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Message |
jstevh@msn.com
science forum Guru
Joined: 21 Jan 2006
Posts: 951
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 Posted: Wed Jun 21, 2006 2:18 am Post subject:
SF: Simpler factoring idea, but does it work?
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After yet another failure with what I call surrogate factoring, where
this time I had been doing some basic algebra wrong, I sat down to
think about it all for a while, and considered that I was quite
reasonably just going in circles, using equations to try and factor
that could only give one answer.
So I started thinking about equations that could give two.
It didn't take long till I was concentrating on:
S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))
Here the square roots mean that expression can't just factor S, which
is what I call the surrogate, as something else is being factored as
well, but how do I get that something else to be a target composite?
After I posted that equation and started talking about it, a Tim Peters
worked out details following my instructions, but unfortunately, he is
a dedicated, um, "crank" buster you might call it, who spends his time
trying to shoot down my ideas, so when he worked out the equations, and
got to something useable, he promptly began throwing up distracting
posts meant to show it was useless.
However, oddly enough, his results can be used quite simply, where the
first thing is to use some of his equations, to introduce a target
composite, which I call T.
You introduce T using
(k_1*k_4 + k_2*k_3) / (k_1*k_4 - k_2*k_3) = T
Multiply both sides by
k_1*k_4 - k_2*k_3
to get
(k_1*k_4 + k_2*k_3) = T (k_1*k_4 - k_2*k_3)
and just subtract the left from the right to get
0 = (T-1)* k_1*k_4 - (T+1)*k_2*k_3
So
(T-1)* k_1*k_4 = (T+1)*k_2*k_3
And you have
(k_1*k_4)/(k_2*k_3) = (T+1)/(T-1)
so k_1 and k_4 are integer factors of T+1, and k_2 and k_3 are integer
factors of T-1.
Easy. Just like that you're most of the way to using the equations.
That gives a finite set of possibles for the k's.
For instance, with
So, for instance if T=15, you have
(k_1*k_4)/(k_2*k_3) = 16/14 = 8/7
so k_1*k_4 = 8, and k_2*k_3 = 7
and one possible setup then is
k_1 = 2, k_4 = 4, k_2 = 7, k_3 = 1
so plug those into
S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))
and you need S, and can use any square you like, so it's easiest just
to use
x=y=1, and take the positive result of the square roots to get
S = (2 + 7)(1 + 4) = 45
which promptly factors it, but I'll continue, as that's just one
possibility, so you have to check for it.
If S were coprime to T, then you now use factors of S, where with
S = g_1*g_2
you have
k_1*sqrt(x) + k_2*sqrt(y) = g_1
k_3*sqrt(x) + k_4*sqrt(y) = g_2
and you just find find squares for x and y that will work to give you
g_1 and g_2, and here's where that second solution from the square
roots comes in, as with each set of squares for x and y that will work,
you just change the sign of one of the square roots, to get the shadow
factorization.
That's it. Remarkably simple, as you go for the hidden factorization.
For instance, still using x=y=1, with my simple example, now take the
negative of ONE of the square roots:
S = (2 - 7)(1 + 4) = 55
I guess T=15 is too dinky of an example as it just keeps factoring it,
no matter what you do, but at least that still shows the basic idea
here.
To recap, I looked at an expression
S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))
that can't just be the factorization of a single number because of the
use of square roots. I posted about it and a Tim Peters worked through
some analysis following instructions I gave, which gives up the simple
equation
(k_1*k_4 + k_2*k_3) / (k_1*k_4 - k_2*k_3) = T
which can be solved to get
(k_1*k_4)/(k_2*k_3) = (T+1)/(T-1)
so you can find the k's based on the factors of T+1 and T-1, to plug
back into the original equations and get an S, and then you use the
factors of S to find squares that are solutions for x and y, and then
you do the remarkable trick of just switching signs one at a time to
get to the hidden factorization.
But does it work beyond toy examples like factoring 15?
Unfortunately, as I said, Tim Peters is a hostile when it comes to my
research, so he promptly busied himself trying to obscure use of the
equations, while claiming that they don't work. You can look at recent
threads I've created on sci.crypt and sci.math to see him at work.
Some of you might be shocked by such behavior, as, hey, it's the
factoring problem.
But consider for YEARS Peters and people like him have been shadowing
my posts working to convince people that my research is useless.
I assure you that he has worked in this way many times.
He is only doing what he has always been doing, so there is nothing new
in his behavior.
So the question is an open one. Does this method work?
