# all positive integers satisfying the following series

Find all positive integers $$n>1$$ such that the following series is divisible by $$n$$: $$1^n + 2^n + 3^n + \ldots + (n-1)^n.$$

I started with considering separate cases for even and odd values of $$n$$ but couldn't make any progress, and on searching I found following conjecture:

If $$n$$ is a prime, then $$n \mid 1^{n-1}+2^{n-1} + \ldots + (n-1)^{n-1} + 1,$$

but I am not able to understand the proof. It seems the above is by fermat

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Suppose $$n$$ is odd. Consider the sum modulo $$n$$. Then $$(n - k)^n \equiv (-k)^n \equiv (-1)^n k^n \equiv -k^n.$$ So, each term of the sum pairs uniquely with another term of the sum which is its negative. So, when $$n$$ is odd, the sum should be $$0$$ modulo $$n$$, i.e. divisible by $$n$$.
Otherwise, consider the case where $$n$$ is even. Then $$n = 2^km$$ for some odd $$m$$, and positive integer $$k$$. Consider the sum modulo $$2^k$$. Note that $$k \le n$$, hence if $$r$$ is an even number between $$1$$ and $$n - 1$$, then $$2^k \mid r^n$$.
Otherwise, if $$r$$ is an odd number in the range $$1$$ to $$n - 1$$. Then $$r^n$$ is odd, and there are an odd number of such terms. So, adding all the terms modulo $$2^k$$, we obtain an odd number, implying that the sum cannot be $$0$$ modulo $$2^k$$, and hence it can't be $$0$$ modulo $$n$$.
Therefore, the sum is divisible by $$n$$ if and only if $$n$$ is odd.