y)$ satisfies $x^2-2y^2=1$ and $(x^{\prime },y^{\prime })$ satisfies ${x{}^{\prime }}^2-2{y{}^{\prime }}^2=-1$, then

so that $(x{x}^{\prime }+2y{y}^{\prime },y{x}^{\prime }+x{y}^{\prime })=(x,y)\ast (x^{\prime },y^{\prime })$ is a solution of $X^2-2Y^2=-1$.

We define by induction $(x_n,y_n)$ with

Then $(x_n,y_n)$ is a solution of $x^2-2y^2=-1$ for all $n\in \mathbb{N}$, with

This gives the solutions

Moreover, for all $n\in \mathbb{N}$, $n\ne 0$, $x_n>0,y_n>0$, thus $x_{n+1}=3x_n+4y_n>3x_n>x_n$ (and $x_1=7>x_0=1$). Therefore the sequence ${(x_n)}_{n\in \mathbb{N}}$ is strictly increasing, and so the solutions $(x_n,y_n)$ are distinct. The Diophantine equation $x^2-2y^2=-1$ has infinitely many solutions. □