I can assure you that the math society that has busied itself ignoring
my research for years is in no hurry to acknowledge a result of mine
like this one--if it does work--as then they would be completely
overturned, now wouldn't they?
And what is your security against theirs?
So they wait, seeing if the "crackpot" label will hold, and they wait,
to see if no one can tell if these ideas will work or not, and if they
do work, they wait until there is a disaster big enough for the world
to care about the truth.
And then, finally, they will be able to wait, no more.
James Harris |
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willo_thewisp@hotmail.com
science forum addict
Joined: 04 Mar 2006
Posts: 88
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| Posted: Wed Jun 21, 2006 3:54 am Post subject:
Re: SF: Simpler factoring idea, but does it work?
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jstevh@msn.com wrote:
| Quote: |
S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))
and you need S, and can use any square you like, so it's easiest just
to use
x=y=1, and take the positive result of the square roots to get
S = (2 + 7)(1 + 4) = 45
But you *can't* just choose to take the positive result of the |
square roots! The square roots, as you know, have two values,
and it is *absolutely impossible* to choose the positive ones.
After all, what if aliens on the Planet Contrary chose to take
the negative ones? What then? |
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Tim Peters
science forum Guru
Joined: 30 Apr 2005
Posts: 426
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| Posted: Wed Jun 21, 2006 4:26 am Post subject:
Re: JSH: SF: Simpler factoring idea, but does it work?
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[added "JSH:" to subject, spared sci.crypt and alt.math]
[jstevh@msn.com, comes out of retirement in less than a day  ]
| Quote: | After yet another failure with what I call surrogate factoring, where
this time I had been doing some basic algebra wrong, I sat down to
think about it all for a while, and considered that I was quite
reasonably just going in circles, using equations to try and factor
that could only give one answer.
So I started thinking about equations that could give two.
It didn't take long till I was concentrating on:
S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))
Here the square roots mean that expression can't just factor S, which
is what I call the surrogate, as something else is being factored as
well, but how do I get that something else to be a target composite?
After I posted that equation and started talking about it, a Tim Peters
worked out details following my instructions, but unfortunately, he is
a dedicated, um, "crank" buster you might call it, who spends his time
trying to shoot down my ideas, so when he worked out the equations, and
got to something useable, he promptly began throwing up distracting
posts meant to show it was useless.
|
Including a rigorous proof that it was in fact 100% useless -- although some
variations could be salvaged, and those appeared to be equivalent to the
random-gcd algorithm.
| Quote: | However, oddly enough, his results can be used quite simply, where the
first thing is to use some of his equations, to introduce a target
composite, which I call T.
|
You posted the same stuff yesterday; see my reply yesterday.
| Quote: | You introduce T using
(k_1*k_4 + k_2*k_3) / (k_1*k_4 - k_2*k_3) = T
Multiply both sides by
k_1*k_4 - k_2*k_3
to get
(k_1*k_4 + k_2*k_3) = T (k_1*k_4 - k_2*k_3)
and just subtract the left from the right to get
0 = (T-1)* k_1*k_4 - (T+1)*k_2*k_3
So
(T-1)* k_1*k_4 = (T+1)*k_2*k_3
And you have
(k_1*k_4)/(k_2*k_3) = (T+1)/(T-1)
so k_1 and k_4 are integer factors of T+1, and k_2 and k_3 are integer
factors of T-1.
Easy. Just like that you're most of the way to using the equations.
That gives a finite set of possibles for the k's.
For instance, with
So, for instance if T=15, you have
(k_1*k_4)/(k_2*k_3) = 16/14 = 8/7
so k_1*k_4 = 8, and k_2*k_3 = 7
and one possible setup then is
k_1 = 2, k_4 = 4, k_2 = 7, k_3 = 1
so plug those into
S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))
and you need S, and can use any square you like, so it's easiest just
to use
x=y=1, and take the positive result of the square roots to get
S = (2 + 7)(1 + 4) = 45
which promptly factors it, but I'll continue, as that's just one
possibility, so you have to check for it.
|
Are you unable to find an example that shows something interesting?
Let's try a small but not utterly trivial one off the top of my head, T =
143 = 11*13. Pick, say,
T+1 = 144 = 36 * 4 = k1*k4
T-1 = 142 = 2 * 71 = k2*k3
and pick x=y=1. Then S = (36 + 2)*(4 + 71) = something irrelevant, and
gcd(-36+2, T) = gcd(36-2, T) = gcd(-4+71, T) = gcd(4-71, T) = 1
That's the usual outcome in a non-utterly-trivial example. As below, you
_can_ always find choices that work, but if you don't know T's factorization
before you begin you have no way to find choices that are _likely_ to work.
| Quote: | If S were coprime to T, then you now use factors of S, where with
S = g_1*g_2
you have
k_1*sqrt(x) + k_2*sqrt(y) = g_1
k_3*sqrt(x) + k_4*sqrt(y) = g_2
and you just find find squares for x and y that will work to give you
g_1 and g_2, and here's where that second solution from the square
roots comes in, as with each set of squares for x and y that will work,
you just change the sign of one of the square roots, to get the shadow
factorization.
