### Saturday, February 26, 2005

## JSH: Part of the point, math society

So like I said, surrogate factoring is a

Now, just for the sake of argument, let's say it can.

Well it's a decent idea, and if it can be made to work then there is an enormous impact on the entire world, but my thesis is that math society is corrupt, so you people will sit back, and passively hope that it just can't work, or if it can, that no one will figure it out!

That's part of the point, you're not rational as a group!

People believe all kinds of things about mathematicians now, like they just won't believe that mathematicians as a

I know you can cause I feel very certain I have multiple proofs, for instance, that Wiles did not prove Fermat's Last Theorem.

And they don't believe that you can ignore major mathematical proofs, when I know you can.

I know you can because surrogate factoring is my

But people don't yet get it, yet.

There's nothing like a situation where everyone does get it, for people to learn to drop their beliefs.

And here, everyone will get it.

The idea is out there. There is ample time to inform people who trust in society to efficiently distribute important information.

That's how you can trust medical doctors to mostly not harm you with medical information.

And how you can trust your mechanices to fix your car, or try to trust the car manufacturers to recall when there's a problem, like exploding gas tanks.

Similarly, people, I say naively, believe that they can trust math experts i.e. mathematicians, to inform them of important mathematical issues—like a concept that could affect their life savings.

But I fear that their belief is just wrong and that you people as a

I'm telling the world you

You deliberately act to hide, as a group, discoveries made by amateur mathematicians, I think to preserve your social order against the inherent destabilizing reality that people you consider outsiders, strangers, can make major mathematical discoveries.

And so one small group has the power to to greatly harm many much larger groups, and remind people that sometimes trust can be misplaced.

The Hammer is here.

[A reply to someone who asked if James was only drunk or under the influence of other drugs.]

No, I'm desperate. I guess I didn't help matters much either.

The proof is trivial. The method is easily proven to have to work.

Someone can exploit this and then what?

All the chatter in the world won't make a difference then, and whatever meager satisfaction some of you get from making fun of me will not compare to the consequences.

Math is real, whether you believe in it or not.

People are out there trusting. Trusting that there are smart people watching out for them, watching out for some big result.

But you people still can't get past insulting me.

**concept**so the particulars aren't as important as whether or not the concept can be made to work practically.Now, just for the sake of argument, let's say it can.

Well it's a decent idea, and if it can be made to work then there is an enormous impact on the entire world, but my thesis is that math society is corrupt, so you people will sit back, and passively hope that it just can't work, or if it can, that no one will figure it out!

That's part of the point, you're not rational as a group!

People believe all kinds of things about mathematicians now, like they just won't believe that mathematicians as a

**group**can claim an argument is correct, even when it's not.I know you can cause I feel very certain I have multiple proofs, for instance, that Wiles did not prove Fermat's Last Theorem.

And they don't believe that you can ignore major mathematical proofs, when I know you can.

I know you can because surrogate factoring is my

**fourth**major mathematical discovery.But people don't yet get it, yet.

There's nothing like a situation where everyone does get it, for people to learn to drop their beliefs.

And here, everyone will get it.

The idea is out there. There is ample time to inform people who trust in society to efficiently distribute important information.

That's how you can trust medical doctors to mostly not harm you with medical information.

And how you can trust your mechanices to fix your car, or try to trust the car manufacturers to recall when there's a problem, like exploding gas tanks.

Similarly, people, I say naively, believe that they can trust math experts i.e. mathematicians, to inform them of important mathematical issues—like a concept that could affect their life savings.

But I fear that their belief is just wrong and that you people as a

**group**will do that groupthink thing, and just sit on your hands, hoping later to claim it wasn't your fault, and how could you know?I'm telling the world you

**had**to know, and that later you will just lie like you're lying now, as your actions must be conscious on many levels.You deliberately act to hide, as a group, discoveries made by amateur mathematicians, I think to preserve your social order against the inherent destabilizing reality that people you consider outsiders, strangers, can make major mathematical discoveries.

And so one small group has the power to to greatly harm many much larger groups, and remind people that sometimes trust can be misplaced.

The Hammer is here.

[A reply to someone who asked if James was only drunk or under the influence of other drugs.]

No, I'm desperate. I guess I didn't help matters much either.

