$ \left\{\begin{array}{l} mx +2y = m +1\\ 2x +my = 2m -1\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} 2mx +4y = 2m +2\\ 2mx +m^2y = 2m^2 -m\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} (m^2-4)y=2m^2-3m-2\\ 2mx +m^2y = 2m^2 -m\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} y=\dfrac{2m^2-3m-2}{m^2-4}\\ x=\dfrac{2m-1-my}{2}\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} y=\dfrac{(m-2)(2m+1)}{(m-2)(m+2)}\\ x=\dfrac{2m-1-my}{2}\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} y=\dfrac{2m+1}{m+2}\\ x=\dfrac{m-1}{m+2}\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} y=\dfrac{2m+4-3}{m+2}\\ x=\dfrac{m+2-3}{m+2}\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} y=2-\dfrac{3}{m+2}\\ x=1-\dfrac{3}{m+2}\end{array} \right. $
Để phương trình có nghiệm $(x;y)$ nguyên
$\Rightarrow (m+2) \in Ư(3)\\ \Rightarrow m \in \{-5;-3;-1;1\}$
