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.

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 $\mathbb{Z}_p = \{ 1, a, a^2, \ldots, a^{p-2} \}$.


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*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.
