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Topic: (u-1)^n = u^n -1 (mod n^3) more than two solutions?
Replies: 24   Last Post: Jul 15, 2014 2:04 AM

 Messages: [ Previous | Next ]
 robersi Posts: 2,410 From: NNY Registered: 8/18/09
Re: (u-1)^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 UTC-4, 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?
>
>
>
>
> 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

(u-1)^n = u^n -1

has at most only two solutions}

then this proves FLT unless 2^n = 2 mod n^3.

Date Subject Author
7/12/14 robersi
7/12/14 Brian Q. Hutchings
7/12/14 robersi
7/12/14 quasi
7/12/14 robersi
7/14/14 robersi
7/14/14 quasi
7/14/14 robersi
7/14/14 quasi
7/14/14 quasi
7/14/14 robersi
7/14/14 robersi
7/14/14 quasi
7/14/14 quasi
7/14/14 robersi
7/14/14 quasi
7/15/14 robersi
7/15/14 quasi
7/15/14 robersi
7/15/14 quasi
7/15/14 quasi
7/13/14 quasi
7/14/14 Timothy Murphy
7/14/14 James Waldby
7/13/14 robersi