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Given pairwise distinct numbers $c_1, c_2, \dots c_n \in \mathbb{C} \setminus \{0\}$, does the system of equations $$\frac{6}{c_k} + \sum_{i \ne k} \frac{2}{c_k - c_i} = \sum_{i = 1}^n \frac{1}{c_k - x_i}, \ k = 1,2,3, \dots, n$$ have a nontrivial solution $(x_1, x_2, \dots, x_n) \in \mathbb C^n$?

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Why do you ask? –  Igor Rivin Jul 31 '13 at 0:44
    
I was originally thinking about this question: Given $h(x) = \prod_{i = 1}^n (x - c_i) \in \mathbb C[x]$ with $c_i \ne c_j \forall i \ne j$ Can we find $p(x) \in \mathbb C[x]$ nontrivial and $\deg(p) \leq n$ such that $$(6h'(x) + xh''(x))p(x) = xh'(x)p'(x)$$ And the existence of such $p(x)$ can be guaranteed if I can solve the posted one. –  Zhaoning Yang Jul 31 '13 at 1:18
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But then it's just $p'/p = 6/x + h''/h'$, so $p = c x^6 h'$ and the only nontrivial solutions have degree $n+5$. –  Noam D. Elkies Jul 31 '13 at 4:25
    
Sorry, I did not state the problem correctly, Given $h(x) = \prod_{i = 1}^n (x - c_i)$ where $c_i \in \mathbb C$ pairwise disjoint, Is that possible to find a $p(x) \in \mathbb C[x] \And \deg(p) \leq n$ such that $$S(x) = \bigl( 6h'(x) + xh''(x) \bigl)p(x) - xh'(x)p'(x)$$ has zeros at $x = c_1, c_2, \dots, c_n$ –  Zhaoning Yang Jul 31 '13 at 13:00
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