Show $x^2$ + $y^2$ + $z^2$ = $2xyz$ has no non-trivial solutions using infinite descent. I want to show that the equation
$x^2$ + $y^2$ + $z^2$ = $2xyz$ has no non-trivial solutions using infinite descent. 
This question has been solved with infinite descent by showing that $x,y,z$ are all even and we can infinitely find smaller solutions: 
Integer solutions of the equation $x^2+y^2+z^2 = 2xyz$
Can it be shown instead by assuming you have the least positive solution and then finding a smaller solution.
I let ($a,b,c$) be the least positive solution with gcd(a,b,c) = 1, and I want to find a smaller solution.
I have that exactly one of $a,b,c$ must be even because if two were even, then the left hand side would be odd while the right hand side is even. And if all three are even, then gcd($a,b,c$) $\ge$ 2.  
So we can assume $a$ is even. I've tried considering the equation mod $a,b,c,2$ but couldn't seem to find a way to construct a smaller solution. 

$x^2+y^2+z^2=2xyz...........................(1)$
in this equation $RHS$ is even .then, $LHS$ is also even.
w.l.o.g, $z$ is even $\exists z_1 $ such that $z= 2 z_1$ and $x,y$ are in same parity.
substituting $z= 2 z_1$ in given equation
$x^2+y^2+4z_1^2=4xyz_1 ........................(2)$
$x^2+y^2=4xyz_1-4z_1^2$
$RHS$ is multiple of 4 .then ,$LHS$ must be multiple of 4
therefore $x$ and $y$ are even.
let $x=2x_1$ and $y=2y_1$
substituting  $x=2x_1$ and $y=2y_1$ in $(2)$ we get,
$4x_1^2+4y_1^2+4z_1^2=16x_1y_1z_1 $
$\implies x_1^2+y_1^2+z_1^2=4x_1y_1z_1 $
$LHS$ is multiple of $4$.then,$RHS$ is also multile  of $4$
w.l.o.g,$z_1$ is even, let $z_1=2z_2$
$\implies x_1^2+y_1^2+4z_2^2=8x_1y_1z_2 $
in similar argument .We can find three sequence  of infinite integers 
$$x_1 \gt x_2 \gt x_3 \dots \gt x_n$$
$$y_1 \gt y_2 \gt y_3 \dots \gt y_n$$
$$z_1 \gt z_2 \gt z_3 \dots \gt z_n$$
that can solve the equation $x^2+y^2+z^2=2xyz$.
by using infinite descent we can prove
 $x^2+y^2+z^2=2xyz$ has no solution greater than $0$.
