FLT from Mochizuki's proof of abc - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-25T08:18:01Z http://mathoverflow.net/feeds/question/108277 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108277/flt-from-mochizukis-proof-of-abc FLT from Mochizuki's proof of abc unknown (yahoo) 2012-09-27T19:11:59Z 2012-10-05T05:44:36Z <p>If Mochizuki's proof of abc is correct, why would this provide a new proof of FLT?</p> <p>Edit: In proof of asymptotic FLT, does Mochizuki claim a specific value of n and if so what is this value?</p> http://mathoverflow.net/questions/108277/flt-from-mochizukis-proof-of-abc/108278#108278 Answer by Igor Rivin for FLT from Mochizuki's proof of abc Igor Rivin 2012-09-27T19:20:23Z 2012-09-27T19:20:23Z <p>See <a href="http://mathworld.wolfram.com/abcConjecture.html" rel="nofollow">http://mathworld.wolfram.com/abcConjecture.html</a></p> http://mathoverflow.net/questions/108277/flt-from-mochizukis-proof-of-abc/108279#108279 Answer by J. H. S. for FLT from Mochizuki's proof of abc J. H. S. 2012-09-27T19:21:06Z 2012-10-05T05:44:36Z <p>Let us suppose that $x^{n}+y^{n}= z^{n}$ with $x, y,$ and $z$ relatively prime. By the <em>abc</em> 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 <em>asymptotic</em> version of FLT. Nevertheless, if we have explicit information regarding the constant in the <em>abc</em> conjecture, we could determine explicit bounds for $n$. </p> http://mathoverflow.net/questions/108277/flt-from-mochizukis-proof-of-abc/108281#108281 Answer by Joël for FLT from Mochizuki's proof of abc Joël 2012-09-27T19:33:02Z 2012-09-27T19:33:02Z <p>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.</p> <p>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.</p> <p>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.</p> <p>[1] <a href="http://mathoverflow.net/questions/106560/philosophy-behind-mochizukis-work-on-the-abc-conjecture" rel="nofollow">http://mathoverflow.net/questions/106560/philosophy-behind-mochizukis-work-on-the-abc-conjecture</a></p>