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Is it true / is there any way to prove that, for a given $A\in \mathbb{N}$, and $N\in \mathbb{N}, N >2$ ($N=2014$ in my example) there is at most one solution for $0\leq a_1\leq a_2 \leq \dots \leq a_N$, all in $\in \mathbb{Z}_+$ (positive integers or $0$, they do not have to be pairwise different) such that that

$$A^N=\sum_{i=1}^N a_i^N$$

If there are any conditions on $A$ and $N$ so that the solution is non-existent or unique, I would be interested to know them.

N.B. This is a follow-up on question 965318

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Yulia V
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    For $N=3$ we have $12^3+1^3+0^3=10^3+11^3+0^3$ – Hagen von Eitzen Oct 14 '14 at 14:45
  • @HagenvonEitzen: yes but this sum is not a cube of an integer number (there is no $A$ such that this sum equals to $A^3$) – Yulia V Oct 14 '14 at 14:48
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    If I'm not mistaken, how about $a_1=A,a_i=0$ for $i\not=1$ ? – mathlove Oct 14 '14 at 15:00
  • @mathlove: yes, this is would be one solution; one solution can exist for sure. My question was about the existance of 2 or more different solutions. Probably I should have specified that permutations of $a_i$ are not considered to create multiple solutions; I will edit the question. – Yulia V Oct 14 '14 at 15:03
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    $2^3 + 17^3 + 40^3 = 6^3 + 32^3 + 33^3 = 41^3$. – Dan Shved Oct 14 '14 at 15:11
  • @DanShved: thanks! Do you think you can find a generic formula for such combinations? Do you think it is untrue for all $N$? – Yulia V Oct 14 '14 at 15:12
  • @Yulia I don't think so :) I found this example (and several others) by brute force on my computer. So I don't really have any insight to share. – Dan Shved Oct 14 '14 at 15:14
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    @DanShved: might be a bit tricky to do it for $N=2014$... – Yulia V Oct 14 '14 at 15:14
  • When the numbers $N=2014....$ we always come to the need to solve an algebraic equation of large degrees. And as this problem around? May you know the formula for the solutions of these equations? – individ Oct 14 '14 at 17:38

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