FLT from Mochizuki's proof of abc [closed]

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?

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closed as too localized by Igor Rivin, Andrés Caicedo, Lee Mosher, Vidit Nanda, Steven GubkinSep 27 '12 at 19:34

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Enough with Mochizuki + abc, already! If the proof is accepted, then we will all celebrate. Until then, this is fairly pointless. – Igor Rivin Sep 27 '12 at 19:21
I downvoted this because it took about 30 seconds to find the answer, starting from the Wikipedia page on the abc conjecture. – user5117 Sep 27 '12 at 19:28
but unknown yahoos want to know! – Joel Dodge Sep 27 '12 at 19:30
Igor, while I am sure many of us are getting fed up with the rampant speculatory questions regarding Mochizuki and ABC, maybe the people asking are not regular enough on MO (or meta-MO) to be familiar with the ongoing debates about acceptability of such questions. This is not a crime. My impression is that this question is independent of Mochizuki's work and simply asks for why ABC implies FLT. Since this information is easily available on the internet, the question should be closed for not being research level rather than in annoyance at other recent ABC questions. – Vidit Nanda Sep 27 '12 at 19:30
@Vel: I did answer, though (a) this particular @unknown is obviously not a complete MO novice (judging by reputation) and (b) I do agree re the easy availability of the info. – Igor Rivin Sep 27 '12 at 19:58

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 $n$.

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You need 3 + 3eps not 1 + eps when you bound the product; from your inequality you could prove there are only finitely many primitive Pythagorean triples. (What is perhaps more interesting is that while abc will yield finiteness of solutions for all n=> 4, it won't for n=3, which is not unexpected since heuristically/analytically one could actually expect an infinitude of solutions for n=3.) – user9072 Oct 4 '12 at 16:39
@quid: That was obviously a typo... Thanks. – José Hdz. Stgo. Oct 5 '12 at 5:46
So, with Dimitrov's new preprint at arxiv.org/pdf/1601.03572.pdf , what are the consequences for FLT? Does one get a reasonable estimate for the exponent? – ThiKu Jan 19 at 19:03

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.

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