ratio of residues to non residues What is the asymptotic limit of the ratios of $m^{th}$ power residues to $m^{th}$ power non-residues $\mod n$ as $n\to \infty$, if it exists? And when does $n=$ the number of residues $\mod n$ (e.g. when $m=3$  $n\equiv 3$ or $2\mod 3$)?
 A: Even if we make the restriction that $n$ goes to infinity along the primes, the limit does not exist if $m>2$. 
So for clarity I will use $p$ instead of $n$.  The following result is fairly easily proved using the fact that every prime has a primitive root.
Theorem: The number of incongruent $m$-th power residues of $p$ is $\dfrac{p-1}{\gcd(m,p-1)}$.
If for example $m=4$, we can find arbitrarily large primes for which the $\gcd$ is $2$ (in which case the the ratio is $1$ to $1$), and arbitrarily large primes for which the $\gcd$ is $4$, in which case the ratio of residues to non-residues is $1$ to $3$.
The same idea can be used for arbitrary $m>2$. By appropriate choice of $p$, we can alter $\gcd(p-1,m)$ and thereby alter the ratio. That infinitely many primes that do the job exist follows from Dirichlet's Theorem on primes in arithmetic progressions.   
Suppose first that $m\ge 3$ is odd. Then we can find infinitely many primes $p$ such that (a) $\gcd(p-1,m)=1$, and also infinitely many primes $p$ such that (b) $\gcd(p-1,m)=m$. To prove (a), use the infinitely many primes of the form $mk+2$. To prove (b), use the infinitely many primes of the form $mk+1$. The ratio of residues to non-residues taken along the primes of type (a) is quite different from the ratio as taken along the primes of type (b). 
Suppose next that $m>2$ is even, say $m=2l$. Then again using Dirichlet's Theorem, we can find infinitely many primes $p$ such that (a) $\gcd(p-1,2l)=2l$. We can also find infinitely many primes such that (b) $\gcd(p-1,2l)=2$. The ratios of $m$-th power residues to non-residues are different constants for these two infinite collections of primes. 
