# The reciprocal condition of an equality

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?

• 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

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.

• 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. – HYL Nov 6 '11 at 19:27
• 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