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 !
 A: For a start:
Let $z=xy+1$, then $$z(z+x+1)=a^2\implies \boxed{z^2+z(x+1)-a^2=0}$$
Since discriminant is perfect square we have $$d^2 =(x+1)^2+4a^2$$ which means that (Pythagorean triples) $$d=k\cdot (u^2+v^2),\;\; x+1 =k\cdot(u^2-v^2),\;\; a = k\cdot uv$$ for some positive integers $u,v,k$ and $\gcd(u,v)=1$. Since $x+1>0$ we have $u>v$. Then (solving boxed equation) $$xy+1 =k\cdot {v^2-u^2\pm (u^2+v^2)\over 2}$$
Since $z$ is positive we have only $+$ choise, so we have $xy+1 = kv^2$ so $y = {kv^2-1\over k(u^2-v^2)-1}$


 ??? Since for every integer $m$ we have $ \gcd(mk+1,k)=1$ thus $k(u^2-v^2)-1\mid k(2v^2+u^2) \implies k(u^2-v^2)-1\mid 2v^2+u^2$ and so
 $$k(u^2-v^2)-1\leq  2v^2+u^2$$

So we have only finite possibilities for $k$ ...

A: Disclaimer: This is only a partial answer. I may complete this later.
Let $x$, $y$ and $z$ be three positive integers such that
$$(xy+x+2)(xy+1)=z^2,\tag{1}$$
The greatest common divisor $d$ of the two factors divides $1+x$ and $1-y$, so we have
$$x=ud-1\qquad\text{ and }\qquad y=vd+1,$$
for coprime integers $u$ and $v$, and $z=wd$. Plugging this into $(1)$ then yields
$$w^2=\big(uvd+2u-v\big)\big(uvd+u-v\big).$$
Now the two factors on the right hand side are coprime, and hence they are both perfect squares, so
$$uvd+2u-v=a^2\qquad\text{ and }\qquad uvd+u-v=b^2,$$
for some positive integers $a$ and $b$ with $a>b$. Then $u=a^2-b^2$ and so
$$v=\frac{2b^2-a^2}{d(a^2-b^2)-1}.$$
As $v$ is positive this shows that $a^2<2b^2$.
Conversely, if $a$ and $b$ are two coprime positive integers with $b<a<\sqrt{2}b$ and $d$ is any positive integer such that
$$\frac{2b^2-a^2}{d(a^2-b^2)-1},$$
is also an integer, then the integer $x$ and $y$ given by
$$x=d(a^2-b^2)-1\qquad\text{ and }\qquad y=d\frac{2b^2-a^2}{d(a^2-b^2)-1}+1=\frac{db^2-1}{d(a^2-b^2)-1},$$
satisfy
$$(xy+x+2)(xy+1)=(dab)^2.$$
A: $(xy+x+2)(xy+1) = a^2$
$\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!
