### Friday, August 31, 2007

## JSH: Surrogate factoring, periodic behavior

Having completed better analysis on surrogate factoring I found the equations that explain a periodic behavior at least one person has noted in posts, where for a given k and n, if you find a prime factor p of your target T with that n, then you will find other solutions by adding multiples of p to n.

Two of the equations determining that behavior are

Cw = n + (k + 2xr_1*p_1)( k + 2xr_2*p_2) - ((k + 2xr_1*p_1)( k + 2xr_2*p_2) - 2k^2)/T

and

w = k + 2xr_2*p_2 mod T

where if the second equation is true for a given n, then you will have a solution to the surrogate factoring equations at that n, but that is an only if. There C doesn't matter but is just some non-zero integer, as w just needs to be any factor of the right side—which is an integer I should note as the T must divide through—for which the second condition is met.

That is the primary decision relation that determines if a surrogate factorization can work or not.

Remember the surrogate factorization involves factoring a target composite T by solving

(x+k)^2 = y^2 + 2k^2 + nT

where the primary question has been, how do you pick k and n?

If they are picked correctly then some solution for x and y will also be a solution for

x^2 = y^2 mod p

where p is a prime factor of T.

[A reply to someone who asked why should James' method process have a high probability of working.]

Yes, questions, the mark of true researchers and human beings in general, as human curiosity is such a wonderful thing.

We wonder why and in looking for answers humanity finds new things.

So yeah, like I mentioned in another thread, the question in my mind for some time has been how so many of you seem to lack basic human curiosity.

Does the idea work at all? If not, why not? If so, how?

Learning begins with questions.

Now I have worked for years at answering questions presented by an idea, which was, could you factor one number with another, and I kept at it despite derision and insults from people like you.

You are the jocks of the schoolyard who tease that strange little boy who is so fascinated with his books.

Whether you wanted to be or not, or thought you hated those people growing up, that is your behavior against me and always has been. Maybe you hated them growing up because you wanted to BE them, and given the slightest excuse they are who you became.

You are the cruel jocks picking on the kid you call nothing.

And I am the genius.

Two of the equations determining that behavior are

Cw = n + (k + 2xr_1*p_1)( k + 2xr_2*p_2) - ((k + 2xr_1*p_1)( k + 2xr_2*p_2) - 2k^2)/T

and

w = k + 2xr_2*p_2 mod T

where if the second equation is true for a given n, then you will have a solution to the surrogate factoring equations at that n, but that is an only if. There C doesn't matter but is just some non-zero integer, as w just needs to be any factor of the right side—which is an integer I should note as the T must divide through—for which the second condition is met.

That is the primary decision relation that determines if a surrogate factorization can work or not.

Remember the surrogate factorization involves factoring a target composite T by solving

(x+k)^2 = y^2 + 2k^2 + nT

where the primary question has been, how do you pick k and n?

If they are picked correctly then some solution for x and y will also be a solution for

x^2 = y^2 mod p

where p is a prime factor of T.

[A reply to someone who asked why should James' method process have a high probability of working.]

Yes, questions, the mark of true researchers and human beings in general, as human curiosity is such a wonderful thing.

We wonder why and in looking for answers humanity finds new things.

So yeah, like I mentioned in another thread, the question in my mind for some time has been how so many of you seem to lack basic human curiosity.

Does the idea work at all? If not, why not? If so, how?

Learning begins with questions.

Now I have worked for years at answering questions presented by an idea, which was, could you factor one number with another, and I kept at it despite derision and insults from people like you.

You are the jocks of the schoolyard who tease that strange little boy who is so fascinated with his books.

Whether you wanted to be or not, or thought you hated those people growing up, that is your behavior against me and always has been. Maybe you hated them growing up because you wanted to BE them, and given the slightest excuse they are who you became.

You are the cruel jocks picking on the kid you call nothing.

And I am the genius.

### Tuesday, August 28, 2007

## JSH: Mathematical intuition, surrogate factoring equations

Having come off a tremendous research effort that has given key answers to important questions with surrogate factoring, I find myself still curious about the math world's ability to ignore a factoring idea that has many indications that it should be possible to get it to succeed!

Consider that I have

y^2 = x^2 mod T

and

k = 2x mod T

to get

(x+k)^2 = y^2 + 2k^2 + nT

when I finally to to the explicit, and it's like, if the first and second are assumed to be true to get the last, then how can it now work to factor a target composite T?

Part of the answer is that the math can get to the same place in a different way using:

y^2 = x^2 mod a

and

k = 2x mod b

to get

(x+k)^2 = y^2 + 2k^2 + nT

as long as 'a' and 'b' are coprime to T, but that doesn't explain why the math would CHOOSE to do so, as why would it care?

What difference does it make to the mathematics if it satisfies:

y^2 = x^2 mod T and k = 2x mod T

or

y^2 = x^2 mod a and k = 2x mod b

where 'a' and 'b' are coprime to T, to get to

(x+k)^2 = y^2 + 2k^2 + nT?

