Răspuns:
[tex](x,y)\in\{(2,-2),(3,-1),(1,0)\}[/tex]
Explicație pas cu pas:
[tex]x=\dfrac{2+5y}{2+3y},\:y\ne -2/3\:(y\in\Bbb{Z})[/tex].
[tex]x\in\Bbb{Z}, y\in\Bbb{Z}\implies (2+3y)\mid (2+5y)[/tex].
[tex]2+3y\mid 2+5y\implies 2+3y\mid 3(2+5y)=6+15y\\2+3y\mid 2+3y\implies 2+3y\mid 5(2+3y)=10+15y\\[/tex].
Scazand, obtinem: [tex]2+3y\mid 10+15y-6-15y=4\iff 2+3y\mid 4[/tex].
In concluzie, [tex]2+3y\in\mathcal{D}_4=\{-4,-2,-1,1,2,4\}\iff y\in\{-2, -4/3, -1, -1/3, 0, 2/3\}[/tex].
Dar [tex]y\in\Bbb{Z}\implies y\in\{-2,-1,0\}[/tex].
Cazul 1. [tex]y=-2[/tex]. Ecuatia devine: [tex]8-4x=0\implies x=2\in\Bbb{Z}[/tex]. O solutie este, deci, [tex](x_1, y_1)=(2,-2)[/tex].
Cazul 2. [tex]y=-1[/tex]. Ecuatia devine: [tex]3-x=0\implies x=3\in\Bbb{Z}[/tex]. O solutie este, deci, [tex](x_1, y_1)=(3,-1)[/tex].
Cazul 3. [tex]y=0[/tex]. Ecuatia devine: [tex]x-1=0\implies x=1\in\Bbb{Z}[/tex]. O solutie este, deci, [tex](x_1, y_1)=(1,0)[/tex].
Solutii: {(2,-2),(3,-1),(1,0)}.
