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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 ?

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  • $\begingroup$ A sentence beginning "Prove that" is not a question. As to the problem itself, you might try to find an odd prime that divides the LHS, but does not divide $m(m+1)$, d Bernoulli polynomials do look relevant. $\endgroup$ Jan 18, 2016 at 10:14
  • $\begingroup$ Thanks Geoff for the suggestion. I have attempted that approach but seem not to be getting anything. Perhaps you could help ? $\endgroup$
    – favoured
    Jan 18, 2016 at 10:22
  • $\begingroup$ I do not know a precise answer myself, but I think reading about a little Bernoulli polynomials should give some insights. I think that there should be no problem when $m$ is even. $\endgroup$ Jan 18, 2016 at 10:29
  • $\begingroup$ @Fedor Petrov, for $m=2$, we have $1^n + 2^n = 3^n$, which is clearly not possible for any $n\geq 2$. $\endgroup$
    – favoured
    Jan 18, 2016 at 10:32
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    $\begingroup$ Also posted to m.se, math.stackexchange.com/questions/1616671/… – please, don't do that. $\endgroup$ Jan 18, 2016 at 11:49

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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.

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