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Let $n\geq 3$ be an odd number and $r_1, r_2, \ldots, r_{2n}$ be pairwise distinct integers. Suppose that $(X-r_1)(X-r_2)\ldots(X-r_{2n}) = X^{2n} + \sum_{i=1}^{2n}a_{i}X^{2n-i}$. Can we have $a_1 = a_3 = a_5 = \ldots = a_{n-2} = 0$?

The motivation of the question above is to know whether or not the "non trivial" solution of this diophantine equation exists: $$r_{1}^{2k+1}+\ldots+r_{2n}^{2k+1}=0$$ for all $0\leq k\leq \frac{n-3}{2}$.

Since $r_{1}^{2k+1}+\ldots+r_{2n}^{2k+1}=0$ is equivalent to $r_{1}^{2k+1}+\ldots+r_{n}^{2k+1}=(-r_{n+1})^{2k+1}+\ldots+(-r_{2n})^{2k+1}=0$, "non trivial" solution means $r_i$ aren't all zero and $\{r_1, \dots, r_n\} \neq \{-r_{n+1}, \dots, -r_{2n}\} $ for all rearrangement of $r_1, r_2, \ldots, r_{2n}$

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  • $\begingroup$ You should write out a couple example polynomials, I think the indices $i$ for which $a_i=0$ may need to be adjusted. $\endgroup$ – abiessu May 30 '18 at 12:42
  • $\begingroup$ I added $n\geq 3$ in the question. Is that what you mean? Some cases for n small could be easy, I only want to know in general case, that why I forgot this condition. $\endgroup$ – 曾靖國 May 30 '18 at 12:55
  • $\begingroup$ They can. Put all coefficients equal to zero. The polynomial is $X^{2n}$, which has all of its roots integer. $\endgroup$ – user565560 May 30 '18 at 13:06
  • $\begingroup$ If we add the condition "all the roots of $P$ are some distinct integers", does such $P$ exists? $\endgroup$ – 曾靖國 May 30 '18 at 14:01
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    $\begingroup$ Where does the question come from or what motivates it? Just curiosity? $\endgroup$ – punctured dusk May 30 '18 at 14:16
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Let $n=3$. Take such integers, so that $r_1+r_2+\dots+r_6=0$. For example, $-6, -2, -1, 2, 3, 4$. Then we'll have $a_1=0$, so it's possible. Now, is it possible for every odd $n\geq 3$ ?

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