<s>[INST] 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.
 [/INST] 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 $p \nmid xz+1$.
</s>