3A. Determine the integer values of the parameter $a$ for which the solutions of the system

$$
\left\{\begin{array}{l}
a(x+y-1)=x-2 y-1 \\
a(x-y-3)=x+2 y-3
\end{array}\right.
$$

are integers.
Solution. The given system is equivalent to the system:

$$
\left\{\begin{array}{l}
(a-1) x+(a+2) y=a-1 \\
(a-1) x-(a+2) y=3 a-3
\end{array}\right.
$$

i.e., to the system

$$
\left\{\begin{array}{l}
2(a-1) x=4 a-4 \\
2(a+2) y=-2 a+2
\end{array}\right.
$$

If $a \neq 1$, then $x=2$, and if $a=1$, then $x \in \mathbb{Z}$.

If $a=-2$, there is no solution for $y$, and thus no solution for the system. If $a \neq-2$, we have $y=\frac{-a+1}{a+2}=-1+\frac{3}{a+2}$.

Since the solutions must be integers, it follows that $(a+2) \mid 3$, so $a+2 \in\{ \pm 1, \pm 3\}$, from which it follows that $a \in\{-5,-3,-1,1\}$. We obtain the required solutions as:

$$
\begin{array}{ll}
a=-5: & x=2, y=-2 \\
a=-3: & x=2, y=-4 \\
a=-1: & x=2, y=2 \\
a=1: & x \in \mathbb{Z}, y=0
\end{array}
$$