Exercise 7.8.3 ($x^2 -d y^2 = -1$ has no solution if $d \equiv 3 \pmod 4$)

Proof. If $d\equiv 3(\mathit{mod}4)$, then $x^2-d{y}^2\equiv x^2+y^2(\mathit{mod}4)$. Moreover $x^2\equiv 0$ or $1(\mathit{mod}4)$, thus $x^2+y^2\equiv 0,1$ or $2(\mathit{mod}4)$, therefore $x^2+y^2\not\equiv 3(\mathit{mod}4)$. This shows that $x^2-d{y}^2\not\equiv -1(\mathit{mod}4)$. A fortiori the solution $x^2-d{y}^2=-1$ has no integer solution.

The equation $x^2-d{y}^2=-1$ has no integer solution if $d\equiv 3(\mathit{mod}4)$. □