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.
 A: 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 $p|(x+y)$.
A: It is known that $\mathbb{Z}_p^{\times} = \{ 1, 2, \ldots, p-1 \}$ with multiplication $\operatorname{mod} p$ is cyclic, so there is $a \in \mathbb{Z}_p^{\times}$ such that $\mathbb{Z}_p = \{ 1, a, a^2, \ldots, a^{p-2} \}$.


*

*Every $a^{2j}$ is obviously a square, since $a^{2j} = (a^j)^2$.

*Suppose $a^k$ is a square, i.e. $a^k = b^2$ for some $b \in \mathbb{Z}_p^{\times}$. But $b = a^j$ for some $j$, hence $a^k = a^{2j}$, so $k \equiv 2j \pmod{p-1}$. But $p-1$ is even, so $k$ is even.


We thereby see that $a^k$ is a square if and only if $k$ is even. Now suppose $-1 = a^i$ and $2 = a^j$ and both are not squares. By the above both $i$ and $j$ are odd, so $i+j$ is even, therefore $-2 = a^{i+j}$ is a square.
A: If $-1$ is not a square modulo $p$ then $p\equiv3\pmod 4$ and so
we can write $p=4k+3$. Let $a=2^{k+1}$. Then $a^4=2^{p+1}\equiv 2^2=4\pmod p$ by Fermat's little theorem. So $a^2\equiv2$ or $a^2\equiv-2\pmod p$.
If $2$ isn't a square modulo $p$ then $-2$ must be.
