Find all positive integers $x,y$ such that $\frac{x^2-1}{xy+1}$ is a non-negative integer. QUESTION: Find all positive integers $x,y$ such that $\frac{x^2-1}{xy+1}$ is a non-negative integer.
ATTEMPT (AND SOME SOLUTIONS): So, for a positive integer $x$, $x^2-1\ge0$. In the case $x^2-1=0$ i.e $x=1$ (since it's a positive integer), we get 
$\frac{x^2-1}{xy+1}=0$ which is a non-negative integer, regardless of $y$. So, one solution is $(x,y)=(1,k)$ where $k$ is a positive integer. 
Now, let's say $y=1$. Then $\frac{x^2-1}{xy+1}=\frac{x^2-1}{x+1}=x-1$ which is always a non-negative integer so $(x,y)=(k,1)$ is also a solution. However, I don't know how to find the other or prove that those are the only ones. 

$x+1=xy+1 ⇒ y=1$ then x can get any value greater than 0.
B: $x-1=xy+1 ⇒ x(y-1)=-2 ⇒ x=2, y=0$ are only possible solutions. 
