Đáp án: $(x,y)\in\{(-1, 14), (25, -12), (-2, -12), (-28, 14)\}$
Giải thích các bước giải:
Ta có:
$x^2+2xy+x+1+3y=15$
$\to x^2+2xy+x+3y=14$
$\to x^2+x+y(2x+3)=14$
$\to x^2+x-14=-y(2x+3)$
$\to x^2+x-14\quad\vdots\quad 2x+3$
$\to 4(x^2+x-14)\quad\vdots\quad 2x+3$
$\to 4x^2+4x-56\quad\vdots\quad 2x+3$
$\to (4x^2-9)+(4x+6)-53\quad\vdots\quad 2x+3$
$\to (2x+3)(2x-3)+2(2x+3)-53\quad\vdots\quad 2x+3$
$\to 53\quad\vdots\quad 2x+3$
$\to 2x+3\in U(53)$ vì $x\in Z$
$\to 2x+3\in\{1, 53 ,-1,-53\}$
$\to x\in\{-1, 25, -2, -28\}$
$\to y\in\{14, -12, -12, 14\}$ Vì $x^2+x-14=-y(2x+3)$
$\to (x,y)\in\{(-1, 14), (25, -12), (-2, -12), (-28, 14)\}$
