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When $k=1$ this is trivial, and when $k=2$ the solutions correspond to the nonzero integral points on the elliptic curve:

$$y^2=\frac{x(x+1)(2x+1)}{6}.$$

And the Wolfram|Alpha says there are only four nonzero integral points on it ($x=1$ and $x=24$), hence two solutions. I am looking for an (algebro-geometric) proof for this fact, thanks! Also, are there any general results known for the higher $k$ cases?

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    What is an "algebro-geometric" proof, exactly?2012-06-09
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    The name of this problem is the '*[cannonball problem](http://mathworld.wolfram.com/CannonballProblem.html)*'. References in this link may help you find out what you are searching for.2012-06-09
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    @Phira, I don't have a concrete idea for this, sorry for the vague.2012-06-09
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    @sos440, thank you for the link!2012-06-09
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    Reading about the relation between [Bernoulli numbers](http://en.wikipedia.org/wiki/Bernoulli_number#Sum_of_powers) and your problem, might interesting: You can rephrase your problem to $$ \sum_{k=1}^n k^m = 1^m + 2^m + \cdots + n^m = S_m(n) = {1\over{m+1}}\sum_{k=0}^m {m+1\choose{k}} B_k\; n^{m+1-k}=M^m. $$2012-06-09
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    Some work has been done on the case where $m=n+1$: http://arxiv.org/abs/1011.29562012-06-09
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    Partial (general) result: If $m=-2x$, then $\lim_{n\to \infty} \sum_{k=1}^n k^m=0$ (trivial roots of the [Riemann $\zeta$ function](http://en.wikipedia.org/wiki/Riemann_zeta_function)). OK, you assume $m>0$, but you never wrote it ;-). AND, also true, it's not $0^{\color{red}{-2x}}$.2012-06-09
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    I suspect the general problem is discussed in Guy, Unsolved Problems In Number Theory, but I don't have access to the book right now.2012-06-09
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    I suggest this question be migrated to mathoverflow. It does seem like it is research level.2012-06-10

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