Prove $\{x \mid x^2 \equiv 1 \pmod p\}=\{1, -1\}$ for all primes $p$ One way to prove that $\{x \mid x^2 \equiv 1 \pmod p\}=\{1, -1\}$ is to use the fact that $\{1, -1\}$ is the only subgroup of the cyclic group of primitive residue classes modulo $p$ that has the order 2. Is there a more basic way to prove this?

You can use group theory method.
$(\Bbb Z_{p}^{*}, \otimes_{p})$ is a cyclic group of an even number, so just $\varphi(2)=1$ element has order $2$, so the equation $x^2=1$ has only two solutions identity and that unique element of order $2$.

You can use number theory method.
$x$ satisfies the given equation if and only if $p|x^2-1=(x-1)(x+1)$ so if we assume $$x\in\{0,1,2,\dots,p-1\}$$
then the only possible answers are $1$ and $p-1$.
