For which natural numbers $a,b,p$ with $a>b>0$ do we have $$ \left(\sum_{k=1}^nk^a\right)^p = \sum_{k=1}^nk^b $$ for all $n\in\mathbb N$? This is famously true for $(a,b,p)=(1,3,2)$, but looking at lists of sums of powers shows no other examples. How many "solutions" $(a,b,p)$ are there and what are they?
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http://math.stackexchange.com/questions/1292636/other-variation-of-nicomachuss-theorem – mathlove Aug 22 '16 at 07:29
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See this – Jyrki Lahtonen Aug 22 '16 at 07:30
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@JyrkiLahtonen, I hadn't seen that one, but it seems to be only a partial version of this one ($a=1$). – Joonas Ilmavirta Aug 22 '16 at 07:46
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@mathlove, questions like this are hard to find... Should we close this one as a duplicate? – Joonas Ilmavirta Aug 22 '16 at 07:47
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True, Joonas. Not a perfect match. The technique is still similar. – Jyrki Lahtonen Aug 22 '16 at 07:56
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@JoonasIlmavirta: All I can say is that your question is a partial version of the linked question where $p$ is rational. – mathlove Aug 22 '16 at 07:59
1 Answers
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There are no other examples.
For large $n$ we have1 $$ \sum_{k=1}^nk^b=\frac1{b+1}n^{b+1}+O(n^b). $$ Looking at leading order terms only gives the constraints $p(a+1)=b+1$ (for the power) and $(a+1)^p=b+1$ (for the coefficient). In particular, this gives $p(a+1)=(a+1)^p$, or $p=(a+1)^{p-1}$.
Since $a\geq1$, we have $p=(a+1)^{p-1}\geq2^{p-1}$. But $p\geq2^{p-1}$ only holds for $p\leq2$. (In other words, $\sqrt[p-1]{p}$ is not an integer when $p\geq3$.) It is now easy to see that there is no solution if $a\geq2$. The case $p=1$ (with $a=b$) is uninteresting and was excluded in the question, so the only option is $p=2$ and $a=1$. This leaves $b=3$.
(I didn't find the answer anywhere, so I decided to ask and answer myself. I hope others agree that this is a fun observation.)
1 One can argue this by comparing the sum to an integral or by using Faulhaber's formula.
Joonas Ilmavirta
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