Đáp án: $ (x,y)\in\{(2,5), (3,3), (0,-1), (-1,-1)\}$
Giải thích các bước giải:
Ta có:
$xy-x^2-1=y$
$\to xy-y=x^2+1$
$\to y(x-1)=x^2+1$
$\to x^2+1\quad\vdots\quad x-1$ vì $x,y\in N$
$\to x^2-1+2\quad\vdots\quad x-1$
$\to (x-1)(x+1)+2\quad\vdots\quad x-1$
$\to2\quad\vdots\quad x-1$
$\to x-1\in\{1,2,-1,-2\}$
$\to x\in\{2,3,0,-1\}$
$\to y\in\{5, 3, -1,-1\}$ vì $y(x-1)=x^2+1\to y=\dfrac{x^2+1}{x-1}$
$\to (x,y)\in\{(2,5), (3,3), (0,-1), (-1,-1)\}$
