Given positive integers $m_1,...,m_n$, is it possible to solve the following equation system over the field of complex numbers?

$$m_1x_1+\cdots+m_nx_n=0$$ $$m_1x_1^2+\cdots+m_nx_n^2=0$$ $$\cdots$$ $$m_1x_1^{n-1}+\cdots+m_nx_n^{n-1}=0$$ $$x_1x_2\cdots x_n=1.$$

When $n=1,2,3,4$, I can find a formula for the solutions. But I can not do it for $n>4$. Also, when $m_1=m_2=...=m_n=1$, the solutions can be written as


where $\zeta$ is a $n$-th primitive root of unity and $\sigma$ is an element of the symmetric group $S_n$.

For other concrete examples, the Mathematica numerical computation shows that the number of solutions to this equation system would be $n!$. Does anybody have an idea to prove it?

In general, if we do not assume $m_1,...,m_n$ are positive integers, what is the condition on $m_1,...,m_n$ such that this equation system has a solution?

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    $\begingroup$ For $m_1=\ldots=m_n=1$, $(\zeta^{\sigma(1)},\ldots,\zeta^{\sigma(n)})$ is a solution only when $n$ is odd. For $n$ even, $(\zeta)(\zeta^2)\ldots(\zeta^n)=-1$. $\endgroup$ Nov 27, 2013 at 15:45
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    $\begingroup$ Do you by any chance want $x_1^{m_1}...x_n^{m_n}=1$? $\endgroup$ Nov 27, 2013 at 16:40
  • $\begingroup$ I don’t know if it’s of any use, but if $f(z)$ denotes the polynomial $(1-x_1z)^{m_1}\cdots(1-x_nz)^{m_n}$, then the first $n-1$ equations are equivalent to the condition that the logarithmic derivative $f'(z)/f(z)$ has a root of order at least $n-1$ at the origin (this follows easily enough from the Cauchy integral formula). $\endgroup$ Nov 27, 2013 at 17:59

2 Answers 2


I believe this follows using Vandermonde matrices. First I will prove a proposition that gives a bit more than what the OP asks.

For every integer $n \geq 2$, for every integer $r=1,\dots,n$, define $$ F_r(X_1,\dots,X_n) = F_{n,r}(m_1,\dots,m_n;X_1,\dots,X_n) = m_1X_1^r + \dots + m_nX_n^r.$$ Let $P\subset \mathbb{C}$ be a nonempty subset that is stable under addition and which does not contain $0$, e.g., the set of positive integers.

Proposition $Q_n$. For every integer $n\geq2$, for every choice of $m_1,\dots,m_n$ in $P$, every nonzero solution $(A_1,\dots,A_n)\neq (0,\dots,0)$ of $$ F_{n,1}(m_1,\dots,m_n;X_1,\dots,X_n)=\dots=F_{n,n-1}(m_1,\dots,m_n;X_1,\dots,X_n)=0$$ is not a solution of $F_{n,n}(m_1,\dots,m_n;X_1,\dots,X_n) = 0$, it has no coordinate equal to $0$, and it has no repeated coordinates.

Proof This will be proved by induction on $n$. It is straightforward to prove $Q_2$, the base case of the induction. Basically it boils down to the fact that none of $m_1$, $m_2$ nor $m_1+m_2$ can equal $0$.

Now by way of induction, assume that $n>2$ and that $Q_m$ holds for all integers $2\leq m < n$. Let $(A_1,\dots,A_n)$ be a nonzero solution of the system $F_1=\dots=F_{n-1}=0$. If any $A_i$ is zero, then after permuting the coordinates and $(m_1,\dots,m_n)$, we may assume that $A_n=0$. Then $(A_1,\dots,A_{n-1})$ is a nonzero solution of the system $$ F_{n-1,1}(m_1,\dots,m_{n-1};X_1,\dots,X_{n-1}) = 0,\dots ,$$ $$ F_{n-1,n-2}(m_1,\dots,m_{n-1};X_1,\dots,X_{n-1}) = 0,$$ and which also solves the equation $F_{n-1,n-1}=0$. This contradicts the induction hypothesis. Thus, no $A_i$ equals $0$.