The proof is trivial. The method is easily proven to have to work.

Someone can exploit this and then what?

All the chatter in the world won't make a difference then, and whatever meager satisfaction some of you get from making fun of me will not compare to the consequences.

Math is real, whether you believe in it or not.

People are out there trusting. Trusting that there are smart people watching out for them, watching out for some big result.

But you people still can't get past insulting me.

### Wednesday, February 16, 2005

## Critics. How many have there been?

Who knows. Over the years I've seen so many postings from people who seem to think they gain something by stepping up to criticize, as if they're playing to some imaginary cheering crowd.

They come and go.

Years back, some posters would decide they'd made major points and do celebratory posts, or a couple of them would congratulate each other on how brilliant they thought they were with some cutting and negative remark.

They have come and gone.

I have ideas that I get excited about, and ones that worry me, like surrogate factoring, where I make certain extraordinary statements, which I attempt to justify mathematically.

In response I see some people who challenge those assertions, at times successfully, and a LOT of others who seem to believe that in this world, a good performance can be about just cutting someone else down.

Yeah, I guess there are comedians who seem to get away with nothing but personal attacks, but you actually have to produce something to keep people's interest.

People who just maliciously attack me in posts either do two things, end up just eventually stopping, and being forgotten, or turn into the weird almost ghoulish crew who stalk my postings, replying obsessively, as if they have lost the ability to not do so.

I say the simple thing is to let this be Usenet.

If you don't like my postings do what most people do with things they don't like: turn away

But don't come out, obsessively seek out my posts, reply to them energetically, and then claim to not like my postings, while you try to make a name for yourself with attack postings, as not only is it silly, it does not work.

You don't make a name for yourself.

You make a bigger name for me, that's all, so I can make posts like this one, and it be justified.

Now ask yourself, who else on sci.math or sci.crypt can make a post like this one at all, and pull it off?

Despite any critics comments to the contrary, I'm the only one.

They come and go.

Years back, some posters would decide they'd made major points and do celebratory posts, or a couple of them would congratulate each other on how brilliant they thought they were with some cutting and negative remark.

They have come and gone.

I have ideas that I get excited about, and ones that worry me, like surrogate factoring, where I make certain extraordinary statements, which I attempt to justify mathematically.

In response I see some people who challenge those assertions, at times successfully, and a LOT of others who seem to believe that in this world, a good performance can be about just cutting someone else down.

Yeah, I guess there are comedians who seem to get away with nothing but personal attacks, but you actually have to produce something to keep people's interest.

People who just maliciously attack me in posts either do two things, end up just eventually stopping, and being forgotten, or turn into the weird almost ghoulish crew who stalk my postings, replying obsessively, as if they have lost the ability to not do so.

I say the simple thing is to let this be Usenet.

If you don't like my postings do what most people do with things they don't like: turn away

But don't come out, obsessively seek out my posts, reply to them energetically, and then claim to not like my postings, while you try to make a name for yourself with attack postings, as not only is it silly, it does not work.

You don't make a name for yourself.

You make a bigger name for me, that's all, so I can make posts like this one, and it be justified.

Now ask yourself, who else on sci.math or sci.crypt can make a post like this one at all, and pull it off?

Despite any critics comments to the contrary, I'm the only one.

### Sunday, February 13, 2005

## Surrogate factoring, room for error?

Well, considering the direct evidence that I may be wrong, it seems to me that the simplest explanation would be at the start:

yx^2 + Ax - M^2 = 0

and

yz^2 + Az - j^2 = 0

as that's an assertion of a truth.

Essentially I'm saying that there exists these two quadratics related in that given way, but if in fact no rationals exist such that all conditions can be met for a given natural M and j then that statement is false.

If it is true, then all else follows in the basic argument I've given.

So then, can that initial start be false?

Well I can solve out y to get

z^2 (M^2 - Ax) = x^2 (j^2 - Az)

and then I can solve for A to get

A = (z^2 M^2 - j^2 x^2)/xz(z-x)

so A is not as arbitrary as I imagined.

Hmmm…A again, where it's not clear that it's just settable, but then again, x and z are

Switching to rationals for x and z, with z = a/b and x = c/d, I get

A = (a^2 d^2 M^2 - j^2 b^2 c^2)/ac(ad - bc)

so there must exist integers a, b, c and d such that A is an integer, or it's a fraction.

