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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 12:43 AM

On Tuesday, July 15, 2014 12:35:33 AM UTC-4, quasi wrote:
> robersi730 wrote:
>

> >
>
> >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
>
>
>
> No, I don't accept that.
>
>
>
> Your proof of the above only works if n _doesn't_ divide abc.
>
>
>
> quasi

a^p = -b^p mod (p^p)

if p divides c

a=-b mod p by F Little T

a = -b + pq

and not

a = -b + p^2q'

*note p does not divide q

binomial expansion yield from the former

a^p = -b^p mod (p^2 not p^3)

a + b = 0 mod p^2

and

a^p + b^p = a + b mod p^2

when p|c

(p doesn't divide ab *said w.o.l.o.g.)

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