Consider the following Diophantine equation

$$z^3 = 3(x^3 +y^3+2xyz)$$

Is there any elementary proof for the non-solubility in non-zero integers for this Diophantine equation, where the absolute value of $x, y$ and $z$ are pairwise coprime integers?

I have proved the impossibility of solution of this Diophantine equation in non-zero integers, but that took quite a few pages and I consider this too long. I hope for a much more elementary and shorter proof for this puzzle.

Hint: It should be noted that is also true for the non-availability of the similar form (replacing coefficient 3 by 1) as the following:

$$z^3=x^3+y^3+2xyz,$$ but this case is very simple to prove, where $x, y\;\&\; z$ are nonzero integers.

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    $\begingroup$ Could you just describe your proof in short in the question? $\endgroup$ Apr 4, 2016 at 9:47
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    $\begingroup$ Evidently the only solutions are $(t,-t,0).$ Thus the only primitive solutions are $(1,-1,0)$ and $(-1,1,0).$ Where did you get this problem? There is a repeat of it, about 12 hours before this comment. Here: math.stackexchange.com/questions/1731616/… $\endgroup$
    – Will Jagy
    Apr 7, 2016 at 18:47
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    $\begingroup$ I would like credible and/or official sources for the question. $\endgroup$
    – Will Jagy
    Apr 7, 2016 at 18:54
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    $\begingroup$ The problem leads to finding rational points on the curve $3U^3 +3V^3 +6UV−1=0$, which is birationally equivalent to the elliptic curve $Y^2 +Y=X^3 −270X−1708$ with trivial torsion and rank zero, so it has no rational points. Thus, the original equation has no solutions, except those with z=0. $\endgroup$
    – duje
    Apr 7, 2016 at 19:37
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    $\begingroup$ @duje thank you. It turns out that, if the coefficient $3$ is replaced by other integers, solutions are possible, I put a computer search at math.stackexchange.com/questions/1731616/… Mysterious that the ratios $9$ and $27$ are readily found. $\endgroup$
    – Will Jagy
    Apr 7, 2016 at 22:28

1 Answer 1


The problem leads to finding rational points on the curve $3U^3+3V^3+6UV-1 = 0$, which is birationally equivalent to the elliptic curve $Y^2+Y = X^3-270X-1708$ with trivial torsion and rank zero, so it has no rational points.

See also Theorem 1 in 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. arxiv

  • $\begingroup$ Let us wait and hope for more shorter and elementary proof, also the reference date is 2003 for this similar topic, where as you say in 2009, what is correct date?, since this had been discussed in 2007 in details at sci.math $\endgroup$ Apr 9, 2016 at 5:32
  • $\begingroup$ The reference from 2009 is copied from MathSciNet. Versions on the arxiv are from 2003: arxiv.org/abs/math/0309474 $\endgroup$
    – duje
    Apr 9, 2016 at 7:16
  • $\begingroup$ As indicated in science-bbs.com/121-math/19a7e45249b8ba6c.htm there is some discussion of this and similar equations in Mordell's book Diophantine Equations, around pages 78 and 130. $\endgroup$
    – duje
    Apr 9, 2016 at 7:58
  • $\begingroup$ That is me in the link!, but the problem there is without the coefficient 3, which was too simple and direct conclusion of impossible solution, but the link you provide isn't the original and show very less content, but in other threads it shows how simple the proof was $\endgroup$ Apr 9, 2016 at 14:40

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