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Timeline for answer to Effect of abc conjecture on Fermat's Last Theorem by user9072

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Dec 14, 2018 at 1:10 comment added Đào Thanh Oai @DietrichBurde Please see mathoverflow.net/questions/303141
May 17, 2013 at 20:13 comment added user9072 @Dietrich Burde: I think even for $n \ge 6$, as the inequality is strict.
May 17, 2013 at 20:11 history edited user9072 CC BY-SA 3.0
monir addition in view of an edit to the question
May 17, 2013 at 17:52 comment added Dietrich Burde If we would know that $C_1=1$ for $\epsilon =1$ (and no counterexample is known to that), then FLT would follow for exponents $n\ge 7$.
May 17, 2013 at 17:49 comment added user9072 You are welcome. Yes in some sense one can consider so to say "two limits" (in $z$ and in $n$) and thus there are in some sense different versions. Yet the finiteness of all solutions (under n> 3) contains all, as if there are only finitely many in total then there is a largest $n$ and a largest $z$ and so on. And, while I read it differently at first, thus my edit, I think the version you link to actually is meant to assert the finiteness of the set of all solutions , ie couples $(z,n)$ and thus contains Lang's.
May 17, 2013 at 17:41 vote accept Favst
May 17, 2013 at 17:41 comment added Favst Thanks, quid. So there are two asymptotic versions of FLT which none of my sources cared to distinguish!
May 17, 2013 at 17:27 history edited user9072 CC BY-SA 3.0
added "srongest possible" comment
May 17, 2013 at 17:22 history edited user9072 CC BY-SA 3.0
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May 17, 2013 at 17:16 history answered user9072 CC BY-SA 3.0