Prove that the equation $x^3+2y^3+4z^3=9w^3$ has no solution $(x,y,z,w)\neq (0,0,0,0)$

Question

Prove that the equation $x^3+2y^3+4z^3=9w^3$ has no solution $(x,y,z,w)\neq (0,0,0,0)$

So reduced $\mod 2$ I have $x^3\equiv w^3$, so if $x$ is odd, $w$ is odd and if $x$ is even, $w$ is even.

But I'm not really sure what else I should do.

I tried factoring the $2$ to get $x^3+2(y^3+2z^3)=9w^3$.

Reducing $\mod 3$ I get $x^3+2y^3+z^3=0$.

Answer 1

I guess integer solutions are meant.

Solve it by reducing $\,\mod 9$.

Answer 2

Suppose there is an integer solution $(x, y, z, w)$ for your equation, then we have $$ x^3+2y^2+4z^3\equiv 0 \pmod 9$$

However, $0^3\equiv 3^3\equiv 6^3\equiv 0 \pmod 9$, $1^3\equiv 4^3\equiv 7^3\equiv 1 \pmod 9$, $2^3\equiv 5^3\equiv 8^3\equiv -1 \pmod 9$. So this implies there exists $a, b, c \in \{-1,0,1\}$ such that $a+2b+4c\equiv 0 \pmod 9$. By enumerating all possible combinations of $a$, $b$, and $c$, we see that $a=b=c=0$ is necessary. This means $x$, $y$, $z$ are multiples of three. So is true for $w$, for if $x=3k$, $y=3l$ and $z=3m$, the original equation implies $$ 27(k^3+2l^3+4m^3)=9w^3 $$ Hence $3$ divides $w^3$, and by the fact that $3$ is a prime number we have $3$ divides $w$.

If non-zero solutions exist, let $(x_0, y_0, z_0, w_0)$ be one of them such that $|x|+|y|+|z|+|w|$ is the smallest. Then we see that $(x_0/3, y_0/3, z_0/3, w_0/3)$ is another non-zero integer solution. But $|x_0/3|+|y_0/3|+|z_0/3| + |w_0/3|< |x_0|+|y_0|+|z_0| + |w_0|$ holds, a contradiction.