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?