Surrogate factoring, out of the box
jstevh_at_msn.com
Date: 01/29/05
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Date: 29 Jan 2005 12:09:31 -0800
Well I'll admit that I've been feeling a bit depressed the last couple
of days, as I had calculations showing at least 50% success with a
rational x, and then I checked thoroughly and found that my method gave
a LOT of rational x's, and wasn't factoring with most of them.
And then I realized that for most cases it gives you an x that has your
target itself as the factor.
I puzzled over that result, and realized that the math was too
efficient in searching through rational solutions, so that it could,
most of the time, find solutions such that the target itself was the
factor of x, hence my problem.
So I squared the target, and it factored.
Why? You need to understand quadratic residues to understand why.
Basically the probability was too high that it could get quadratic
residues for both of my factors, so by squaring them, I forced the math
to look for solution for those factors squared, making it more likely
that it would fail for at least one, and it did.
I have verified that the value of A is mostly irrelevant, except with
regard to evens, as either you need A=8, or maybe you might do well by
making A a fraction and using A=1/2 or A=1/4 or A=1/8, which I say as I
haven't checked it, as I now just use A=8.
That issue has to do with allowing rational solutions based on
balancing out evens versus odds with
yx^2 + Ax - M^2 = 0
as M is usually odd, so if you don't make A even, and at least 8, then
you'll get situations where y and x can't be rational, as they want to
be odd, but if A is odd, then you have three odds trying to add up to
0, which can't work.
I'm still puzzling over the quadratic residues a bit, as it's not to
the prime, as I first thought, but to the prime squared. So like if 5
is a factor of M, then the quadratic residue result has to do with
d_1^2 = w_1'w_2 mod 25
not mod 5, as I thought originally.
Oh, the quadratic residue result is
d_1^2 = w_1' w_2 mod g^2
where g is a prime factor of your target, and d_1 is what is picked for
you by the math as it checks through all integers to see if a d_1
exists.
w_1 w_2 = j^2 T
and I'm using w_1' for the modular inverse of w_1.
So yeah, the factorizations of j AND T play a role.
That's the profoundly fascinating feature of this method, as infinity
itself is checked for results!
And that is rigorously proven.
I just didn't realize that more than likely it will find d_1, out of
infinity.
Well, infinity is kind of big...
So yeah, if you try to factor something not prime, and get no factors,
square it.
I've updated the Yahoo! site:
http://groups.yahoo.com/group/sufactor/
and I'm going to work on settling down the probabilities, as I find it
curious.
So even I got stuck thinking inside of the box, as I kept puzzling over
why I didn't get what I wanted--a solution--versus paying attention to
this flood of rational x's that had M^2 itself as a solution, in
defiance of my earlier quadratic residues results, which was just off.
Fun. Math is great.
James Harris
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