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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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    $\begingroup$ Enough with Mochizuki + abc, already! If the proof is accepted, then we will all celebrate. Until then, this is fairly pointless. $\endgroup$
    – Igor Rivin
    Sep 27, 2012 at 19:21
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    $\begingroup$ I downvoted this because it took about 30 seconds to find the answer, starting from the Wikipedia page on the abc conjecture. $\endgroup$
    – user5117
    Sep 27, 2012 at 19:28
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    $\begingroup$ but unknown yahoos want to know! $\endgroup$
    – Joel Dodge
    Sep 27, 2012 at 19:30
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    $\begingroup$ 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. $\endgroup$ Sep 27, 2012 at 19:30
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    $\begingroup$ @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. $\endgroup$
    – Igor Rivin
    Sep 27, 2012 at 19:58

3 Answers 3

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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] Philosophy behind Mochizuki's work on the ABC conjecture

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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. Unfortunately, this establishes only an asymptotic version of FLT. Nevertheless, if we had explicit information regarding the implied constant in the abc conjecture, we could in principle determine explicit upper bounds for those $n$'s for which the abc conjecture doesn't settle FLT.

UPDATE (October 26th, 2021). In the abstract of "Explicit Estimates in Inter-universal Teichmuller Theory", Mochizuki et. al. we read this:

In the final paper of a series of papers concerning inter-universal Teichmüller theory, Mochizuki verified various numerically non-effective versions of the Vojta, ABC, and Szpiro Conjectures over number fields. In the present paper, we obtain various numerically effective versions of Mochizuki’s results... These numerically effective versions imply effective diophantine results such as an effective version of the ABC inequality over mono-complex number fields [i.e., the rational number field or an imaginary quadratic field] and effective versions of conjectures of Szpiro. We also obtain an explicit estimate concerning “Fermat’s Last Theorem” (FLT)—i.e., to the effect that FLT holds for prime exponents greater than $1.615\cdot 10^{14}$—which is sufficient to give an alternative proof of the first case of FLT. In the second case of FLT, if one combines the techniques of the present paper with a recent estimate due to Mihăilescu, then the lower bound $1.615 \cdot 10^{14}$ can be improved to $3.35\cdot 10^{9}$.

If I understand correctly, this paper was uploaded to Mochizuki's homepage four months ago. I dared to update my reply because I consider that the lines above give a definite answer to the OP's second question (regardless of what the generalized opinion about IUTT is).

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  • $\begingroup$ 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.) $\endgroup$
    – user9072
    Oct 4, 2012 at 16:39
  • $\begingroup$ @quid: That was obviously a typo... Thanks. $\endgroup$ Oct 5, 2012 at 5:46
  • $\begingroup$ 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? $\endgroup$
    – ThiKu
    Jan 19, 2016 at 19:03
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See http://mathworld.wolfram.com/abcConjecture.html

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