It is conjectured by Artin that for every number $g$ which is not $-1$ and not a square number, there is an infinite number of $m$ such that $g$ is a primitive root modulo $m$.
How to find $m$ efficiently given primitive root $g$?
The reverse problem that given modulo $m$, find the primitive root $g$ can be solved faster than brute force with random trying with checking $g^{(m-1)/d} \neq 1$ ( $d\mid m$ ).
But I can't find something like that for this problem.