`x;y\in Z`
`\qquad x^2-2xy+3x-y=20`
`<=>x^2+3x-20=2xy+y`
`<=>x^2+3x-20=y(2x+1)`
`<=>y={x^2+3x-20}/{2x+1}`
`y\in Z=>{x^2+3x-20}/{2x+1}\in Z`
`=>{2(x^2+3x-20)}/{2x+1}\in Z`
`=>{2x^2+x+5x-40}/{2x+1}\inZ`
`=>{x(2x+1)+5x-40}/{2x+1}\in Z`
`=>x+{5x-40}/{2x+1}\in Z`
Vì `x\in Z=>{5x-40}/{2x+1}\in Z`
`=>{10x-80}/{2x+1}\inZ`
`=>{5(2x+1)-85}/{2x+1}\in Z`
`=>5- {85}/{2x+1}\in Z`
`=>{85}/{2x+1}\in Z`
`=>(2x+1)\in Ư(85)={-85;-17;-5;-1;1;5;17;85}`
`=>2x\in {-86;-18;-6;-2;0;4;16;84}`
`=>x\in {-43;-9;-3;-1;0;2;8;42}`
Ta có bảng sau:
$\begin{array}{|c|c|c|}\hline x&-43&-9&-3&-1&0&2&8&42\\\hline y=\dfrac{x^2+3x-20}{2x+1}&-20&-2&4&22&-20&-2&4&22\\\hline\end{array}$
Vậy các cặp giá trị $(x;y)$ nguyên thỏa đề bài là `(-43;-20);(-9;-2);(-3;4);(-1;22);(0;-20);(2;-2);(8;4);(42;22)`
