- In a message dated 16/12/03 17:08:42 GMT Standard Time,

hillcino368@... writes:

> I have been trying to prove and searching (Eg., the entire Caldwell Suite

Your Theorem 1 is a special case of a more general theorem, which will now be

> et

> al) for a proof of the

> following theorem I conjured by inspection.

>

> Theorem I

> Let p and q be primes. if q divides N=x^p - (x-1)^p for some integer x,

> then p divides q-1

> and q = 2kp+1 for some integer k. If N is prime then N-1 = q-1 and the

> theorem still holds.

>

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.

Proof:

As q | N, a^p = b^p mod q.

.

Let m be the _least_ power such that a^m = b^m mod q.

Then, since a^p = b^p mod q, p must be a multiple of m.

[It is easy to show that if p is not a multiple of m, then we get a

contradiction to m being the _smallest_ power,

as r = p mod m also satisfies a^r = b^r mod q and r < m.]

m <> 1, as a-b <> 0 mod q by assumption.

Thus, since p is prime and a multiple of m, p = m.

By Fermat's Little Theorem:

a^(q-1) = b^(q-1) mod q.

So, by the same reasoning as outlined above, (q-1) is a multiple of m,

i.e. (q-1) is a multiple of p.

As q and p are odd, (q-1) is even, so:

q = 2*k*p + 1.

Q.E.D.

-Mike Oakes

[Non-text portions of this message have been removed] - Mike Oakes proved
> Theorem:

are

> Let p and q be odd primes, and q divide N=a^p-b^p, where a and b

> arbitrary positive integers except for the condition

Can we also prove the case for N=a^p+b^p?

> a <> b mod q.

> Then q = 2*k*p + 1, for some integer k > 0.

>

of m,

> Proof:

> As q | N, a^p = b^p mod q.

> .

> Let m be the _least_ power such that a^m = b^m mod q.

> Then, since a^p = b^p mod q, p must be a multiple of m.

>

> [It is easy to show that if p is not a multiple of m, then we get a

> contradiction to m being the _smallest_ power,

> as r = p mod m also satisfies a^r = b^r mod q and r < m.]

>

> m <> 1, as a-b <> 0 mod q by assumption.

>

> Thus, since p is prime and a multiple of m, p = m.

>

> By Fermat's Little Theorem:

> a^(q-1) = b^(q-1) mod q.

>

> So, by the same reasoning as outlined above, (q-1) is a multiple

> i.e. (q-1) is a multiple of p.

Cino

>

> As q and p are odd, (q-1) is even, so:

> q = 2*k*p + 1.

>

> Q.E.D.

>

> -Mike Oakes

- 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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