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.

Mod $p$, there is an equal number of quadratic residues and nonresidues (ignoring 0 of course). Therefore, the index of $(\mathbb{Z}_p^\times)^2$ in $\mathbb{Z}_p^\times$ is 2, and so the quotient is isomorphic to $\{\pm 1\}$, as this is the only group of order 2.
Edited to only use words from group theory:
Note that $\mathbb{Z}_p^\times$ is cyclic of order $p-1$.  We'll show that the subgroup $(\mathbb{Z}_p^\times)^2$, which is the set of squares in $\mathbb{Z}_p^\times$, has index 2. Then the quotient is $\{\pm 1\}$, since this is the only group of order 2.
Let $a$ be a generator for $\mathbb{Z}_p^\times$.
Then the elements of $\mathbb{Z}_p^\times$ are $a,a^2,\ldots,a^{p-1}=1$.
If $k$ is even, then $a^k = (a^{k/2})^2$ is a square.
However if $k$ is odd, then $a^k$ is not a square, for if $a^k = b^2$ for some $b \in \mathbb{Z}_p^\times$ and $k$ odd, then $b= a^j$ for some $j$, and so $a^k = a^{2j}$. But then $k -2j\equiv 0$ mod $p-1$, which is impossible since $k-2j$ is odd.
Since $p-1$ is even, there are the same number of elements of $\mathbb{Z}_p^\times$ which are even and odd powers of $a$, so $(\mathbb{Z}_p^\times)^2$ has index 2.
