4
$\begingroup$

Suppose $Q_1,..., Q_n$ are polynomials in $x_1,x_2,...,x_{2n-1}, x_{2n}$ such that $$\tag{1}Q_1 x_1+Q_2 x_3+\cdots_+Q_n x_{2n-1}=0,\\ Q_1 x_2+Q_2 x_4+\cdots_+Q_n x_{2n}=0.$$ I wonder if it is always true that all $Q_1, \cdots, Q_{n}$ vanishes.

It is easy to prove the case when $n=1$. So I want to prove it by using induction, but I am stuck.

Edit: I can solve the case for $n=1,2$. For $n=1$, $(1)$ implies that $Q_1x_1=0$ and $Q_1x_2=0$, which implies that $Q_1=0$. For $n=2$, $(1)$ implies that $$Q_1x_1+Q_2x_3=0\mbox{ and }Q_1x_2+Q_2x_4=0.\tag{2}$$ As XTL pointed out, multiply the first equation by $x_2$ and multiply the second equation by $x_1$ and take the difference, we obtain $Q_2(x_2x_3-x_1x_4)=0$, which implies that $Q_2=0$. Put it back to $(2)$, we get $Q_1=0$.

  • 0
    Does it work for $n=2$? I think I see the line of reasoning for $n=1$ ($Q_1(x_1-x_2) = 0 \to Q_1 = 0$ if $x_1 \neq x_2$) but it doesn't seem to fall out so cleanly even for $n=2$? – 2017-01-03
  • 0
    Yes, maybe that's the reason why I am stuck in the induction. Maybe I should think about the case for $n=2$ first. – 2017-01-03
  • 2
    For $n=2$, it's very easy, multiply the first equation by $x_2$ and multiply the second equation by $x_1$. Finish by taking their difference. This trick does not work for $n=3$ – 2017-01-03

1 Answers 1

3

Why would that be true? For take $n\ge 3$, and $Q_3=\ldots = Q_n=1$. Then you can solve for $Q_1$ and $Q_2$ in the system (Cramer rule) to get them as rational fractions. Now multiply all the $Q_i$ by a common denominator.