Yeah, but I

So, rationals is kind of a big set, there must exist a, b, c and d for any natural A chosen.

That means the start of my proof is valid, and

yx^2 + Ax - M^2 = 0

and

yz^2 + Az - j^2 = 0

must be true for some naturals M, j and A, with rational y, x and z.

Though I'm still convinced that if M is odd, j needs to be odd.

Mystery continues…why don't implementations work?

Where's the mistake?

Yup. And I had the reason correct at the beginning as if j is even when M is odd then both quadratics cannot necessarily be true, with rational nonzero x, y and z.

So that breaks things at the start of the argument, so nothing else follows.

A proof begins with a truth. The start of the proof is the set of quadratics for

If M is odd while j is even, then that start is not valid, as then no rational nonzero x, y and z exist.

Neat.

yx^2 + Ax - M^2 = 0

and

yz^2 + Az - j^2 = 0

as that's an assertion of a truth.

Essentially I'm saying that there exists these two quadratics related in that given way, but if in fact no rationals exist such that all conditions can be met for a given natural M and j then that statement is false.

If it is true, then all else follows in the basic argument I've given.

So then, can that initial start be false?

Well I can solve out y to get

z^2 (M^2 - Ax) = x^2 (j^2 - Az)

and then I can solve for A to get

A = (z^2 M^2 - j^2 x^2)/xz(z-x)

so A is not as arbitrary as I imagined.

Hmmm…A again, where it's not clear that it's just settable, but then again, x and z are

**rationals**, so more analysis is needed.Switching to rationals for x and z, with z = a/b and x = c/d, I get

A = (a^2 d^2 M^2 - j^2 b^2 c^2)/ac(ad - bc)

so there must exist integers a, b, c and d such that A is an integer, or it's a fraction.

Yeah, but I

**can**just choose an A, and then get rational x, y and z.So, rationals is kind of a big set, there must exist a, b, c and d for any natural A chosen.

That means the start of my proof is valid, and

yx^2 + Ax - M^2 = 0

and

yz^2 + Az - j^2 = 0

must be true for some naturals M, j and A, with rational y, x and z.

Though I'm still convinced that if M is odd, j needs to be odd.

Mystery continues…why don't implementations work?

Where's the mistake?

Yup. And I had the reason correct at the beginning as if j is even when M is odd then both quadratics cannot necessarily be true, with rational nonzero x, y and z.

So that breaks things at the start of the argument, so nothing else follows.

A proof begins with a truth. The start of the proof is the set of quadratics for

**rational**x, y and z.If M is odd while j is even, then that start is not valid, as then no rational nonzero x, y and z exist.

Neat.

## I was right, surrogate factoring proof

Last weekend I came across a simpler way to describe surrogate factoring with a couple of quadratics and solutions for x and z, which I promptly programmed—and it didn't seem to work.

I puzzled over that a bit, and posted the equations:

yx^2 + Ax - M^2 = 0

and

yz^2 + Az - j^2 = 0

where A, j, and M are integers greater than 0 chosen, where M is the target to be factored, and you find that you can use T, where

T = M^2 - j^2

and substituting out y to solve for x and z gives

x = z(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)

and

z = x(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)

and you

But, as I said, I tried the idea last weekend, and I didn't see it work, and also a poster named Rick Decker claimed to have a counter-example.

However, I was missing something easy, as consider A (yes, I've screwed this up before but here's the correct argument) as it's

Now then, for any rational x and z, there must exist some A, such that

Ax and Az

are integers, right? Like if x = 29343/3947387492 then

A = 3947387492, will force Ax to be an integer, equal to 29343.

Well, look at

x = z(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)

and

z = x(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)

and notice then that an A can always be chosen such that Ax and Az are integers.

So what, right?

Well, multiply both equations by A, and you get

Ax = Az(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)

and

Az = Ax(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)

so that now you have integers inside and out!!!

And looking at

sqrt((Az - 2M^2)^2 - 4TM^2)

it MUST be true that for some integer Az, M is factored.

In retrospect, with my tests last weekend, I, um, assumed that A=1, as I just figured it didn't matter, and that's probably why they didn't work, I guess, or also I might not have done plus and minus.

