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

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 $$x^3+2y^3+z^3=0$$.

• $x=y=z=1\,$ and $\,w\,=\,\left(\frac79\right)^{\frac13}$ Dec 11, 2019 at 7:07

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.

I guess integer solutions are meant.

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

• So I get that $x^3+2y^3+4z^3\equiv 0$ so then $x$ should be even. and $w$ is then even. Dec 11, 2019 at 7:52
• By working $\,\mod 9,\,$ with relatively little effort you will get the whole enchilada. Dec 11, 2019 at 9:44
• Wlod, I see two comments of yours referring to the whole enchilada, one says an hour ago, one says 11 hours ago. Try some of the ways of refreshing your screen/browser/cache... Dec 11, 2019 at 20:49
• @WillJagy, thank you. I've removed the later repetitive comment. My computer or connection works very poorly (it's so annoying and makes my Internet life miserable -- I guess, this must be the point of that virtual joker). Dec 12, 2019 at 8:44
• The equation has nontrivial solutions mod $9$. For instance, $(3,3,3,1)$.
– lhf
Dec 13, 2019 at 12:03