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Topic:
a+b=c mod n^2 for a^n+b^n=c^n
Replies:
6
Last Post:
Jul 15, 2014 2:01 AM



robersi
Posts:
2,410
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NNY
Registered:
8/18/09


Re: a+b=c mod n^2 for a^n+b^n=c^n
Posted:
Jul 15, 2014 12:20 AM


On Monday, July 14, 2014 11:00:21 PM UTC4, quasi wrote: > robersi730 wrote: > > > > > >proof here > > > > > ><https://sites.google.com/site/thetalkingfly/n21.pdf> > > > > Here's my attempt to state and prove a clear version of what > > I think Simon Robert's paper actually shows. > > > > Proposition: > > > > If a^p + b^p + c^p = 0, where > > > > (1) p is an odd prime > > (2) a,b,c are pairwise coprime nonzero integers > > (3) p does not divide any of a,b,c > > > > then > > > > a^p = a (mod p^2) > > b^p = b (mod p^2) > > c^p = c (mod p^2) > > > > Proof (by Simon Roberts, edited by quasi): > > > > Since c is nonzero, a + b is nonzero. > > > > Since p is odd, (a + b)  (a^p + b^p). > > > > Let > > > > S = a + b > > > > T = (a^p + b^p)/(a + b) > > > > Since S*T = c^p and p doesn't divide c, p doesn't divide > > either of S,T. > > > > Claim gcd(S,T) = 1. > > > > Suppose otherwise. > > > > Let q be a common prime factor of S,T. > > > > S = 0 (mod q) > > > > => b = a (mod q) > > > > Then > > > > T = (a^p + b^p)/(a + b) > > > > = sum((a^(p1k))*(b^k), k = 0..(p1)) > > > > = sum(((1)^k)*(a^(p1k))*((a)^k), k = 0..(p1)) (mod q) > > > > = sum((a^(p1)), k = 0..(p1)) (mod q) > > > > = p*(a^(p1)) (mod q) > > > > But then > > > > T = 0 (mod q) > > > > => p*(a^(p1)) = 0 (mod q) > > > > => (q = p) or qa. > > > > If q = p, then > > > > qT and Tc^p > > > > => qc^p > > > > => qc > > > > => pc, > > > > contrary to hypothesis. > > > > If qa then qb (since qS), contradiction since a,b are > > coprime. > > > > Hence gcd(S,T) = 1, as claimed. > > > > Since S,T are coprime > > > > S*T = c^p, and c nonzero > > > > => S = s^p and T = t^p for some nonzero integers s,t. > > > > By Fermat's little Theorem > > > > a^p = a (mod p) > > > > b^p = b (mod p) > > > > Hence > > > > a^p + b^p = a + b (mod p) > > > > => ST = S (mod p) > > > > [Since p doesn't divide c, > > Sc^p => p doesn't divide S] > > > > => T = 1 (mod p) > > > > => t^p = 1 (mod p) > > > > => t = 1 (mod p) [by Fermat's little Theorem] > > > > => t = 1 + u*p for some integer u > > > > => t^p = (1 + u*p)^p > > > > => t^p = 1 (mod p^2) > > > > [all terms of the expansion of (1 + u*p)^p > > are multiples of p^2 except the constant term] > > > > => T = 1 (mod p^2) > > > > => (a^p + b^p)/(a + b) = 1 (mod p^2) > > > > => a^p + b^p = a + b (mod p^2) I apologise in advance, but right here I have
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' 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.) > > > > By symmetry of a,b,c, we have > > > > [eq1] a^p + b^p = a + b (mod p^2) > > > > [eq2] b^p + c^p = b + c (mod p^2) > > > > [eq3] c^p + a^p = c + a (mod p^2) > > > > Adding eq1,eq2,eq3 and dividing by 2 (since p is odd) yields > > > > [eq4] a + b + c = 0 (mod p^2) > > > > Then > > > > eq3 + eq1  eq4 => a^p = a (mod p^2) > > > > eq1 + eq2  eq4 => b^p = b (mod p^2) > > > > eq2 + eq3  eq4 => c^p = c (mod p^2) > > > > This completes the proof. > > > > Remarks: > > > > (1) This result applies to case (1) of FLT. > > > > (2) All the logic of the proof is due to Simon Roberts. > > > > (3) My only contribution was to try to make a clear statement > > of what I think Simon Roberts' logic actually proved, and to > > try to make the proof more readable. > > > > (3) Prior to Wiles' Theorem, was this result previously known? > > > > quasi



