Tarwyn,

A couple of things... first of all, I made a mistake in writing out that theorem. Instead of saying

If p and q are distinct primes, p > q, and (a,p) = 1, then a^(p-1) == 1 (mod
pq)

I should have said

If p and q are distinct primes, p > q, and (a,p) = 1, then a^p == a (mod
pq).

(I'm using == for "is congruent to")

This works for some "trivial examples" like
p=5, q=2: a^5 == a (mod 10) (my brother's example).
p=7, q=3: a^7 == a (mod 21)
...

HOWEVER... even though it worked for enough trivial examples to make me confident enough to send it to you as a conjecture, right after I sent that
e-mail, I found a counter-example!

p=7, q=5: a^7 =/= a (mod 35)... (but a^13 == a!)

So... let me CAREFULLY restate the problem here :) ...

Take the congruence

a^n == a (mod m)

The question is in two parts:

1) For what values of m does there exist an n > 1 that works for all integers a?

We know that when m is prime, n = m. But when m is composite, sometimes there's an n, and sometimes there's not. What determines when such an n exists?

2) Can we come up with a function or algorithm that we can use to find n given an m?

There... try that. :)

As for the theorem he sent... is he basically saying that there is no F(x) that will be equivalent to P(n) for ALL n's? It's definitely interesting, and I'd like to play with it sometime, but most of what used to be my free math time has been eaten up by a graduate Real & Complex Analysis course I'm taking this year.

Cheers! :)

§§ There is your explanation Mike... §§

J J J

J J J