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Topic: n does not divide abc
Replies: 4   Last Post: Jul 15, 2014 2:18 AM

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robersi

Posts: 116
From: NYS
Registered: 8/18/09
Re: n does not divide abc
Posted: Jul 15, 2014 1:23 AM
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On Tuesday, July 15, 2014 12:39:32 AM UTC-4, quasi wrote:
> robersi730 wrote:
>

> >
>
> >Assume a^n + b^n = c^n where a,b,c are non zero integers
>
> >and n is an odd prime.
>
> >
>
> >and the gcd(a,b,c)=1
>
> >
>
> >also assume
>
> >
>
> >a^n-a=0 (mod n^2)
>
> >b^n-b=0 (mod n^2)
>
> >c^n-c=0 (mod n^2)
>
>
>
> You can't assume the above congruences.
>
>
>
> Your claimed proof of those congruences only works if n _doesn't_
>
> divide abc.
>
>
>
> quasi


It does work. I showed a+b=0 mod p^2
and of course a^p+b^p=0 mod p^2.

This result can be added to your edited version nicely.
As for the clarity of this addition I dunno.

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 p|c

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



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