Đáp án:
`(x;y)\in {(1;-3);(-1;-1);(1;0);(-1;2)}`
Giải thích các bước giải:
`\qquad (2x+y)(y+1)=x+1`
`<=>2xy+2x+y^2+y=x+1`
`<=>y^2+2xy+2x-1=x-y`
`<=>2xy+x=-y^2-y+1`
`<=>(2y+1)x=-y^2-y+1`
`<=>x={-y^2-y+1}/{2y+1}` `(y\in ZZ=>2y+1\ne 0)`
Để `x\in ZZ`
`=>{-y^2-y+1}/{2y+1}\in ZZ`
`=>(-y^2-y+1)\ \vdots\ (2y+1)`
`=>(2y^2+2y-2) \vdots\ (2y+1)`
`=>y(2y+1)+y-2 \vdots\ (2y+1)`
Vì `y(2y+1) \vdots\ (2y+1)`
`=>(y-2) \vdots\ (2y+1)`
`=>(2y-4) \vdots\ (2y+1)`
`=>(2y+1-5) \vdots\ (2y+1)`
`=>5 \vdots\ (2y+1)`
`=>(2y+1)\in Ư(5)={-5;-1;1;5}`
`=>2y\in {-6;-2;0;4}`
`=>y\in {-3;-1;0;2}`
$\\$
`\qquad x={-y^2-y+1}/{2y+1}`
+) `TH: y=-3=>x={-9+3+1}/{-6+1}=1`
+) `TH: y=-1=>x={-1+1+1}/{-2+1}=-1`
+) `TH: y=0=>x={-0-0+1}/{0+1}=1`
+) `TH: y=2=>x={-4-2+1}/{4+1}=-1`
$\\$
Vậy các cặp số nguyên `(x;y)` thỏa đề bài là:
`(1;-3);(-1;-1);(1;0);(-1;2)`
