Let's consider the following diophantine system:
$$(\mathscr S)\quad\begin{cases} 2x+y\in a\mathbb{Z} \\ x-2y\in a\mathbb{Z},\end{cases}$$
where $a\in\mathbb Z$ is a large integer.
We know that $x,y$ are integers such that
$$x,y\in\{0,\ldots,a\}\qquad(\star)$$
and such that
$$(x,y)\notin\{(0,0),(0,a),(a,0),(a,a)\}.\qquad(\star\star)$$
The question.
Can we show that $(\mathscr S)$ has no such integers solutions?
What I tried.
We can rewrite the system with matrices:
$$MX=aK$$
where
$$M=\begin{pmatrix} 2 & 1 \\ 1 & -2\end{pmatrix},$$
$$X=\begin{pmatrix} x \\ y\end{pmatrix},$$
and
$$K=\begin{pmatrix} k \\ k'\end{pmatrix}$$
where $k\in\{0,1,2\}$ and $k'\in\{-1,0\}$ since $(\star)$ and $(\star\star)$.
We have $\det M=-5$, but I don't know how to proceed from here...
Any help or leads would be much appreciated.
If you add twice the first to the second, then subtract twice the second from the first, you get $$(\mathscr S)\quad\begin{cases} 5x\in a\mathbb{Z} \\ 5y\in a\mathbb{Z},\end{cases}$$ We can solve it for $a=5,$ finding $$x=1,\ y=3\\ x=2,\ y=1\\ x=3,\ y=4\\ x=4,\ y=2$$ as a solutions. For $a$ a multiple of $5$ you can multiply these solutions by $\frac a5$. If $a$ is not a multiple of $5$ the first lines show there is no solution.