That's it. Remarkably simple, as you go for the hidden factorization.
|
It is indeed simple ;-)
| Quote: | For instance, still using x=y=1, with my simple example, now take the
negative of ONE of the square roots:
S = (2 - 7)(1 + 4) = 55
I guess T=15 is too dinky of an example as it just keeps factoring it,
no matter what you do, but at least that still shows the basic idea
here.
To recap, I looked at an expression
S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))
that can't just be the factorization of a single number because of the
use of square roots.
|
That claim doesn't make sense, but I'm not going to challenge it because it
doesn't change that this method doesn't work well. I'll only note that if
you could understand _why_ that claim doesn't make sense, you'd be close to
understanding why this method has no legs to stand on.
| Quote: | I posted about it and a Tim Peters worked through some analysis
following instructions I gave, which gives up the simple equation
(k_1*k_4 + k_2*k_3) / (k_1*k_4 - k_2*k_3) = T
which can be solved to get
(k_1*k_4)/(k_2*k_3) = (T+1)/(T-1)
so you can find the k's based on the factors of T+1 and T-1, to plug
back into the original equations and get an S, and then you use the
factors of S to find squares that are solutions for x and y, and then
you do the remarkable trick of just switching signs one at a time to
get to the hidden factorization.
But does it work beyond toy examples like factoring 15?
|
Provided k_1 and k_2 are coprime (if they're not, you can divide both by
their gcd to make them coprime), and T is composite, there are an infinite
number of <x, y> pairs for which:
1 < gcd(k_1 * -sqrt(x) - k_2 * sqrt(y), T) < T
That statement is true regardless of how k_1, k_2, x and y are chosen (i.e.,
this much has nothing to do with the specific equations you're using).
Alas, there's no reason to hope that "your way" of picking them is better
than picking them at random.
| Quote: | Unfortunately, as I said, Tim Peters is a hostile when it comes to my
research, so he promptly busied himself trying to obscure use of the
equations, while claiming that they don't work.
|
It was you who suggested using a derived difference of squares, and for the
specific way you suggested it was straightforward to prove that a
non-trivial factor could _never_ be found. I (and Rick Decker) found
variations for you that worked better than your idea, but nobody (including
you) found a variation of the differences-of-squares idea of any real use.
| Quote: | You can look at recent threads I've created on sci.crypt and sci.math
to see him at work.
|
Of course I usually cut sci.crypt from repiles.
| Quote: | Some of you might be shocked by such behavior, as, hey, it's the
factoring problem.
|
If proof shocks you, you should take up a different hobby.
| Quote: | But consider for YEARS Peters and people like him have been shadowing
my posts working to convince people that my research is useless.
|
Your factoring methods to date _have_ been useless. Why shoot the
messenger? Or do you believe you've thrown away dozens of useful factoring
methods to get to this one? LOL.
The results would be the same if you did the algebra (provided you did it
correctly). Blaming me for that your methods haven't worked makes you look
like a crank.
| Quote: | I assure you that he has worked in this way many times.
He is only doing what he has always been doing, so there is nothing new
in his behavior.
|
Nor yours. You make wild claims, I investigate some of them, and then you
make a fool of yourself whining, moaning, bitching, threatening and
slandering because you don't like the results. From a day to a year later,
so far you've always been forced to agree that each method in fact didn't
work well. I don't often object to your bad behavior because making a fool
of yourself doesn't actually injure me. I don't know why you _want_ to do
that to yourself, and sincerely suggest you'd be happier if you didn't, but
in the end your reputation and honor are yours to throw away.
| Quote: | So the question is an open one. Does this method work?
|
AFAICT, it works exactly as well as random-gcd, but is more complicated. If
I'm wrong about that, it will be possible for you to demonstrate it.
| Quote: | I can assure you that the math society that has busied itself ignoring
my research for years
|
Actually, taking the time to prove results about your claims is nothing at
all like ignoring them. You're just crabby because your factoring claims
_so far_ have been shot down. Sorry, but this one isn't going to fare
better. If that pisses you off, find a better method. Trying to blame
others for the failure of the methods is crackpot behavior.
| Quote: | is in no hurry to acknowledge a result of mine like this one--if it
does work--as then they would be completely overturned, now wouldn't
they?
