robersi
Posts:
901
From:
Northern New York
Registered:
8/18/09


Re: (u1)^n = u^n 1 (mod n^3) more than two solutions?
Posted:
Jul 15, 2014 12:00 AM


On Monday, July 14, 2014 11:48:21 PM UTC4, quasi wrote: > >robersi730 wrote: > > >>quasi wrote: > > >>robersi730 wrote: > > >>> > > >>>also I can prove n doesn't divide the product abc > > >> > > >>Not believable (unless you plan to invoke Wiles' Theorem). > > > > > >I can, I did, and I can. > > > > So let me get this straight ... > > > > Are you claiming that you have an elementary proof that the > > equation > > > > a^p + b^p = c^p > > > > where > > > > p is an odd prime > > a,b,c are pairwise coprime nonzero integers > > p divides abc > > > > has no solutions? > > > > In other words, you claim to have an elementary proof that > > case (2) of FLT is impossible? > > > > Your pdf article (from 2011) definitely fails to achieve that > > goal. > > > > If you think you have a proof, let's see it. > > > > quasi
I am only claiming I had the logic for
a + b = c mod n^2 a^n = a mod n^2 b^n = b mod n^2 c^n = c mod n^2
regardless of whether n divides abc or not
and then logic for the proof
n does not divide abc
and then showing {if
(u1)^n = u^n 1
has at most only two solutions}
then this proves FLT unless 2^n = 2 mod n^3.

