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What are all the triples $(x,y,z) \in \Bbb Z^3$ with $$x^3+y^3+z^3=29, x \geq y \geq z$$ ? We find immediately $(3,1,1)$, but are there other? According to this question, it could be a difficult problem. There might be useful references treating of this equation.

Thank you!

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    $\begingroup$ Are these numbers assumed to be positive? $\endgroup$ – Dr. Sonnhard Graubner Jan 16 at 8:56
  • $\begingroup$ @Dr.SonnhardGraubner : for me (and for almost all people, I guess) $\Bbb Z$ includes negative numbers. $\endgroup$ – Alphonse Jan 16 at 8:58
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    $\begingroup$ A second solution is $(4,-2,-3)$, a third $(18,13,-20)$, $(235,-69,-233)$ and so on... $\endgroup$ – Raymond Manzoni Jan 16 at 9:31
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    $\begingroup$ From what I've seen this is normally called the "sum of three cubes" problem which is a difficult problem. Only $x^3+y^3+z^3=k^3,2k^3$ have parameterized solutions (hence infinite solutions) I think. For the rest it is not known whether is solution set is finite. The latest paper I have seen comes from the Elliptic curve technique Newer sum of three cubes. Another older reference would be this. $\endgroup$ – Yong Hao Ng Jan 16 at 9:32
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    $\begingroup$ In addition to the 3 solutions given by @RaymondManzoni, there are $-7584, 7552, 1765$; $-11195, 11190, 1234$; $22734, -20075, -15410$; $ 104860, -100464, -51803$; $-146415, 144115, 52609$; $386839, -379133 -150237,\dots$. $\endgroup$ – Rosie F Jan 17 at 11:08

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