System of quadratic equations over field of size 2 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.

Let $K=\mathbb{F}_2$. There are $2^n+2^m-1$ trivial solutions ($(X,0)$ or $(0,Y)$). We consider the non-trivial solutions. Let $f:K^n\rightarrow K$ be random linear and $f_1$ be its restriction to $K^n\setminus\{0\}$. We assume that $X,Y$ are random. Then $prob(f(X)=1)=\dfrac{2^{n-1}}{2^n-1}\approx 1/2$ when $n$ is great enough. Note that each of our $k$ equation has the form $f_1(X)g_1(Y)=0$; thus $prob(f_1(X)g_1(Y)=0)\approx 3/4$. Here the total equation is $F=0$ where $F:K^{n}\setminus\{0\}\times K^{m}\setminus\{0\}\rightarrow K^k$. If $Z$ is the random variable "number of non-trivial solutions", then the average of $Z$ is $E(Z)\approx(\dfrac{3}{4})^k2^{n+m}$. Finally $E(Z)=1$ when $k=\dfrac{\ln(2)}{\ln(4/3)}(n+m)\approx 2.4094(n+m)$.
There are several methods for solving $F(X,Y)=0$. 
1 The brute force one. Test all possible value of $X,Y$; its complexity is  $2^{m+n}(m+n)k$.


*The decomposition in subsets of $S=\{1,\cdots,k\}$; if $T\subset S$, then assume $ f_i(X)=0$ when $i\in T$ and $g_j(X)=0$ when $j\notin T$. Its complexity is in  $O(2^k (\inf(k^3,n^3)+kn)(\inf(k^3,m^3)+km))$.

*Calculate a Grobner basis. Let $k=\lambda (n+m)$. When $\lambda \approx 2.4094$, the complexity is in $O(2^{n+m})$; when $k$ increases, the complexity quickly decreases.
Conclusion: when $\lambda<1$ use 2. When $1<\lambda<3$ use 1. When $\lambda>3$ use 3.
Anyway it seems that this problem is NP. It is known that to solve a system of equations of degree $2$ in $K$ is a NP problem (in the numbers of equations and unknowns).
