Is there an easy way to determine when this fractional expression is an integer? For $x,y\in \mathbb{Z}^+,$ when is the following expression an integer? 
$$z=\frac{(1-x)-(1+x)y}{(1+x)+(1-x)y}$$
The associated Diophantine equation is symmetric in $x, y, z$, but I couldn't do anything more with that. I tried several factoring tricks without luck. The best I could do was find three solutions such that $0<x\le y\le z$. They are: $(2,5,8)$, $(2,4,13)$ and $(3,3,7)$.
The expression seems to converge pretty quickly to some non-integer between 1 and 2.

Since $$ \frac{(1-x)-(1+x)y}{(1+x)+(1-x)y} = \frac{ xy+x+y-1}{xy-x-y-1} = 1 + \frac{2(x+y) }{xy-x-y-1} $$ 
and $ 2x+2y < xy - x -y - 1 $ if $ 3(x+y) < xy - 1 .$ Suppose $ x\leq y$, then $ 3(x+y) \leq 6y \leq xy-1 $ if $ x\geq 7. $ So all solutions must have $0\leq x< 7 $ so it is reduced to solving $7$ simpler Diophantine equations. 
If $x=0 $ then $ \displaystyle z= 1 - \frac{2y}{y+1}$ so the only solutions are $ (0,0,1)$ and $ (0,1,0).$
If $x=1$ then $ \displaystyle z=  -y$ so $(1,m,-m)$ is a solution for $ m\geq 1.$ 
If $x=2$ then $ \displaystyle z = 1 + \frac{4+2y}{y-3}$ which is an integer for $y=1,2,4,5,8,13.$ 
I will leave you to find the others. Each of the cases are now simple Diophantine equations.
