I am working on system of quadratic equations.
\begin{cases} (\alpha_1^1x_1+\ldots+\alpha_n^1x_n)(\beta_1^1y_1+\ldots+\beta_m^1y_m)=0\\ \ldots \\ (\alpha_1^kx_1+\ldots+\alpha_n^kx_n)(\beta_1^ky_1+\ldots+\beta_m^ky_m)=0\\ \end{cases}
where $\alpha_i^j$ and $\beta_i^j$ are constants in $\mathbb F_2=\{0,1\}$ and $x_i,y_j$ are unknowns in $\mathbb F_2$.
I need help on these problems:
1) If I get another equation of same form $(\alpha_1^0x_1+\ldots+\alpha_n^0x_n)(\beta_1^0y_1+\ldots+\beta_m^0y_m)=0$, how can I check that if I add it to the system it will change number of solutions. If the equations were linear I can check if it is linearly dependent from vectors in the system. But for quadratic case I have no idea how to do it.
2) What is minimal number of equations one must choose so that the solution of system is trivial. By trivial solution I mean all vectors that first $n$ or last $m$ coordinates are 0. For linear case again it is well known that if number of variables is $n$, one should choose at lease $n$ equations to get only trivial solution.
Update:
I understood that in order to have only trivial solution one must choose at least $n+m-1$ equations and for $n=m=2$ I found a system with $3$ equation which doesn't have nontrivial solutions.
\begin{cases} x_1y_1=0\\ x_2y_2=0\\ (x_1+x_2)(y_1+y_2)=0\\ \end{cases}
But $n+m-1$ bound is not tight. For $n=3$ and $m=2$ I have found only $5$ equation system. Solution with $4$ equations doesn't exist.
Finding exact number of equations for general case is hard problem so I am trying to find it only for the case $m=2$.
Looks like a system with $\lceil \frac{3n}{2}\rceil$ equation is enough but I can's show that solution with lower number of equations doesn't exist. I am giving bounty to anyone who can help on that problem.