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 $r_1, r_2, \ldots, r_{2n}$
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 $n\geq 3$ ?