And what is your security against theirs?
So they wait, seeing if the "crackpot" label will hold, and they wait,
to see if no one can tell if these ideas will work or not, and if they
do work, they wait until there is a disaster big enough for the world
to care about the truth.
And then, finally, they will be able to wait, no more.
James Harris
|
From all evidence, I pay more _serious_ attention to your methods than you
do. In fact, that's why I discover whether they work long before you do.
This method is no exception. |
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mensanator@aol.compost
science forum Guru
Joined: 24 Mar 2005
Posts: 826
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| Posted: Wed Jun 21, 2006 4:40 am Post subject:
Re: SF: Simpler factoring idea, but does it work?
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jstevh@msn.com wrote:
| Quote: | After yet another failure with what I call surrogate factoring, where
this time I had been doing some basic algebra wrong, I sat down to
think about it all for a while, and considered that I was quite
reasonably just going in circles, using equations to try and factor
that could only give one answer.
So I started thinking about equations that could give two.
It didn't take long till I was concentrating on:
S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))
Here the square roots mean that expression can't just factor S, which
is what I call the surrogate, as something else is being factored as
well, but how do I get that something else to be a target composite?
After I posted that equation and started talking about it, a Tim Peters
worked out details following my instructions, but unfortunately, he is
a dedicated, um, "crank" buster you might call it, who spends his time
trying to shoot down my ideas, so when he worked out the equations, and
got to something useable, he promptly began throwing up distracting
posts meant to show it was useless.
However, oddly enough, his results can be used quite simply, where the
first thing is to use some of his equations, to introduce a target
composite, which I call T.
You introduce T using
(k_1*k_4 + k_2*k_3) / (k_1*k_4 - k_2*k_3) = T
Multiply both sides by
k_1*k_4 - k_2*k_3
to get
(k_1*k_4 + k_2*k_3) = T (k_1*k_4 - k_2*k_3)
and just subtract the left from the right to get
0 = (T-1)* k_1*k_4 - (T+1)*k_2*k_3
So
(T-1)* k_1*k_4 = (T+1)*k_2*k_3
And you have
(k_1*k_4)/(k_2*k_3) = (T+1)/(T-1)
so k_1 and k_4 are integer factors of T+1, and k_2 and k_3 are integer
factors of T-1.
Easy. Just like that you're most of the way to using the equations.
That gives a finite set of possibles for the k's.
For instance, with
So, for instance if T=15, you have
(k_1*k_4)/(k_2*k_3) = 16/14 = 8/7
so k_1*k_4 = 8, and k_2*k_3 = 7
and one possible setup then is
k_1 = 2, k_4 = 4, k_2 = 7, k_3 = 1
so plug those into
S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))
and you need S, and can use any square you like, so it's easiest just
to use
x=y=1, and take the positive result of the square roots to get
S = (2 + 7)(1 + 4) = 45
which promptly factors it, but I'll continue, as that's just one
possibility, so you have to check for it.
If S were coprime to T, then you now use factors of S, where with
S = g_1*g_2
you have
k_1*sqrt(x) + k_2*sqrt(y) = g_1
k_3*sqrt(x) + k_4*sqrt(y) = g_2
and you just find find squares for x and y that will work to give you
g_1 and g_2, and here's where that second solution from the square
roots comes in, as with each set of squares for x and y that will work,
you just change the sign of one of the square roots, to get the shadow
factorization.
That's it. Remarkably simple, as you go for the hidden factorization.
For instance, still using x=y=1, with my simple example, now take the
negative of ONE of the square roots:
S = (2 - 7)(1 + 4) = 55
I guess T=15 is too dinky of an example as it just keeps factoring it,
no matter what you do, but at least that still shows the basic idea
here.
To recap, I looked at an expression
S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))
that can't just be the factorization of a single number because of the
use of square roots. I posted about it and a Tim Peters worked through
some analysis following instructions I gave, which gives up the simple
equation
(k_1*k_4 + k_2*k_3) / (k_1*k_4 - k_2*k_3) = T
which can be solved to get
(k_1*k_4)/(k_2*k_3) = (T+1)/(T-1)
so you can find the k's based on the factors of T+1 and T-1, to plug
back into the original equations and get an S, and then you use the
factors of S to find squares that are solutions for x and y, and then
you do the remarkable trick of just switching signs one at a time to
get to the hidden factorization.
