Does Mochizuki's proof of abc conjecture gives an upper bound for the quality of a triple?

Asked Nov 16 '15 at 08:37

Active Mar 15 '16 at 20:41

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The quality of a triple $(a,b,c)$ of coprime positive integers with $a + b = c$ is defined as $$q(a,b,c) := \frac{\log(c)}{\log(\mathrm{rad}(abc))}.$$ Then $$a+b = c = \mathrm{rad}(abc)^{q(a,b,c)}.$$
The triple with the highest known quality is $(2, 3^{10} \cdot 109, 23^5)$, with $q(a,b,c) = 1.6299...$
If the abc conjecture is true then the quality of any triple is bounded (expected by $2$).

Assuming Mochizuki's proof correct, does it gives an explicit upper bound for the quality of a triple?

edited Nov 16 '15 at 15:30

GH from MO

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asked Nov 16 '15 at 08:37

Sebastien Palcoux

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Somewhat related: http://mathoverflow.net/a/106399 – Nov 16 '15 at 12:31
3

The role of (many!) Belyi maps in the proof (if correct...) is probably a substantial obstacle to extracting any explicit numerical bounds for theoretical situations. – nfdc23 Nov 16 '15 at 14:41
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The answer is no, according to Mochizuki. – Felipe Voloch Dec 18 '15 at 11:36

Does Mochizuki's proof of abc conjecture gives an upper bound for the quality of a triple?

0 Answers0