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[PrimeNumbers] Re: If q divides a^p +b^p then p divides q-1

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  • mikeoakes2@aol.com
    In a message dated 05/03/04 09:26:14 GMT Standard Time, ... Yes. Nowhere is the positivity of a or b used in the proof (so the Theorem should not have required
    Message 1 of 4 , Mar 5, 2004
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      In a message dated 05/03/04 09:26:14 GMT Standard Time,
      hillcino368@... writes:


      > Mike Oakes proved
      > > Theorem:
      > > Let p and q be odd primes, and q divide N=a^p-b^p, where a and b
      > are
      > > arbitrary positive integers except for the condition
      > > a <> b mod q.
      > > Then q = 2*k*p + 1, for some integer k > 0.
      >
      > Can we also prove the case for N=a^p+b^p?
      >

      Yes.
      Nowhere is the positivity of a or b used in the proof (so the Theorem should
      not have required this, in fact).
      Putting b => -b gives your case (since p is odd).

      -Mike Oakes



      [Non-text portions of this message have been removed]
    • cino hilliard
      ... But of course! Thanks. I also noticed that if p is not prime then the prime factors of p divide q-1. This applies to even numbers also. Does the even case
      Message 2 of 4 , Mar 5, 2004
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        >In a message dated 05/03/04 09:26:14 GMT Standard Time,
        >hillcino368@... writes:
        >
        >
        > > Mike Oakes proved
        > > > Theorem:
        > > > Let p and q be odd primes, and q divide N=a^p-b^p, where a and b
        > > are
        > > > arbitrary positive integers except for the condition
        > > > a <> b mod q.
        > > > Then q = 2*k*p + 1, for some integer k > 0.
        > >
        > > Can we also prove the case for N=a^p+b^p?
        > >
        >
        >Yes.
        >Nowhere is the positivity of a or b used in the proof (so the Theorem
        >should
        >not have required this, in fact).
        >Putting b => -b gives your case (since p is odd).
        >
        >-Mike Oakes
        >
        But of course! Thanks.

        I also noticed that if p is not prime then the prime factors of p divide
        q-1. This applies to even numbers also. Does the even case follow from
        a^(2p) - b^(2p) = (a^p-b^p)(a^p+b^p) =
        q1*q2*q3*...?

        Cino

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