<s>[INST] Find $x ,y$ satisfying : $x,y$ are $2$ positive integers and $(xy+x+2)(xy+1)$ is a perfect square. Find $x ,y$ satisfying : $x,y$ are $2$ positive integers and $(xy+x+2)(xy+1)$ is a perfect square.
I have solved this problem and will post the solution as soon as possible. Hope everyone can check my solution! Thanks very much !
 [/INST] $(xy+x+2)(xy+1) = a^2$
$\Rightarrow (2xy+x+3)^2 = (2a)^2 +(x+1)^2 $
Because $2 | 2a$  , there exist $3$ positive integers $m , n,d$ satisfying :
$(m,n) = 1$ and  :  $2a = 2dmn ; x+1 = d(m^2-n^2) ; 2xy+x+3 = d(m^2+n^2)$
$\Rightarrow xy+1 = dn^2 $
$\Rightarrow y = \frac{dn^2-1}{d(m^2-n^2)-1} $
$\Rightarrow \frac {y-1}{d} = \frac {2n^2-m^2}{d(m^2-n^2)-1} $
It is easy to see that $d|(y-1)$ , so $\frac {2n^2-m^2}{d(m^2-n^2)-1} \in \mathbb N^+ $
Let  $\frac {2n^2-m^2}{d(m^2-n^2)-1} = k $
Let $ m = \frac{e+f}{2} , n =\frac{e-f}{2}$.
$\Rightarrow k = \frac{e^2+f^2-6ef}{4def-4}$
$\Rightarrow e^2 -ef(6+4kd)+f^2+4k = 0 $
Let $e_1$ is the smallest solution of the above equation .
We have $e_1 + e_2 = (6+4kd)f$ and $e_1 e_2 = f^2+4k >0 $
So $e_2$ is also a solution of the above equation, so $e_1 \le e_2$
$\Rightarrow f^2 \le (e_1-1)(e_2-1) $
$\Rightarrow f(6+4kd) \le 4k+1 $ ( a contradiction )
So there is no $(x,y)$ satisfying the problem.
I just finished it, so no one has checked it yet! Hope you guys can check if I'm wrong!
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