--- 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

been instead in Q[x].

> 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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