Suppose we have the complex numbers $x_1,x_2,\dots,x_N$ and that we know that $$x_1+\dots+x_N =0,$$ $$x_1^2+\dots+x_N^2 =0,$$ $$\dots$$ $$x_1^N+\dots+x_N^N =0.$$ Does this imply that all $x_m=0$? There should be enough equations to get a unique solution. I have proven it in the case $N=2$, but going further requires a lot of algebra, and I'm wondering if it can be seen in an easier way than solving the system of equations by brute force.
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AFAICT, if your numbers are 1, $i$, -1, and $-i$, you get a sum of zero for any power. I haven't bothered to demonstrate this rigorously, though. – bob.sacamento Aug 28 '19 at 22:36
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1@bob.sacamento The sum of their fourth powers is equal to $4$. – José Carlos Santos Aug 28 '19 at 22:39
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1Hint: Vectors $(x_i,x_i^2,...,x_i^N)$ , $i \in {1,...,N}$ are linearly independent for different $x_i$ – Dominik Kutek Aug 28 '19 at 22:42
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Yes. By the Newton-Girard formulas the fact that $p_1=p_2=\ldots=p_N=0$ implies the fact that $e_1=e_2=\ldots=e_N=0$, so the variables $x_1,\ldots,x_N$ are roots of the polynomial $z^N$, i.e. they are all zero.
A celebrated consequence is the following: if $M$ is a $N\times N$ real symmetric matrix such that $\text{Tr}(M)=\text{Tr}(M^2)=\ldots=\text{Tr}(M^N)=0$, then $M=0$.
Jack D'Aurizio
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