13. Giải hệ ta có [tex]\left\{\begin{matrix} x=\frac{2m^2+5m-1}{2m^2+m+1}\\ y=-\frac{2(2m^2+m)}{2m^2+m+1} \end{matrix}\right.[/tex]
Để y nguyên thì [tex]2(2m^2+m)\vdots 2m^2+m+1\Leftrightarrow 2(2m^2+m+1)-2\vdots 2m^2+m+1\Rightarrow 2\vdots 2m^2+m+1\Rightarrow 2m^2+m+1\in \left \{ 1;2 \right \}[/tex]
+ [tex]2m^2+m+1=1\Rightarrow m(2m+1)=0\Rightarrow m=0[/tex]
+ [tex]2m^2+m+1=2\Rightarrow 2m^2+m-1=0\Rightarrow (m+1)(2m-1)=0\Rightarrow m=-1[/tex]