But does it work beyond toy examples like factoring 15?
Unfortunately, as I said, Tim Peters is a hostile when it comes to my
research, so he promptly busied himself trying to obscure use of the
equations, while claiming that they don't work. You can look at recent
threads I've created on sci.crypt and sci.math to see him at work.
Some of you might be shocked by such behavior, as, hey, it's the
factoring problem.
But consider for YEARS Peters and people like him have been shadowing
my posts working to convince people that my research is useless.
I assure you that he has worked in this way many times.
He is only doing what he has always been doing, so there is nothing new
in his behavior.
So the question is an open one. Does this method work?
I can assure you that the math society that has busied itself ignoring
my research for years is in no hurry to acknowledge a result of mine
like this one--if it does work--as then they would be completely
overturned, now wouldn't they?
And what is your security against theirs?
So they wait, seeing if the "crackpot" label will hold, and they wait,
to see if no one can tell if these ideas will work or not, and if they
do work, they wait until there is a disaster big enough for the world
to care about the truth.
And then, finally, they will be able to wait, no more.
|
Boy, your un-retirement was quicker than Michael Jordan's.
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Proginoskes
science forum Guru
Joined: 29 Apr 2005
Posts: 2593
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| Posted: Wed Jun 21, 2006 4:52 am Post subject:
Re: SF: Simpler factoring idea, but does it work?
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jstevh@msn.com wrote:
I thought you were going to quit posting. You _did_ post that
"baseball" thread, right? The one where you said:
| Quote: | I'm at the end of my run. Mathematics is a young man's game, so far,
and hopefully, it'll be a young person's game soon enough, as it's past
time for a woman to step up, and I am looking for her.
But for now, it's a young man's game and as I'm past 35 I am well past
my prime, and my time is over.
[...]
I am retired. The spirit has died. The Muse has left me. I don't
feel any more discovery is left in these bones.
|
Were you lying then?
[Back to the OP:]
| Quote: | After yet another failure with what I call surrogate factoring, where
this time I had been doing some basic algebra wrong, I sat down to
think about it all for a while, and considered that I was quite
reasonably just going in circles, using equations to try and factor
that could only give one answer.
So I started thinking about equations that could give two.
It didn't take long till I was concentrating on:
S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))
Here the square roots mean that expression can't just factor S, which
is what I call the surrogate, as something else is being factored as
well, but how do I get that something else to be a target composite?
After I posted that equation and started talking about it, a Tim Peters
worked out details following my instructions, but unfortunately, he is
a dedicated, um, "crank" buster you might call it, who spends his time
trying to shoot down my ideas, so when he worked out the equations, and
got to something useable, he promptly began throwing up distracting
posts meant to show it was useless.
However, oddly enough, his results can be used quite simply, where the
first thing is to use some of his equations, to introduce a target
composite, which I call T.
You introduce T using
(k_1*k_4 + k_2*k_3) / (k_1*k_4 - k_2*k_3) = T
Multiply both sides by
k_1*k_4 - k_2*k_3
to get
(k_1*k_4 + k_2*k_3) = T (k_1*k_4 - k_2*k_3)
and just subtract the left from the right to get
0 = (T-1)* k_1*k_4 - (T+1)*k_2*k_3
So
(T-1)* k_1*k_4 = (T+1)*k_2*k_3
And you have
(k_1*k_4)/(k_2*k_3) = (T+1)/(T-1)
so k_1 and k_4 are integer factors of T+1, and k_2 and k_3 are integer
factors of T-1.
Easy. Just like that you're most of the way to using the equations.
|
The only problem is that you now need to factor T+1 and T-1. (Granted,
you can show that 2 is a factor of one and 4 a factor of the other, but
the complete factorization is needed here.) And if you're going to use
this to factor T-1, you need to know the factors of (T-1)+1 = T, so
you're going in circles after all.
| Quote: | That gives a finite set of possibles for the k's.
For instance, with
So, for instance if T=15, you have
(k_1*k_4)/(k_2*k_3) = 16/14 = 8/7
so k_1*k_4 = 8, and k_2*k_3 = 7
and one possible setup then is
k_1 = 2, k_4 = 4, k_2 = 7, k_3 = 1
so plug those into
S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))
and you need S, and can use any square you like, so it's easiest just
to use
x=y=1, and take the positive result of the square roots to get
S = (2 + 7)(1 + 4) = 45
which promptly factors it,
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Er. the object was to factor T, _the target_, not the surrogate.
