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Evidently the problem is to show this is impossible: $$\frac{z^3}{x^3+y^3+2xyz}=3$$$(x,y,z\in \mathbb N)$

Posted as

Why there isn't any solution in positive integers for $z^3 = 3(x^3 +y^3+2xyz)$?

How to prove if this equation provides an integral solution divisible by $3$?

A complete solution was given in comment by duje after my answer, elliptic curves.

ADDED: it seems one of the people asking has been wondering about similar problems for years:

http://www.science-bbs.com/121-math/19a7e45249b8ba6c.htm

I keep asking for the source, no joy. One possible reason for that is that this is part of some sort of contest, probably online but not necessarily in English. Oh, there seem to be a fair number of contests that are aimed at programmers but have explicit mathematical content.

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closed as off-topic by choco_addicted, Shailesh, JonMark Perry, Surb, BLAZE Apr 9 '16 at 11:08

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  • $\begingroup$ I found this paper on the arxiv arxiv.org/pdf/math/0309474v4.pdf $\endgroup$ – duje Apr 8 '16 at 22:19
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    $\begingroup$ Are you sure that meta is the right place for this question? $\endgroup$ – user26857 Apr 8 '16 at 22:19
  • $\begingroup$ Nathan, Joseph Amal(6-BARC-REP) Revisiting Fermat's last theorem for exponent 3. (English summary) Indian J. Math. 51 (2009), no. 2, 379–390. $\endgroup$ – duje Apr 8 '16 at 22:25
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    $\begingroup$ @duje thank you, so the problem really can be done at the same level as exponent 3 in FLT, which is in many beginning number theory books. There is also a chance of a genuinely easy trick with all variables restricted positive (as was stated in both postings), making for a contest problem. $\endgroup$ – Will Jagy Apr 8 '16 at 22:32

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