robersi
Posts:
685
From:
Northern New York
Registered:
8/18/09


Re: (u1)^n = u^n 1 (mod n^3) more than two solutions?
Posted:
Jul 14, 2014 10:38 PM


On Monday, July 14, 2014 10:22:38 PM UTC4, rober...@gmail.com wrote: > On Monday, July 14, 2014 9:35:04 PM UTC4, quasi wrote: > > > quasi wrote: > > > > > > >robersi730 wrote: > > > > > > >>quasi wrote: > > > > > > >>>robersi730 wrote: > > > > > > >>>> > > > > > > >>>>Let there be a non zero integer solution to a^n+b^n=c^n > > > > > > >>>>where n is an odd prime. > > > > > > >>>> > > > > > > >>>>Also let a,b,c be such that none is divisible by n. > > > > > > >>> > > > > > > >>>Ok so far (for case 1 of FLT) > > > > > > >>> > > > > > > >>>>And let a+b=c (mod n^2) > > > > > > >>> > > > > > > >>>That's a pretty strong assumption. > > > > > > >>> > > > > > > >>>By Fermat's little Theorem, it's automatic that > > > > > > >>> > > > > > > >>> a + b = c (mod n) > > > > > > >>> > > > > > > >>>but I don't see how to prove > > > > > > >>> > > > > > > >>> a + b = c (mod n^2) > > > > > > >>> > > > > > > >>>Can you prove the above from the prior hypotheses? > > > > > > >>> > > > > > > >>>If not, any contradiction you might get would only prove > > > > > > >>>that a hypothetical nontrivial solution a,b,c would have > > > > > > >>>to be such that a + b != c (mod n^2). > > > > > > >> > > > > > > >>Yea actually I can prove > > > > > > >> > > > > > > >> a^n = a mod n^2 > > > > > > >> b^n = b mod n^2 > > > > > > >> c^n = c mod n^2 > > > > > > > > > > > > > >Sorry, I looked at the pdf of your claimed proof. > > > > > > > > > > > > > >The proof has fatal flaws. > > > > > > > > > > > > Parts of your proof are correct, so there may actually be > > > > > > a nontrivial result that can be salvaged from that mess. > > > > > > > > > > > > If I have time, I'll try to post a readable version of what > > > > > > I think your work actually shows. > > > > > > > > > > > > quasi > > > > What do you think it actually shows? nevermind
> > > > What are the flaws? > nevermind > > > If you can write it better be my guest. > I'll be second author > > > Mess? yes Sound? YES! nevermind
Simon

