In $F_p$ (p prime and odd) one of $-1,2,-2$ is a square

I was reading the solution to a problem by Hagen von Eitzen in this post, and they used an argument like this:

[in $F_p$] If $p$ is odd and neither $−1$ nor $2$ is a square, then their product $−2$ is a square

and I'm really confused why! I hope this is not a dumb question which the answer to is obvious. Can someone explain?

Solution 1:

Quadratic reciprocity is one answer, but you also can prove it from first principles. We know that the multiplicative group of any finite field is cyclic. Let $x$ be a generator of the field's multiplicative group. If $-1=x^k$ and $2=x^m$ are both odd powers of $x$ (as they must be if neither is a square), then their product must be $x^{k+m}$, where $k+m=2n$ is even because it is the sum of two odd numbers. In that case, $-2=(x^n)^2$ is a square.

Note that this proof establishes the result for any finite field, not just those of prime order.

Solution 2:

Quadratic reciprocity and the cyclic nature of $\mathbb{F}_q^{\times}$ are ways to answer this, and there is at least one more way that is even more first-principles, which does not require knowing the cyclic structure of $\mathbb{F}_q^{\times}$.

Define $S$ as the set of nonzero squares in $\mathbb{F}_q$. There is a two-to-one map from $\mathbb{F}_q^{\times}\to S$, with $\{h,-h\}\mapsto h^2$. And this map is onto, since any square $s\in S$ has a preimage $\{\pm\sqrt{s}\}$. So:

1. $\left\lvert S\right\rvert=\frac12\left\lvert\mathbb{F}_q^{\times}\right\rvert$. And it follows that the nonsquares, $S^c$, also have $\left\lvert S^c\right\rvert=\frac12\left\lvert\mathbb{F}_q^{\times}\right\rvert$.

At this point, we have $S$, a subgroup of a finite group, with index $2$. So $\mathbb{F}_q^{\times}/S\cong C_2$, and multiplying two nonsquares together must produce a square. But for a more direct proof that doesn't use quotient group theory, read on.

Fix a nonsquare $b$. Multiplication by $b$ on the set $S$ is injective since this is all happening within a field. The image is a subset of $S^c$ (the nonsquares) because if $by^2=z^2$, then $b=(z/y)^2$, a contradiction. By (1) we know that $|S^c|=|S|$, so the injection is really a bijection:

2. Multiplication by $b$ is a bijection from $S$ to $S^c$.

Consider multiplication by $b$ on the set $S^c$. The image cannot contain a nonsquare, because if $bc=d$ for nonsquares $c,d$, then also $by^2=d$ because of (2). So $bc=by^2\implies c=y^2$, a contradiction. So multiplication by $b$ is an injection from $S^c$ into $S$. That is, multiplying $b$ by a nonsquare yields a square. Since $b$ was an aribitrary nonsquare:

3. The product of two nonsquares is a square.

So even if $-1$ and $2$ are both nonsquares, then by (3), $-1\cdot 2=-2$ will be a square.