>In a message dated 05/03/04 09:26:14 GMT Standard Time,
> > 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?
>Nowhere is the positivity of a or b used in the proof (so the Theorem
>not have required this, in fact).
>Putting b => -b gives your case (since p is odd).
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) =
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