Given the conditions above, find when $x$, $y$, $z$ satisfy below: $ (x^2-1)(y+1)=\dfrac{z^2+1}{y-1}$ 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.
 A: I solve this Diophantine equation with the assumption that $x$, $y$, and $z$ can be any integers (not necessarily positive and not necessarily unequal). Rewrite the equation as
$$\left(x^2-1\right)\,\left(y^2-1\right)=z^2+1\,.$$
Since the largest power of $2$ that divides $z^2+1$ is $2$ (i.e., $4\nmid z^2+1$ for all $z\in\mathbb{Z}$) and $8\mid t^2-1$ for any odd integer $t$, we conclude that both $x$ and $y$ must be even.
If either $x$ or $y$ is nonzero, say, $x\neq 0$, then $x^2-1$ is a positive integer congruent to $3$ modulo $4$.  Therefore, a prime natural number $p\equiv 3\pmod{4}$ divides $x^2-1$.  This means $p\mid z^2+1$ as well, but we know this is impossible as $-1$ is not a quadratic residue modulo $p$.  Hence, the only possible solution in $\mathbb{Z}$ must come from $(x,y)=(0,0)$, which then produces $(x,y,z)=(0,0,0)$.
A: 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?
A: The only solution is $x = y = z = 0$.
Suppose there were a solution with x, y, and z not all zero.  Choose the maximum $k$ with $2^k | gcf(x, y, z)$.  So $x = 2^k m$, $y = 2^k n$, and $z = 2^k p$, and $m$, $n$, and $p$ are not all even.
$(xy)^2 = x^2 + y^2 + z^2$
$(2^k m 2^k n)^2 = (2^k m)^2 + (2^k n)^2 + (2^k p)^2$
$(4^k mn)^2 = 4^k m^2 + 4^k n^2 + 4^k p^2$
$16^k (mn)^2 = 4^k m^2 + 4^k n^2 + 4^k p^2$
$4^k (mn)^2 = m^2 + n^2 + p^2$
But the last equation only has solutions (mod 4) when m, n, and p are all even, contradicting the choice of k.
