Showing that $x^3+2y^3+4z^3=2xyz$ has no integer solutions except $(0,0,0)$. Let $x,y,z\in \mathbb{Z}$ satisfy the equation:
$$
x^3+2y^3+4z^3=2xyz
$$
How do I prove that the only solution is  $x=y=z=0$?
 A: If following equation has a non-trivial solution
$$x^3 + 2y^3 + 4z^3 = 2xyz\tag{*1}$$
there must be one whose $x^2 + y^2 + z^2$ is minimal. Let $(x_1,y_1,z_1)$ be such a minimal non-trivial solution. We have
$$x_1^3 + 2y_1^3 + 4z_1^3 = 2x_1y_1z_1 \implies x_1^3 \equiv 0 \pmod 2 \implies 2|x_1$$
Let $(x_2,y_2,z_2) = (y_1,z_1,\frac{x_1}{2})$, we have
$$8z_2^3 + 2x_2^3 + 4y_2^3 = 4z_2x_2y_2\quad\iff\quad
x_2^3 + 2y_2^3 + 4z_2^3 = 2x_2y_2z_2$$
This means $(x_2,y_2,z_2)$ is again a solution for $(*1)$. Since $(x_1,y_1,z_1)$ is minimal, we have
$$x_1^2 + y_1^2 + z_1^2 \le x_2^2 + y_2^2 + z_2^2 \implies x_1^2 \le z_2^2 = \frac14 x_1^2 \implies x_1 = 0$$ 
This leads to $2y_1^3 + 4z_1^3 = 0$. Since $(x_1,y_1,z_1)$ is assumed to be non-trivial, $z_1 \ne 0$ and this implies $\sqrt[3]{2} = \left|\frac{y_1}{z_1}\right|$ is a rational number! 
This is absurd and hence the original assumption on existence of non-trivial solution for $(*1)$ is simply wrong!
A: We have $x^3 +2y^3+4z^3=2xyz$ so x is even and $8x_1^3+2y^3+4z^3=4x_1yz$ $\implies$
$4x_1^3+y^3+2z^3=2x_1yz$ so y is even and $4x_1^3+8y_1^3+2z^3=4x_1y_1z$
$\implies$ $2x_1^3+4y_1^3+z^3=2x_1y_1z$ so z is even and
$2x_1^3+4y_1^3+8z_1^3=4x_1y_1z_1$ $\implies$ $x_1^3+2y_1^3+4z_1^3=2x_1y_1z_1$.
Consequently, the procedure can be repeated indefinitely and, by infinite descent (Fermat), there are no solution distinct of (0,0,0).Obviously no matter if we consider negative integers.
