Prove that if $p$ divides $a^2+b^2$ , then $ p \equiv \ 1 \pmod 4$.
I know that both $a$ and $b$ are coprimes with $p$ but i dont know what to do next.
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Here $(ab^{-1})^2\equiv-1\pmod4$. This implies $p\equiv1\pmod4$. See here. – Jyrki Lahtonen Jun 04 '20 at 18:27
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$b$ is invertible $\bmod p$, i.e., there is $c$ with $bc\equiv 1\pmod p$. Then $(ac)^2+1=c^2(a^2+b^2)\equiv 0\pmod 4$. Can you take it from here?
Hagen von Eitzen
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How did we get to the mod 4 there? And i am afraid i still can't see what to do. – Tom Jun 04 '20 at 17:22