TITLE: convex polytope integer points
QUESTION [3 upvotes]: is there a simple proof for the following lemma:
An unbounded convex polytope (defined by linear constraints) has either zero integer points or infinite many integer points.

REPLY [5 votes]: Edit: the following argument is true when the polyhedron $P$ is the intersection of finitely many linear constraints: $a_1 x_1 + \ldots + a_n x_n \leq b$.

The coefficient of linear constraints must be rational. Otherwise, your lemma will not be true (e.g. consider $\sqrt{2}x_1-\sqrt{3}x_2=0$).
Since your polyhedron $P \subset \mathbb{R}^n$ is unbounded and defined by rational linear constraints, it has a rational recession direction, i.e. 
there exists $ 0 \neq d \in \mathbb{Q}^n$ such that $x+\lambda d \in P $,
for all $x \in P$ and $ \lambda  \geq 0 $. 
Therefore, if $x_0\in P \cap \mathbb{Z}^n $ then $x_0 +\lambda d \in P \cap \mathbb{Z}^n $ for infinitely many values of $\lambda$.
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