Suppose that a linear system of inequalities $Ax \le b$, where $A\in Z^{m\times n}$ and $b\in Z^m$ have integral coefficients, has an infinite number of integral solutions $x$. 


Can one conclude that there is a ray $\{x+tp\mid t\ge 0\}$ containing infinitely many integral solutions? (If one drops the integrality condition, the answer is clearly yes.)



On linear integer inequalities with infinitely many solutions
Let the constraints be numbered 1 to $m$. Let the $i$th constraint be $a^{(i)}\cdot x\le b$. Let $S$ be the set of solutions. Inductively refine the constraints as follows: either there are $\limsup\_{x\in S, x\to\infty} a^{(1)}\cdot x=-\infty$ or not. If $\limsup\_{x\in S,x\to\infty} a^{(1)}\cdot x=c^{(1)}>-\infty$, then replace the system with the stronger constraint, $a^{(1)}\cdot x=c^{(1)}$. It still has infinitely many integer solutions. 


Now consider the second constraint. If the $\limsup$ is finite, replace it by an equality etc. 


In the end, you have a set of equalities, and a set of inequalities. For the inequalities, the inner products tend to $-\infty$ as the solution goes to $\infty$. In particular, you can find two solutions where for each equality, then inner product is the same, but for each inequality, the inner product of the first is larger than the inner product of the second. 


Call these $x$ and $y$. Now $x+n(y-x)$ is an infinite family of integer solutions satisfying all of the (stronger) constraints. 


