`x+xy+y=1`

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

`(y+1)(x+1)=2`

Mà: `x,y∈Z`

Nên: `x+1,y+1∈Ư(2)={-2;-1;1;2}`

Ta có bảng giá trị sau:

\begin{array}{|c|c|c|}
\hline \text{x+1}&\text{-2}&\text{-1}&\text{1}&\text{2}\\
\hline \text{y+1}&\text{-1}&\text{-2}&\text{2}&\text{1}\\
\hline \text{x}&\text{-3}&\text{-2}&\text{0}&\text{1}\\
\hline \text{y}&\text{-2}&\text{-3}&\text{1}&\text{0}\\\hline\end{array}

Vậy các cặp số nguyên `(x;y)` cần tìm là:

`(-3;-2),(-2;-3),(0;1),(1;0)`