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.

If our field is of order $2^k$, then the map $x \mapsto x^2$ is injective and thus bijective since the field is finite. Therefore every element is a square.
When the field has odd order, consider the equivalence relation on the nonzero elements defined by $x \sim y \Leftrightarrow x^2 = y^2$. The number of nonzero squares is the number of equivalence classes of this relation. Each equivalence class contains exactly two elements, so half of the nonzero elements are squares. 
Intuitively, in the list of squares of nonzero elements
$$a_1^2,\ a_2^2,\ a_3^2,\ \ldots$$
we get repetitions when $a_i^2 = a_j^2 \Leftrightarrow a_i = a_j$ or $a_i = -a_j$. Since $a_i \neq -a_i$ for each $a_i$, rearranging and relabeling the list gives
$$b_1^2, (-b_1)^2, b_2^2, (-b_2)^2, b_3^2, (-b_3)^2, \ldots$$
where $b_i \neq \pm b_j$. Again, the number of different elements in this list is half the number of nonzero elements.
