# The solutions of a polynomial system

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 $n$.

• This seems harder than the previous question, where the answer was that $a_i\neq 0, b_j\neq 0$ for every solution. In particular, there are some choices of $m$, $n$, $r_i$ such that some solutions have some $a_i$ 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. – Jason Starr Jan 19 '14 at 20:15
• @JasonStarr I think it is difficult to count the number of solutions. Do you have any idea to prove the existence of such solutions? – Zhihua Chang Jan 20 '14 at 8:40

Here is an observation which does not solve your problem, but translates it into one which has a large literature. Note that the $k$-th elementary symmetric function can be expressed as a polynomial with rational coefficients in the first $k$ power symmetric functions. For example, $e_2(x_i) = \Sigma(x_ix_j | i < j) = 1/2[(\Sigma(x_i))^2 - \Sigma(x_i^2)] = 1/2[(p_1(x_i))^2 - p_2(x_i)]$. This implies in your problem that if $Q(z) = \Pi (z - x_i)^{r_i} = z^a + q_1z^{a-1} + q_2z^{a-2} + \cdots$ (where $a = r_1 + \cdots + r_k$), and $S(z) = \Pi(z - y_i) = z^n + s_1z^{n-1} + s_2z^{n-2} + \cdots$, then $q_i = s_i$ for $i = 1, \dots , m+n-1$. Equivalently, $Q(z) - z^{a-n}S(z)$ has degree $\leq a - (m+n)$. This is a situation where the much generalized Mason-Stothers Theorem applies.
For a polynomial $F(z)$ over $\mathbb{C}$, let $h(F)$ be the number of distinct roots of $F$, and let $d(F)$ be the degree of $F$. Mason-Stothers Theorem: If $Q(z) + T(z) + V(z) = 0$, where $Q,T,V$ are complex polynomials with no common root, then $h(Q) + h(T) + h(V) \geq max(d(Q),d(T),d(V)) + 1$.
The theorem was first proved in W. W. Stothers, Polynomial Identities and Hauptmoduln, Quart. J. Math. Oxford 32 (1981), 349-370. It can be proved in a few lines by elementary means, but most of Stothers' paper uses deeper methods from the theory of automorphic forms to study the boundary case where $h(Q) + h(T) + h(V) = max(d(Q),d(T),d(V)) + 1$. In the boundary case he gives examples of choices of root multiplicities for $Q,T,V$ for which there are solutions and examples for which there are no solutions. I couldn't tell if his examples include any of your polynomial systems. Since his method requires looking at finite index subgroups of a free group of rank 2, I expect that more examples could be analyzed with today's computers.
When there is a solution to one of your polynomial systems, let $Q(z),S(z)$ be as in the first paragraph above, and let $T(z) = -z^{a-n}S(z), V(z) = -Q(z) + z^{a-n}S(z) = -Q(z) - T(z)$. Then $h(Q) = m$, $h(T) = n + 1$, $a - (m +n) \geq d(V) \geq h(V)$, $d(Q) = a$, and Mason-Stothers implies that $a+1 = m + (n + 1) + (a - (m + n)) \geq h(Q) + h(T) + h(V) \geq d(Q) + 1 = a + 1$, so all inequalities become equalities, which is the boundary case.