(For those of you who pay attention to certain other "posters", you'll
recognize this as Archimedes Plutonium's favorite technique: Change the
problem until it becomes trivial enough to solve.)
| Quote: | but I'll continue, as that's just one
possibility, so you have to check for it.
If S were coprime to T, then you now use factors of S, where with
S = g_1*g_2
you have
k_1*sqrt(x) + k_2*sqrt(y) = g_1
k_3*sqrt(x) + k_4*sqrt(y) = g_2
and you just find find squares for x and y that will work to give you
g_1 and g_2, and here's where that second solution from the square
roots comes in, as with each set of squares for x and y that will work,
you just change the sign of one of the square roots, to get the shadow
factorization.
That's it. Remarkably simple, as you go for the hidden factorization.
For instance, still using x=y=1, with my simple example, now take the
negative of ONE of the square roots:
S = (2 - 7)(1 + 4) = 55
I guess T=15 is too dinky of an example as it just keeps factoring it,
no matter what you do, but at least that still shows the basic idea
here.
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No, it doesn't. As far as I can tell, the only thing that the k_i's say
about T is that
T = (k_1*k_4 + k_2*k_3) / (k_1*k_4 - k_2*k_3) = (2*4 + 7*1)/(2*4 -
7*1)
= 2*4 + 7*1,
(since 2*4-7*1 = 1) which isn't a factorization of T. Nowhere have you
deduced that
15 = 3*5 or 5*3, which are the only non-trivial (integer)
factorizations of 15.
| Quote: | To recap, I looked at an expression
S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))
that can't just be the factorization of a single number because of the
use of square roots. I posted about it and a Tim Peters worked through
some analysis following instructions I gave, which gives up the simple
equation
(k_1*k_4 + k_2*k_3) / (k_1*k_4 - k_2*k_3) = T
which can be solved to get
(k_1*k_4)/(k_2*k_3) = (T+1)/(T-1)
so you can find the k's based on the factors of T+1 and T-1, to plug
back into the original equations and get an S, and then you use the
factors of S to find squares that are solutions for x and y, and then
you do the remarkable trick of just switching signs one at a time to
get to the hidden factorization.
But does it work beyond toy examples like factoring 15?
|
It doesn't even factor 15; see above.
--- Christopher Heckman
P.S. And now for the "I'll get my revenge one day" part of every JSH
post:
| Quote: | Unfortunately, as I said, Tim Peters is a hostile when it comes to my
research, so he promptly busied himself trying to obscure use of the
equations, while claiming that they don't work. You can look at recent
threads I've created on sci.crypt and sci.math to see him at work.
Some of you might be shocked by such behavior, as, hey, it's the
factoring problem.
But consider for YEARS Peters and people like him have been shadowing
my posts working to convince people that my research is useless.
I assure you that he has worked in this way many times.
He is only doing what he has always been doing, so there is nothing new
in his behavior.
So the question is an open one. Does this method work?
I can assure you that the math society that has busied itself ignoring
my research for years is in no hurry to acknowledge a result of mine
like this one--if it does work--as then they would be completely
overturned, now wouldn't they?
And what is your security against theirs?
So they wait, seeing if the "crackpot" label will hold, and they wait,
to see if no one can tell if these ideas will work or not, and if they
do work, they wait until there is a disaster big enough for the world
to care about the truth.
And then, finally, they will be able to wait, no more.
James Harris |
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Mike
science forum addict
Joined: 14 Oct 2005
Posts: 50
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| Posted: Wed Jun 21, 2006 5:13 am Post subject:
Re: SF: Simpler factoring idea, but does it work?
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In article <1150865572.908937.137510@c74g2000cwc.googlegroups.com>, CCHeckman@gmail.com says...
| Quote: |
jstevh@msn.com wrote:
The only problem is that you now need to factor T+1 and T-1. (Granted,
you can show that 2 is a factor of one and 4 a factor of the other, but
the complete factorization is needed here.) And if you're going to use
this to factor T-1, you need to know the factors of (T-1)+1 = T, so
you're going in circles after all.
Ahh, but by dividing out the 2 and 4 you have reduced the difficult task of factoring one 200 digit number with the |
much easier task of factoring one 199.57 digit number and one 199.13 digit number. Repeat this process approximately
230 times and the problem is vastly simplified into one of factoring 1.7x10^69 numbers, _none_ of which are over 100
digits long!
Mike |
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blown cap
science forum beginner
Joined: 21 Jun 2006
Posts: 1
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| Posted: Wed Jun 21, 2006 5:55 am Post subject:
Re: JSH: SF: Simpler factoring idea, but does it work?
