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I apologise for all these questions of Fermat's Last Theorem, but I am fascinated by the topic, even if I cannot understand all of it.

I must admit that I am not well versed in the language of modular forms or elliptic equations, but they seem quite complicated to me.

However, while reading Simon Singh's book "Fermat's Last Theorem", I found one particular bit very interesting.

I immediately thought that there was an easier way to prove Fermat's Last Theorem. It was to do with Falting's Theorem and the geometrical representations of equations like $x^n + y^n = 1$.

I quote: "Faltings was able to prove that, because these shapes always have more than one hole, the associated Fermat equation could only have a finite number of whole number solutions."

Surely, now all that is needed is to prove that a Fermat equation has infinite solutions.

Suppose we take the original equation: $$A^3 + B^3 = C^3$$ Surely we can find infinite solutions to this by doubling $A, B, C$ $$(2A)^3 + (2B)^3 = (2C)^3$$ $$8A^3 + 8B^3 = 8C^3$$ $$A^3 + B^3 = C^3$$ Surely this means there are infinitely many solutions to these equations. But now we have a contradiction, so therefore our original assumption, that $A^3 + B^3 = C^3$ has solutions, is false.

Who can point out my error as this seems a very simple step to take from Falting's to Fermat's. And surely that step wouldn't have taken years to take, especially for Andrew Wiles.

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    $\begingroup$ Note that your FLT equation has 3 variables but the equations to which Faltings' theorem apply have 2 variables. If you divide everything by $C$ you'll find that you haven't generated new solutions after all. $\endgroup$ – Qiaochu Yuan Jul 22 '18 at 22:04
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    $\begingroup$ Singh's statement as quoted by you is inaccurate. It should say that Falting's work implies that the Fermat equation has only a finite number of primitive solutions (i.e., solutions in which the variables have no common factor). $\endgroup$ – Rob Arthan Jul 22 '18 at 22:25
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    $\begingroup$ @RobArthan Thank you. That makes sense. $\endgroup$ – Roskiller Jul 23 '18 at 8:45
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    $\begingroup$ You have to be careful with these popular books; they often say something that's almost right. I remember from reading it that he gives axioms of arithmetic in an appendix, but forgets to say $0\ne 1$. So the axioms boil down to "numbers form a field, which I forgot to mention should also be nontrivial". $\endgroup$ – J.G. Jul 23 '18 at 9:16
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This is a nice idea but the quote is about the equation $$x^n+y^n=\color{red}1$$So, assume that $$A^3+B^3=C^3$$We get $$\left(\frac AC\right)^3+\left(\frac BC\right)^3=1$$

And also indeed $$(2A)^3+(2B)^3=(2C)^3$$ And we again get$$\left(\frac {2A}{2C}\right)^3+\left(\frac {2B}{2C}\right)^3=\left(\frac AC\right)^3+\left(\frac BC\right)^3=1$$ Which is the same solution as before.

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The original question was posited in terms of exponent $n$, but argued in terms of exponent $3$. In the particular case $n=3$, the argument presuming that one solution might generate infinite solutions is not without merit. Although I discovered this relationship some time ago, it may be the case that others have come across it as well; I have not seen it reported anywhere else, however.

Assume integers $a,b,c$ could be found that satisfy $a^3+b^3=c^3$. Then it is the case that the integers $$x=a(c^3+b^3), y=c(b^3-a^3), z=b(c^3+a^3)$$ satisfy the equation $x^3+y^3=z^3$.

The new numbers $x,y,z$ are not simple multiples of $a,b,c$; thus, I expect the process to generate further sets of triples could be carried on indefinitely.

Examples of such solutions abound when one (or more) of each set of numbers is allowed to be real, rather than integral. For example, $a=2, b=3, c=\sqrt[3]{35}$ yields $x=124, y=19\sqrt[3]{35}, z=129$.

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    $\begingroup$ Hmm, looking backwards (from $x,y,z$ towards $a,b,c$) - wouldn't that allow an "infinite descent"-argument? $\endgroup$ – Gottfried Helms Jul 23 '18 at 17:06
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    $\begingroup$ @Gottfried Helms The problem with descent (which I have thought about) is that even if I assume that $x,y,z$ are integers, I have not been able to show that the smaller derived numbers $a,b,c$ would also be integers. They might be rational, supporting an integral solution, but that integral solution would not then necessarily be smaller than the starting $x,y,z$. Hence, there might not be any descent. I'm not saying it's impossible, but I haven't been able to crack it. $\endgroup$ – Keith Backman Jul 23 '18 at 17:31

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