If $d \equiv 3 \pmod 4 $ then $x^2 − dy^2 = −1$ has no solutions in positive integers $x, y$. In this case, I have come to the conclusion that $x^2 \equiv 0,1 \pmod 4$, $y^2 \equiv 0,1 \pmod 4$ but I am not sure what this means for $dy^2$ and how this won't have any solutions for the above equation. Can someone clarify this?

$d\equiv3\equiv-1\pmod4$, so $x^2-dy^2\equiv x^2+y^2\pmod 4$.
Since $x^2\equiv0$ or $1$ and $y^2\equiv 0$ or $1$, $x^2+y^2\equiv 0, 1, $ or $2$, but not $3$ ($\equiv-1$).
