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Suppose $m_{1}^{h}+\cdots m_{k}^{h}=n_{1}^{h}+\cdots n_{k}^{h}$ for $h=1,\dots ,k$, where $0 positive integer. How do one show that the natural number $n_{v}$ must be equal (in some order) to the numbers $m_{v}$? Thanks.

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If the power sums are the same, then the elementary symmetric functions are the same, so the integers on either side are zeros of the same polynomial of degree $k$, hence, equal (up to order).

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    Thanks, I didn't know this idea before.2012-10-02