- Patrick Capelle wrote:

> It would better to ask :

You may have misunderstood this "limit".

> Do all the prime-generating polynomials of degree 3 (without absolute

> value) give less than 40 positive and distinct prime numbers ?

>

> This question is related to the problem of the eventual existence of a

> limit.

> There is already a limit for the quadratic polynomials of Euler's form.

If c = 2, 3, 5, 11, 17, or 41, then x^2-x+c is prime for x = 0..c-1.

It has been proved there are no other such c values.

See http://mathworld.wolfram.com/LuckyNumberofEuler.html

(That page incorrectly says x = 0..c-2 but it makes no

difference for the solutions).

However, there may be large c for which x^2-x+c is prime

for x = 0..41 or longer. Finding such a c is infeasible.

> If we do an extension for the degree n, we could ask :

Probably not. Heuristics support this plausible conjecture:

> is there a limit, for each degree n, in the number of positive and

> distinct prime numbers given by the prime-generating polynomials of

> degree n ?

For any n>0 and k, there is a polynomial of degree n which

gives distinct positive primes for the first k integers.

It is trivial to show that for any n and p, there are infinitely many

polynomials of degree n which never have a prime factor <= p.

Consider x^n-x+c. x=0 and x=1 give the same value.

Then for any prime q, there must be at least one value x^n-x

cannot take modulo q.

Choose c (mod q) to ensure that q never divides x^n-x+c.

There is no polynomial which gives primes for infinitely many

consecutive integers - except for a constant polynomial f(x) = p ;-)

--

Jens Kruse Andersen - --- gordon_as_number <gordon_as_number@...> wrote:
> Only a comment.

First you need to specify the ring in which you're working.

>

> If you want to prove a polynomials P(x) is prime,

You do not do so.

One cannot assume Z[x] given that many of the examples we've seen have

been instead in Q[x].

> isnt enough to determine the assoiated roots.

Bases ought to be irrelevant. If they're not, you're doing something wrong.

>

> E.g. P(x)=x^2+2

>

> The number N is P(x) which can base expanded in

> some base b as

>

> N=\sum_{i=0}^m c_i b^i .

> Determining the roots and the coefficients I claim

Order(funtion(N)) only makes sense if you let N vary.

> is order \ln^3(N), which is a rough estimate here

> because I didnt let N vary

You've reached that level of illucidity again...

> and I didnt include what

Yup - illucid. Polynomials can be evaluated, but their value is

> the value of m is. So you have to polynomials evaluated

> in the different bases b

independent of any base.

> with the same coefficients c_i.

The c_i, as you introduced them, are not coefficients.

> The number of real c.c. pairs (x-\alpha_i)*(x-\alpha_j)^\star

With no firm foundation to build on, you've completely lost me now.

> = integer or (x-\alpha_i) an integer span the prime factors.

No point in continuing.

Phil

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