The number of elements which are squares in a finite field.

Question

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.

Answer 1

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 $f(F^*)$.

Answer 2

The multiplicative group of a finite field is cyclic.

Answer 3

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.

Answer 4

Another way to prove it, way less elegant than Dustan's but perhaps slightly more elementary: let $$a_1,a_2,...,a_{q-1}$$ be the non zero residues modulo $\,q\,,\,q$ an odd prime . Observe that $\,\,\forall\,i\,,\,\,a_i^2=(q-a_i)^2 \pmod q\,$ , so that all the quadratic residues must be among $$a_1^2\,,\,a_2^2\,,...,a_m^2\,\,,\,m:=\frac{q-1}{2} $$

Note that $\,\,\forall\,1\leq i,j\leq m\,:$$$a_i+a_j=0\Longrightarrow a_i=-a_j=q-a_i\Longrightarrow$$$$\Longrightarrow a_i-a_j=q=0$$

Both left most equalities above would lead us to $\,a_i=a_j=0\,$, which is absurd.

Finally, we prove that not two of the above $\,\,(q-1)/2\,\,$ elements are equal. The following's done modulo $\,q$:$$a_i^2=a_j^2\Longrightarrow (a_i-a_j)(a_i+a_j)=0\Longrightarrow a_i-a_j=0$$since we already showed that $\,\,a_i+a_j\neq 0$