`\qquad x^2+3xy-y=7`
`<=>x^2-7+3xy-y=0`
`<=>x^2-7+y(3x-1)=0`
`<=>x^2-7=-y(3x-1)`
`=>y={-(x^2-7)}/{3x-1}`
Vì `y\in Z=> {-(x^2-7)}/{3x-1}\in Z`
`=>(x^2-7)\ \vdots\ (3x-1)`
`=>3(x^2-7)\ \vdots\ (3x-1)`
`=>3x^2-x+x-21\ \vdots\ (3x-1)`
`=>x(3x-1)+x-21\ \vdots\ (3x-1)`
Vì `x(3x-1)\ \vdots\ (3x-1)`
`=>(x-21)\ \vdots\ (3x-1)`
`=>3(x-21)\ \vdots\ (3x-1)`
`=>3x-1-62\ \vdots\ (3x-1)`
Vì `(3x-1)\ \vdots\ (3x-1)`
`=>62\ \vdots\ (3x-1)`
`=>(3x-1)\in Ư(62)={-62;-31;-2;-1;1;2;31;62}`
`=>3x\in {-61;-30;-1;0;2;3;32;63}`
`=>x\in{{-61}/3;-10;{-1}/3;0;2/ 3; 1;{31}/3;21}`
Vì `x\in Z=>x\in {-10;0;1;21}`
Ta có bảng sau:
$\begin{array}{|c|c|c|}\hline x&-10&0&1&21\\\hline y=\dfrac {-(x^2-7)}{3x-1}&3&-7&3&-7\\\hline\end{array}$
Vậy cặp số nguyên `(x;y)` thỏa đề bài là $(-10;3);(0;-7);(1;3);(21;-7)$
