robersi
Posts:
2,007
From:
NNY
Registered:
8/18/09


Re: (u1)^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 UTC4, 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)
contradiction. therefore
a + b = 0 mod p^2
and
a^p + b^p = a + b mod p^2
when pc
(p doesn't divide ab *said w.o.l.o.g.)

