How do I show that $x^\frac{q-1}{2} = -1$ if $x$ is nonsquare for a finite field $k$ with order $q$

Question

I have managed to show the opposite, that for square $x$ that $x^\frac{q-1}{2} = 1$, and I have shown that precisely half the elements of $k^x$ are squares but I think I am missing some obvious part of the puzzle to show this last part.

Answer 1

Let $q$ be an odd prime power.

For all non-zero $x$, we know $(x^{\frac{q-1}2})^2=1$, hence $x^{\frac{q-1}2}=\pm1$. You already know all the $x\in(\mathbb F_q^\times)^2$ are solutions to $T^{\frac{q-1}2}=1$. The set $(\mathbb F_q^\times)^2$ has size $\frac{q-1}2$, since $a^2=b^2$ if and only if $a=\pm b$.

Now, for any $x\in\mathbb F_q^\times\backslash(\mathbb F_q^\times)^2$ we have $x^{\frac{q-1}2}=-1$. Indeed, if $x^{\frac{q-1}2}=1$, then the equation $T^{\frac{q-1}2}=1$ has degree $\frac{q-1}2$ but has at least $\frac{q+1}2$ solutions $(\mathbb F_q^\times)^2\sqcup\{x\}$, which is absurd.