Show that for $p \neq 2$ not every element in $\mathbb{Z}/p\mathbb{Z}$ is a square. Show that for $p\neq2$, not every element in $\mathbb{Z}/p\mathbb{Z}$ is a square of an element in $\mathbb{Z}/p\mathbb{Z}$. (Hint: $1^2=(p-1)^2=1$. Deduce the desired conclusion by counting).
So far I have that $1=p^2-2p-1\Rightarrow p^2-2p=0\Rightarrow p^2=2p$, but I don't know where to go from here. I also don't fully understand what it means to deduce it by counting.

Because $\mathbb{Z}/p\mathbb{Z}$ is finite the map $f \colon \mathbb{Z}/p\mathbb{Z} \to \mathbb{Z}/p\mathbb{Z}$, $x \mapsto x^2$ is surjective (i.e. every element is a square) if and only if it is injective. But for $p \neq 2$ we have $-1 \neq 1$ with $f(-1) = 1 = f(1)$.
