Exercise 7.8.7 ($x^2 - (k^2 - 1)y^2 = -1$ has no solutions in integers)

Proof. We observe that $(9,1)$ is a solution of $x^2-80y^2=1$, and it is the smallest positive solution. By Problem 1, if $x^2-80y^2=-1$ was solvable, the smallest positive solution $(x_1,y_1)$ of $x^2-80y^2=-1$ would satisfy

But then $2x_1{y}_1=1$, so $2\mid 1$. This contradiction shows that $x^2-80y^2=-1$ has no solution.

Generalization. If $k=0$, then $x^2-(k^2-1)y^2=x^2+y^2=-1$ has no solution.

Moreover $k^2-1$ is not a perfect square, otherwise $k^2-1=l^2$ for some integer $l$, so $1=k^2-l^2=(k-l)(k+l)$, thus $k-l=k+l=𝜀$, where $𝜀=\pm 1$. Therefore $l=0$ and $k=\pm 1$. In this case, $x^2-(k^2-1)y^2=-1$ is equivalent to $x^2=-1$, which has no solution.

So we may suppose that $|k|>1$, and then $k^2-1$ is not a perfect square, so $\sqrt{k^2-1}$ is irrational.

Note that $(|k|,1)$ is a positive solution of he equation $x^2-(k^2-1)y^2=1$. Every other positive solution $(x,y)$ satisfies $1\le y$, hence $(|k|,1)$ is the smallest positive solution.

By Problem 1, if $x^2-(k^2-1)y^2=-1$ was solvable, the smallest positive solution $(x_1,y_1)$ of $x^2-(k^2-1)y^2=-1$ would satisfy

But the irrationality of $\sqrt{k^2-1}$ implies that $2x_1{y}_1=1$, so $2\mid 1$. This contradiction shows that $x^2-(k^2-1)y^2=-1$ is not solvable.

For any integer $k$, $x^2-(k^2-1)y^2=-1$ has no solution in integers. □