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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

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robersi

Posts: 116
From: NYS
Registered: 8/18/09
Re: (u-1)^n = u^n -1 (mod n^3) more than two solutions?
Posted: Jul 14, 2014 10:38 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On Monday, July 14, 2014 10:22:38 PM UTC-4, rober...@gmail.com wrote:
> On Monday, July 14, 2014 9:35:04 PM UTC-4, 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 non-trivial 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


Date Subject Author
7/12/14
Read (u-1)^n = u^n -1 (mod n^3) more than two solutions?
robersi
7/12/14
Read Re: (u-1)^n = u^n -1 (mod n^3) more than two solutions?
Brian Q. Hutchings
7/12/14
Read Re: (u-1)^n = u^n -1 (mod n^3) more than two solutions?
robersi
7/12/14
Read Re: (u-1)^n = u^n -1 (mod n^3) more than two solutions?
quasi
7/12/14
Read Re: (u-1)^n = u^n -1 (mod n^3) more than two solutions?
robersi
7/14/14
Read Re: (u-1)^n = u^n -1 (mod n^3) more than two solutions?
robersi
7/14/14
Read Re: (u-1)^n = u^n -1 (mod n^3) more than two solutions?
quasi
7/14/14
Read Re: (u-1)^n = u^n -1 (mod n^3) more than two solutions?
robersi
7/14/14
Read Re: (u-1)^n = u^n -1 (mod n^3) more than two solutions?
quasi
7/14/14
Read Re: (u-1)^n = u^n -1 (mod n^3) more than two solutions?
quasi
7/14/14
Read Re: (u-1)^n = u^n -1 (mod n^3) more than two solutions?
robersi
7/14/14
Read Re: (u-1)^n = u^n -1 (mod n^3) more than two solutions?
robersi
7/14/14
Read Re: (u-1)^n = u^n -1 (mod n^3) more than two solutions?
quasi
7/14/14
Read Re: (u-1)^n = u^n -1 (mod n^3) more than two solutions?
quasi
7/14/14
Read Re: (u-1)^n = u^n -1 (mod n^3) more than two solutions?
robersi
7/14/14
Read Re: (u-1)^n = u^n -1 (mod n^3) more than two solutions?
quasi
7/15/14
Read Re: (u-1)^n = u^n -1 (mod n^3) more than two solutions?
robersi
7/15/14
Read Re: (u-1)^n = u^n -1 (mod n^3) more than two solutions?
quasi
7/15/14
Read Re: (u-1)^n = u^n -1 (mod n^3) more than two solutions?
robersi
7/15/14
Read Re: (u-1)^n = u^n -1 (mod n^3) more than two solutions?
quasi
7/15/14
Read Re: (u-1)^n = u^n -1 (mod n^3) more than two solutions?
quasi
7/13/14
Read Re: (u-1)^n = u^n -1 (mod n^3) more than two solutions?
quasi
7/14/14
Read Re: (u-1)^n = u^n -1 (mod n^3) more than two solutions?
Timothy Murphy
7/14/14
Read Re: (u-1)^n = u^n -1 (mod n^3) more than two solutions?
James Waldby
7/13/14
Read Re: (u-1)^n = u^n -1 (mod n^3) more than two solutions?
robersi

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