On the congruence $1^m+2^m+\cdots +m^m\equiv n(\mathrm{mod}m)$ with $n|m$

Abstract

We show that if the congruence above holds and $n\mid m$, then the quotient $Q:=m/n$ satisfies ${\sum }_{p\mid Q}\frac{Q}{p}+1\equiv 0(\mathrm{mod}Q)$, where $p$ is prime. The only known solutions of the latter congruence are $Q=1$ and the eight known primary pseudoperfect numbers $2,6,42,1806,47058,2214502422,52495396602,$ and $8490421583559688410706771261086$. Fixing $Q$, we prove that the set of positive integers $n$ satisfying the congruence in the title, with $m=Qn$, is empty in case $Q=52495396602$, and in the other eight cases has an asymptotic density between bounds in $(0,1)$ that we provide.

13 pages, 1 table; introduced proofs of Lemma 5 and Theorem 5; notational changes (${\mathfrak{M}}_Q\to {\mathfrak{N}}_Q$) and some typos corrected

Cited by in corpus (2)

• A Carlitz-von Staudt type theorem for finite rings
• Computing solutions to the congruence $1^n+2^n+\cdots +n^n\equiv p(\mathrm{mod}n)$