### Sunday, January 23, 2005

## JSH: Nearly done

There has been a lot of verbiage flying back and forth related to my theory and method for factoring that I call surrogate factoring.

The mathematics though is surprisingly simple, while checking it thoroughly can test you to your limits. It's some of the best kind of mathematics to consider as elementary methods are shown to show surprising results never conceived of before.

At the heart of the theory are two simple quadratics:

yx^2 + Ax - j^2 = T

and

yz^2 + Az - j^2 = 0,

where all are to be rational, while A, j and T are also integers.

The primary question is, do rational non-zero x, y, z exist?

So what does that have to do with factoring?

Well j and T are chosen such that M^2 = j^2 + T, where M is the number you're trying to factor, and that means that from the first equation you have

x(yx + A) = M^2

so x is a factor of M, but it can be a fraction, so you concentrate on its numerator to see if that gives a prime factor of M.

So why M^2 and not M? Well that's where the theory starts pushing you, as I tried M, and found out that mathematically it didn't give me something I could work with easily, but M^2 did.

If you let b_1 b_2 = -j^2, and f_1 f_2 = T, it is easy to prove that

x = (b_1 f_2 + b_2 f_1 - 2j^2)/A

meaning that x is defined by the factors of j^2 and T, though it is a factor of M^2, which is why this is surrogate factoring. The surrogates are factored to try and factor M.

It's a brilliant idea. No matter what any poster says in reply, what I've already shown is simply brilliant, as I have x, a factor of M, defined by factors of numbers other than M, and that is the start.

The first problem you face is that b_1, b_2, f_1 and f_2 may not be integers, but are in the field of rationals.

My paper goes over methods that put f_1 and f_2 into integers, while allowing the mathematics to *choose* b_1 and b_2 so that they are still in the field of rationals, but you get the mathematics selecting them out of infinity.

Yup, that's why I call it a super sieve, as in picking b_1 and b_2 for you, the algebra checks against all rationals, the entire set, and that's an infinite set!

Now we are beyond brilliant into the arena of almost impossible to imagine, with a technique for factoring, which loops through the entire field of rationals in searching for a solution.

So must x reveal a non-trivial factor of M?

The answer, amazingly enough, depends on quadratic residues!!!

The mathematics requires only elementary methods, but in the space of a few paragraphs I've talked about looking for a prime factor of some target M, where the answer depends on the factorization of numbers other than M, and you pick one of those numbers, while letting the algebra factor the other, going through the entire field of rationals to pick a solution!

All of that is easy to prove, and not only did I prove it in a paper, I wrote a program that does it.

See http://groups.yahoo.com/group/sufactor/

I'm a person not just making claims--I have the full demonstration of everything I say here, and I've worked it out, and put it out for people to consider.

So, how can that be possible? How can what I say here be true, and the mathematical establishment not pay attention?

There's something wrong with them. I have other big results and they've tried to ignore those, and I've talked about some of them here where people have usually lied about my work.

It's weird. But it's the reality.

With "pure math" people can lie. That's just a fact. If some group of mathematicians write gibberish, and declare it to be a proof of something, then basically that goes over, if they are considered to be mathematicians of note.

Sure, mathematicians say that's not the way it is, but that's the way it is.

And if an amateur mathematician makes major discoveries, it goes the other way.

Mainstram mathematicians just band together and ignore their work, like they will try to do with my latest discovery, which they also try to do because they're very stupid.

The factoring problem is quite important in the real world.

It's not "pure math", so some of those people you respect and admire may soon be in jail, hated worldwide, and villified by people who will not be able to comprehend their behavior.

I find it hard to understand myself.

You do not believe that will happen.

It does not matter.

It will happen.

There are probably only a few days before many of your heroes are cast down, humiliated, ripped from their positions, and put up on public display as enemies of humanity itself, as people too blind to act in the best interests of society, and too dumb to realize they wouldn't get away with it.

You do not believe but it will happen, and the foundations of your society will not just be shaken--they will be shattered--as I promised years ago.

The end was never in doubt. You cannot betry the truth, and you cannot block mathematics.

You betrayed mathematics itself, so you will be destroyed by it.

It is the just solution, the most logical, the most rational one, and one of absolute perfection.

Those who betrayed the field of mathematics, who sullied that field with their lies, and their belief that social rules could win that they could make up "truth" and get away with it are about to learn just how powerful of a field it truly is as a lesson that humanity will never forget.

No one will be able to forget.

You will learn this time.

You will learn, never to make these mistakes again, and the lesson will be complete.

The mathematics though is surprisingly simple, while checking it thoroughly can test you to your limits. It's some of the best kind of mathematics to consider as elementary methods are shown to show surprising results never conceived of before.

