We know that there are at most finitely many counter-examples to Beal's conjecture, assuming Mochizuki's proof of the abc conjecture is valid. Specifically, a coprime solution to the Beal equation: x^p + y^q = z^r with p, q, r >= 3 must satisfy the following inequality (since p, q and r cannot all be equal to 3 by the special case of FLT established by Euler): 1/p + 1/q + 1/r <= 1/3 + 1/3 + 1/4 = 11/12 Now define a = x^p, b = y^q, c = z^r, so we obtain: a + b = c with a,b,c coprime. Then rad(abc) <= xyz = a^(1/p) b^(1/q) c^(1/r) < c^(1/p + 1/q + 1/r) <= c^(11/12). Hence, infinitely many counter-examples to Beal's conjecture implies that we can find infinitely many abc triples with c >= rad(abc)^(12/11), which contradicts the abc conjecture. Sincerely, Adam P. Goucher
Sent: Saturday, October 25, 2014 at 1:25 PM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Announced proof of Beal's conjecture (I)
On the other hand, see http://www.coolissues.com/mathematics/Beal/beal.htm
<< Beal's Conjecture is disproved for the same reasons Fermat's Last Theorem is proved. >>
Simple when you know how, ennit ... WFL
On 10/25/14, Dan Asimov <dasimov@earthlink.net> wrote:
I haven't looked at this, but it's freely downloadable here:
< http://www.scirp.org/journal/APM/ >.
--Dan
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