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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. –  Arturo Magidin Nov 7 '11 at 4:49
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1 Answer 1

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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Thank you. I've slightly modified my question, now I'd like to find all sets $\{a_1,\ldots,a_n\}$ such that this equality holds. –  Hsueh-Yung Lin Nov 6 '11 at 19:27
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Why is this getting many downvotes? I can remove the answer, but I think it might help others who want to answer the complete modified question. –  Listing Nov 6 '11 at 19:39
    
I don't get the downvotes unless they think you are not answering the question. But the question has changed and your answer is a good one for the original. I wish people would leave a comment for every downvote (or at least a comment upvote if their sentiment is already expressed.) But I wouldn't sweat it-you have more upvotes (including mine) than downvotes for this answer. –  Ross Millikan Nov 7 '11 at 4:48
    
@Ross: I think the downvotes came early, and the upvotes came later, partly in response to Listing's question in the comments. –  Mike Spivey Nov 7 '11 at 4:50
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