## Re: prime generating quadratic conjecture

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• ... 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
Message 1 of 14 , Apr 1, 2006
--- In primenumbers@yahoogroups.com, Wed March 29, 2006, "Jens Kruse
Andersen" <jens.k.a@...> wrote:
> Heuristics support this plausible conjecture:
> 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
Message 2 of 14 , Apr 2, 2006
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.
• ... First you need to specify the ring in which you re working. You do not do so. One cannot assume Z[x] given that many of the examples we ve seen have been
Message 3 of 14 , Apr 2, 2006
--- gordon_as_number <gordon_as_number@...> wrote:
> Only a comment.
>
> If you want to prove a polynomials P(x) is prime,

First you need to specify the ring in which you're working.
You do not do so.
One cannot assume Z[x] given that many of the examples we've seen have

> 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 .

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

> 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

Order(funtion(N)) only makes sense if you let N vary.
You've reached that level of illucidity again...

> and I didnt include what
> the value of m is. So you have to polynomials evaluated
> in the different bases b

Yup - illucid. Polynomials can be evaluated, but their value is
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
> = integer or (x-\alpha_i) an integer span the prime factors.

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

Phil

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