I've read the following fact on my number theory textbook, there's no proof on the book of such result, I tried working it out on my own but I'm kinda lost, the lemma is the following:
Given two integers $a,b$ and a prime of the form $p=4q+3$, $p|a^2+b^2$ $\iff p|a$ and $p|b$
I would post my work but I doubt it would be of any help since I don't feel I got anywhere, I tried analyzing $x^2+y^2\equiv0 \pmod p$ since $x^2\equiv-y^2$ looked somewhat promising to me, the only thing I noticed is that $x^2 \neq-1\pmod p$ since $p\equiv3\pmod4$
As suggested by Ethan Bolker in the comments, continuing my line of reasoning we have: $$x^2\equiv-y^2\pmod p$$ If $p$ divides $y$ we're done, otherwise suppose $p$ doesn't divide $y$, then $p$ can't divide $y^{-1}$ (its modular inverse) either since if it did: $$y\cdot y^{-1}\equiv y\cdot0\equiv0\neq1\pmod p$$ Thus we can multiply both sides of the first conguence by $(y^{-1})^2$ without changing the mod since $(y^{-1},p)=1$, thus:$$x^2\cdot (y^{-1})^2\equiv(x\cdot y^{-1})^2\equiv-1\pmod p$$ Which is a contradiction because $k^2\equiv -1 \pmod p \iff p\equiv 1 \pmod4$