Given equation over $\mathbb F_2$: $$x_1x_2x_3+x_4x_5x_6=0$$
It has $50$ solutions. Let $N$ be set of solutions. If we give some linear dependencies to variables we will get cosets (linear subspace + a vector) that are inside $N$. For example $x_1=0,\ x_4+x_5=1$ will give coset of dimension $4$ that lies on $N$. I have checked using computer that there is not coset of dimension $4$ that is in $N$ and contains $111111$ vector. I need to prove it formally.
Any help for proving this or some books, articles about such problems will be appreciated.
Assume that such a 4-dimensional vector space $V$ exists. Consider the 3-dimensional subspace $U\subset \Bbb{F}_2^6$ consisting of those vectors whose three first coordinates all vanish. Because $\dim V+\dim U=4+3>6$ these two subspaces intersect non-trivially. Thus there exists a non-zero vector $v$ of the form $$ v=(0,0,0,v_4,v_5,v_6)\in V\cap U. $$ Check what happens at the point $(1,1,1,1,1,1)+v$.