Fermat's Last Theorem and Faltings' Theorem.

Solution 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.

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