Re: Basically a sieve method, relation to quantum

jstevh_at_msn.com
Date: 01/26/05


Date: 25 Jan 2005 15:16:31 -0800

Michael Brown wrote:
> jstevh@msn.com wrote:
> > Michael Brown wrote:
> >> Michael Brown wrote:
> >> [...]
> >>> How do you get the original primes from the above factorisation?
Or
> >>> does j have to be specially chosen?
> >>
> >> From what I can gather, it's supposed to go something like:
> >> b_i = factors of j^2 ( = 16 in this case)
> >> f_i = factors of T (product squared minus j^2)
> >>
> >> b_1 = product of some subset of b_i
> >
> > It's a rational, typically a fraction.
>
> OK, this wasn't made clear in your previous posts. I was basing my
> assumptions off the post where you said "f_1, f_2, b_1 and b_2 are
given
> integers".
>

My mistake then, as b_1 and b_2 are not typically integers, while f_1
and f_2 typically are.

> >> b_2 = -j^2 / b1
> >
> > It's a rational, typically a fraction.
>
> Presumably still given by the relation?
>

Well, I don't bother looking at the b's, though you can calculate them
if you wish, and I guess for research purposes they'd be of interest.

The point of my paper is in letting the mathematics find them, use
them, and give you what you want, factors of M, without you having to
bother where you can't really do anything anyway, as you can't check
infinity.

> >> f_1 = product of some subset of f_i
> >> f_2 = T / f_1
> >
> > They are both integers.
>
> I'm assuming this is an agreement with the definitions of f_1 and
f_2.
>

Yes.

> >> A = some integer (not sure how to calculate this)
> >
> > It's value doesn't usually matter though it can intefere.
> >
> > Safest is to just set it to 1.
> >
> >> Then a possible factor is:
> >> (b_1 f_2 + b_2 f_1 + 2 j^2) / A
> >>
> >> Since I'm not sure how to calculate A, I just calculated each
> >> possibility mod each of the original primes to see if it was zero.
> >> No such combination
> >
> > There's now way you calculated each possible as the b's are
fractions.
> >
> > Are you saying you iterated through an infinite set?
>
> This is why you get abused on usenet. It was clear from above that I
was
> treating b_1 and b_2 as integer, so obviously I was iterating through
a very
> finite set. If you want to be taken seriously by anyone, usenet or
> otherwise, you should cut back on the "you are an idiot" style
comments,
> especially if it was your mistake in the first place (like you saying
that
> b_1 and b_2 were integers).
>
>

Go away. I don't need your attitude as I get enough from lots of other
people.

I don't need you.

I like talking out the math, but I HATE how full of yourselves some of
you are.

It's like you're all so sensitive or something.

I'm not interested in worrying about hurting your feelings or not.

If you can't take my style, just walk away.
You will not be missed.

James Harris



Relevant Pages

  • Usenet problem?
    ... Usenet has been all but dead most of this ... and is down to a fraction of its normal activity as of Monday ...
    (comp.publish.prepress)
  • Re: Basically a sieve method, relation to quantum
    ... > It's a rational, typically a fraction. ... this wasn't made clear in your previous posts. ... This is why you get abused on usenet. ... Michael Brown ...
    (sci.math)
  • Re: Basically a sieve method, relation to quantum
    ... > It's a rational, typically a fraction. ... this wasn't made clear in your previous posts. ... This is why you get abused on usenet. ... Michael Brown ...
    (sci.crypt)
  • Re: The "Oh no not another Burpday Moot" Thread
    ... summat astonishing like Google Earth, ... can't manage a broswer-based interface for usenet that's ... even a fraction as good as the text-only jobbie I had a ...
    (uk.rec.sheds)
  • Re: For sale: HP Integrity rx2600 parts (various)
    ... anyone harvests usenet for email addresses. ... few, maybe up to ten, spam a day. ... increasing fast enough that the fraction that went to old ones would ...
    (comp.os.vms)