$x^3=2y^3+4z^3$ in the set of integers

Question

If $x, y, z$ are integers solve: $$x^3=2y^3+4z^3$$

Answer 1

Just found the solution.

First notice that $(x,y,z)=(0,0,0)$ is a solution. If only one of $x,y,z$ is $0$ then we get a contradiction. If two of $x,y,z$ are equal to $0$ then we get the above solution.

Let's assume that none of $x,y,z$ is equal to $0$. In that case: $$x=2^ka$$ $$y=2^mb$$ $$z=2^nc$$ where $$a,b,c,k,m,n\in\mathbb{Z}$$ $$k,m,n\ge0$$ and $a,b,c$ are odd numbers

From the equation $x^3$ is even, so $x$ is even. The equation becomes: $$(2\cdot 2^{k-1}a)^3=2y^3+4z^3$$ $$4(2^{k-1}a)^3=y^3+2z^3$$ In the same way y and z are even and we get $(2^{k-1}a)^3=2(2^{m-1}b)^3+4(2^{n-1}c)^3$. Repeating the same until one of $k,m,n$ gets eliminated we conclude that one of $a,b,c$ is an even number, hence a contradiction.