"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?
On
$x^2-y^2=2xyz \implies (x+y)(x-y)=2xyz$
$LHS$ is multiple of $2$ .then, $RHS$ must be multiple of $2$
$x$ and $y$ are in same parity.
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 $0$.
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 $0$.
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 $x$.