Conclude: $-3 \in K$ is a square if and only if $q \equiv 1 \bmod 3$.

Question

Let $p \neq 2,3$ be a prime number, $q=p^n$ for an $n \in \mathbb{N}$ and $K$ a field with $q$ elements. Prove the following:

$-3 \in K$ is a square if and only if $q \equiv 1 \bmod 3$.

My idea:

We can write: $x^2\equiv-3 \bmod q$

But what next?

Answer 1

The relation $q \equiv 1 \bmod 3$ holds iff $3 \mid q-1$ iff the (cyclic!) group $\mathbb{F}_q^{\times}$ has an element of order $3$, i.e. there is some $a \in \mathbb{F}_q$ such that $a^3=1$ and $a \neq 1$. In other words, $x^3-1$ needs to have a root $\neq 1$ in $\mathbb{F}_q$.

If $K$ is any field of characteristic $\neq 2$, the roots of $x^3-1 = (x-1)(x^2+x+1)$ in $\overline{K}$ are given by $x=1$ and $x = \dfrac{-1 \pm \sqrt{-3}}{2}$ (quadratic formula), the latter of which are $\neq 1$ if $\mathrm{char}(K) \neq 3$. Therefore, it has a root $\neq 1$ in $K$ iff $\sqrt{-3} \in K$ and $\mathrm{char}(K) \neq 3$.

A very similar proof shows $q \equiv 1 \bmod 4 \iff \sqrt{-1} \in \mathbb{F}_q$ and $q$ odd.