That is, the square roots give a positive and a negative solution, and I don't remember if I checked both.

In any event, the proof that the method has to work is easy. There must exist integers Ax and Az, as shown above.

[A reply to someone who askd James to inform the newsgroup when it turns out that his method won't work.]

Huh? Do you believe in algebra, or not?

Given

yx^2 + Ax - M^2 = 0

yz^2 + Az - j^2 = 0

and

T = M^2 - j^2

you must also have

x = z(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)

and

z = x(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)

and, you can multiply both equations by A, to get

Ax = Az(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)

Az = Ax(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)

where for any rational x and z there must be an integer A such that both Ax and Az are integers, so looping through

So, I can't be wrong, or, algebra is wrong.

Now that's a trivially easy proof. And I want to emphasize that it is a proof as I've seen weird stuff from sci.math'ers where proofs are just discounted, as if they don't matter.

But they do. A mathematical proof gives absolute certainty.

Surrogate factoring must work. There is no room for doubt.

[A reply to someone who said to have missed the proof that the method works in polynomial time.]

Sounds like

It's a trivially easy proof. End of story.

If you wish to betray that you're either not bothering to read it, or can't comprehend VERY basic algebra, then ok.

Reply again, and so betray.

[A reply to someone who asekd James to state the therem that he think he proved.]

And there you have it.

I've seen this mindless behavior go on for

Yup, hundreds of posts, where the fact that an actual proof has been given and shown is lost in the shuffle.

What motivates these people?

I think their behavior is instinctive.

Easily this thread could in a few hours be over a hundred posts long, despite my having given a very basic argument, using very basic algebra, at the level that teenagers learn.

Oh, and note that the posters so far in this thread are obsessive repliers.

The "Last Danish Pastry" is notable for starting a flame webpage against me some years back. Yes, years back.

"Jose Carlos Santos" is a current regular flamer.

The other two are as well.

And they will reply, reply, reply.

Notice that reasoning doesn't work with them.

I puzzled over that a bit, and posted the equations:

yx^2 + Ax - M^2 = 0

and

yz^2 + Az - j^2 = 0

where A, j, and M are integers greater than 0 chosen, where M is the target to be factored, and you find that you can use T, where

T = M^2 - j^2

and substituting out y to solve for x and z gives

x = z(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)

and

z = x(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)

and you

**should**be able to always factor M, by iterating through factors of Tj^2 to get the full set of of integer solutions for Ax.But, as I said, I tried the idea last weekend, and I didn't see it work, and also a poster named Rick Decker claimed to have a counter-example.

However, I was missing something easy, as consider A (yes, I've screwed this up before but here's the correct argument) as it's

**chosen**arbitrarily.Now then, for any rational x and z, there must exist some A, such that

Ax and Az

are integers, right? Like if x = 29343/3947387492 then

A = 3947387492, will force Ax to be an integer, equal to 29343.

Well, look at

x = z(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)

and

z = x(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)

and notice then that an A can always be chosen such that Ax and Az are integers.

So what, right?

Well, multiply both equations by A, and you get

Ax = Az(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)

and

Az = Ax(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)

so that now you have integers inside and out!!!

And looking at

sqrt((Az - 2M^2)^2 - 4TM^2)

it MUST be true that for some integer Az, M is factored.

In retrospect, with my tests last weekend, I, um, assumed that A=1, as I just figured it didn't matter, and that's probably why they didn't work, I guess, or also I might not have done plus and minus.

That is, the square roots give a positive and a negative solution, and I don't remember if I checked both.

In any event, the proof that the method has to work is easy. There must exist integers Ax and Az, as shown above.

[A reply to someone who askd James to inform the newsgroup when it turns out that his method won't work.]

Huh? Do you believe in algebra, or not?

Given

yx^2 + Ax - M^2 = 0

yz^2 + Az - j^2 = 0

and

T = M^2 - j^2

you must also have

x = z(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)

and

z = x(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)

and, you can multiply both equations by A, to get

Ax = Az(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)

Az = Ax(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)

where for any rational x and z there must be an integer A such that both Ax and Az are integers, so looping through

**all**possible integer Ax and Ay as required by the square roots MUST factor M.So, I can't be wrong, or, algebra is wrong.

