Given the conditions above, find when $x$, $y$, $z$ satisfy below: $ (x^2-1)(y+1)=\dfrac{z^2+1}{y-1}$

Question

Let $x,y,z \in \mathbb{Z^+}$ and $x \neq y \neq z$.

Given the conditions above, find when $x$, $y$, $z$ satisfy below:

$$ (x^2-1)(y+1)=\frac{z^2+1}{y-1}\,.$$

What I did was I factored the numerator to

$$(x+1)(x-1)(y+1)=\frac{z^2+1}{y-1}\,,$$

but I am having trouble figuring out how to isolate the variables. I tried some values with trial and error and wasn't able to get any.

Answer 1

Multiply both sides by $y-1$, expand, and simplify. You then have several relatively easy ways to attack the question. Is that enough to go on?