| Topic: |
Science > Physics |
| User: |
"JSH" |
| Date: |
16 Jan 2008 09:51:58 PM |
| Object: |
JSH: Synopsis post |
Kind of a quicker post to cover the big picture while the previous one
was a long one:
Short of it is that I found that you could approach factoring through
a two equation system:
x^2 = y^2 + pr_1
z^2 = y^2 + nT
while mathematicians have traditionally used a single equation system:
x^2 = y^2 + nT or often given as x^2 = y^2 mod T.
where in both cases T is the target to factor.
With the two equation system I found mathematical laws represented by
a set of congruence relations:
z = (2a)^{-1} (1 + 2a^2)k mod p, k^2 = (a^2+1)^{-1}(nT) mod p
and
y = (1+2a^2)^{-1} z mod p, or y = -(1+2a^2)^{-1} z mod p.
I pointed out in my prior long posting that you can verify for
yourself quite a few things by considering a simple example with n=1,
T=21, as y = 2 is a solution, as, of course 5^2 = 2^2 + 21, and
possible values for pr_1 go out to infinity with the start of the
sequence being useful for understanding important facts:
pr_1 = (5)(1) = 5, pr_1 = (6)(2) = 12, pr_1 = (7)(3) = 21,
pr_1 = (8)(4) = 32, pr_1 = (9)(5) = 45,
pr_1 = (10)(6) = 60, and pr_1 = (11)(7) = 77.
You can add 4 to any of those to see you get another square and see
why you will have results for any prime p.
The congruence relations are trivially derived. The idea that a two
equation system will give you more information than just one is not a
complicated one. And the proof of the existence of solutions is easy.
Mathematicians just used one equation to study a problem that is best
handled by using two.
In a way it is a profoundly cool thing that there were these
underlying mathematical laws all along controlling things that people
just didn't know anything about, but the sense of satisfaction is
muted by the potential impact of the result.
However, it is the reality now that the information is known.
I HAVE been notifying mathematicians directly by email, by posting on
math newsgroups, and I have submitted a paper to a major mathematical
journal. But so far to no avail.
James Harris
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| User: "James" |
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| Title: Re: Synopsis post |
17 Jan 2008 12:10:54 PM |
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"JSH" <jstevh@gmail.com> wrote in message
news:8ba3760d-142e-47e5-9b71-1f91d30dbc61@j20g2000hsi.googlegroups.com...
Kind of a quicker post to cover the big picture while the previous one
was a long one:
Short of it is that I found that you could approach factoring through
a two equation system:
x^2 = y^2 + pr_1
z^2 = y^2 + nT
while mathematicians have traditionally used a single equation system:
x^2 = y^2 + nT or often given as x^2 = y^2 mod T.
where in both cases T is the target to factor.
With the two equation system I found mathematical laws represented by
a set of congruence relations:
z = (2a)^{-1} (1 + 2a^2)k mod p, k^2 = (a^2+1)^{-1}(nT) mod p
and
y = (1+2a^2)^{-1} z mod p, or y = -(1+2a^2)^{-1} z mod p.
I pointed out in my prior long posting that you can verify for
yourself quite a few things by considering a simple example with n=1,
T=21, as y = 2 is a solution, as, of course 5^2 = 2^2 + 21, and
possible values for pr_1 go out to infinity with the start of the
sequence being useful for understanding important facts:
pr_1 = (5)(1) = 5, pr_1 = (6)(2) = 12, pr_1 = (7)(3) = 21,
pr_1 = (8)(4) = 32, pr_1 = (9)(5) = 45,
pr_1 = (10)(6) = 60, and pr_1 = (11)(7) = 77.
You can add 4 to any of those to see you get another square and see
why you will have results for any prime p.
The congruence relations are trivially derived. The idea that a two
equation system will give you more information than just one is not a
complicated one. And the proof of the existence of solutions is easy.
Mathematicians just used one equation to study a problem that is best
handled by using two.
so that was it, all those mathematicians, all that time, and NOBODY thought
of using more than ONE algebraic equation.
In a way it is a profoundly cool thing that there were these
underlying mathematical laws all along controlling things that people
just didn't know anything about, but the sense of satisfaction is
muted by the potential impact of the result.
However, it is the reality now that the information is known.