Now that's a trivially easy proof. And I want to emphasize that it is a proof as I've seen weird stuff from sci.math'ers where proofs are just discounted, as if they don't matter.

But they do. A mathematical proof gives absolute certainty.

Surrogate factoring must work. There is no room for doubt.

[A reply to someone who said to have missed the proof that the method works in polynomial time.]

Sounds like

**you**missed algebra in school.It's a trivially easy proof. End of story.

If you wish to betray that you're either not bothering to read it, or can't comprehend VERY basic algebra, then ok.

Reply again, and so betray.

[A reply to someone who asekd James to state the therem that he think he proved.]

And there you have it.

I've seen this mindless behavior go on for

**hundreds**of posts.Yup, hundreds of posts, where the fact that an actual proof has been given and shown is lost in the shuffle.

What motivates these people?

I think their behavior is instinctive.

Easily this thread could in a few hours be over a hundred posts long, despite my having given a very basic argument, using very basic algebra, at the level that teenagers learn.

Oh, and note that the posters so far in this thread are obsessive repliers.

The "Last Danish Pastry" is notable for starting a flame webpage against me some years back. Yes, years back.

"Jose Carlos Santos" is a current regular flamer.

The other two are as well.

And they will reply, reply, reply.

Notice that reasoning doesn't work with them.

### Wednesday, February 09, 2005

## Factoring problem solution

I've been working to figure out the particulars of my approach to surrogate factoring focusing on getting solutions for y, and then using those to get solutions for x, as it's a little complicated.

It turns out there's an easier way, where you focus on x at the start, and actually solve out y, and at a key point you do an inversion to get the full solution set, solving the problem.

Notice the mathematics is very easy. The factoring problem had never been proven to be difficult, but had been supposed to be difficult because no one knew of an easy answer!

Well, here it is.

Take the two quadratics

yx^2 + Ax - M^2 = 0

and

yz^2 + Az - j^2 = 0

where A, j, and M are integers greater than 0 chosen, where M is the target to be factored, and you find that you can use T, where

T = M^2 - j^2

and substituting out y to solve for x and z gives

x = z(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)

and

z = x(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)

where you have the problem that x may be a fraction, and its denominator is what's important to determine.

So, invert it. Let x = 1/Aw, then

z = 1(-1/w +/- sqrt((1/w - 2j^2)^2 + 4Tj^2))/(2wM^2 - 2)

and focusing on the square root I have

sqrt((1 - 2wj^2)^2 + 4Tw^2 j^2))/w^2

and focusing on the square root again, multiplying it out, gives

sqrt(1 - 4 j^2 w + 4j^4 w^2 + 4Tj^2 w^2)

and collecting with respect to w, gives

sqrt(4(j^4 + Tj^2)w^2 - 4A^2 j^2 w + 1)

which is

sqrt(4j^2 (j^2 + T)w^2 - 4 j^2 w + 1)

and since M^2 = j^2 + T, that is

sqrt(4j^2 M^2 w^2 - 4j^2 w + 1)

and completing the square inside, gives

sqrt(4j^2 M^2 w^2 - 4j^2 w + j^2/M^2 + 1 - j^2 /M^2)

which is

sqrt((2j M w - j/M)^2 + (M^2 - j^2)/M^2)

which is

sqrt(j(2M^2 w - 1)^2 + T)/M^2

and focusing on the square root again, you have

sqrt(j(2M^2 w - 1)^2 + T)

showing that the factorization is dependent on T.

The inversion is what's necessary to finish out.

Since x = 1/Aw, if x is a fraction like

x = x_num/x_denum, where x_num and x_denum are integers, then

w = x_denum/A x_num

so you can easily recover x from the solution to w, and you can let A have different values but it's easier to just let A=1, as you don't gain anything using other values.

That means the solution is almost trivially easy, and not hard to implement.

The belief that factoring was a hard problem had nothing to do with mathematical reality, but simply human wants and needs.

It turns out there's an easier way, where you focus on x at the start, and actually solve out y, and at a key point you do an inversion to get the full solution set, solving the problem.

Notice the mathematics is very easy. The factoring problem had never been proven to be difficult, but had been supposed to be difficult because no one knew of an easy answer!

