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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.

Bassam Karzeddin
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    Could you just describe your proof in short in the question? – TheRandomGuy Apr 04 '16 at 09:47
  • @Dhruv I never thought that this question might be very difficult to answer, but the idea is somewhat clear, it is obvious here that $z$ is divisible by 3, then you conclude more that it must be divisible by 9, then continue..., actually writing the proof in mathematical language would take me more time to prepare and learn all the notations first! – Bassam Karzeddin Apr 06 '16 at 16:45
  • I've verified it till 1000 and there seem to be no solutions. – TheRandomGuy Apr 07 '16 at 14:08
  • I have derived that $x$, $y$ and $z$ are $1, 8, 0 \pmod 9$ and exactly one of $x$ and $y$ is even. Which implies that $z$ is odd. – TheRandomGuy Apr 07 '16 at 14:12
  • Just mention how you go about the proof without any notations. It would give us some ideas. Just mention how you go through it. – TheRandomGuy Apr 07 '16 at 14:22
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    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: http://math.stackexchange.com/questions/1731616/how-to-prove-if-this-equation-provides-an-integral-solution-divisible-by-3 – Will Jagy Apr 07 '16 at 18:47
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    I would like credible and/or official sources for the question. – Will Jagy Apr 07 '16 at 18:54
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    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. – duje Apr 07 '16 at 19:37
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    @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 http://math.stackexchange.com/questions/1731616/how-to-prove-if-this-equation-provides-an-integral-solution-divisible-by-3 Mysterious that the ratios $9$ and $27$ are readily found. – Will Jagy Apr 07 '16 at 22:28
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    see http://meta.math.stackexchange.com/questions/22988/anyone-know-the-contest-this-is-from – Will Jagy Apr 08 '16 at 19:12
  • This is still eligible for bounty, but strangely not allowing me to do so, despite my credit can cover the bounty of 50 credit only!? – Bassam Karzeddin May 15 '16 at 15:14

1 Answers1

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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

duje
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  • 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 – Bassam Karzeddin Apr 09 '16 at 05:32
  • The reference from 2009 is copied from MathSciNet. Versions on the arxiv are from 2003: http://arxiv.org/abs/math/0309474 – duje Apr 09 '16 at 07:16
  • As indicated in http://www.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. – duje Apr 09 '16 at 07:58
  • 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 – Bassam Karzeddin Apr 09 '16 at 14:40