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Let $\{ r_1, \dots, r_n \}$ and $\{ s_1, \dots, s_m \}$ be two sets of positive integers. If $\sum_{i = 1}^{n} r_{i} = \sum_{i = 1}^{m} s_{i}$ and $\prod_{i = 1}^{n}(r_i + 1) = \prod_{i = 1}^{m} (s_i + 1)$, does it necessarily follow that $n = m$? More generally, given the equalities, does it follow that there is a permutation $\pi \in S_{n}$ such that $\pi( \{ r_1, \dots, r_n \}) = \{ s_1, \dots, s_m \}$?

Perhaps symmetric functions play a non-trivial role here.

Update: As stated, neither question above can be answered in the affirmative (See Prof. Israel's counter-example). Let $e_k(x_1, \dots, x_n)$ denote the $k^{\text{th}}$-elementary symmetric polynomial of $n$ indeterminates.

Is there a positive integer $l < n$ such that if $e_{k}(r_1, \dots, r_n) = e_{k}(s_1, \dots, s_m)$ for $0 \leqslant k \leqslant l$ and $\prod_{i = 1}^{n}(r_i + 1) = \prod_{i = 1}^{m} (s_i + 1)$, then $n = m$ and $e_{k}(r_1, \dots, r_n) = e_{k}(s_1, \dots, s_m)$ for $0 \leqslant k \leqslant n$, or more generally, there is a permutation $\pi \in S_{n}$ such that $\pi( \{ r_1, \dots, r_n \}) = \{ s_1, \dots, s_m \}$?

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2 Answers 2

No, it's not true. For example, try $\{1, 2, 14\}$ and $\{8, 9\}$

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Thank you for the counter-example. I posted a new follow-up question. –  user02138 Aug 14 '12 at 4:55
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For any $\ell$, there are disjoint sets $\{{r_1,\dots,r_n\}}$ and $\{{s_1,\dots,s_n\}}$ of positive integers such that $$\sum r_i^k=\sum s_i^k,\quad 1\le k\le\ell$$ Information can be found under the headings "multigrade equations", or "Tarry-Escott problem". It follows that the polynomials $\prod(x-r_i)$ and $\prod(x-s_i)$ will have identical coefficients from $x^n$ down to $x^{n-\ell}$, which is to say, the first $\ell$ elementary symmetric functions will agree. It's conjectured that one can take $n$ as small as $\ell+1$, but only proved that one can take $n$ to be on the order of $\ell^2$.

I know this doesn't account for the condition on $\prod(r_i+1)$, but there are so many solutions to the multigrade equation I suspect that at worst this would push $n$ up by 1.

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