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Considering the cubic case of Fermat’s Last Theorem, I make the following claim:

Proposition: The Diophantine equation $$ X^3 + Y^3 = Z^3 \tag{$\star$} $$ has a finite number of primitive [and non-trivial] integer solutions $(x,y,z)$.

Is there a simple and elementary way to prove this statement?

Note: I am aware of Wiles’s proof of FLT in the general case, and the infinite descent proof of the cubic case by Euler et al., and the Mordell–Weil theorem, etc. I’m just curious, independent of those things, whether this weaker proposition has a simple and elementary proof.

Kieren MacMillan
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    So you mean really different, since if $x,y,z$ is a solution so is $(ax,ay,az)$. – André Nicolas Mar 27 '16 at 19:32
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    You never know. You should be aware of Selmer's example, $3x^3 + 4 y^3 + 5 z^3.$ There are no local obstructions to this being zero, however it cannot be zero with integers $x,y,z$ unless all three are zero. http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/selmerexample.pdf – Will Jagy Mar 27 '16 at 19:34
  • Related: http://math.stackexchange.com/questions/662313 ? – Watson Mar 27 '16 at 19:38
  • @AndréNicolas: Thanks — I edited the question and Proposition accordingly. Hope it more clearly and accurately represents what I'm asking. – Kieren MacMillan Mar 27 '16 at 19:41
  • @WillJagy: Nice! I look forward to absorbing that. – Kieren MacMillan Mar 27 '16 at 19:42
  • @Watson: Yes, that question includes the infinite descent proof (originally by Euler, with corrections/completions by a few subsequent mathematicians), as mentioned in the body of my question. I'm asking a slightly different question than the full cubic case of FLT. – Kieren MacMillan Mar 27 '16 at 19:44
  • We can assume $X$ and $Y$ are relatively prime WLOG. Let them have a common factor $p$. $$X^3 + Y^3 =Z^3 \Longleftrightarrow (px)^3 + (py)^3 = (pz)^3 \Longleftrightarrow x^3 + y^3 = z^3$$ – TheRandomGuy Mar 28 '16 at 17:35
  • @Dhruv: Okay… and then? – Kieren MacMillan Mar 28 '16 at 19:29
  • Kieren MacMillan, $x^3+y^3=z^3$

    Suppose $z=x+d_1$ and $y=x+d_2$ also $y>x$ ⇒ $d_2<d_1$. Substituting in equation we get:

    $x^3 +3x^2(d_2-d_1) +3x(d_2^2 -d_1^2)-d_1^3=0$

    Now If you claim equation $x^3+y^3=z^3$ have a solution then you have to prove the equation $x^3 +3x^2(d_2-d_1) +3x(d_2^2 -d_1^2)-d_1^3=0$ has a solution. You can either use standard algorithm for cubic equations or reasoning to show this. I think this could be a simple elementary method.

     – sirous Dec 18 '17 at 06:56
  • @sirous: I think you mean $x^3+3x^2(d_2−d_1)+3x(d_2^2−d_1^2)+d_2^3-d_1^3=0$, right? – Kieren MacMillan Dec 21 '17 at 14:28
  • Kieren MacMilan, you are right, sorry about the mistake. Meanwhile I used this algorithm to find a general formula for finding Pythagorean triples. – sirous Dec 23 '17 at 06:28
  • Kieren MacMilan, We can write: $z^3-y^3=x^3$ or $(z-y)(z^2 + zy + y^2)= x . x^2$ suppose $x=z-y$, it contradicts $x^2=z^2+zy+y^2$. we may assume $z-y=1$. among $z, y$ couples only $z=8$, $y=7$ give one term, i.e. $8^3-7^3=13^2$. For others usually the terms are more than one. One way is to search whether there exist more such triples and if so $x$ is a cube?. – sirous Dec 24 '17 at 08:39
  • @sirous: What justification do we have for assuming $z-y=1$? – Kieren MacMillan Dec 27 '17 at 13:15
  • We have to assume $x^3=z^3-y^3$ or $x^2=z^3-y^3$ and $x =1$. This is another possibility when assuming $z-y=x$ contradicts other factor. – sirous Dec 27 '17 at 17:19
  • @sirous: Why can’t $z-y=x_1^3$ and $z^2+yz+y^2=x_2^3$, where $x=x_1x_2$? – Kieren MacMillan Dec 28 '17 at 03:18
  • The equation is supposed to have only one solution for each parameter. – sirous Dec 30 '17 at 13:16
  • @sirous: Sounds like you’re mixing up "solution" with "integer". In my one solution, I simply factor $x=x_1x_2$ where $x_1, x_2 > 1$. – Kieren MacMillan Dec 30 '17 at 15:42

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