The number of elements which are squares in a finite field. Meanwhile reading some introductory notes about the projective special linear group $PSL(2,q)$ wherein $q$ is the cardinal number of the field; I saw:

....in a finite field of order $q$, the number of elements ($≠0$) which are squares is $q-1$ if $q$ is even number and is $\frac{1}{2}(q-1)$ if $q$ is a odd number..." .

I can see it through $\mathbb Z_5$ or $GF(2)$. Any hints for proving above fact? Thanks.

Let $F^*$ be the multiplicative group of the field $F$. Find the kernel of the squaring homomorphism $f: F^* \to F^*$, $f(x) = x^2$. Use that to find the order of the image $f(F^*)$.
