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

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  • 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 1 of 4 , Mar 5, 2004
      >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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