I want to show that for every prime $p$ and $n \in \mathbb Z$:
$$n^2 \equiv 1 \mod p \implies n \equiv \pm 1 \mod p$$
I think I already got a solution for odd $n \,\,\,(n := 2k + 1)$:
$$ n^2 \equiv 1 \mod p \\ \implies (2k + 1)^2 \equiv 1 \mod p \\ \implies 4k(k+1) \equiv 0 \mod p \\ \implies 4k \equiv 0 \mod p \text{ or } k + 1 \equiv 0 \mod p \\ \implies k \equiv \pm 1 \mod p \\ \implies n \equiv \pm 1 \mod p $$
But I'm stuck for even $n$. Maybe this differentiation between odd and even $n$ isn't even necessary at all and there is a more simple solution. Any help is welcome.