tìm pt nghiệm nguyên x^2 +x(y-2)+3-y=0

Giải thích các bước giải:

Ta có:
$x^2+x(y-2)+3-y=0$

$\to x^2+xy-2x+3-y=0$

$\to x^2-2x+3=y-xy$

$\to 1-2x+x^2+2=y(1-x)$

$\to (1-x)^2+2=y(1-x)$ 

Vì $x,y\in Z\to 1-x\in Z$

$\to y(1-x)\quad\vdots\quad 1-x$

$\to (1-x)^2+2\quad\vdots\quad 1-x$

$\to 2\quad\vdots\quad 1-x$

$\to 1-x\in U(2)$

$\to 1-x\in\{1, 2, -1, -2\}$

$\to x\in\{0, 2, -1, 3\}$

Ta có: $(1-x)^2+2=y(1-x)$ 

$\to y=\dfrac{(1-x)^2+2}{1-x}$

$\to y\in\{3, -3, 3, -3\}$

$\to (x,y)\in\{(0, 3), (2, -3), (-1, 3), (3, -3)\}$