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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

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robersi

Posts: 116
From: NYS
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
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On Monday, July 14, 2014 11:00:21 PM UTC-4, 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^(p-1-k))*(b^k), k = 0..(p-1))
>
>
>
> = sum(((-1)^k)*(a^(p-1-k))*((-a)^k), k = 0..(p-1)) (mod q)
>
>
>
> = sum((a^(p-1)), k = 0..(p-1)) (mod q)
>
>
>
> = p*(a^(p-1)) (mod q)
>
>
>
> But then
>
>
>
> T = 0 (mod q)
>
>
>
> => p*(a^(p-1)) = 0 (mod q)
>
>
>
> => (q = p) or q|a.
>
>
>
> If q = p, then
>
>
>
> q|T and T|c^p
>
>
>
> => q|c^p
>
>
>
> => q|c
>
>
>
> => p|c,
>
>
>
> contrary to hypothesis.
>
>
>
> If q|a then q|b (since q|S), 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,
>
> S|c^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 p|c

(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





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