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

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 $f$.