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?

The integers modulo $p$ form a field. A polynomial of degree $n$ over a field has at most $n$ roots in the field.