At the heart of the theory are two simple quadratics:

yx^2 + Ax - j^2 = T

and

yz^2 + Az - j^2 = 0,

where all are to be rational, while A, j and T are also integers.

The primary question is, do rational non-zero x, y, z exist?

So what does that have to do with factoring?

Well j and T are chosen such that M^2 = j^2 + T, where M is the number you're trying to factor, and that means that from the first equation you have

x(yx + A) = M^2

so x is a factor of M, but it can be a fraction, so you concentrate on its numerator to see if that gives a prime factor of M.

So why M^2 and not M? Well that's where the theory starts pushing you, as I tried M, and found out that mathematically it didn't give me something I could work with easily, but M^2 did.

If you let b_1 b_2 = -j^2, and f_1 f_2 = T, it is easy to prove that

x = (b_1 f_2 + b_2 f_1 - 2j^2)/A

meaning that x is defined by the factors of j^2 and T, though it is a factor of M^2, which is why this is surrogate factoring. The surrogates are factored to try and factor M.

It's a brilliant idea. No matter what any poster says in reply, what I've already shown is simply brilliant, as I have x, a factor of M, defined by factors of numbers other than M, and that is the start.

The first problem you face is that b_1, b_2, f_1 and f_2 may not be integers, but are in the field of rationals.

My paper goes over methods that put f_1 and f_2 into integers, while allowing the mathematics to *choose* b_1 and b_2 so that they are still in the field of rationals, but you get the mathematics selecting them out of infinity.

Yup, that's why I call it a super sieve, as in picking b_1 and b_2 for you, the algebra checks against all rationals, the entire set, and that's an infinite set!

Now we are beyond brilliant into the arena of almost impossible to imagine, with a technique for factoring, which loops through the entire field of rationals in searching for a solution.

So must x reveal a non-trivial factor of M?

The answer, amazingly enough, depends on quadratic residues!!!

The mathematics requires only elementary methods, but in the space of a few paragraphs I've talked about looking for a prime factor of some target M, where the answer depends on the factorization of numbers other than M, and you pick one of those numbers, while letting the algebra factor the other, going through the entire field of rationals to pick a solution!

All of that is easy to prove, and not only did I prove it in a paper, I wrote a program that does it.

See http://groups.yahoo.com/group/sufactor/

I'm a person not just making claims--I have the full demonstration of everything I say here, and I've worked it out, and put it out for people to consider.

So, how can that be possible? How can what I say here be true, and the mathematical establishment not pay attention?

There's something wrong with them. I have other big results and they've tried to ignore those, and I've talked about some of them here where people have usually lied about my work.

It's weird. But it's the reality.

With "pure math" people can lie. That's just a fact. If some group of mathematicians write gibberish, and declare it to be a proof of something, then basically that goes over, if they are considered to be mathematicians of note.

Sure, mathematicians say that's not the way it is, but that's the way it is.

And if an amateur mathematician makes major discoveries, it goes the other way.

Mainstram mathematicians just band together and ignore their work, like they will try to do with my latest discovery, which they also try to do because they're very stupid.

The factoring problem is quite important in the real world.

It's not "pure math", so some of those people you respect and admire may soon be in jail, hated worldwide, and villified by people who will not be able to comprehend their behavior.

I find it hard to understand myself.

You do not believe that will happen.

It does not matter.

It will happen.

There are probably only a few days before many of your heroes are cast down, humiliated, ripped from their positions, and put up on public display as enemies of humanity itself, as people too blind to act in the best interests of society, and too dumb to realize they wouldn't get away with it.

You do not believe but it will happen, and the foundations of your society will not just be shaken--they will be shattered--as I promised years ago.

The end was never in doubt. You cannot betry the truth, and you cannot block mathematics.

You betrayed mathematics itself, so you will be destroyed by it.

It is the just solution, the most logical, the most rational one, and one of absolute perfection.

Those who betrayed the field of mathematics, who sullied that field with their lies, and their belief that social rules could win that they could make up "truth" and get away with it are about to learn just how powerful of a field it truly is as a lesson that humanity will never forget.

No one will be able to forget.

You will learn this time.

You will learn, never to make these mistakes again, and the lesson will be complete.

### Tuesday, January 18, 2005

## Factoring problem, solved

I am increasingly certain that I've solved the factoring problem.

That is a bold claim to add to previous bold claims of mine, but here in THIS post I'll give the underlying theory as it's very basic, and direct you to a site where you can download a rough, and somewhat flawed, but still good enough to show the idea, prototype factoring program, which implements some of the theory.

