Okay maybe we ought to not post maths on the board..LoL you should just email me or give this guy my email addy...

What I am looking at reminds me of the euler totient, and of fermats little theorem. Basically Fermats little theorem is
if p a prime number and (a,p) = 1 then a^(p-1) == 1 mod p. I am now going to use group theoretic language, so if you are not familiar with group theory
let me know and I will explain in arithmetic language. The Euler Phi function is defined for all integers n by phi(n) = | { a | a < n , (a,n) = 1 } | , that is, the number if integers less than n and relatively prime to n. If m is any integer and (m,n) = an + bm = 1 , then n is a unit modulo m, that is, n has a multipicative inverse modulo m.

Theorem 1. The collection of all units mod any integer m forms a group under multiplication modulo m.
Theorem 2. The equivalence class of n modulo m is a unit if and only if (n,m) = 1
Corollary: The group of units modulo m is a group of order phi(m) for m any integer
Theorem 3. If G is a finite group of cardinal number k and a is an element of G, then a^k = 1 "the unit in G"
Corollary : If m,n are integers with (m,n) = 1 then n^phi(m) = 1 mod m <-- half an answer
Theorem 4. If a^n == 1 mod m for any n then (a,m) = 1
Thus we have what values of n and what a will work for a given m as a solution to the equation a^n == 1 mod m
unfortunately no values of n,m will work for every integer a, since no integer is relatively prime to all integers.
The following theorems are trivial

1. for all primes p phi(p) = p - 1
2. if (a,b) = 1 then phi(a*b) = phi(a)*phi(b)
3. for all primes p and integers k, phi(p^k) = p^k - p^(k-1) = (p^k)(1 - (1/p))

from these three theorems we obtain, let m be any integer with prime factorization m = (q_1)^(k_1)*...*(q_n)^(k_n)
then phi(m) = m((1 - 1/(q_1))*...*(1 - 1/(q_n)), which is an explicit formula for phi(m) in terms of the prime factorization of m.

I hope this is a complete answer to your question...

I actually typed the theorem I sent incorrectly... I meant to say what you said, which is that no F(x) will be equivalent to P(n) for all n. This is the theorem for which I have a proof, however, I had not considered if F(x) can coincide with P(n) for infinite many n, I do not think that this is possible. I believe that in a few days I can in fact that F(n) can be the n'th prime for only a finite set of integers n. If you will notice the distinction, the second theorem says a great deal more than does the first.. the first mearly states that a polynomial can not produce the prime number sequence, the second gives an idea of how "close" a polynomial can get to producing the prime number sequence, it looks like a far more difficult theorem than the first. Solving the first problem took me a few days even though I a had a correct idea of how to solve it from the beginning. I am not even sure where to begin on this one.

Regards........ Mike