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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 $0<m_{v}\leq q, 0<n_{v}\leq q, q$ 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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up vote 2 down vote accepted

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. – ericc Oct 2 '12 at 7:23

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