TITLE: Prove that $x^{2} \equiv -1 \pmod p$ has no solutions if prime $p \equiv 3 \pmod 4$.
QUESTION [17 upvotes]: Assume: $p$ is a prime that satisfies $p \equiv 3 \pmod 4$
Show: $x^{2} \equiv -1 \pmod p$ has no solutions $\forall x \in \mathbb{Z}$.
I know this problem has something to do with Fermat's Little Theorem, that $a^{p-1} \equiv 1\pmod p$. I tried to do a proof by contradiction, assuming the conclusion and showing some contradiction but just ran into a wall. Any help would be greatly appreciated.

REPLY [2 votes]: It is equivalent to $\nexists n: \frac{n^2+1}{p}\in \mathbb{N}$, which is a case of the sum of squares theorem.