TITLE: How to prove this formula for the Legendre symbol for a finite field
QUESTION [5 upvotes]: Let $\mathbb{F}_q$ be a finite field with $q$ odd, let $x\in\mathbb{F}_q$ and define the Legendre symbol for $\mathbb{F}_q$ as
\begin{equation} \left(\frac{x}{\mathbb{F}_q} \right) = \begin{cases}
 \phantom{-}1 & \text{if $t^2=x$ has a solution $t\in\mathbb{F}_q^*$}\;,\\
 -1 & \text{if $t^2=x$ has no solution $t\in\mathbb{F}_q$}\;,\\
 \phantom{-}0 & \text{if } x=0\;.
 \end{cases} \end{equation}
How do I see that
$$ \left( \frac{x}{\mathbb{F}_q} \right) = x^{(q-1)/2} $$
as elements of $\mathbb{F}_q$?
This is left as an exercise in "Elliptic Curves - Number Theory and Cryptography" by Washington but I need it in a proof and I can't seem to figure it out. 
Any help is greatly appreciated.

REPLY [7 votes]: The same proof as in Euler's criterion works fine for $\mathbb{F}_q$ because $\mathbb{F}_q$ is a field:
If $x\ne 0$, then $x^{q-1}=1$ because of Lagrange's theorem applied to the multiplicative group $\mathbb{F}_q^\times$.
Write $0=x^{q-1}-1=(x^{\frac{q-1}{2}}-1)(x^{\frac{q-1}{2}}+1)$.
The map $x \mapsto x^2$ is homomorphism $\mathbb{F}_q^\times \to \mathbb{F}_q^\times$ whose kernel is $\{\pm1\}$. Therefore, the image has order $\frac{q-1}{2}$ and so there are exactly $\frac{q-1}{2}$ squares; they must be exactly the solutions of $x^{\frac{q-1}{2}}-1=0$.