Showing that if $-1$ and $2$ are not squares in $\mathbb{Z}_p$, then $-2$ is a square I would like to prove that if $-1$ and $2$ are not squares in $\mathbb{Z}_p$, then $-2$ is a square. I have searched for some hints on this, but the answers that I find all involve cosets and quadratic reciprocity. I have also heard that the number of squares in $\mathbb{Z}_P^{\times}$ is $(p-1)/2)$, but I haven't seen a proof of this.
My question is if there is a more direct/basic/elementary way to prove that if $-1$ and $2$ are not squares in $\mathbb{Z}_p$, then $-2$ must be a square.

We don't need to know that $\mathbb{Z}_p^\times$ is cyclic, which I think is a much more advanced fact.  And we certainly don't need quadratic reciprocity.
In fact, all we need is that (for $p$ an odd prime) the number of squares and non-squares modulo $p$ is the same: $(p-1)/2$.  I'll prove this below, and leave the corollary that a non-square times a non-square is a square as an exercise.
It's enough to show that $x\mapsto x^2$ is two-to-one on $\mathbb{Z}_p^\times$.  We know that $x$ and $-x$ are mapped to the same place (and they're never equal), so it's enough to show that $x^2 \equiv y^2 \pmod{p}$ implies $x\equiv y$ or $x\equiv -y$.
But this is straightforward: if $x^2\equiv y^2$ then $p | (x-y)(x+y)$, so either $p|(x-y)$ or $p|(x+y)$.
