<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] 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
$$(xy+x+2)(xy+1)=(dab)^2.$$
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