$y(x-1)=x^{2}+2$
⇔$y(x-1)-(x^{2}-1)=3$
⇔$y(x-1)-(x-1)(x+1)=3$
⇔$(x-1)(y-x-1)=3$
Ta thấy: 3=1.3=3.1=(-1)(-3)=(-3)(-1)
⇒ Xảy ra 4 TH:
TH1:$\left \{ {{x-1=1} \atop {y-x-1=3}} \right.$ ⇔$\left \{ {{x=2} \atop {y=6}} \right.$
TH2:$\left \{ {{x-1=3} \atop {y-x-1=1}} \right.$ ⇔$\left \{ {{x=4} \atop {y=6}} \right.$
TH3:$\left \{ {{x-1=-1} \atop {y-x-1=-3}} \right.$ ⇔$\left \{ {{x=0} \atop {y=-2}} \right.$
TH4:$\left \{ {{x-1=-3} \atop {y-x-1=-1}} \right.$ ⇔$\left \{ {{x=-2} \atop {y=-2}} \right.$
Vậy phương trình có nghiệm: $(x;y)∈${$(2;6);(4;6);(0;-2);(-2;-2)$}
