- This is exactly the case as you described it.If 5

didnt divide k,then we would have a finite covering

set.But since 5/k then the numbers k*2^(4*m+2)+1 have

an algebraic factorization.

Let k=r^4.Then:

r^4*2^(4*m+2)+1=4*(r*2^m)^4+1

=[2*(r*2^m)^2+2*(r*2^m)+1]*[2*(r*2^m)^2-2*(r*2^m)+1]=A*B

I have not proven that A or B are divisible by

infinitely many different primes as m varies ,but

this seems very logical.

I tried to prove it without success by assuming that

there is a finite set of primes[p1,p2,,,pj] that

divide A for every m>1 ,for a specific k.Then i used

as a value for m,m=(p1-1)*(p2-1)*...*(pj-1) and tried

to end up with something absurd.No luck.This is the

most interesting part of my effort.The k's i found

produce numbers k*2^n+1,some of which have factors

belonging to a finite set and some of them have

factors belonging to the set of primes that divide a

polynomial.I dont know what step is next.Your valuable

ideas are welcomed

Pavlos

--- jbrennen <jack@...> wrote:> --- In primenumbers@y..., Pavlos N <pavlos199@y...>

__________________________________________________

> wrote:

> > I managed to construct a Sierpinski multiplier k

> such

> > that k*2^n+1 is always composite for every n>1 and

> > there is no finite set of primes that divide it.I

> will

> > post a more detailed email explaining exactly how

> i

> > worked and other details.For now on i post my

> final

> > result only.The k's belong to the following set of

> > integers:

> > k=(18446744073709551615*a+2039404931861161240)^4

> > where a is a positive integer

>

> This is a really cool finding. I suspected that

> such a multiplier

> (or family of multipliers) was possible, but I never

> thought of

> using the fact that (4*x^4+1) has an algebraic

> factorization. I

> tried looking for k being a perfect cube or a fifth

> power, with

> no luck. Great work. :)

>

> So what we have here is:

>

> k*2^n+1 is of the form 4*x^4+1 if n

> == 2 (mod 4)

>

> k mod 3 == 1 -> 3 | k*2^n+1 if n

> == 1 (mod 2)

> k mod 17 == 16 -> 17 | k*2^n+1 if n

> == 0 (mod 8)

> k mod 257 == 16 -> 257 | k*2^n+1 if n

> == 4 (mod 16)

> k mod 65537 == 16 -> 65537 | k*2^n+1 if n

> == 12 (mod 32)

> k mod 641 == 16 -> 641 | k*2^n+1 if n

> == 28 (mod 64)

> k mod 6700417 == -16 -> 6700417 | k*2^n+1 if n

> == 60 (mod 64)

>

> Putting them together, k*2^n+1 is always composite.

>

> The requirement for k to be divisible by 5 is to

> ensure that

> when n == 2 (mod 4), that k*2^n+1 is not divisible

> by 5 -- if

> it is so divisible, then a finite covering set of

> primes exists.

>

> So all that is left is to prove that for at least

> one k in the

> given group, no covering set S of primes exists such

> that every

> k*2^n+1 is divisible by a member of S. Please let

> us in on this

> aspect of the finding... How are you able to prove

> the

> non-existence of a finite covering set? Or is the

> non-existence

> of a finite covering set only a conjecture?

>

>

>

>

Do You Yahoo!?

Yahoo! Finance - Get real-time stock quotes

http://finance.yahoo.com - Just looked in between trips. Skipping predictable

Brown trash, I saw a really _neat_ thing by

Pavlos Saridis:

To make N=k*2^n+1 always composite [Saridis]:

a) take k=x^4 so that for n=4*s+2 we have an

algebraic factorization

N=(2*y^2-2*y+1)*(2*y^2+2*y+1) with y=2^s*x

b) find a covering set for n != 2 mod 4

c) if desired, make x=0 mod 5, so that p=5

does not cover what you did with Aurifeuille in (a)

Question: What is the smallest k=x^4 for which

this can be done, with and without the grace note (c)?

Back in a few weeks time...

David