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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?

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up vote 9 down vote accepted

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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Note that your first sentence can be strengthened: not only is each $\lambda_j$ a root of $X^n = (-1)^{n+1}\lambda_1\cdots \lambda_n$, but the polynomial $(X-\lambda_1)\cdots(X-\lambda_n)$ is actually equal to the polynomial $X^n - (-1)^{n+1}\lambda_1\cdots \lambda_n$. This latter polynomial factors as $(X-w)(X-w\zeta)\cdots(X-w\zeta^{n-1})$, where $\zeta$ is a primitive $n$th root of unity and $w$ is any $n$th root of $(-1)^{n+1}\lambda_1\cdots \lambda_n$ (note this is true even if $w=0$). This immediately establishes the OP's proposition. – Greg Martin Jan 4 '12 at 1:46
Do you want to just edit yours? It's an observation on what's otherwise your solution.... – Greg Martin Jan 4 '12 at 7:44
@GregMartin, done, thanks! – lhf Jan 4 '12 at 12:00

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