Re: Polynomials and Prime Numbers
- From: "Chip Eastham" <hardmath@xxxxxxxxx>
- Date: 28 Feb 2006 13:24:14 -0800
Maury Barbato wrote:
a
Maury Barbato wrote:
Hello,
I read the following result. There's no rational
polynomial P(x) wich generates all the primes
(we say that a rational polynomial P(x) generates
prime p, if for some n in Z, we have P(n)=p).
Do you know a simple proof of this result?
Thank you for your attention.
Hi, Maury:
This seems trivially false as P(x) = x would then
"generate" prime p when P(p) = p.
Perhaps you have a more restrictive condition
in mind?
regards, chip
What a stupid I am!!!
Surely, the author meant that P(x) generates only
primes
for x in Z! So the right formulation is:
There's no rational polynomial P(x) such that P(z) is
prime for every z in Z, and, for every prime p,
there's
some z in Z such that P(z)=p.
Forgive my embarrassing slip!
My wrong formulation of the problem suggested the
following question to me:
there's a rational polynomial P(x) of degree greater
then 1, such that {P(n): n in Z} contains the set of
prime numbers?
The density of {P(n): n in Z} in [-N,+N] is O(N^-(1-1/d))
where d = degree(P) >= 2. Since this density is
strictly lower than the density of primes, which is
O(1/log(N)), there is no such polynomial, rational
or otherwise.
regards, chip
.
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- From: Maury Barbato
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