Răspuns :

[tex]2x-3y=4=>x= \frac{4+3y}{2} \in N=>y=nr.par=2k\\ (x-2)(y+2)=( \frac{4+3y}{2}-2)(y+2)= \frac{3y}{2} (y+2)= \frac{3 \cdot 2k}{2} (2k+2)=\\ =3k\cdot2(k+1)=6k(k+1)=>(x-2)(y+2)\vdots6[/tex]