- In a message dated 05/03/04 09:26:14 GMT Standard Time,

hillcino368@... writes:

> Mike Oakes proved

Yes.

> > 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 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,

But of course! Thanks.

>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

>

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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