$\left \{ {{mx + y = 2m} \atop {x + my = m +1}} \right.$
⇔ $\left \{ {{m²x + my = 2m²} \atop {x + my = m +1}} \right.$
⇔ $\left \{ {{m²x -x = 2m²-m-1} \atop {x + my = m +1}} \right.$
⇔ $\left \{ {{x=\frac{2m²-m-1}{m²-1}=\frac{2(m-1).(m+0,5)}{(m-1).(m+1)}=\frac{2m+1}{m+1}} \atop {y=2m-mx}} \right.$
⇔ $\left \{ {{x=\frac{2m+1}{m+1}} \atop {y=2m-m.\frac{2m+1}{m+1}}} \right.$
⇔ $\left \{ {{x=\frac{2m+1}{m+1}} \atop {y=\frac{m}{m+1}}} \right.$
Để $x$, $y∈Z$ thì $2m+1⋮m+1$
⇔ $2.(m+1)-1⋮m+1$
⇒ $m+1∈Ư(1)={-1;1}$
⇔ $m∈{-1;0}$
và $m⋮m+1$
⇔ $m+1-1⋮m+1$
⇒ $m+1∈Ư(1)={-1;1}$
⇔ $m∈{-1;0}$
Để hpt có 2 nghiệm phân biệt thì: $m\neq\frac{1}{m}$
⇔ $m\neq±1$
⇒ $m=0$