I HAVE been notifying mathematicians directly by email, by posting on
math newsgroups, and I have submitted a paper to a major mathematical
journal. But so far to no avail.
James Harris
Oh By the Way, How long does it take to do thousands of calculations using
mod T anyway ?
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| User: "Bob Cain" |
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| Title: Re: JSH: Synopsis post |
17 Jan 2008 12:57:48 AM |
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JSH wrote:
I HAVE been notifying mathematicians directly by email, by posting on
math newsgroups, and I have submitted a paper to a major mathematical
journal. But so far to no avail.
Submit an executable procedure here. The proof is in the putting.
Bob
--
"Things should be described as simply as possible, but no simpler."
A. Einstein
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| User: "JSH" |
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| Title: Re: JSH: Synopsis post |
17 Jan 2008 06:56:06 PM |
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On Jan 16, 10:57 pm, Bob Cain <arc...@arcanemethods.com> wrote:
JSH wrote:
I HAVE been notifying mathematicians directly by email, by posting on
math newsgroups, and I have submitted a paper to a major mathematical
journal. But so far to no avail.
Submit an executable procedure here. The proof is in the putting.
Bob
All the information needed is in the original post.
You can pick 'a' and solve for n, so that you have a k. Then it's
just plug and chug all the way through until you get values for z mod
p and y mod p, and then you just check the gcd of your target T, with
z-y mod p and z+y mod p.
Easy.
James Harris
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| User: "Bob Cain" |
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| Title: Re: JSH: Synopsis post |
17 Jan 2008 09:03:29 PM |
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JSH wrote:
On Jan 16, 10:57 pm, Bob Cain <arc...@arcanemethods.com> wrote:
Submit an executable procedure here. The proof is in the putting.
Bob
All the information needed is in the original post.
You can pick 'a' and solve for n, so that you have a k. Then it's
just plug and chug all the way through until you get values for z mod
p and y mod p, and then you just check the gcd of your target T, with
z-y mod p and z+y mod p.
Easy.
James Harris
Put up or shut up. Write an executable procedure. Unwillingness to do so
cannot be viewed as rational or honest and fully justifies dismissing you.
Bob
--
"Things should be described as simply as possible, but no simpler."
A. Einstein
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| User: "JSH" |
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| Title: Re: JSH: Synopsis post |
17 Jan 2008 09:45:48 PM |
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On Jan 17, 7:03 pm, Bob Cain <arc...@arcanemethods.com> wrote:
JSH wrote:
On Jan 16, 10:57 pm, Bob Cain <arc...@arcanemethods.com> wrote:
Submit an executable procedure here. The proof is in the putting.
Bob
All the information needed is in the original post.
You can pick 'a' and solve for n, so that you have a k. Then it's
just plug and chug all the way through until you get values for z mod
p and y mod p, and then you just check the gcd of your target T, with
z-y mod p and z+y mod p.
Easy.
James Harris
Put up or shut up. Write an executable procedure. Unwillingness to do so
cannot be viewed as rational or honest and fully justifies dismissing you.
Bob
What do you think is happening here? Think I'm trying to convince you
of something?
No, I'm trying to convince myself.
Every time I find something I have a pile of people from all over the
world trying to tell me it's nothing.
In response just for breathing room, I defend my ideas, and then work
to figure out if there is anything to them or not.
Looks like there is something here. The two equations do give more
information, of that there is no doubt, but the crucial issue always
has been, can they give enough information that I or someone else can
definitively solve the factoring problem with them.
These do. Talking it out has given me the final answer. That is all.
James Harris
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| User: "James" |
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| Title: Re: JSH: Synopsis post |
17 Jan 2008 10:07:53 PM |
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"JSH" <jstevh@gmail.com> wrote in message
news:fab0fa8e-2ff3-4b09-ab93-8febb22d0f0e@u10g2000prn.googlegroups.com...
On Jan 17, 7:03 pm, Bob Cain <arc...@arcanemethods.com> wrote:
JSH wrote:
On Jan 16, 10:57 pm, Bob Cain <arc...@arcanemethods.com> wrote:
Submit an executable procedure here. The proof is in the putting.
Bob
All the information needed is in the original post.
You can pick 'a' and solve for n, so that you have a k. Then it's
just plug and chug all the way through until you get values for z mod
p and y mod p, and then you just check the gcd of your target T, with
z-y mod p and z+y mod p.
