Find all $n>1$ such that $1^{n} + 2^{n} + 3^{n} +\cdots + (n-1)^{n}$ divisible by $n$. I'm not good at Number Theory so , give elementary answer.
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3Clearly, $n$ is odd is a sufficient condition – lab bhattacharjee May 16 '16 at 13:32
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@lab bhattacharjee pls explain. – mnulb May 16 '16 at 13:33
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See http://math.stackexchange.com/questions/641443/proof-of-anbn-divisible-by-ab-when-n-is-odd – lab bhattacharjee May 16 '16 at 13:34
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@lab bhattacharjee we have to club two-two terms from starting, if I'm not wrong? – mnulb May 16 '16 at 13:36
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2Possible duplicate of Find all $n$ such that $n|1^n + 2^n + 3^n + \cdots + (n-1)^n$ where $n \in \mathbb{Z}^+$. – TheRandomGuy May 17 '16 at 14:15
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If $n=2k+1$ is odd $$\sum_{i=1}^{n-1} i^n=\sum_{i=1}^{k} i^n+(n-i)^n$$ and $i^n+(-i)^n\equiv 0\pmod{n}$. So, the claim holds for odd $n$.
So, suppose $n$ is even. Let $n=2^kl$, where $l$ is odd. Then, as $\phi(2^k)=2^{k-1}|n$, all the odd terms are congruent to $1\bmod{2^k}$. However, $n\ge 2^k\ge k$ implies $2^k$ divides the even terms. So, the whole sum is congruent to $n/2$ modulo $2^k$, i.e. not divisible by $2^k$. Thus, it is not divisible by $n$.
Emre
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