Similarly, if any coordinates are repeated, say $A_{n-1}=A_n$, then $(A_1,\dots,A_{n-1})$ is a nonzero solution of the system, $$ F_{n-1,1}(m_1,\dots,m_{n-2},m_{n-1}+m_n;X_1,\dots,X_{n-1}) = 0, \dots, $$ $$ F_{n-1,n-2}(m_1,\dots,m_{n-2},m_{n-1}+m_n;X_1,\dots,X_{n-1}) = 0,$$ and which also solves the equation $$F_{n-1,n-1}(m_1,\dots,m_{n-2},m_{n-1}+m_n;X_1,\dots,X_{n-1})=0.$$ Again this contradicts the induction hypothesis. Thus, there are no repeated coordinates.

Therefore, let $(A_1,\dots,A_n)$ be an $n$-tuple with no zero coordinates and no repeated coordinates. Then, by the theory of the Vandermonde determinant, there is only the trivial solution $(Y_1,\dots,Y_n)$ of the following linear system, $$ \left\{ \begin{array}{rrrrrrr} Y_1A_1 & + & \dots & + & Y_nA_n & = & 0, \\ Y_1A_1^2 & + & \dots & + & Y_nA_n^2 & = & 0, \\ \vdots & & \ddots & & \vdots & & \vdots \\ Y_1A_i^r & + & \dots & + & Y_nA_n^r & = & 0, \\ \vdots & & \ddots & & \vdots & & \vdots \\ Y_1A_1^n & + & \dots & + & Y_nA_n^n & = & 0. \end{array} \right.$$ Indeed, the determinant of the coefficient matrix is $$ \text{det}([A_s^r]_{1\leq r,s \leq n}) = \pm A_1\cdots A_n \prod_{k<l}(A_l-A_k), $$ which is nonzero since all $A_i$ are nonzero and distinct. Therefore, if $(m_1,\dots,m_n)$ is a solution, then all $m_i$ equal zero. But this contradicts the hypothesis that all $m_i$ lie in $P$, which does not contain $0$. Therefore, for $(A_1,\dots,A_n)$ as above, there is no choice of $(m_1,\dots,m_n)$ in $P$ such that all $F_{n,1} = \dots = F_{n,n-1}=F_{n,n}=0$. Therefore, if $(A_1,\dots,A_n)$ is as above and solves $F_{n,1}=\dots=F_{n,n-1}=0$, then it does not solve $F_{n,n}=0$. Therefore Property $Q_n$ holds. Therefore, for every integer $n\geq 2$, Property $Q_n$ holds by induction on $n$.

End Proof of Proposition $Q_n$.

Combinining the proposition with Francesco's homogenization argument, there are $n!$ solutions of the original system, when counted with multiplicity. The point is that, for Francesco's homogeneous system, $x_0$ cannot equal $0$, or that would force at least one of the coordinates $A_i$ to be zero, contradicting the proposition.

But, in fact, the multiplicity is always $1$. If you form the Jacobian matrix of Francesco's system evaluated at a solution $(A_0,A_1,\dots,A_n)$, it is the following $n\times(n+1)$-matrix, $$ \left[ \begin{array}{rr} -nA_0^{n-1} & * & * & \dots & * \\ 0 & m_1 & m_2 & \dots & m_n \\ 0 & 2m_1A_1 & 2m_2A_2 & \dots & 2m_n A_n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & rm_1A_1^{r-1} & rm_2A_2^{r-1} & \dots & rm_nA_n^{r-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & (n-1)m_1A_1^{n-2} & (n-1)m_2A_2^{n-2} & \dots & (n-1)m_nA_n^{n-2} \end{array} \right]. $$ If you delete any one of the final $n$ columns, then the determinant of the remaining square matrix is, $$ \pm n!m_1\cdots m_n A_0^{n-1}\prod_{k<l} (A'_l-A'_k), $$ where $(A'_1,\dots,A'_{n-1})$ is what you get from $(A_1,\dots,A_n)$ by removing the entry corresponding to the deleted column. By the proposition, this determinant is nonzero. Thus the Jacobian matrix has full rank $n$ at every solution of Francesco's system. Therefore every solution has multiplicity $1$.

In summary, for every choice of $(m_1,\dots,m_n)$ in $P$, there are precisely $n!$ solutions of the original system.