I'll explain why I say some of the theory in just a bit.

My work boils down to the analysis of a system of equations:

a_1 x + b_1 = f_1

a_2 x + b_2 = f_2

a_1 b_2 + a_2 b_2 = A

a_1 a_2 x^2 + Ax + b_1 b_2 = f_1 f_2

where to match the variables in my initial paper

y = a_1 a_2, T = f_1 f_2, and j^2 = - b_1 b_2.

(Here b_1 b_2 is a negative number so j is still an integer.)

I have a rather complex looking solution in the paper, but remarkably you can encompass it by solving the first three equations for key variables:

a_1 = A(b_1 - f_1)/(b_1 f_2 + b_2 f_1 - 2b_1 b_2)

and

x = (b_1 f_2 + b_2 f_1 - 2b_1 b_2)/A

which shows that you can actually factor simply by cycling through the factors of j and T, while my focus in the program is on cycling through the factors of T.

All together you then get the factors of your target M, as

M^2 = j^2 + T, so

a_1 a_2 x^2 + Ax + b_1 b_2 = f_1 f_2

is

a_1 a_2 x^2 + Ax - M^2 = 0

so x is a factor of M.

The idea may seem pathetically simple, but that is because you are programmed to believe that the factoring problem is hard. It is not.

I have just shown you in a few lines how to factor an arbitrary non-zero positive integer M in polynomial time.

Many of you are programmed to believe that mathematics must be abstruse and difficult to work. You are wrong. The factoring solution I've presented here while looking extremely simple can work and is simple because the PROBLEM is simple.

I have demonstration code, which is rough as I threw something together, to test out my own theory, and hopefully to be more convincing. That code implements a VERSION of my own theory, as I don't factor both j and T, but only T, as j gets background factored by the mathematics.

If you watch it factoring and spitting out primes (though it also mislabels some as prime and can't factor some easy numbers...rough version remember) then you can at least see that something is happening here worth investigating further.

In the hopes of convincing

http://groups.yahoo.com/group/sufactor/

It is CRITICAL that the information be taken seriously as soon as possible so that some other method can be used for Internet security, and quickly implemented.

Time is a factor.

That is a bold claim to add to previous bold claims of mine, but here in THIS post I'll give the underlying theory as it's very basic, and direct you to a site where you can download a rough, and somewhat flawed, but still good enough to show the idea, prototype factoring program, which implements some of the theory.

I'll explain why I say some of the theory in just a bit.

My work boils down to the analysis of a system of equations:

a_1 x + b_1 = f_1

a_2 x + b_2 = f_2

a_1 b_2 + a_2 b_2 = A

a_1 a_2 x^2 + Ax + b_1 b_2 = f_1 f_2

where to match the variables in my initial paper

y = a_1 a_2, T = f_1 f_2, and j^2 = - b_1 b_2.

(Here b_1 b_2 is a negative number so j is still an integer.)

I have a rather complex looking solution in the paper, but remarkably you can encompass it by solving the first three equations for key variables:

a_1 = A(b_1 - f_1)/(b_1 f_2 + b_2 f_1 - 2b_1 b_2)

and

x = (b_1 f_2 + b_2 f_1 - 2b_1 b_2)/A

which shows that you can actually factor simply by cycling through the factors of j and T, while my focus in the program is on cycling through the factors of T.

All together you then get the factors of your target M, as

M^2 = j^2 + T, so

a_1 a_2 x^2 + Ax + b_1 b_2 = f_1 f_2

is

a_1 a_2 x^2 + Ax - M^2 = 0

so x is a factor of M.

The idea may seem pathetically simple, but that is because you are programmed to believe that the factoring problem is hard. It is not.

I have just shown you in a few lines how to factor an arbitrary non-zero positive integer M in polynomial time.

Many of you are programmed to believe that mathematics must be abstruse and difficult to work. You are wrong. The factoring solution I've presented here while looking extremely simple can work and is simple because the PROBLEM is simple.

I have demonstration code, which is rough as I threw something together, to test out my own theory, and hopefully to be more convincing. That code implements a VERSION of my own theory, as I don't factor both j and T, but only T, as j gets background factored by the mathematics.

If you watch it factoring and spitting out primes (though it also mislabels some as prime and can't factor some easy numbers...rough version remember) then you can at least see that something is happening here worth investigating further.

In the hopes of convincing

**someone**willing to help I have put put up a Yahoo! Group with my paper and some rudimentary code:http://groups.yahoo.com/group/sufactor/

It is CRITICAL that the information be taken seriously as soon as possible so that some other method can be used for Internet security, and quickly implemented.

Time is a factor.