Why should it care?

There are two things really that jump out at you about my surrogate factoring idea as it's so simple:

Regardless of the failures I had with these equations I needed to understand why they didn't work all the time.

In contrast, I noticed that many of you seemed satisfied with simply being told that they did not work all the time, which presents me with a puzzle, why?

Why weren't any of you more curious than that?

Why wouldn't you demand an explanation for why and wonder if with such relatively easy math, none were given?

That is, where was your human curiosity?

This post is about probing that question, which is a question of your humanity.

Given the information presented here, why wouldn't this bug some of you, like a burr under your saddle until you got an answer?

Where is your human curiosity?

Consider that I have

y^2 = x^2 mod T

and

k = 2x mod T

to get

(x+k)^2 = y^2 + 2k^2 + nT

when I finally to to the explicit, and it's like, if the first and second are assumed to be true to get the last, then how can it now work to factor a target composite T?

Part of the answer is that the math can get to the same place in a different way using:

y^2 = x^2 mod a

and

k = 2x mod b

to get

(x+k)^2 = y^2 + 2k^2 + nT

as long as 'a' and 'b' are coprime to T, but that doesn't explain why the math would CHOOSE to do so, as why would it care?

What difference does it make to the mathematics if it satisfies:

y^2 = x^2 mod T and k = 2x mod T

or

y^2 = x^2 mod a and k = 2x mod b

where 'a' and 'b' are coprime to T, to get to

(x+k)^2 = y^2 + 2k^2 + nT?

Why should it care?

There are two things really that jump out at you about my surrogate factoring idea as it's so simple:

- Why didn't anyone think of this before, as mostly it is about completing the square?
- Why doesn't it always work?

Regardless of the failures I had with these equations I needed to understand why they didn't work all the time.

In contrast, I noticed that many of you seemed satisfied with simply being told that they did not work all the time, which presents me with a puzzle, why?

Why weren't any of you more curious than that?

Why wouldn't you demand an explanation for why and wonder if with such relatively easy math, none were given?

That is, where was your human curiosity?

This post is about probing that question, which is a question of your humanity.

Given the information presented here, why wouldn't this bug some of you, like a burr under your saddle until you got an answer?

Where is your human curiosity?

### Sunday, August 26, 2007

## JSH: Eliminating possibilities

With previous research I could say, ok, it's "pure math" and maybe mathematicians don't want to know the truth or the world to know the truth, but still it was odd when I could step through easy differences with my prime counting function and get this bizarre denial response.

So I went to factoring and have stepped out most of a fairly trivial proof that relies mostly on basic algebra and at this point the easy explanations go away for denial, especially with the financial markets on edge.

That leaves me considering intelligence services and it just defies reason that any US ones would be involved, or even the British, as why?

Clearly someone wants ME to pull the trigger, and the posts I'm seeing are all about me doing the demonstration.

Meanwhile sci.crypt has been flooded with spam messages and recently alt.math.recreational has as well.

But RSA pulled the money from the challenges recently so it seems reasonable to suppose US involvement but not from any of the standard players but maybe some rogue agency pretending to be one of them, and using government credentials to play some stupid corporate execs for fools.

So an organization fully established here in the US making its big play.

Toss in some weirdness on the world markets, and you get a shadow organization with a lot of money and enough connections to make this thing work that has several operatives posting against me to guide me in the direction they want.

And CLEARLY they want me to pull the trigger, program a full factoring solution, and post factorizations that could shatter a lot of things, and be at the point of the knife against the current encryption system.

But why? And why now?

I say, not Russians, and not Chinese. Definitely not the Israelis. Some new big players? Or someone new wanting to be a big player?

[A reply to someone who asked why did James rule out the Chinese.]

A trillion US dollars in reserves and a lot of US treasury notes, I guess.

But am I being naive on this one?

Who has not only the resources and the know-how, but more specifically the psi-ops know-how, in the world?

That narrows things down a bit, but I find it hard to believe this is state sponsored, but then again, who else is smart enough?

Or I am actually dealing with some of the most dedicated yet dumb math people around who not only can't comprehend basic algebra and obvious differences like with my prime counting function but they obsessively need to reply to me in highly directed ways.

Oh yeah, and why focus me on sci.math with the spamming of the other two newsgroups sci.crypt and alt.math.recreational?

That actually makes me think more it is just a bunch of angry math people not thinking straight, as, what's the difference?

Oh, it's about the number of readers?

Mysteries abound here.

Ok, so maybe it is the Chinese, but why wouldn't they just recognize my research and use it to shut-up loudmouth mathematicians in the current top countries who say the Chinese are horrible at math?

Or hey, it could be the Russians or even rogue elements in the CIA or NSA I guess.

Maybe they're just mad.

So I went to factoring and have stepped out most of a fairly trivial proof that relies mostly on basic algebra and at this point the easy explanations go away for denial, especially with the financial markets on edge.

