Exercise 7.8.11 (If $p\mid d,\ p\equiv 3 \pmod 4$, then $x^2 - dy^2 = -1$ is not solvable)

Proof. We suppose that $d$ is divisible by a prime number $p$, $p\equiv 3(\mathit{mod}4)$. Assume for the sake of contradiction that there are integers $x$, $y$ such that $x^2-d{y}^2=-1$. Then $x^2\equiv -1(\mathit{mod}p)$, thus

Therefore $p=2$ or ${(-1)}^{(p-1)∕2}=1$, so $p\equiv 1$ or $2(\mathit{mod}4)$, in contradiction with $p\equiv 3(\mathit{mod}4)$, This contradiction shows that the equation $x^2-d{y}^2=-1$ has no solution in integers. □