Easy.
James Harris
Put up or shut up. Write an executable procedure. Unwillingness to do
so
cannot be viewed as rational or honest and fully justifies dismissing
you.
Bob
What do you think is happening here? Think I'm trying to convince you
of something?
No, I'm trying to convince myself.
Every time I find something I have a pile of people from all over the
world trying to tell me it's nothing.
In response just for breathing room, I defend my ideas, and then work
to figure out if there is anything to them or not.
Looks like there is something here. The two equations do give more
information, of that there is no doubt, but the crucial issue always
has been, can they give enough information that I or someone else can
definitively solve the factoring problem with them.
These do. Talking it out has given me the final answer. That is all.
Bob you need to understand the JSH way-
--that you gotta pick a "a" then solve for "n" then pick a p, then pick a T
then pick a mod then pick a "r" then pick an x and a z, then solve for y and
the others, just a pickin and a solvin, no math to it at tall.
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| User: "Kenneth Doyle" |
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| Title: Re: JSH: Synopsis post |
17 Jan 2008 10:08:02 PM |
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JSH <jstevh@gmail.com> wrote in news:fab0fa8e-2ff3-4b09-ab93-
8febb22d0f0e@u10g2000prn.googlegroups.com:
I defend my ideas, and then work
to figure out if there is anything to them or not.
Ah, well that's your problem right there. Wouldn't it be better to do it
the other way around?
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| User: "Bob Cain" |
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| Title: Re: JSH: Synopsis post |
18 Jan 2008 12:14:41 AM |
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JSH wrote:
What do you think is happening here? Think I'm trying to convince you
of something?
Well, you have been trying for a long time to convince somebody of an ever
shifting something. Otherwise, why all the usenet self publicity?
No, I'm trying to convince myself.
Every time I find something I have a pile of people from all over the
world trying to tell me it's nothing.
Listen, you can easily show them that it's something but you refuse. You are
describing an algorithm and it is not fully specified until it exists in a
procedural form which terminates having yielded the factors, i.e. an executable
program. Showing that it terminates is of course can be difficult and that
might be another problem you want to take a look at.
What you are going to find, should you care to try to fully specify your
algorithm via a programmed procedure, is that there is some step that you don't
actually know how to do in detail. If you study that step carefully you will
find within it the very problem you think you are solving. i.e. that step will
either be just as complex as or is itself exactly the factoring process that you
think you've solved. It is incumbent upon you and you alone to show otherwise
in order to gain the recognition you so dearly desire and have been campaigning
for for such a long time.
Give it a go. If it works and others verify that (trust me, many will try it)
you are a made man.
Bob
--
"Things should be described as simply as possible, but no simpler."
A. Einstein
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| User: "JSH" |
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| Title: Re: JSH: Synopsis post |
17 Jan 2008 09:02:09 PM |
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On Jan 17, 4:56 pm, JSH <jst...@gmail.com> wrote:
On Jan 16, 10:57 pm, Bob Cain <arc...@arcanemethods.com> wrote:
JSH wrote:
I HAVE been notifying mathematicians directly by email, by posting on
math newsgroups, and I have submitted a paper to a major mathematical
journal. But so far to no avail.
Submit an executable procedure here. The proof is in the putting.
Bob
All the information needed is in the original post.
You can pick 'a' and solve for n, so that you have a k. Then it's
just plug and chug all the way through until you get values for z mod
p and y mod p, and then you just check the gcd of your target T, with
z-y mod p and z+y mod p.
Easy.
Might be better to just let n=1, and try a's until you get a k.
The full set of equations are that given
z^2 = y^2 + nT
where T is the target to factor and n is some nonzero integer you
pick, with p a prime you pick:
z = (2a)^{-1} (1 + 2a^2)k mod p,
k^2 = (a^2+1)^{-1}(nT) mod p
and
y = (1+2a^2)^{-1} z mod p, or y = -(1+2a^2)^{-1} z mod p.
That's the set of rules previously unknown and the question is, how do
you factor with them?
If p>sqrt(nT), then z-y mod p or z+y mod p must have for some solution
prime factors of nT that are less than p, because those are their own
residues. For instance, 7(17) = 119, and 7 mod 11 = 7, while 17 mod
11 = 6.
