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I figured out that:

$$\mathop{\LARGEΣ}_{k = 1}^{n}k^3 = \bigg(\mathop{\LARGEΣ}_{k = 1}^{n}k\bigg)^2$$

Are there any other solutions for the following equation, for which $α \neq 3, \ β \neq 2$ and $α \neq β$?

$$\mathop{\LARGEΣ}_{k = 1}^{n}k^α = \bigg(\mathop{\LARGEΣ}_{k = 1}^{n}k\bigg)^β$$

P.S. The Sigmas in the equations look different to how its font usually looks like on the MSE and/or MO, (it's typical look appearing like this $\rightarrow \sum$) because I am using

\mathop{\LARGEΣ}_{k = 1}^{n}

just to test how it looks like, and because of its boldness, I don't mind its appearance.

Edit:

There is another answer that solves my problem here, mentioned by mathlove.

Mr Pie
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2 Answers2

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Since $\displaystyle \left(\sum_{k=1}^nk \right)^\beta = \left( \frac{n(n+1)}2 \right)^\beta\sim\frac{n^{2\beta}}{2^\beta}$ and $\displaystyle \sum_{k=1}^n k^\alpha \sim \frac{n^{\alpha+1}}{\alpha+1}$, two necessary conditions for the equality to hold for all $n$ are

  • $\alpha+1 = 2\beta$

  • $2^\beta=\alpha+1$

This implies $2\beta =2^\beta$, hence $\beta \in \{1,2\}$.

Conversely, $(\alpha, \beta)=(1,1)$ and $(\alpha, \beta)=(3,2)$ are indeed solutions.

Gabriel Romon
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  • Sorry I cannot vote $(+1)$ because I reached my limit....but I can do this $$\color{green}{\checkmark}$$ – Mr Pie Oct 22 '17 at 12:56
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Everything here can be put into semi-closed form. $\sum_{k=1}^n k = n(n+1)/2$ and for $\alpha>3$ odd, $\sum_{k=1}^n k^\alpha = n^2(n+1)^2/2\cdot g(n)$ where $g(n)$ is not of the form $(n(n+1)/2)^\beta$ for any $\beta$. For $\alpha$ even, a similar case occurs. So the only case we have equality is $\alpha,\beta=1$ and the case you mentioned.

  • Why for $g(n)$? What does $g(n)$ mean? – Mr Pie Oct 22 '17 at 12:53
  • You can prove it using an inductive argument. Consider the series $\sum_{i=1}^n (n+1)^k-n^k$ which is telescoping, so more precisely $(n+1)^k-1$. Then apply binomial theorem to both sides and apply induction. This is probably more hectic than the proof in the other post. –  Oct 22 '17 at 13:07