`\qquad yx^2+xy-x+y=3`
`<=>yx^2+xy+y=x+3`
`<=>y(x^2+x+1)=x+3`
`=>y={x+3}/{x^2+x+1}`
Vì `y\in Z=>(x+3)\ \vdots \ (x^2+x+1)`
Vì `x\in Z=>x-2\in Z`
`=>(x-2)(x+3)\ \vdots \ (x^2+x+1)`
`=>(x^2+3x-2x-6)\ \vdots \ (x^2+x+1)`
`=>(x^2+x+1-7)\ \vdots \ (x^2+x+1)`
$\\$
Do `(x^2+x+1)\ \vdots \ (x^2+x+1)`
`=>7\ \vdots \ (x^2+x+1)`
`=>x^2+x+1\in Ư(7)={-7;-1;1;7}`
Ta có:
`x^2+x+1=x^2+2.x. 1/ 2 + 1/ 4 +3/ 4`
`=(x+ 1/ 2)^2+3/ 4\ge 3/ 4>0` với mọi $x$
$\\$
`=>x^2+x+1\in {1;7}`
+) Nếu `x^2+x+1=1`
`<=>x^2+x=0`
`<=>x(x+1)=0`
$⇔\left[\begin{array}{l}x=0\\x+1=0\end{array}\right.$ $⇔\left[\begin{array}{l}x=0\\x=-1\end{array}\right.$
$\\$
`y={x+3}/{x^2+x+1}`
++) $TH: x=0$
`=>y={0+3}/{0^2+0+1}=3` (TM)
`=>(x;y)=(0;3)`
$\\$
++) $TH: x=-1$
`=>y={-1+3}/{(-1)^2-1+1}=2` (TM)
`=>(x;y)=(-1;2)`
$\\$
+) Nếu `x^2+x+1=7`
`<=>x^2+x-6=0`
`<=>x^2-2x+3x-6=0`
`<=>x(x-2)+3(x-2)=0`
`<=>(x-2)(x+3)=0`
$⇔\left[\begin{array}{l}x-2=0\\x+3=0\end{array}\right.$ $⇔\left[\begin{array}{l}x=2\\x=-3\end{array}\right.$
$\\$
`y={x+3}/{x^2+x+1}`
++) $TH: x=2$
`=>y={2+3}/{2^2+2+1}=5/ 7` (loại)
$\\$
++) $TH: x=-3$
`=>y={-3+3}/{(-3)^2-3+1}=0` (TM)
`=>(x;y)=(-3;0)`
$\\$
Vậy nghiệm nguyên của phương trình là:
`(x;y)\in {(0;3);(-1;2);(-3;0)}`
