Fermat's Last Theorem and Faltings' Theorem.

Question

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.

Answer 1

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.

Answer 2

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 $x=124, y=19\sqrt[3]{35}, z=129$.