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I was led by this question to the following problem:

Find $n$ complex numbers $\lambda_1\dots\lambda_n\in\mathbb{C}$ that satisfy

$\begin{align} \sum_i\lambda_i & =0\\ \sum_i\lambda_i^2 & =0\\ \sum_i\lambda_i^3 & =0\\ &\vdots\\ \sum_i\lambda_i^{n-1} & =0\\ \end{align}$

I'm pretty sure that the only solution for these equations is $\lambda_k = w \zeta^k$ where $w$ is some complex number and $\zeta$ is the primitive $n^{\text{th}}$ root of unity. By "only" I mean of course that you can permute the $\lambda_i$'s because the equations are all symmetric polynomials, but that's it.

Is there a proof for this?

1 Answers 1

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By writing the elementary symmetric functions in terms of your power sums using Newton's identities, you get $(X-\lambda_1)\cdots(X-\lambda_n)=X^n+(-1)^n\lambda_1\cdots \lambda_n=X^n-w^n=(X-w)(X-w\zeta)\cdots(X-w\zeta^{n-1})$ where $\zeta$ is a primitive $n$th root of unity and $w$ is any $n$th root of $-(-1)^{n}\lambda_1\cdots \lambda_n$. The result follows at once now.

This solution also follows from the observation that the power sums are homogeneous and hence one can normalize the problem by dividing all $\lambda_i$ by some factor. Choosing the factor such that $(-1)^{n}\lambda_1\cdots \lambda_n=-1$ gives $X^n-1$.

[Thanks to Greg Martin for simplifying my original answer.]

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    @GregMartin, done, thanks!2012-01-04