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 :)


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

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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

$$n^5\equiv-4\implies n^{15}\equiv-64\equiv-2\implies n^{30}\equiv4\mod31$$

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 $31\mid n^5+4$.


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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