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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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    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 $(3k\pm 1)^3\equiv \pm 1\pmod{9}$, $(3k)^3\equiv 0\pmod{9}$. – user236182 Dec 29 '17 at 16:49
  • If $n=0$ then all possible solutions have one of $x,y$ or $z$ equal to $0$ by Fermat's Last Theorem. – Matt B Dec 29 '17 at 16:50
  • @user236182 thanks I've edited now - my point was there are no nontrivial solutions. – Matt B Dec 29 '17 at 16:53
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    It is an open problem to determine whether there exists a solution when $n=33$, see https://mathoverflow.net/a/100324 . – tristan Dec 29 '17 at 16:56
  • see here https://www.google.de/url?sa=t&rct=j&q=&esrc=s&source=web&cd=5&cad=rja&uact=8&ved=0ahUKEwjNo9aE1q_YAhXQJuwKHR0GCLMQFghaMAQ&url=https%3A%2F%2Fpeople.hofstra.edu%2FEric_Rowland%2Fpapers%2FKnown_families_of_integer_solutions_of_x%255E3%2By%255E3%2Bz%255E3%3Dn.pdf&usg=AOvVaw00K0fibj6GVfEmBOPuoarA – Dr. Sonnhard Graubner Dec 29 '17 at 17:04
  • Looks like a duplicate of https://math.stackexchange.com/questions/2584378/parametric-solutions-to-x3y3z3-n – rtybase Dec 29 '17 at 18:05
  • @rtybase Clearly no. That one asks for parametrizations. –  Dec 29 '17 at 18:09
  • @Dr.SonnhardGraubner I cannot open that file, that is, visit that page. –  Dec 29 '17 at 18:14
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    This arXiv paper and this MO post both give some background. – Paul LeVan Dec 29 '17 at 20:22
  • @PaulLeVan Thanks, Honestly, I do not know is it safe to conjecture that there are no solutions for some $n$ that are not of the form $n\equiv \pm 4\pmod{9}$ –  Dec 29 '17 at 20:30

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