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]

>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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