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My factoring method, quadratic residues
| jstevh at msn.com | | Justin | | ošin | | Gib Bogle | | jstevh at msn.com |
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| | From: | jstevh at msn.com | | Subject: | My factoring method, quadratic residues | | Date: | 20 Jan 2005 16:51:03 -0800 |
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 | Well, it looks like my method has a probabilistic component.
It looks through all factors of j^2 T d^2, where j and T are given as
before, and d is ANY integer that will fit, such that with f_1 and f_2
factors of j^2 T d^2,
f_1 = j^2 mod g
AND
f_2 = - j^2 mod g
where g is a prime factor of your target M.
In looping out to use d, the mathematics looks for some k, such that
f_1 k^2 + f_2 d^2 = 0 mod g.
Can any of you figure out the pobabilities on that information?
I'm kind of puzzled by this result though it's easy enough to derive.
The quadratic residues popping up is kind of disappointing, as I'd
hoped to have a perfect closed solution.
Oh well, it's mathematics. No point whining about it, it's not like
it's going to change.
Does this kill this method, or is it just a minor setback?
In any event, no doubt, my factoring idea is interesting mathematics.
I'm still trying to get a handle on it, working as hard as I can,
posting my most arrogant, obnoxious or otherwise, barely getting any
sleep, drinking lots of coffee, and a moderate doses of alcohol, it's
still taking me to my limit.
EXTREME MATHEMATICS at its finest. I know none of you can fully
appreciate the sheer joy I'm experiencing, but at least some of you are
stupid enough to post silly and obnoxious comments in reply to me.
You never learned how I work.
James Harris
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| | From: | Justin | | Subject: | Re: My factoring method, quadratic residues | | Date: | Fri, 21 Jan 2005 10:04:14 +0000 (UTC) |
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| In sci.math jstevh@msn.com wrote:
: Well, it looks like my method has a probabilistic component.
That's an interesting way of putting it.
: In looping out to use d, the mathematics looks for some k, such that
Something's looping out, that's for sure...
: Can any of you figure out the pobabilities on that information?
Can you figure out what the heck you're talking about?
: The quadratic residues popping up is kind of disappointing, as I'd
: hoped to have a perfect closed solution.
As far as we can tell, you *do* have a perfect(ly) closed solution.
: Does this kill this method, or is it just a minor setback?
What method could it possibly kill?
Justin
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| | From: | ošin | | Subject: | Re: My factoring method, quadratic residues | | Date: | Thu, 20 Jan 2005 17:00:58 -0800 |
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| > Does this kill this method, or is it just a minor setback?
I sure hope it kills it. And how can you be set back from zero progress.
Your code never worked!
> In any event, no doubt, my factoring idea is interesting mathematics.
Your factoring idea is neither interesting nor mathematics.
> ... barely getting any
> sleep, drinking lots of coffee, and a moderate doses of alcohol, it's
> still taking me to my limit.
Alcohol? Sounds more like LSD.
> EXTREME MATHEMATICS at its finest.
Yup... must be LSD.
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| | From: | Gib Bogle | | Subject: | Re: My factoring method, quadratic residues | | Date: | Fri, 21 Jan 2005 16:10:29 +1300 |
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| jstevh@msn.com wrote:
> Well, it looks like my method has a probabilistic component.
As in, it probably doesn't work?
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| | From: | jstevh at msn.com | | Subject: | Re: My factoring method, quadratic residues | | Date: | 20 Jan 2005 17:08:05 -0800 |
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| jst...@msn.com wrote:
> Well, it looks like my method has a probabilistic component.
>
> It looks through all factors of j^2 T d^2, where j and T are given as
> before, and d is ANY integer that will fit, such that with f_1 and
f_2
> factors of j^2 T d^2,
>
> f_1 = j^2 mod g
>
> AND
>
> f_2 = - j^2 mod g
>
> where g is a prime factor of your target M.
>
> In looping out to use d, the mathematics looks for some k, such that
>
> f_1 k^2 + f_2 d^2 = 0 mod g.
>
Oh, I just tossed that out there and said it's easy to calculate.
Here's how you do it:
Remember the method I use boils down to using
a_1 x + b_1 = f_1
a_2 x + b_2 = f_2
and
a_1 b_2 + a_2 b_1 = A, as
a_1 a_2 x^2 + A x + b_1 b_2 = f_1 f_2, and
f_1 f_2 - b_1 b_2 = M,
where M is the number you're trying to factor.
First, solve for x, which gives you
x = (b_1 f_2 + b_2 f_1 - 2 b_1 b_2)/A
and let b_1 b_2 = -j^2, to fit with the paper.
And now solve for b_1, to get
b_1 = (-(2j^2 - Ax) +/- sqrt((2j^2 - Ax)^2 + 4j^2T))/2f_2
and now I want x to have g as a factor where g is a prime factor of the
target M.
Um, it should be easy enough to figure out the rest from there.
If not, then you don't know a lot of mathematics.
If necessary maybe later I can fill in the details.
In any event, you can go from there to get the quadratic residue result
in my original post.
James Harris
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