Given $z^2-1\mid x^2z^2-1$, prove $\frac{x^2z^2-1}{z^2-1}$ is never prime, for $x$, $z$ integers such that $x>z>1$. 
Given $z^2-1\mid x^2z^2-1$, prove that $\frac{x^2z^2-1}{z^2-1}$ can never be prime, assuming $x$, $z$ are integers such that $x>z>1$.

So far I have tried taking mod a lot of different numbers, but I cannot find solution. I also tried writing as a quadratic and using quadratic formula, but that doesn't work either. Please help.
 A: Let $p = \frac{x^2z^2-1}{z^2-1} \implies pz^2 - p = x^2z^2 - 1$
So $z^2(x^2 - p) = 1 - p$.
We know $p > 1$. So this implies $p > x^2$.
Now $p$ divides $xz-1$ or $xz+1$ but not both.
But $p > x^2 > x > z \implies p > xz - 1$ and $p > xz + 1$.
which means $p \nmid xz-1$ and $p \nmid xz+1$.
A: $x^2z^2-1=(xz-1)(xz+1)$
$xz-1>z^2-1$
So $x^2z^2-1$ has at least two factors greater than $1$, neither of which can be made equal to $1$ upon division by $z^2-1$. Ergo it is not prime.
A: Answered for my own benefit, as I already had seen the above two answers first******
Note that $x^2z^2-1 = (xz-1)(xz+1)$. Then, as $\frac{x^2z^2-1}{z^2-1} = \frac{(xz-1)(xz+1)}{z^2-1}$ is an integer, it follows that $z^2-1$ can be written as follows: $z^2-1 = ab$ for some positive integers $a$ and $b$, that satisfy the following:

*

*$a|(xz-1)$ say $xz-1=ca$ and


*$b|(xz+1)$ say $xz+1=bd$.
[One of $a,b$ may be $1$]. But then, as $xz-1 > z^2-1$ [because $x>z$], it follows that $c$ as in 1. above must be an integer greater than $1$, and similarly, $d$ as in 2. above must be an integer greater than $1$. But then that gives:
$$\frac{x^2z^2-1}{z^2-1} = \frac{(xz-1)(xz+1)}{z^2-1} = \frac{(ca)(db)}{ab} = cd,$$ where $c$ and $d$ are both integers greater than $1$.
Thus, as $cd$ is clearly compositive for such $c$ and $d$, it follows that $\frac{x^2z^2-1}{z^2-1}=cd$ cannot be prime.
In general, the following is true:

Let $A,B,C$ be positive integers such that both $\frac{AB}{C}$ is integral, and $C<\min\{A,B\}$. Then $\frac{AB}{C}$ is always not prime.

