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It is well known that $\left(\sum_{k=1}^n k\right)^2 = \sum_{k=1}^n k^3.$

Now given $n$ integers $a_1,a_2,\ldots,a_n > 0$, is it possible to show that if
$\left(\sum_{k=1}^n a_k\right)^2 = \sum_{k=1}^n a_k^3,$ then $\{a_1,\ldots,a_n\} = \{1,\ldots,n\}$?

If it is false, for which $a_1,\ldots,a_n$ does this equality hold?

Thanks for your responses.

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    If you modify your question in such a way as to render already-written answers incorrect, then you invite undeserved down-votes on an innocent answerer. It might be better to indicate that you are making changes and/or additions to the question, so that readers will know that answers may be addressing an old version.2011-11-07

1 Answers 1

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No, because

$(2+2)^2=4^2=16=2^3+2^3$

This is even false for every $n\neq1$ because you can choose $a_k=n$ and get:

$\left(\sum _{k=1}^n n\right){}^2=n^4=\sum _{k=1}^n n^3$

Besides it holds for all permutations of $\{1,2,\ldots,n\}$ of course, yet it remains to show that there is no other possibility to choose the $a_k$.

Edit: For $n=4$ we also have for example

$(2+2+4+4)^2=2^3+2^3+4^3+4^3$

$(1+2+2+4)^2=1^3+2^3+2^3+4^3$

So this suggests there are more non-trivial tuples that satisfy your equation, I am not sure if there is some closed form for the solution set.

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    @Ross: I think the downvotes came early, and the upvotes came later, partly in response to Listing's question in the comments.2011-11-07