# Uniqueness of $a_1, a_2, \dots, a_N \in \mathbb{Z}_+ : A^N=\sum_{i=1}^N a_i^N$ given $A\in\mathbb{N},N>2$

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

• 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
• If I'm not mistaken, how about $a_1=A,a_i=0$ for $i\not=1$ ? – mathlove Oct 14 '14 at 15:00
• $2^3 + 17^3 + 40^3 = 6^3 + 32^3 + 33^3 = 41^3$. – Dan Shved Oct 14 '14 at 15:11
• @DanShved: might be a bit tricky to do it for $N=2014$... – Yulia V Oct 14 '14 at 15:14