Well, here it is.

Take the two quadratics

yx^2 + Ax - M^2 = 0

and

yz^2 + Az - j^2 = 0

where A, j, and M are integers greater than 0 chosen, where M is the target to be factored, and you find that you can use T, where

T = M^2 - j^2

and substituting out y to solve for x and z gives

x = z(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)

and

z = x(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)

where you have the problem that x may be a fraction, and its denominator is what's important to determine.

So, invert it. Let x = 1/Aw, then

z = 1(-1/w +/- sqrt((1/w - 2j^2)^2 + 4Tj^2))/(2wM^2 - 2)

and focusing on the square root I have

sqrt((1 - 2wj^2)^2 + 4Tw^2 j^2))/w^2

and focusing on the square root again, multiplying it out, gives

sqrt(1 - 4 j^2 w + 4j^4 w^2 + 4Tj^2 w^2)

and collecting with respect to w, gives

sqrt(4(j^4 + Tj^2)w^2 - 4A^2 j^2 w + 1)

which is

sqrt(4j^2 (j^2 + T)w^2 - 4 j^2 w + 1)

and since M^2 = j^2 + T, that is

sqrt(4j^2 M^2 w^2 - 4j^2 w + 1)

and completing the square inside, gives

sqrt(4j^2 M^2 w^2 - 4j^2 w + j^2/M^2 + 1 - j^2 /M^2)

which is

sqrt((2j M w - j/M)^2 + (M^2 - j^2)/M^2)

which is

sqrt(j(2M^2 w - 1)^2 + T)/M^2

and focusing on the square root again, you have

sqrt(j(2M^2 w - 1)^2 + T)

showing that the factorization is dependent on T.

The inversion is what's necessary to finish out.

Since x = 1/Aw, if x is a fraction like

x = x_num/x_denum, where x_num and x_denum are integers, then

w = x_denum/A x_num

so you can easily recover x from the solution to w, and you can let A have different values but it's easier to just let A=1, as you don't gain anything using other values.

That means the solution is almost trivially easy, and not hard to implement.

The belief that factoring was a hard problem had nothing to do with mathematical reality, but simply human wants and needs.

### Tuesday, February 08, 2005

## JSH: Easy math, easy solution

I am talking about quadratics people:

yx^2 + Ax - M^2 = 0

and

yz^2 + Az - j^2 = 0

and solving those equations to get

x = z(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)

and

z = x(-Ax +/- sqrt((Ax + 2j^2)^2 - 4Tj^2))/(2M^2 - 2Ax)

is the kind of thing most of you should have been able to do when you were twelve years old.

The variables x and z are connected to each other while depending on factorizations of numbers other than the number of which they are a factor.

Think of it as a clever trick if it will help you.

But somehow I managed to get x a factor of M constrained by the factorization of Tj^2, where T = M^2 - j^2.

It's trivial to figure that out.

It's trivial to see A, and see that if you have some solution for x that is a fraction where A=1, then A can absorb the denominator of x, and you have a rational solution.

It's trivial to see that proves a closed form.

Now then, I say you people are worse than frauds, you are selfish, and stupid and willing to let other people suffer for your mistakes and failures.

I say you don't care about the hundreds of billions of dollars flowing as we speak around the world dependent on an idea that I've shown to be flawed with a couple of quadratics!!!

And I say that if other people die because of you being frauds then you should die with them.

I say that if people are harmed because you are liars then you should suffer harm in equal measure.

I say that you will face the consequences of refusing to acknowledge the truth, of refusing to be real mathematicians, of refusing to show social responsibility not in years, not in months, but in days.

You people are beneath contempt. Even now you still have a chance to do the right thing, to act for the benefit of others, to do SOMETHING other than lie.

You are corrupt from surface to core.

I say we deserved better, but you are what we got. Well I fear you like poverty, misery, and death as that's what you're setting the world up for on a large scale.

I have called some of you evil directly, and it seems to me that you are working to prove that you are evil.

The problem can be handled if recognized. But if you sit on your damn hands, then the consequences are HUGE.

It's trival math in many ways. You cannot plausibly say you didn't understand.

