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Find all $n$ such that $$n|1^n + 2^n + 3^n + \cdots + (n-1)^n$$ where $n \in \mathbb{Z}^+$.

I don't know how to start. $n = 3, 5$ are simple solutions. Induction seems strange since the divisor is changing every time. How do I go about it?

Thanks.

TheRandomGuy
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All odd numbers $n$ are solutions to this. Take $n=2k+1$ and see what remainder the first and last terms give when divided by $n$ (it's $1^{2k+1}$ and $(-1)^{2k+1} = -1^{2k+1}$). Then see what remainder the 2nd and the 2nd last terms give (it's $2^{2k+1}$ and $(-2)^{2k+1}=-2^{2k+1}$) and so on.

peter.petrov
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  • Yes, and $n$ even? – almagest May 04 '16 at 12:02
  • When $n$ is even then $n$ is not a solution. The proof shouldn't be too hard. Let me think about it. – peter.petrov May 04 '16 at 12:02
  • I got what you meant. For odd $n$ we can pair them to form divisors of $n$. I tried proving that there exist no solutions for even $n$ and it finally comes down to prove that $$n \nmid 1^n\cdot 2 + 2^n \cdot 2 + \cdots + (n/2 - 1)^n \cdot 2 + (n/2)^n$$ Have any idea how I should do this? – TheRandomGuy May 04 '16 at 12:36
  • Yeah, it comes down to that. No idea yet. I think you missed $\cdot 2$ in the last term. – peter.petrov May 04 '16 at 12:37
  • @peter.petrov No. I intentionally left that. You see for an even $n$, the $n/2$ will be the one which only comes once and has no other number whose $^k$ is equal to it $\pmod n$. – TheRandomGuy May 04 '16 at 14:50