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I'm stuck if $m$ is not a prime or has a single prime divider (Then using Eisenstein's criterion), e.g, $m=4$ and $p=5$.

Any suggestions?

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    $\begingroup$ I think we have a perfect duplicate somewhere. This is the best match I've found so far. $\endgroup$ – Jyrki Lahtonen Apr 6 '19 at 15:38
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If $x^p-m=A(x)B(x)$ with $\deg(A)=a$ and $\deg(B)=b$, then $|A(0)|=|m|^{a/m}$ and $|B(0)|=|m|^{b/m}$ (using roots-coefficients relations). This should lead to a contradiction when you decompose $m$ in prime factors.

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