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"Tim Peters" <tim.one@comcast.net> wrote in message
news:ZY6dnaOzVPofVwXZnZ2dnUVZ_t-dnZ2d@comcast.com...
| Quote: | [added "JSH:" to subject, spared sci.crypt and alt.math]
|
Tim, I must correct you. You should have typed [added "JSH:" to subject,
spared sci.crypt and alt.math, then, replied point by point, making me yet
another dumbass troll feeder]
Please get it right in the future.
Thanks. |
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Tim Peters
science forum Guru
Joined: 30 Apr 2005
Posts: 426
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| Posted: Wed Jun 21, 2006 6:46 am Post subject:
Re: JSH: SF: Simpler factoring idea, but does it work?
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[spared sci.crypt and alt.math]
[blown cap]
| Quote: | "Tim Peters" <tim.one@comcast.net> wrote in message
news:ZY6dnaOzVPofVwXZnZ2dnUVZ_t-dnZ2d@comcast.com...
[added "JSH:" to subject, spared sci.crypt and alt.math]
Tim, I must correct you. You should have typed [added "JSH:" to subject,
spared sci.crypt and alt.math, then, replied point by point, making me yet
another dumbass troll feeder]
Please get it right in the future.
|
Not needed -- the last part goes without saying ;-)
BTW, why did you add the x-posts to sci.crypt and alt.math back again? For
someone who doesn't want to play what he believes is a troll game, that's
incongruously trollish behavior  . |
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Tim Peters
science forum Guru
Joined: 30 Apr 2005
Posts: 426
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| Posted: Wed Jun 21, 2006 7:45 am Post subject:
Re: JSH: SF: Simpler factoring idea, but does it work?
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[added "JSH:" to subject, spared sci.crypt and alt.math]
[JSH]
| Quote: | ...
And you have
(k_1*k_4)/(k_2*k_3) = (T+1)/(T-1)
so k_1 and k_4 are integer factors of T+1, and k_2 and k_3 are integer
factors of T-1.
Easy. Just like that you're most of the way to using the equations.
|
[Proginoskes]
You really shouldn't make that claim :-)
| Quote: | is that you now need to factor T+1 and T-1. (Granted, you can show
that 2 is a factor of one and 4 a factor of the other, but the complete
factorization is needed here.)
|
Why? Finding 2 and 4 is already enough to find k_i that work. For that
matter, k_1 = T+1, k_2 = T-1, k_3 = k_4 = 1 is one of several (trivially
different) ways to find k_i that work without finding any proper factors of
anything.
This one is most like last year's surrogate factoring algorithms, where
James would _like_ to believe that any way whatsoever will work. Also like
last year's, it's not true that any way will work. Then layers of search
get added, one at a time, as it's shown that "oops! that doesn't always work
either". He has to find that out for himself, though.
| Quote: | And if you're going to use this to factor T-1, you need to know the
factors of (T-1)+1 = T, so you're going in circles after all.
|
Spoken too much like a mathematician  It would be a legitimate advance
if an efficient way were known to factor T _given_ full factorizations for
T+1 and T-1. For example, many RSA-class composites would fall with ease if
such a way were known.
Besides, as you just pointed out, even if he does need full factorizations,
he at worst (in number of bits) needs to factor (T-1)/2 or (T+1)/2, and
adding one to those doesn't land back on T.
| Quote: | ...
x=y=1, and take the positive result of the square roots to get
S = (2 + 7)(1 + 4) = 45
which promptly factors it,
Er. the object was to factor T, _the target_, not the surrogate.
|
He meant that gcd(2+7, 15) = 3 and gcd(1+4, 15) = 5, i.e. that the factors
of S can be used to "promptly factor" T as well in this case. That's
unusual.
| Quote: | ...
For instance, still using x=y=1, with my simple example, now take the
negative of ONE of the square roots:
S = (2 - 7)(1 + 4) = 55
I guess T=15 is too dinky of an example as it just keeps factoring it,
no matter what you do, but at least that still shows the basic idea
here.
No, it doesn't. As far as I can tell, the only thing that the k_i's say
about T is that
T = (k_1*k_4 + k_2*k_3) / (k_1*k_4 - k_2*k_3) = (2*4 + 7*1)/(2*4 - 7*1)
= 2*4 + 7*1,
(since 2*4-7*1 = 1) which isn't a factorization of T. Nowhere have you
deduced that 15 = 3*5 or 5*3, which are the only non-trivial (integer)
factorizations of 15.
|
James is just being unclear again. What he meant to say is that all of:
2-7
-2+7
1-4
-1+4
are non-trivial factors of 15. You don't even need to take a gcd here.
