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What is a good test for identifying cubic non-residues/residues and higher power non-residues/residues modulo a prime $R$ in terms of computational complexity?

Given $M$ and $N$, is there a good way to find a prime $R$ such that $M$ is cubic non-residue modulo $R$ and $N$ is cubic residue modulo $R$?

What is a good test for identifying cubic non-residues/residues and higher power non-residues/residues modulo a prime $R$ in terms of computational complexity?

Given $M$ and $N$, is there a good way to find a prime $R$ such that $M$ is cubic non-residue modulo $R$ and $N$ is cubic residue modulo $R$?

Update after David Speyer's answer: Does the probability estimate hold good if $R$ is restricted to be between $M$ and $N$ and $M < 2N$ or $N < 2M$?

What is a good test for identifying cubic non-residues/residues and higher power non-residues/residues modulo a prime $R$ in terms of computational complexity?

Given $M$ and $N$, is there a good way to find a prime $R$ such that $M$ is cubic non-residue modulo $R$ and $N$ is cubic residue modulo $R$?

Let $a_{i}$ for $i=1,2,3,4$ be cubic non-residues modulo prime $R$. Let $M$ and $N$ be integers.

$(1)$ If $M$ is cubic non-residue and $N$ is cubic residue modulo $R$, then if $(a_{1}a_{2}a_{3}a_{4})^{2}MN$ is a cubic residue, then is it true that atleast one of $a_{i}^{2}M$ is a cubic residue?

$(2)$ If $M$ and $N$ are cubic non-residues modulo $R$ with $MN$ a cubic non-residue, then if $(a_{1}a_{2}a_{3}a_{4})MN$ is a cubic residue, then is it true that atleast one of $a_{i}MN$ is a cubic residue? Do we need $MN$ to be cubic non-residue to conclude atleast one of $a_{i}MN$ is a cubic residue?

What is a good test for identifying cubic non-residues/residues and higher power non-residues/residues modulo a prime $R$ in terms of computational complexity?

Given $M$ and $N$, is there a good way to find a prime $R$ such that $M$ is cubic non-residue modulo $R$ and $N$ is cubic residue modulo $R$?

What is a good test for identifying cubic non-residues/residues and higher power non-residues/residues modulo a prime $R$ in terms of computational complexity?

Let $a_{i}$ for $i=1,2,3,4$ be cubic non-residues modulo prime $R$. Let $M$ and $N$ be integers.

$(1)$ If $M$ is cubic non-residue and $N$ is cubic residue modulo $R$, then if $(a_{1}a_{2}a_{3}a_{4})^{2}MN$ is a cubic residue, then is it true that atleast one of $a_{i}^{2}M$ is a cubic residue?

$(2)$ If $M$ and $N$ are cubic non-residues modulo $R$ with $MN$ a cubic non-residue, then if $(a_{1}a_{2}a_{3}a_{4})MN$ is a cubic residue, then is it true that atleast one of $a_{i}MN$ is a cubic residue? Do we need $MN$ to be cubic non-residue to conclude atleast one of $a_{i}MN$ is a cubic residue?

What is a good test for identifying cubic non-residues/residues and higher power non-residues/residues modulo a prime $R$ in terms of computational complexity?

Given $M$ and $N$, is there a good way to find a prime $R$ such that $M$ is cubic non-residue modulo $R$ and $N$ is cubic residue modulo $R$?

Let $a_{i}$ for $i=1,2,3,4$ be cubic non-residues modulo prime $R$. Let $M$ and $N$ be integers.

$(1)$ If $M$ is cubic non-residue and $N$ is cubic residue modulo $R$, then if $(a_{1}a_{2}a_{3}a_{4})^{2}MN$ is a cubic residue, then is it true that atleast one of $a_{i}^{2}M$ is a cubic residue?

$(2)$ If $M$ and $N$ are cubic non-residues modulo $R$ with $MN$ a cubic non-residue, then if $(a_{1}a_{2}a_{3}a_{4})MN$ is a cubic residue, then is it true that atleast one of $a_{i}MN$ is a cubic residue? Do we need $MN$ to be cubic non-residue to conclude atleast one of $a_{i}MN$ is a cubic residue?