Q:

Number of $x \in [0,1]^n$ such that Ax integer

If $A$ is an integer matrix in $\mathbb{Z}^{m \times n}$, why is there only a finite number of vectors $x \in [0,1]^n$ such that $Ax$ is a vector of integers?

A:

The claim is true if and only if $\ker A=\{0\}$.

Let $\ker A=\{0\}$. Then necessarily $m\ge n$ and there is $B\in \mathbb R^{n,m}$ such that $BA=I_n$.
Since multiplication with $A$ is a linear and bounded map, the image $A([0,1]^n)$ of $[0,1]^n$ is bounded, so $A([0,1]^n) \cap \mathbb Z^m$ is a finite set.
Then the set of vectors $x\in [0,1]^n$ with $Ax\in \mathbb Z^m$ is precisely
$$
B \left( A([0,1]^n) \cap \mathbb Z^m \right),
$$
which is finite as well.
Let now $y \in \ker A$, $y\ne 0$. Since $A$ is an integer matrix, it holds $Ax\in \mathbb Z^m$ for all $x\in \{0,1\}^n$.
Now define $x$ as 
$$
x_i = \begin{cases} 1 & \text{ if } y_i<0,\\
0 & \text{ if } y_i\ge0.
\end{cases}$$
Then for all $t\ge0$ small enough, $x+ty\in [0,1]^n$ and $A(x+ty)=Ax\in \mathbb Z^m$.

