In $GF(q)$, are there the same number of quadratic residues as quadratic nonresidues? I know that for $\mathbb Z_p^*$ (the multiplicative group of a field with $p$ elements where $p$ is a prime), there are $(p-1)/2$ quadratic residues, and thus $(p-1)/2$ quadratic nonresidues. We can see this by considering $f:\mathbb Z_p^* \longrightarrow \mathbb Z_p^*$, defined by $f(\bar x)=\bar x^2$, where $\bar x$ is a least non-negative residue. But what about for the general finite field with $q$ elements $GF(q)$, where $q$ is a power of $p$? 
Are there the same number of quadratic residues as nonresidues? How would I go about showing this?
 A: The argument that works when $q$ is an odd prime  works equally well in a $q$-element field where $q$ is odd. 
For any non-zero $x$ in our field, let $f(x)=x^2$. Note that $f(x)=f(-x)$. We show that conversely if $f(t)=x^2$, then $t=\pm x$. For if $t^2=x^2$, then $t^2-x^2=0$, so $(t-x)(t+x)=0$. Since we are working in a field, it follows that $t=x$ or $t=-x$. Since $q$ is odd, we have $x\ne -x$.
Thus the function $f$ is two-to-one, and the result follows in the same way as in the case $q$ is an odd prime. 
A: For the multiplicative group of a finite field $ K $, the mapping $ f(x) = x^2 $ is an endomorphism whose kernel is $ \{ 1, -1 \} $, which means the image $ N $ of $ f $ is a subgroup with index 2, which means the quotient $ K^{\times}/N $ has $ 2 $ elements. Now, let $ h : K^{\times} \to K^{\times}/N $ be the quotient epimorphism. This epimorphism maps every quadratic residue in $ K $ to $ N $, and since it cannot map a nonresidue to $ N $, it maps it to the other coset of $ N $. The result follows from the fact that the cosets of a subgroup have the same cardinality.
Note: If $ K $ has characteristic $ 2 $ then $ 1 = -1 $, so the kernel of $ f $ really has only one element. Every element is a quadratic residue in these fields.
