Suppose that $n \in \mathbb Z$ and that we want to determine for which $n$ there exists at least one solution $(x,y,z) \in \mathbb Z^3$ so that $x^3+y^3+z^3=n$

What is known about this?

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    $\begingroup$ If $n\equiv \pm 4\pmod{9}$, then there are no solutions because $x^3\equiv \pm 1, 0\pmod{9}$, $\forall x\in\mathbb Z$ because, e.g., by Binomial theorem $(3k\pm 1)^3\equiv \pm 1\pmod{9}$, $(3k)^3\equiv 0\pmod{9}$. $\endgroup$ – user236182 Dec 29 '17 at 16:49
  • $\begingroup$ If $n=0$ then all possible solutions have one of $x,y$ or $z$ equal to $0$ by Fermat's Last Theorem. $\endgroup$ – Matt B Dec 29 '17 at 16:50
  • $\begingroup$ @user236182 thanks I've edited now - my point was there are no nontrivial solutions. $\endgroup$ – Matt B Dec 29 '17 at 16:53
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    $\begingroup$ It is an open problem to determine whether there exists a solution when $n=33$, see mathoverflow.net/a/100324 . $\endgroup$ – tristan Dec 29 '17 at 16:56
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    $\begingroup$ This arXiv paper and this MO post both give some background. $\endgroup$ – Paul LeVan Dec 29 '17 at 20:22

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