Edit. In fact, the argument above holds so long as $(m_1,\dots,m_n)$ is a sequence of complex numbers such that no "subset sum" equals $0$, i.e., $m_{i_1} + \dots + m_{i_q} \neq 0$ for every $1\leq i_1 < \dots < i_q \leq n$. Since Francesco's homogeneous system defines a closed subscheme $Z$ of the product of projective spaces $\mathbb{P}^{n-1} \times \mathbb{P}^n$ with homogeneous coordinates $([m_1,\dots,m_n],[X_0,X_1,\dots,X_n])$, and since the projection $\pi_1:Z\to \mathbb{P}^{n-1}$ is finite, étale of degree $n!$ on the complement of the union of hyperplanes $Z(m_{i_1} + \dots + m_{i_q})$, then presumably the branch locus is precisely this union of hyperplanes.

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    $\begingroup$ Umm, a most excellent answer, but why is this community wiki? $\endgroup$
    – Igor Rivin
    Nov 27, 2013 at 17:31
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    $\begingroup$ @IgorRivin: "Why is this community wiki?" Why not? $\endgroup$ Nov 27, 2013 at 18:45
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    $\begingroup$ Well, it's obviously your choice, but it seems that this does not fit into any of the community wiki categories (it is a thorough answer to a non-CW question). So, I was wondering if you have some deep reason (like "reputation points are the root of all evil"). $\endgroup$
    – Igor Rivin
    Nov 27, 2013 at 21:33
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    $\begingroup$ @IgorRivin: Yes, reputation points are the root of all evil :) $\endgroup$ Nov 27, 2013 at 22:33

This is not a complete solution but it seems to me that it can be a good starting point. I will prove that the system has $n!$ distinct solutions for a general choice of the coefficients $m_i$.

All your equations are homogeneous in the $x_i$, except the last one. Homogenize adding a variable $x_0$; then all equations remain unchanged except for the last one, which becomes $$x_1x_2 \cdots x_n-x_0^n=0.$$ Now you can see the solutions of your system as the intersections of $n$ hypersurfaces in $\mathbb{P}^n(\mathbb{C})$, with homogeneous coordinates $[x_0: \ldots :x_n]$. If these hypersurfaces intersect transversally, then Bézout Theorem implies that there are only a finite number of solutions in the projective space, and this number is the product of the degrees of the equations, namely $1\cdot 2\cdot 3 \cdots \ n= n!$.

On the other hand, for $m_1=m_2 = \ldots =m_n=1$ you find exactly $n!$ solutions. So you can conclude that for a general choice of the coefficients $(m_1, \ldots, m_n) \in \mathbb{C}^n$ your system has exactly $n!$ distinct solutions in $\mathbb{P}^n(\mathbb{C})$.

Finally, these solution will be solutions of the original system if and only if all of them are outside the hyperplane at infinity $x_0=0$. This means that for any solution all the coordinates $x_1,\ldots, x_n$ must be nonzero. Again, you know that this happens in the case $m_1=m_2 = \ldots =m_n=1.$

So we can conclude that your system has exactly $n!$ solutions for a general choice of the coefficients $(m_1, \ldots, m_n) \in \mathbb{C}^n$, where general means that the coefficients can be chosen outside a Zariski closed subset $Z$ of $\mathbb{C}^n$.

In particular the general $(m_1, \ldots, m_n) \in \mathbb{N}^n$ does not belong to $Z$, hence for a general choice of the positive integers $(m_1, \ldots, m_n)$ your system has precisely $n!$ distinct solutions.

  • $\begingroup$ Why is it true that the general $(m_1, \dots, m_n)$ do not belong to $Z?$ Is it a general fact that $\mathbb{Z}^n$ is Zariski dense? $\endgroup$
    – Igor Rivin
    Nov 27, 2013 at 17:04
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    $\begingroup$ Yes, $\mathbb Z^n$ is Zariski dense. This can be seen by induction and the fact that $\mathbb Z$ (as an infinite set) is not Zariski dense in $\mathbb C$. $\endgroup$ Nov 27, 2013 at 17:14
  • $\begingroup$ @PeterMueller: In final sentence, "is not Zariski dense " --> "is Zariski dense". $\endgroup$ Nov 27, 2013 at 18:11
  • $\begingroup$ @Jason Starr: Yes, of course, thanks. Too silly that one cannot edit comments. Deleting and writing a new one isn't a solution unless the order doesn't matter. $\endgroup$ Nov 27, 2013 at 18:38

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