Đáp án: $ (x,y)\in\{(0,3), (-2,5)\}$
Giải thích các bước giải:
Ta có:
$x^2(y-5)+x+y-3=0$
$\to x^2(y-5)+x+2+(y-5)=0$
$\to x^2(y-5)+(y-5)=-(x+2)$
$\to (x^2+1)(y-5)=-(x+2)$
Vì $x,y\in Z$
$\to x+2\quad\vdots\quad x^2+1$
$\to (x+2)(x-2)\quad\vdots\quad x^2+1$
$\to x^2-4\quad\vdots\quad x^2+1$
$\to x^2+1-5\quad\vdots\quad x^2+1$
$\to 5\quad\vdots\quad x^2+1$
$\to x^2+1\in U(5)$
$\to x^2+1\in\{1,5\}$
$\to x^2\in\{0,4\}$
$\to x\in\{0,2,-2\}$
Lại có $(x^2+1)(y-5)=-(x+2)\to y=\dfrac{-(x+2)}{x^2+1}+5$
$\to y\in\{3, \dfrac{21}{5}, 5\}$
Do $x,y\in Z\to (x,y)\in\{(0,3), (-2,5)\}$
