I'm trying to prove the following:
If $p$ is an odd prime, then exactly half the elements of $\mathbb{U}_p$ are squares.
Where $\mathbb{U}_p$ refers to the units of $\mathbb{Z}_p$. It seems somewhat intuitive that no more than half could be squares, but I'm struggling to prove that exactly half are squares. Any ideas/hints?
Hint $$f: \mathbb{U}_p \to \mathbb{U}_p \\ f(x)=x^2$$ is a group homomorphism with $\ker(f)=\{ \pm 1 \}$.
Use the fact that $$ \mathbb{U}_p /\ker(f) \simeq \mbox{Im}(f)$$
If $p$ is prime, then $(\Bbb{Z}/p\Bbb{Z})^\times$ is cyclic of order $(p-1)$, which means it has an even number of elements. Pick a generator $q \in (\Bbb{Z}/p\Bbb{Z})^\times$, so that $(\Bbb{Z}/p\Bbb{Z})^\times = \{ q, q^2, ..., q^{p-1} = 1 \}$. The squares in $(\Bbb{Z}/p\Bbb{Z})^\times$ are then the even powers of $q$, and there are $(p-1)/2$ of them. QED.
The equation $x^2=a^2$ with $a \neq 0$ has exactly two solution in $\mathbb U_p$, $x=a$ and $x=-a$ (which are different if $p > 2$, a condition you forgot to mention).
So if you square all $p-1$ elements of $\mathbb U_p$, you get each of the squares exactly twice, so there are exactly $\frac{p-1}2$ of them.
