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This is a continuation of my previous question. Suppose $Q_1,..., Q_n$ are polynomials in $x_1,x_2,...,x_{2n-1}, x_{2n}$ such that $$\tag{1}Q_1 x_1+Q_2 x_3+\cdots +Q_n x_{2n-1}=0,\\ Q_1 x_2+Q_2 x_4+\cdots +Q_n x_{2n}=0$$ with the extra condition that $$\tag{2}Q_i(x_1,...,\underbrace{0}_{(2j-1)-\mbox{th}},\underbrace{0}_{2j-\mbox{th}},...,x_{2n})=0\mbox{ for all }j\neq i.$$ I wonder if it is always true that all $Q_1, \ldots, Q_{n}$ vanishes.

As shown in my previous question, it is true for $n=1,2$ without the extra condition $(2)$. In the answer, a counterexample is constructed without condition $(2)$.

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    What is the context in which you encounter this problem? or as a generalization to what? – 2017-01-06
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    Condition $(2)$ seems a little weird to me. What is the intuition behind of it? – 2017-01-06

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Take the counterexample from the previous question. Multiply each $Q_i$ by the monomial $\prod_{j=1}^{2n} x_j$. (Or just the even indices $\prod_{j=1}^n x_{2j}$. They now satisfy condition (2).

It seems like you only took the counterexample at face value and haven't given much thought to the general line of argument, which shows that counterexamples are plentiful because you're dealing with an underdetermined system of 2 equations in $n$ unknowns.

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    I would guess that you have a particular application in mind already that is the source of condition (2). Why don't you just share the original motivation instead of asking variants of the same question? – 2017-01-06
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    You are right. I am thinking of the question of solving the system of first order PDE, which reduces to proving some properties of polynomials. I will share it in the new question. Thanks. – 2017-01-06