robersi
Posts:
2,398
From:
NNY
Registered:
8/18/09


Re: a+b=c mod n^2 for a^n+b^n=c^n
Posted:
Jul 15, 2014 1:45 AM


On Tuesday, July 15, 2014 1:20:59 AM UTC4, quasi wrote: > robersi730 wrote: > > > > > >a^p = b^p mod (p^p) if p divides c > > > > Yes, if pc then > > > > a^p + b^p + c^p = 0 > > > > => a^p = b^p (mod p^p) > > > > >a=b mod p by F Little T > > > > Right. > > > > >a = b + pq > > > > Right. > > > > >and not a = b + (p^2)q, p does not divide q > > > > No. > > > > In the equation > > > > a = b + pq > > > > you can't claim that p does not divide q.
This is the assumption that leads to a contradiction. > > > > >binomial expansion yield from the former > > > > > >a^p = b^p mod p^2 not p^3 > >
> > Wrong. Right, wrong contradiction based on the assumption. > > > > Under the conditions of the problem, if pc, it can shown > > that (p^(p1))(a + b). > > yes I know. but p^2(a+b) will suffice > > quasi

