Giải thích các bước giải:
`x^2 +xy-2016x-2017y-2018=0`
`<=>x^2+xy+x-2017x-2017y-2017-1=0`
`<=>x(x+y+1)-2017(x+y+1)=1`
`<=>(x-2017)(x+y+1)=1 (1)`
Vì x; y là số nguyên nên (1) `<=>`\(\left[ \begin{array}{l}\left \{ {{x-2017=1} \atop {x+y+1=1}} \right. \\\left \{ {{x-2017=-1} \atop {x+y+1=-1}} \right.\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}\left \{ {{x=2018} \atop {y=-2018}} \right. \\\left \{ {{x=2016} \atop {y=-2018}} \right.\end{array} \right.\)
Vậy `(x ; y) ∈ {(2018; -2018);(2016; -2018)}`