That's also unusual, although James doesn't know that yet because this is
apparently the only example he's tried, and he's not thinking critically
about what's actually going on (see my sci.math reply for more on that).
| Quote: | ...
P.S. And now for the "I'll get my revenge one day" part of every JSH
post:
|
Well, if he stuck to the math, more people might take him seriously, and if
nothing else he's consistent about doing all he can to ensure that never
happens |
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marc.t.davies@gmail.com
science forum addict
Joined: 31 May 2006
Posts: 52
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| Posted: Wed Jun 21, 2006 8:44 am Post subject:
Re: SF: Simpler factoring idea, but does it work?
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|
| Quote: | Ahh, but by dividing out the 2 and 4 you have reduced the difficult task of factoring one 200 digit number with the
much easier task of factoring one 199.57 digit number and one 199.13 digit number. Repeat this process approximately
230 times and the problem is vastly simplified into one of factoring 1.7x10^69 numbers, _none_ of which are over 100
digits long!
|
We have a winner! |
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Justin
science forum Guru Wannabe
Joined: 28 Apr 2005
Posts: 204
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| Posted: Wed Jun 21, 2006 11:31 am Post subject:
Re: SF: Simpler factoring idea, but does it work?
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In sci.math jstevh@msn.com wrote:
: S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))
Every time you start one of these threads I hope you're actually going to
factor something, but again you've managed to blather on and on without
actually doing anything.
Justin |
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Denis Feldmann
science forum addict
Joined: 23 Apr 2006
Posts: 87
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| Posted: Wed Jun 21, 2006 12:46 pm Post subject:
Re: JSH: SF: Simpler factoring idea, but does it work?
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Tim Peters a écrit :
| Quote: | [added "JSH:" to subject, spared sci.crypt and alt.math]
[cut the beginning, because...
James is just being unclear again. What he meant to say is that all of:
2-7
-2+7
1-4
-1+4
are non-trivial factors of 15. You don't even need to take a gcd here.
That's also unusual, although James doesn't know that yet because this is
apparently the only example he's tried, and he's not thinking critically
about what's actually going on (see my sci.math reply for more on that).
|
....I suddenly made a HUGE discovery. You are (correctly, I guess) making
James an actualisation of the ideal intuitionnistic mathematician
(Brouwer-like description) : no mathematical result is a theorem before
*he* has proved (or, in James case, at least suspected) it to be true.
So, for instance, those pesky RSA numbers could well be prime, which
explains he doesn't even try to factorize them. This is also the reason
you are lying (i.e. saying deliberate untruths): your results are not
true yet for James, so you are lying *now*. Later, when/if he comes to
the conclusion they are true, it will not change the fact that you were
lying now (and as you will not bother to repeat those results then, you
will have been lying forever ; note that trying to prevent that by
repeating them would not work either, as stupidly repeating true results
*after* James has proved them would be at best plagiarizing, and trying
to hinder further research by laying heavy on James' previous unlucky
attempts)
| Quote: |
...
P.S. And now for the "I'll get my revenge one day" part of every JSH
post:
Well, if he stuck to the math, more people might take him seriously, and if
nothing else he's consistent about doing all he can to ensure that never
happens
|
At one time, he described his strategy as one of *purposefully*
destroying his credibility so that when he comes with his HUGE results,
nobody will believe him, allowing him to show the res(t of the world
what a bunch of corrupted and phonies we all are. Phase 1 of this master
plan , up to now, is a complete success
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marc.t.davies@gmail.com
science forum addict
Joined: 31 May 2006
Posts: 52
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| Posted: Wed Jun 21, 2006 1:12 pm Post subject:
Re: SF: Simpler factoring idea, but does it work?
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Give the man his dues. He's managed to factor the number 15! He's up to
2**4-1, he'll be cracking those RSA composites in no time. |
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marc.t.davies@gmail.com
science forum addict
Joined: 31 May 2006
Posts: 52
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| Posted: Wed Jun 21, 2006 1:14 pm Post subject:
Re: JSH: SF: Simpler factoring idea, but does it work?
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| Quote: | Phase 1 of this master plan , up to now, is a complete success
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Oh, it hurts. It hurts! |
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david.florman@att.net
science forum beginner
Joined: 21 Jun 2006
Posts: 1
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| Posted: Wed Jun 21, 2006 3:22 pm Post subject:
Re: SF: Simpler factoring idea, but does it work?
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This is an idle question, as I have no intention of attempting to
break RSA nos. Does anyone try to deconstruct these nos from the
integer end? |
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