About the congruence $\sum_{k=1}^n k^{f(n)} \equiv 0 \textrm{(mod n)} $

Abstract

In this paper we characterize, in terms of the prime divisors of $n$, the pairs $(k,n)$ for which $n$ divides $\sum_{j=1}^n j^{k}$. As an application, we study the sets $\mathcal{M}_f :=\{n: n \textrm{divides} \sum_{j=1}^n j^{f(n)} \}$ for some choices of $f$.