Prove that $\mathbb{Z}_p^{\times}/(\mathbb{Z}_p^{\times})^2$ is isomorphic to $\{\pm1\}$. Prove that $\mathbb{Z}_p^{\times}/(\mathbb{Z}_p^{\times})^2$ is isomorphic to $\{\pm1\}$, where $p$ is a prime integer.

I'll write $\mathbf F_p$ for $\mathbf Z/p\mathbf Z$. We need to assume that $p$ is odd, since all elements of $\mathbf F_2$ are squares. Maybe the easiest way to see the result is by looking at the squaring map $\mathbf F_p^* \to \mathbf F_p^*$, $x \mapsto x^2$. Since multiplication is abelian this map is a group homomorphism, and its image is the set of squares. You'd like to show that the kernel has order $2$. If $x$ is one of the integers $\{1, 2, \ldots, p - 1\}$ and $x^2 - 1$ is divisible by the prime $p$, then what?
