<s>[INST] Find all $x,y$ so that $\dfrac{x+y+2}{xy-1}$ is an integer. I am trying to find the integers $x,y$ so that 
$\dfrac{x+y+2}{xy-1}$ is an integer. 
What I have done:
I suppose there exists $t$ such that $$t=\dfrac{x+y+2}{xy-1}$$ where $xy\neq 1$ then consider the following scenarios:
$$x=y$$ $$x>y>0$$  $$x>0>y$$ ,etc.
This approach  helps me find the solutions but it is very long. Any simpler method?
 [/INST] For this to happen we need $|x+y+2|\geq |xy-1|$ or $x+y+2=0$.
If $x+y+2=0$ we get solutions $(a,-a-2),(-a-2,a)$.
otherwise we have $|x+y|+2\geq |xy|-1\rightarrow |x|+|y|+3\geq |x||y|$.
Notice that if $x$ or $y$ is $0$ it holds trivially.
We now find all possible values $n,m> 0$ so that $n+m+3\geq nm\iff \frac{n+3}{n-1}\geq m$
If $n=1$ any $m$ works.
If $n=2$ we get $m\leq 5$
If $n=3$ we get $m \leq 3$
If $n=4$ we get $m\leq 2$
If $n=5$ we get $m\leq 2$.
If $n\geq 6$ we get $m=1$.
So we first analyse the case when $|x|=1$.
If $x=1$ we get $y-1$ divides $y+3$, so $y-1$ divides $(y+3)-(y-1)=4$, hence $y-1$ can be $-4,-2,-1,1,2$ or $4$. It is easy to see all of them work. So we get $(1,-3),(1,-1),(1,0),(1,2),(1,3),(1,5)$.
If $x=-1$ we get $-(y-1)$ divides $y-1$. So $y+1$ divides $y-1$ and hence $y+1$ divides $y+1-(y-1)=2$. So $y+1$ can be $-2,-1,1,2$. It is easy to see all of them work. So we get $(-1,-3),(-1,-2),(-1,0),(-1,1)$.
Since $x$ and $y$ play symmetric roles in the equation, this gives all solutions with $|x|=1$ or $|y|=1$.
Now we analyse when $|x|=2$.
If $x=2$ we get $2y-1$ divides $y+4$, we need only try with $y=-5,-4,-3,-2,0,2,3,4,5$. We see only $(2,-4),(2,0),(2,2),(2,5)$ work.
If $x=-2$ we get $-2y-1$ divides $y$, so $2y+1$ divides $y$, and so $2y+1$ divides $2y+1-2(y)=1$. So we only try with $2y+1=1$ and $2y+1=-1$. So $(-1,-1)$ and $(-1,1)$ are solutions.
So we only need to check when $|x|,|y|>2$.
If $|x|=3$ we need only check if $|y|=3$. There are four cases, and out of these only $(3,3)$ works.
This finishes all the cases.
(I'll try to make this a bit cleaner when I get home).
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