# Integer solutions to $x^3+y^3+z^3=29$

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!

• Are these numbers assumed to be positive? – Dr. Sonnhard Graubner Jan 16 at 8:56
• @Dr.SonnhardGraubner : for me (and for almost all people, I guess) $\Bbb Z$ includes negative numbers. – Alphonse Jan 16 at 8:58
• A second solution is $(4,-2,-3)$, a third $(18,13,-20)$, $(235,-69,-233)$ and so on... – Raymond Manzoni Jan 16 at 9:31
• 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. – Yong Hao Ng Jan 16 at 9:32
• In addition to the 3 solutions given by @RaymondManzoni, there are $-7584, 7552, 1765$; $-11195, 11190, 1234$; $22734, -20075, -15410$; $104860, -100464, -51803$; $-146415, 144115, 52609$; $386839, -379133 -150237,\dots$. – Rosie F Jan 17 at 11:08