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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 1:45 AM
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On Tuesday, July 15, 2014 1:20:59 AM UTC-4, quasi wrote:
> robersi730 wrote:
>

> >
>
> >a^p = -b^p mod (p^p) if p divides c
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>
>
> Yes, if p|c then
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>
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> a^p + b^p + c^p = 0
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>
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> => a^p = -b^p (mod p^p)
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>
>

> >a=-b mod p by F Little T
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>
>
> Right.
>
>
>

> >a = -b + pq
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>
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> Right.
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>
>

> >and not a = -b + (p^2)q, p does not divide q
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> No.
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> In the equation
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>
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> a = -b + pq
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> you can't claim that p does not divide q.


This is the assumption that leads to a contradiction.
>
>
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> >binomial expansion yield from the former
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> >
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> >a^p = -b^p mod p^2 not p^3
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>


>
> Wrong.

Right, wrong contradiction based on the assumption.
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> Under the conditions of the problem, if p|c, it can shown
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> that (p^(p-1))|(a + b).
>
>

yes I know.
but p^2|(a+b) will suffice
>
> quasi





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