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Problem: find all integers $n$ such as $$31|n^5+4$$ Solution I have no idea how to solve it. It is the same as solving $n^5+4\equiv0 \text{ (mod } 31)$, but that's as far as I got.

I will appreciate any hint :)

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closed as off-topic by John B, heropup, Semiclassical, Stefan Mesken, Chris Godsil Mar 10 '16 at 1:33

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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John B, heropup, Semiclassical, Stefan Mesken, Chris Godsil
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  • $\begingroup$ For one thing, since $2^5\equiv 1\mod 31$, if $n$ is one solution, then so is $2n$ (and $4n,8n$ and $16n$). This is not as surprising as one might think, since a solvable fifth degree equation "should" have five solutions. The only surprising thing is that the roots are so nicely related. $\endgroup$ – Arthur Mar 9 '16 at 21:36
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$$n^5\equiv-4\implies n^{15}\equiv-64\equiv-2\implies n^{30}\equiv4\mod31$$

But $n^{30}\equiv1$ mod $31$ for all $n\not\equiv0$ mod $31$ (and $n^5\not\equiv-4$ if $n\equiv0$). So there are no integers $n$ such that $31\mid n^5+4$.

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To expand on Barry’s answer, perhaps, let me say that you’re asking whether $-4$ is a fifth power. But something in this cyclic group of order $30$ is a fifth power if and only if its sixth power is $1$. And $(-4)^6\equiv4\pmod{31}$, so the anser is No, $-4$ isn’t a fifth power.

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