solutions of $x^2\equiv 1 \pmod p $

Question

If p is a prime, show that the only solutions of $x^2\equiv 1 \pmod p $ are $x=1$ and $x\equiv -1 \pmod p$. (from herstein's abstract algebra chapter2 section4 lagrange's theorem problem 15, this section is unfamiliar to me, and the questions are all damned hard!)

Answer 1

Hint: $x^2 = 1$ if and only if $(x-1)(x+1)=0$.

Answer 2

If a prime divides a product of integers then it divides at least one of the factors.

If $x$ is a solution of $x^2\equiv 1\pmod p$, apply this factlet to $(x-1)(x+1)$.