- --- In primenumbers@yahoogroups.com, Wed March 29, 2006, "Jens Kruse

Andersen" <jens.k.a@...> wrote:> Heuristics support this plausible conjecture:

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

> gives distinct positive primes for the first k integers.

The zoo of prime numbers did not accustom us to so much simplicity and

absence of special constraints.

In spite of similar formal aspects, we are far here from the idea that

for any k, there is a polynomial of degree n which gives distinct

positive primes for the first k integers.

Paradoxical situation : we have a lot of difficulties to find out one

quadratic polynomial giving more than 40 distinct positive prime

numbers , and during this time one can imagine the existence of a

quadratic polynomial P(x) giving distinct and positive prime numbers

when x goes from 0 to 10^10 ...

Regards,

Patrick Capelle. - Only a comment.

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

isnt enough to determine the assoiated roots.

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

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

because I didnt let N vary and I didnt include what

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

in the different bases b with the same coefficients c_i.

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

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

Arent there theorems that tell you when polynomials have

non-integer root pairs so that you can make a statement

towards their construction. I think there are. There

should be large classes of integers N of the same degree

and with the same coefficients c_j but with varying base;

I think these can be constructed in ln^3 N_{max} time with

N_{max} being the largest integer considered.

The construction I refer to is given in Fast Factoring

by Dr. Gordon Chalmers. - --- 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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