Có: `a= x^2 + xy - 2012x - 2013y - 2014`

Lại có: `a=0`

 `=> x^2 + xy + x - 2013x - 2013y - 2014=0`

`x^2 + xy + x - 2013x - 2013y - 2013-1=0`

`(x^2 + xy + x) - (2013x - 2013y - 2013)=1`

`x(x+y+1) - 2013(x+y+1)=1`

`(x+y+1)(x-2013)=1`

Vì `x,y in Z` nên `(x+y+1); (x-2013) in Z`

`=> (x+y+1); (x-2013) in Ư(1)`

`=> (x+y+1); (x-2013) in {+-1}`

Ta có bảng sau:

\begin{array}{|c|c|c|}\hline \text{x-2013}&\text{-1}&\text{1}\\\hline \text{x+y+1}&\text{1}&\text{-1}\\\hline \text{x}&\text{2012}&\text{2014}\\\hline \text{y}&\text{-2012}&\text{-2016}\\\hline\end{array}

Vậy các cặp số nguyên `(x;y)` tm là : `(2012;-2012) ; (2014;-2016)`