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

José María Grau; Antonio M. Oller-Marcén

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arXiv · 1304.2678

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

José María Grau·Antonio M. Oller-Marcén

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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 $f$.

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José María Grau, Antonio M. Oller-Marcén. 2013-04-09. About the congruence $\sum_{k=1}^n k^{f(n)} \equiv 0 \textrm{(mod n)} $. https://arxiv.org/abs/1304.2678

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