Does $1^n + 2^n + \cdots + m^n$ divide $(1+2+ \cdots +m)^n$ for any even integers $m, n\geq 2$ ?.


For $n\leq 4$, the solution easily follows from the relevant identities. For $n\geq 6$, i suspect that Bernoulli polynomials might be the best approach, but haven't found how ?



On the divisibility of a certain power sum
Let $n$ be even, fixed, large.


Sketch of proof that $\{m|m>0, 1^n+\dots+m^n \mbox{ divides } (1+\dots+m)^n \}$ has zero density in $\mathbb{N}$.


Let $f(n)$ be the degree of the largest irreducible factor of $B\_{n+1}(x)/(n+1)$ in $\mathbb{Q}[x]$. Let $S$ be the set of primes $p$ for which this factor splits into linear factors mod $p$. $1^n+\dots+m^n$ divides $(1+\dots+m)^n$ only if $m+1$ mod $p$ for each $p\in S$ is in one of the at most $p-(f(n)-2)$ classes mod $p$ for which $B\_{n+1}(m+1)/(n+1)$ is not $0$ mod $p$ or both $B\_{n+1}(m+1)/(n+1)$ and $(B\_2(m+1)-1/6)/2$ are $0$ mod $p$. Since $f(n)>2$ (check literature or prove it), and $S$ has positive density in the primes by Chebotarev's density theorem, the density of these $m$ is zero.


