Question

If Mochizuki's proof of abc is correct, why would this provide a new proof of FLT?

Edit: In proof of asymptotic FLT, does Mochizuki claim a specific value of n and if so what is this value?

Answer 1

The two answers of J.H.S and Igor Rivin don't really answer the question : they prove FLT only for an exposant $n$ large enough. Since Mochizuki's version of ABC is effective, one could certainly find an effective bound for $n$, and maybe prove by hand the remaining $n$. But as they stand, they are incomplete.

Here is a short proof that Mochizuku implies Fermat. If the proof is correct, since it is also incorrect (cf. the second answer to [this question] by Vasselin Dimitrov), it implies trivially Fermat't last theorem as well as its contrary.

I guess I should explicit my point: it is prematurate to ask precise questions on the consequence of Mochizuki's proof. As for vague philosophical question, as the one given in the link above, that have a problem too : they are vague. Let us at least wait for the MO community to accept or not Vasselin Dimitrov's very interesting answer, which moreover is exactly about the effective version of ABC conjecture needed to answer the PO's question, before asking such questions.

[1] http://mathoverflow.net/questions/106560/philosophy-behind-mochizukis-work-on-the-abc-conjecture

Answer 2

Let us suppose that $x^{n}+y^{n}= z^{n}$ with $x, y,$ and $z$ relatively prime. By the abc conjecture, $|x^{n}|\ll |xyz|^{1+\epsilon}$, $|y^{n}|\ll |xyz|^{1+\epsilon}$ and $|z^{n}|\ll |xyz|^{1+\epsilon}$. Therefore, $|xyz|^{n}\ll |xyz|^{3+\epsilon}$ which implies that for $|xyz|>1$, $n$ is bounded. So, what we actually have is a proof of an asymptotic version of FLT. Nevertheless, if we have explicit information regarding the constant in the abc conjecture, we could determine explicit bounds for $n$.

Answer 3

See http://mathworld.wolfram.com/abcConjecture.html