And for that, if you do nothing, you may die, and far sooner than you think.

yx^2 + Ax - M^2 = 0

and

yz^2 + Az - j^2 = 0

and solving those equations to get

x = z(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)

and

z = x(-Ax +/- sqrt((Ax + 2j^2)^2 - 4Tj^2))/(2M^2 - 2Ax)

is the kind of thing most of you should have been able to do when you were twelve years old.

The variables x and z are connected to each other while depending on factorizations of numbers other than the number of which they are a factor.

Think of it as a clever trick if it will help you.

But somehow I managed to get x a factor of M constrained by the factorization of Tj^2, where T = M^2 - j^2.

It's trivial to figure that out.

It's trivial to see A, and see that if you have some solution for x that is a fraction where A=1, then A can absorb the denominator of x, and you have a rational solution.

It's trivial to see that proves a closed form.

Now then, I say you people are worse than frauds, you are selfish, and stupid and willing to let other people suffer for your mistakes and failures.

I say you don't care about the hundreds of billions of dollars flowing as we speak around the world dependent on an idea that I've shown to be flawed with a couple of quadratics!!!

And I say that if other people die because of you being frauds then you should die with them.

I say that if people are harmed because you are liars then you should suffer harm in equal measure.

I say that you will face the consequences of refusing to acknowledge the truth, of refusing to be real mathematicians, of refusing to show social responsibility not in years, not in months, but in days.

You people are beneath contempt. Even now you still have a chance to do the right thing, to act for the benefit of others, to do SOMETHING other than lie.

You are corrupt from surface to core.

I say we deserved better, but you are what we got. Well I fear you like poverty, misery, and death as that's what you're setting the world up for on a large scale.

I have called some of you evil directly, and it seems to me that you are working to prove that you are evil.

The problem can be handled if recognized. But if you sit on your damn hands, then the consequences are HUGE.

It's trival math in many ways. You cannot plausibly say you didn't understand.

And for that, if you do nothing, you may die, and far sooner than you think.

## JSH: Time is not on your side

To me the basic surrogate factoring method, where you factor T and also j is not an elegant solution.

However, it's the easiest to show, so I posted the equations, and I've shown I'm right.

For even a beginning math student, the math is basically trivial.

You people, however, don't seem to have a clue what is going on.

Right now, someone can exploit the theory, and factor some really huge numbers.

My guess is that the theory can be developed with a distributed solution to the point where you could factor a 2048 bit number in milliseconds.

It's anybody's guess how long it would take develop the theory and implementiaon to that point, but I'd guess about two months.

Right now, it probably would take someone a couple of days with the BASIC theory and some powerful workstations to factor a 2048 bit number, just with the very basic and inelegant set of equations I've posted the last two days.

The more elegant solution I'm working on involves just factoring T.

And it's so elegant it allows a recursive solution, which potentially could factor a 4048 bit number on a single pc in a few hours.

You people are worrying about social crap, when you can DO THE MATH, and see that solutions are now being found that can end the world as you know it.

The surrogate factoring method in and of itself is just knowledge, and it's really important knowledge, which would have an impact anyway, but if it gets developed on the fringes and is acknowledged by the mainstream only when something really bad happens, then I assure you the world as you know it, the world of today, will be gone, replaced by who know's what.

Time is not on your side.

Play your social games, the ones I know so well, and ask yourselves some serious questions, how did you get here?

Soon enough events will unfold very rapidly. I fear your decision, so I suggest to those who listen that they enjoy the world, whatever their piece of it may be, as much as they can over the next few days, as soon enough, it will pass away, thanks to people who call themselves "mathematicians".

The tests were to be mathematical, and they were, given to those who called themselves mathematicians.

You have ignored the three previous, but cannot ignore the final one.

It is the final test for those who rejected the first three.

You were given three chances.

You failed them all.

So now, The Hammer is here, and with it, the end of days.

The world will be destroyed, and then remade, as foretold.

You will be lost, with your children, and then there will be others, and one day they will be tested, and will pass, but that is another story.

However, it's the easiest to show, so I posted the equations, and I've shown I'm right.

For even a beginning math student, the math is basically trivial.

You people, however, don't seem to have a clue what is going on.

Right now, someone can exploit the theory, and factor some really huge numbers.

My guess is that the theory can be developed with a distributed solution to the point where you could factor a 2048 bit number in milliseconds.

