## In an integral domain, there are at most two square roots of 1

Prove that if $R$ is an integral domain and $x^2 = 1$ for some $x \in R$, then $x = 1$ or $x = -1$.

If $x^2 = 1$, then $x^2 - 1 = 0$. Evidently, then, $(x-1)(x+1) = 0$. Since $R$ is an integral domain, we must have $x-1 = 0$ or $x+1 = 0$; thus $x = 1$ or $x = -1$.