This started off as a joke post of mine on a Facebook Group called "Bad Maths that Gives the Right Answer", in which I pulled a Fermat and claimed that the last bit of the proof was too long to post. But the "proof" that I posted raised some questions that intrigued me, so I will repost it here and then raise my questions. Here we go ...

I will start with the supposition that Fermat’s celebrated conjecture is false i.e., that there do in fact exist non-zero integer values of $x$, $y$, $z$ , and $n>2$ such that

$$x^n+y^n=z^n$$

The degenerate case of $y=x$ will be ignored, since $2^{1/n}$ is irrational for $n>1$ , proving the conjecture for this trivial instance. Accordingly, since the ordering of $x$ and $y$ above is arbitrary, the fraction $r=y/x$ which is by definition rational, can be constrained to $0<r<1$ . Substitution into the supposition above, leads to

$$\frac{x}{z}=(1+r^n)^{-\frac{1}{n}}$$

The LHS is (again, by definition) rational. Proof of Fermat’s conjecture therefore, requires proving that the RHS cannot be rational for $n>2$, for any $r\in \mathbb{Q}$. With the Generalised Binomial Theorem, we have

$$ (1+r^n)^{-\frac{1}{n}} = \sum_{k=0}^{\infty} {{-n^{-1}}\choose{k}}r^{kn} = \sum_{k=0}^{\infty} \frac{(-n^{-1})_k}{k!}\prod_{j=1}^{kn}r $$

which converges absolutely, since $|r|<1$ ($y\neq x$), and the Pochhammer Symbol for the Binomial Coefficient is

$$(-n^{-1})_k=-n^{-1}(-n^{-1}-1)(-n^{-1}-2)\cdots(-n^{-1}-k+1)$$

Note that for finite $k$ , each term of the series is a finite product of rationals and hence rational. However, if it can be proved that in the limit $k\rightarrow \infty$, the product of the infinite products always yields an irrational number for $n>2$ and $r\in \mathbb{Q}$, then Fermat’s conjecture is at once proved. Moreover, even if this were not the case, the falsity of the conjecture would still require showing that the infinite sum of rationals provably converges to a rational. For $n=2$, there are at least *some* rational values of $r$ corresponding to the Pythagorean triples, for which the infinite sum does apparently converge to a rational number. Formal proof of Fermat’s conjecture for therefore, requires demonstrating that there are *no* rational values of $r$ for which this is the case.

I will resist the juvenile urge to “Pull a Fermat” again, and claim that I have a proof that is too large to fit here, and instead readily concede that such a proof is well above my capabilities as a lowly engineer. I do have the following questions, however:

1) Is this a viable approach?

2) Has something along these lines been attempted by Number Theorists before?

3) Can a form of the Thue-Siegel-Roth Theorem be used to determine the convergence to a rational/irrational?

I thank you for your indulgence!