The start of finding answers with the congruences is k^2 = (a^2+1)^{-1}
(nT) mod p, where I suggested you just pick some 'a' and then solve
for n, which means you pick k^2 as well, but if, you pick something
easy, like 4, then there is no connection to your target T. So like
if q is the quadratic residue modulo p then
(a^2+1)^{-1}(nT) = q mod p
and with 'a' chosen you have simply enough that n = q(a^2 + 1)T^{-1}
mod p, but now you're picking q, and it makes sense that you pick it
fairly large and not as a perfect square so that T is in the picture,
so you have to search for a minimum n, where n=1 is the best case.
So maybe that's a problem? With a large T there is a bigger and
bigger region to search?
I don't know. Like with 119, though 119 mod 11 = 9. and 9^{-1} mod
11 = 5. If I have a=1, then
n = q(2)(5) mod 11 = 10q mod 11, and I don't see how to get n=1 with a
residue, but I can get n=2, with q=9, so k=3 and I can go from there.
Then
z = (2a)^{-1} (1 + 2a^2)k mod p = (2)^{-1}(3) mod 11 = 7 mod 11.
y = (1+2a^2)^{-1} z mod p = (3)^{-1}(7) mod 11 = 6 mod 11
or y = -(1+2a^2)^{-1} z mod p = -6 mod 11 = 5 mod 11.
Then z-y = 1 mod 11 or 2 mod 11, and z+y = 2 mod 11 or 1 mod 11.
And that didn't work. It gave me the residues modulo 11 for the
trivial factorization 2(119).
Oh wait, there's also k=-3.
z = (2a)^{-1} (1 + 2a^2)k mod p = (2)^{-1}(-3) mod 11 = 4 mod 11.
y = (1+2a^2)^{-1} z mod p = (3)^{-1}(4) mod 11 = 5 mod 11
or y = -(1+2a^2)^{-1} z mod p = -5 mod 11 = 6 mod 11.
Then z-y = 10 mod 11 or 9 mod 11, and z+y = 9 mod 11 or 10 mod 11.
The negative of the previous. Oh well. It just gave the trivial
factorization both times.
James Harris
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| User: "Eric Gisse" |
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| Title: Re: JSH: Synopsis post |
17 Jan 2008 05:54:25 AM |
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On Jan 16, 6:51 pm, JSH <jst...@gmail.com> wrote:
Kind of a quicker post to cover the big picture while the previous one
was a long one:
Short of it is that I found that you could approach factoring through
a two equation system:
x^2 = y^2 + pr_1
z^2 = y^2 + nT
while mathematicians have traditionally used a single equation system:
x^2 = y^2 + nT or often given as x^2 = y^2 mod T.
where in both cases T is the target to factor.
With the two equation system I found mathematical laws represented by
a set of congruence relations:
z = (2a)^{-1} (1 + 2a^2)k mod p, k^2 = (a^2+1)^{-1}(nT) mod p
and
y = (1+2a^2)^{-1} z mod p, or y = -(1+2a^2)^{-1} z mod p.
I pointed out in my prior long posting that you can verify for
yourself quite a few things by considering a simple example with n=1,
T=21, as y = 2 is a solution, as, of course 5^2 = 2^2 + 21, and
possible values for pr_1 go out to infinity with the start of the
sequence being useful for understanding important facts:
pr_1 = (5)(1) = 5, pr_1 = (6)(2) = 12, pr_1 = (7)(3) = 21,
pr_1 = (8)(4) = 32, pr_1 = (9)(5) = 45,
pr_1 = (10)(6) = 60, and pr_1 = (11)(7) = 77.
You can add 4 to any of those to see you get another square and see
why you will have results for any prime p.
The congruence relations are trivially derived. The idea that a two
equation system will give you more information than just one is not a
complicated one. And the proof of the existence of solutions is easy.
Mathematicians just used one equation to study a problem that is best
handled by using two.
In a way it is a profoundly cool thing that there were these
underlying mathematical laws all along controlling things that people
just didn't know anything about, but the sense of satisfaction is
muted by the potential impact of the result.
However, it is the reality now that the information is known.
I HAVE been notifying mathematicians directly by email, by posting on
math newsgroups, and I have submitted a paper to a major mathematical
journal. But so far to no avail.
Start small - explain why sci.physics should care.
James Harris
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