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Given n and q, how to find p so q$\neq$n-th power (mod p)?
Reasonable exceptions allowed on $q$. Example solution: $n=2$.
Suppose $q$ is odd. Let $p$ be so $pq\equiv -1$ (mod 8). Then $q\neq$ 2nd power (mod $p$) is the same as $\left(\frac{q}{p}\right)=-1 …