It's anybody's guess how long it would take develop the theory and implementiaon to that point, but I'd guess about two months.

Right now, it probably would take someone a couple of days with the BASIC theory and some powerful workstations to factor a 2048 bit number, just with the very basic and inelegant set of equations I've posted the last two days.

The more elegant solution I'm working on involves just factoring T.

And it's so elegant it allows a recursive solution, which potentially could factor a 4048 bit number on a single pc in a few hours.

You people are worrying about social crap, when you can DO THE MATH, and see that solutions are now being found that can end the world as you know it.

The surrogate factoring method in and of itself is just knowledge, and it's really important knowledge, which would have an impact anyway, but if it gets developed on the fringes and is acknowledged by the mainstream only when something really bad happens, then I assure you the world as you know it, the world of today, will be gone, replaced by who know's what.

Time is not on your side.

Play your social games, the ones I know so well, and ask yourselves some serious questions, how did you get here?

Soon enough events will unfold very rapidly. I fear your decision, so I suggest to those who listen that they enjoy the world, whatever their piece of it may be, as much as they can over the next few days, as soon enough, it will pass away, thanks to people who call themselves "mathematicians".

The tests were to be mathematical, and they were, given to those who called themselves mathematicians.

You have ignored the three previous, but cannot ignore the final one.

It is the final test for those who rejected the first three.

You were given three chances.

You failed them all.

So now, The Hammer is here, and with it, the end of days.

The world will be destroyed, and then remade, as foretold.

You will be lost, with your children, and then there will be others, and one day they will be tested, and will pass, but that is another story.

### Thursday, February 03, 2005

## Petty mathematicians

I've discovered that today's mathematicians don't care about how important a mathematical discovery is, how dramatic or "beautiful" the proof, if their social agenda is not advanced.

When considering mathematical results, they look FIRST and foremost to social impact.

A side result of this behavior is that there are no amateur mathematicians today of note.

If you look in the history books you see important amateur mathematician, but in today's high tech world where information flows at rapid speeds around the world, there are NO amateur mathematicians of note.

With over six billion people, and advanced educational systems, computers and libraries, expertise distributed on a large scale, today there are no amateur mathematicians of note.

I suggest to you that mathematicians who are high in the social order of that society refuse to note work from outside of their small group.

I've given a dramatic example with my prime counting function and the Wikipedia.

I am an amateur mathematician forced to mainly talk mathematics on Usenet, not becauae I really like Usenet, but because mainstream mathematicians block.

Repeatedly people have told me that I should get a math degree, learn their rules, work to appease mathematicians, and basically kiss ass to get my work accepted.

That in and of itself is an indication of how corrupt the society has become, as the idea of an amateur mathematician forced to become a professional, and follow arbitrary social rules to get correct results accepted should scare you.

There is a role for amateurs in mathematics, while the party line is that mathematics is too hard at this point and too difficult for amateurs to have an impact.

The reality is that today's professional mathematicians simply refuse to acknowledge results from amateur mathematicians, without regard to its value.

When considering mathematical results, they look FIRST and foremost to social impact.

A side result of this behavior is that there are no amateur mathematicians today of note.

If you look in the history books you see important amateur mathematician, but in today's high tech world where information flows at rapid speeds around the world, there are NO amateur mathematicians of note.

With over six billion people, and advanced educational systems, computers and libraries, expertise distributed on a large scale, today there are no amateur mathematicians of note.

I suggest to you that mathematicians who are high in the social order of that society refuse to note work from outside of their small group.

I've given a dramatic example with my prime counting function and the Wikipedia.

I am an amateur mathematician forced to mainly talk mathematics on Usenet, not becauae I really like Usenet, but because mainstream mathematicians block.

Repeatedly people have told me that I should get a math degree, learn their rules, work to appease mathematicians, and basically kiss ass to get my work accepted.

That in and of itself is an indication of how corrupt the society has become, as the idea of an amateur mathematician forced to become a professional, and follow arbitrary social rules to get correct results accepted should scare you.

There is a role for amateurs in mathematics, while the party line is that mathematics is too hard at this point and too difficult for amateurs to have an impact.

The reality is that today's professional mathematicians simply refuse to acknowledge results from amateur mathematicians, without regard to its value.