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


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).

alt text

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    $\begingroup$ It looks the same in the (non-Indian) 6th edition I have (although you have only reproduced the 1st page of a 2-page argument). $\endgroup$ – Gerry Myerson Apr 27 '11 at 5:02
  • $\begingroup$ @Gerry: Yeah, I wasn't able to get the rest, unfortunately. $\endgroup$ – Aryabhata Apr 27 '11 at 6:09

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.

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    $\begingroup$ If $y$ is 3 mod 4 then $y^3$ is 3 mod 4, so I don't think your calculations work. $\endgroup$ – Gerry Myerson Jul 8 '11 at 2:59
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    $\begingroup$ If z is even and x and y odd,if x = 1 mod 4 --> x^3 = 1 mod 4 and y = 3 mod 4 ---> y^3 = 3 mod 4, then x^3 + y^3 = 4=0 mod 4 as well as 3z^2 = 0 mod 4. So there can be solutions. You cannot exclude solutions this way. $\endgroup$ – user55514 Jan 17 '13 at 5:27

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