<s>[INST] 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  $x=y=z=0$?
 [/INST] 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.
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