Đáp án: $(x,y)\in\{(1,1), (1,2), (2, 1), (2, 3), (3,2 )\}$
Giải thích các bước giải:
Không mất tính tổng quát giả sử $x\le y$
Nếu $x=1\to x+1=2$
Vì $x+1\quad\vdots\quad y$
$\to 2\quad\vdots\quad y$
$\to y\in\{1,2\}$
$\to (x,y)\in\{(1,1), (1,2)\}$
Nếu $x\ge 2\to 2\le x\le y$
Ta có:
$\begin{cases}x+1\quad\vdots\quad y\\y+1\quad\vdots\quad x\end{cases}$
$\to (x+1)(y+1)\quad\vdots\quad xy$
$\to xy+x+y+1\quad\vdots\quad xy$
$\to x+y+1\quad\vdots\quad xy$
$\to \dfrac{x+y+1}{xy}\in Z$
$\to \dfrac1y+\dfrac1x+\dfrac1{xy}\in Z$
Mà $2\le x\le y$
$\to \dfrac1y+\dfrac1x+\dfrac1{xy}\le\dfrac12+\dfrac12+\dfrac14=\dfrac54$
Do $x,y\in Z^+\to \dfrac1y+\dfrac1x+\dfrac1{xy}>0$
$\to 1=\dfrac1y+\dfrac1x+\dfrac1{xy}$
$\to 1\le\dfrac1x+\dfrac1x+\dfrac1{2x}=\dfrac5{2x}$
$\to 2x\le 5$
$\to x\le 2.5$
Mà $x\ge 2, x\in N$
$\to x=2$
$\to \dfrac12+\dfrac1y+\dfrac1{2y}=1\to y=3$
