Proving the existence of integer solutions to linear inequalities Let $b_k\in\mathbb{Z}^n$ for $1\le k\le m$ for some $m,n$. 
I wish to prove the existence of two vectors $x,y\in\mathbb{Z}^n$ such that for all $k$, $b_k\cdot x\ne 0$ and $b_k\cdot y\ne 0$ with $\cdot$ being the usual scalar product.
Moreover, I wish to find $a\in \{0,1\}^n$ such that $a\cdot x=0$ and $a\cdot y\ne 0$, in the case where $m < 2^n-2$.
So far I only thought of very hand-wavy proofs without a good insight. I hope that there's some systematic way to deal with such questions.
 A: You obviously need to assume the $b_k$ are nonzero; I suppose this was an oversight. The question does not change if you replace some $b_k$ by a scalar multiple that lies in $\def\Z{\Bbb Z}\Z^n$, so I'll suppose in addition that for every $k$ the coordinates of $b_k$ have no nontrivial common factor to all of them at once (divide by their $\gcd$ to obtain this).
Here is one way to show that the union of finitely many hyperplanes $\{\, x\in \Z^n \mid x\cdot b_k=0\,\}$ cannot be all of$~\Z^n$, if $n>0$. Let $p$ be a prime larger than the number of hyperplanes considered. Let $f:\Z^n\to(\Z/p\Z)^n$ be defined by reduction of all coordinates modulo$~p$. The image of the hyperplane defined by $b_k$ is contained in $\{\, x\in (\Z/p\Z)^n \mid x\cdot f(b_k)=0\,\}$; since $f(b_k)\neq0$ due to the rescaling of the $b_k$ this is a hyperplane in $(\Z/p\Z)^n$, and has $p^{n-1}$ elements. By the choice of$~p$, the union of the images of all hyperplanes has less than $p\times p^{n-1}=p^n$ elements, so it does not fill all of $(\Z/p\Z)^n$. Taking pre-images by the surjective map$~f$, the hyperplanes did not fill up all of$~\Z^n$. (Taking $p$ prime is not really necessary, but it facilitates the counting.)
Now for answer to the question. The condition $m<2^n-2$ forces $n>1$. Take $a\neq0$ an element of $\{0,1\}^n$ that does not occur among the $b_k$ after rescaling, which is possible since $m\leq 2^n-2$. Now $y$ has to be chosen in the complement of the $m+1$ hyperplanes defined by the $b_k$ and $a$, and this is possible by the above. As for $x$, it has to be chosen inside the hyperplane$~H$ defined by $a$, which is of dimension $n-1>0$, and within$~H$ in the complement of the $m$ hyperplanes of$~H$ defined by the $b_k$; since none of the $b_k$ is (a scalar multiple of) $a$ these are indeed hyperplanes of$~H$. As a $\Z$-module $H\cong\Z^{n-1}$, and images of the hyperplanes of$~H$ are defined by appropriate vectors of$~\Z^{n-1}$. Therefore the above again allows to conclude that $x$ can be chosen. There are no conditions that make the choices of $x$ and $y$ dependent in any way, once $a$ is fixed.
