# Three variable, third degree Diophantine equation

I haven’t found any useful method to solve the following problem: Prove that if $x,y,z\in\mathbb{Z}$ and $x^3+y^3=3z^3$ then $xyz=0$.

Source: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=382377

#### Solutions Collecting From Web of "Three variable, third degree Diophantine equation"

The technique used to prove that $x^3 + y^3 + z^3 = 0$ has no non-trivial solutions in $\mathbb{Z}(\sqrt{-3})$ is also applicable to showing that $x^3 + y^3 = 3z^3$ has no non-trivial solutions (in $\mathbb{Z}(\sqrt{-3})$).

In fact, this appears (with proof) in section 13.5 as Theorem 232 in the excellent book, “An Introduction to the Theory of Numbers”, by Hardy & Wright, 5th Edition (I have the Indian Edition, so might be a bit different from yours).

Here is a snapshot I managed to scrape (though the notation is quite old, and you would need parts of the rest of the book to make sense of it).

Suppose $x$, $y$, and $z$ are relatively prime integers. Then, either $z$ is even or, say, $x$ is even.

If $z$ is even, $x^3+y^3\equiv 2 \mod{4}$, whereas $3z^3 \equiv 0 \bmod{4}$.

If $x$ is even, $x^3+y^3\equiv 1 \mod{4}$, whereas $3z^3 \equiv 3 \bmod{4}$.

In both cases, the LHS is not equal to the RHS. Therefore, the Diophantine equation admits no integer triplet.