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Let $k>1$ be a positive integer.

Question 1. Does there exist infinitely many positive integers $n$ such that $$n \, | \, (1^n + 2^n + \cdots + k^n)?$$

And, more generally:

Question 2. Does there exist infinitely many positive integers $n$ such that $$n \,| \, (\epsilon_11^n + \epsilon_22^n + \cdots + \epsilon_kk^n),$$ where $\epsilon_i \in \{-1,1\}$?

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    $\begingroup$ there is no $n>1$ which divides $2^n-1$, so the generalization is not always true $\endgroup$
    – Fedor Petrov
    Commented Feb 20, 2021 at 14:56
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    $\begingroup$ For $k=3$ it is true; note that for the expression $3^n\pm 2^n-1$ we can take $n=2^j$ and for the remaining one $3^n-2^n+1$ we can take $n=2\cdot 3^j$. $\endgroup$
    – Vlad Matei
    Commented Feb 20, 2021 at 16:11

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Yes (for the first question). Assume at first that $k$ is even, and let $p$ be a prime divisor of $k+1$. Choose $n=p^s$. I claim that $a^n+(k+1-a)^n$ is divisible by $n$ for all $a=1,2,\ldots,k/2$. This is clear if $p$ divides $a$. If $a$ and $p$ are coprime, we get $$a^n+(k+1-a)^n=a^n-(a-k-1)^n$$ is divisible by $p^{s+1}$ which is called Lifting the Exponent Lemma and may be proved by induction in $s$: $$a^{pn}-(a-k-1)^{pn}=\left(a^{n}-(a-k-1)^{n}\right)(a^{n(p-1)}+\ldots+(a-k-1)^{n(p-1)}),$$ the second bracket is divisible by $p$ since all $p$ summands are congruent modulo $p$.

If $k$ is odd, take $n=p^s$ where $p$ is any prime divisor of $k$. The last summand $k^n$ is divisible by $n$, for the previous summands use the even case.

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