Re: Polynomials and Prime Numbers


Maury Barbato wrote:

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 a
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!

Actually, just the statement "There's no rational polynomial P(x) such
that P(z) is
prime for every z in Z" is true, with the proviso that P(x) is not a
constant. Here's a hint. Prove that

P(x + y*P(x)) is divisible by P(x) for every pair of integers x
and y.

.

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