Let $m>0,n\geqslant0$ be integers and $r_1,\ldots,r_m$ be $m$ positive integers. I am considering the polynomial system

\begin{eqnarray*} r_1\cdot x_1+\cdots+r_m\cdot x_m&=&y_1+\cdots+y_n,\\ r_1\cdot x_1^2+\cdots+r_m\cdot x_m^2&=&y_1^2+\cdots+y_n^2,\\ &\vdots&\\ r_1\cdot x_1^{m+n-1}+\cdots+r_m\cdot x_m^{m+n-1}&=&y_1^{m+n-1}+\cdots+y_n^{m+n-1}. \end{eqnarray*}

I want to prove that if $$r_1+r_2+\cdots+r_m\geqslant m+n,$$ Then the polynomial system has at least one solution $(a_1,\ldots,a_m,b_1,\ldots,b_n)$ such that $$a_i\neq0, b_j\neq0$$ for all $i=1,\ldots,m,j=1,\ldots,n$.

A special case is that if all $r_i\geqslant n+1$ for $i=1,\ldots,m$. Then the problem has been solved in the answers to my previous question (cf. The solutions of a system of polynomials).

I have verified using a computer that this is true in all possibilities with $m+n\leqslant8$. But I still have no idea to prove the general case stated above. Does anybody have any idea to solve this problem? Furthermore, is it possible to express the solutions using $r_1,\ldots,r_m$ and $n$, or express the number of such solutions using $r_1,\ldots,r_m$ and $n$.

everysolution. In particular, there are some choices of $m$, $n$, $r_i$ such thatsomesolutions havesome$a_i$ or $b_j$ equal to $0$. So now I don't see any strategy short of counting the number of solutions in coordinate hyperplanes (together with multiplicities) and showing that these do not account for all $(m+n-1)!$ solutions. $\endgroup$ – Jason Starr Jan 19 '14 at 20:15