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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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    $\begingroup$ For $N=3$ we have $12^3+1^3+0^3=10^3+11^3+0^3$ $\endgroup$ Oct 14, 2014 at 14:45
  • $\begingroup$ @HagenvonEitzen: yes but this sum is not a cube of an integer number (there is no $A$ such that this sum equals to $A^3$) $\endgroup$
    – Yulia V
    Oct 14, 2014 at 14:48
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    $\begingroup$ If I'm not mistaken, how about $a_1=A,a_i=0$ for $i\not=1$ ? $\endgroup$
    – mathlove
    Oct 14, 2014 at 15:00
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    $\begingroup$ $2^3 + 17^3 + 40^3 = 6^3 + 32^3 + 33^3 = 41^3$. $\endgroup$
    – Dan Shved
    Oct 14, 2014 at 15:11
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    $\begingroup$ @DanShved: might be a bit tricky to do it for $N=2014$... $\endgroup$
    – Yulia V
    Oct 14, 2014 at 15:14

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