I know for any pairwise co-prime integers $x,y,z$ that $$x^3+y^3+z^3\neq 0$$ $$x^3+y^3+3z^3\neq 0$$ Do we also have $$x^3+y^3+2z^3\neq 0?$$ Any Hints?
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Ok! But you should change the title please! – Alex Silva Jan 24 '17 at 07:57
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1Trivial answers as mentioned below: $(x, y, z) \in {(a,-a,0), (a,a,-a)}$. In these cases clearly $x,y,z$ are not coprime so these don't qualify. – TMM Jan 24 '17 at 07:58
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What would you suggest? – Jan 24 '17 at 07:58
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2... have any pairwise co-prime integer solutions? – Alex Silva Jan 24 '17 at 07:59
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http://www.artofproblemsolving.com/community/c3046h1056636_diophantine_equation_3rd_degree – individ Jan 24 '17 at 11:53
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There are no non-trivial solutions. A proof that there are no non-trivial solutions of the closely related equation:
$$x^3+y^3=2z^3$$
is given in Sierpinski's Elementary Theory of Numbers, Chapter 2 page 79, which may be accessed here (which I found via here). Then it just needs to be noted that a solution of either one of these equations would provide a solution of the other by change of sign of $z$.
Adam Bailey
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There should be also a proof by Euler in L. Euler, Algebra 2 (1770), Art. 247; French transl., 2 (1774) 355-360; Opera Omnia I (1) 491 – Sil Mar 21 '22 at 23:06
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Then there is also On the equation $x^3+y^3=2z^3$ by A. Wakulicz 1957, can be viewed online here – Sil Mar 22 '22 at 11:18