Proving $x^2-y^2 = 2xyz$ has no integer solutions greater than 0. "Prove that there are no integers $x,y, z >0$ such that $x^2-y^2=2xyz$. Solve by infinite descent."
I've been able to solve this problem myself just by playing around with the algebra, but I have no idea how to solve it with infinite descent. I recognize that the $xy$ term on RHS is probably important somehow, but I can't figure much beyond that. Can someone help? 
 A: Maybe you are expected to argue like this.
Assume you have a solution of
$$\tag{eq}
x^{2} - y^{2} = 2 x y z
$$
with $x$ as small as possible.
Since the right-hand side is positive, we have $x > y > 0$.
Thus $x > 1$, so that there is a prime $p$ dividing $x$.
Since $p$ divides $x^{2}$ and $2 x y z$, we have that $p$ divides $y^{2}$, and thus $p$ divides $y$.
Dividing (eq) by $p^{2}$, we obtain
$$
\left( \frac{x}{p} \right)^{2} - \left( \frac{y}{p} \right)^{2}
=
2 \cdot \frac{x}{p} \cdot \frac{y}{p} \cdot z,
$$
so that
$$
\frac{x}{p}, \frac{y}{p}, z
$$
is another solution with
$$
\frac{x}{p} < x,
$$
contradicting the minimality of $x$.
A: $x^2-y^2=2xyz \implies (x+y)(x-y)=2xyz$
$LHS$ is multiple of $2$ .then, $RHS$ must be multiple of $2$
$x$ and $y$ are in same parity.
CASE 1
If both  are even.let $x=2x_1$ and $y=2y_1$
then,
$(2x_1+2y_1)(2x_1-2y_1)=4(x_1+y_1)(x_1-y_1)$
$ 4(x_1+y_1)(x_1-y_1)=8x_1y_1z$
$\implies (x_1+y_1)(x_1-y_1)=2x_1y_1z$
by similar argument  we can find sequence of infinitely many integers
$$x_1 \gt x_2 \gt x_3 \gt\dots \gt x_n$$
$$y_1 \gt y_2 \gt y_3 \gt\dots \gt y_n$$
by using infinite descent,$x^2-y^2=2xyz$ has no even solutions greater than $0$.
CASE 2
if both are odd $(x+y)(x-y)$ is multiple of $4$ .then, $RHS$ is multiple of $4$ . $z$ is even . let, $z=2z_1$ ,$x=2x_1+1$ and $y=2y_1+1$
$(2x_1+2y_1+2)(2x_1-2y_1)=4xyz_1$
$(x_1+y_1+1)(x_1-y_1)=xyz_1$
$(x_1+y_1)(x_1-y_1)=xyz_1 +y_1-z_1$
$(x_1+y_1)(x_1-y_1)$ is multiple of $4$
by similar argument  we can find sequence of infinitely many integers
$$x_1 \gt x_2 \gt x_3 \gt\dots \gt x_n$$
$$y_1 \gt y_2 \gt y_3 \gt\dots \gt y_n$$
by using infinite descent,$x^2-y^2=2xyz$ has no even solutions greater than $0$.