That leaves me considering intelligence services and it just defies reason that any US ones would be involved, or even the British, as why?

Clearly someone wants ME to pull the trigger, and the posts I'm seeing are all about me doing the demonstration.

Meanwhile sci.crypt has been flooded with spam messages and recently alt.math.recreational has as well.

But RSA pulled the money from the challenges recently so it seems reasonable to suppose US involvement but not from any of the standard players but maybe some rogue agency pretending to be one of them, and using government credentials to play some stupid corporate execs for fools.

So an organization fully established here in the US making its big play.

Toss in some weirdness on the world markets, and you get a shadow organization with a lot of money and enough connections to make this thing work that has several operatives posting against me to guide me in the direction they want.

And CLEARLY they want me to pull the trigger, program a full factoring solution, and post factorizations that could shatter a lot of things, and be at the point of the knife against the current encryption system.

But why? And why now?

I say, not Russians, and not Chinese. Definitely not the Israelis. Some new big players? Or someone new wanting to be a big player?

[A reply to someone who asked why did James rule out the Chinese.]

A trillion US dollars in reserves and a lot of US treasury notes, I guess.

But am I being naive on this one?

Who has not only the resources and the know-how, but more specifically the psi-ops know-how, in the world?

That narrows things down a bit, but I find it hard to believe this is state sponsored, but then again, who else is smart enough?

Or I am actually dealing with some of the most dedicated yet dumb math people around who not only can't comprehend basic algebra and obvious differences like with my prime counting function but they obsessively need to reply to me in highly directed ways.

Oh yeah, and why focus me on sci.math with the spamming of the other two newsgroups sci.crypt and alt.math.recreational?

That actually makes me think more it is just a bunch of angry math people not thinking straight, as, what's the difference?

Oh, it's about the number of readers?

Mysteries abound here.

Ok, so maybe it is the Chinese, but why wouldn't they just recognize my research and use it to shut-up loudmouth mathematicians in the current top countries who say the Chinese are horrible at math?

Or hey, it could be the Russians or even rogue elements in the CIA or NSA I guess.

Maybe they're just mad.

## JSH: How could they?

Is there any rational explanation for why mathematicians around the world would ignore important methods that give a short proof of Fermat's Last Theorem, or a key finding in the prime numbers field, or a new way to factor?

Is it at all possible that the world has waited for leadership from the US and Britain?

I'm tackling factoring to stop the madness, but the question remains, how has the madness survived?

What crucial stupidity has motivated the entire world of mathematicians to sit back for over five years, even when a math journal imploded, they sat back?

How could they?

What can explain such profound idiocy on a worldwide scale?

Any answers?

Why is the math world, a world of dense, stupid, fools?

Any answer?

I'm looking at finishing this in the worst possible way, and I want an answer for why it was necessary.

Are you all just fools? Or is there something else? Something you thought you knew, perhaps?

Some lie told to you by people in this country who convinced you that lying could win?

I think so. I think many of you were convinced that the group always won and that it was pointless to fight such powerful people.

So they dragged you down with them, and you get to share in the same consequences.

Is it at all possible that the world has waited for leadership from the US and Britain?

I'm tackling factoring to stop the madness, but the question remains, how has the madness survived?

What crucial stupidity has motivated the entire world of mathematicians to sit back for over five years, even when a math journal imploded, they sat back?

How could they?

What can explain such profound idiocy on a worldwide scale?

Any answers?

Why is the math world, a world of dense, stupid, fools?

Any answer?

I'm looking at finishing this in the worst possible way, and I want an answer for why it was necessary.

Are you all just fools? Or is there something else? Something you thought you knew, perhaps?

Some lie told to you by people in this country who convinced you that lying could win?

I think so. I think many of you were convinced that the group always won and that it was pointless to fight such powerful people.

So they dragged you down with them, and you get to share in the same consequences.

## Pathetically simple answer

For years I've pursued an approach to the factoring problem that I call surrogate factoring, where you factor one number I call the surrogate in order to factor the target.

And last year I finally realize that my difficulties had come from lack of simplification, so last August I sat down and thoroughly thought about what I meant by surrogate factoring and out popped only two equations:

x^2 = y^2 mod T and k^2 = 2xk mod T

where now I've simplified even further and write as

y^2 = x^2 mod T

and

k = 2x mod T

as then it's easier to see that the second one has to exist, if y and x exist, and then it's easier to show multiplying it by k, and adding to the top one:

y^2 + k^2 = x^2 + 2xk mod T, so y^2 + 2k^2 = (x+k)^2 mod T, and reverse to get

(x+k)^2 = y^2 + 2k^2 mod T, and go explicit with

(x+k)^2 = y^2 + 2k^2 + nT

and you get x and y from factoring 2k^2 + nT, and they may be the answers to your congruence of squares

x^2 = y^2 mod T

that factor T.

I had that LAST YEAR in August after that time pondering the question and I began thinking off and on about when this should work while talking about it to mathematicians, and I informed the NSA.

Yup, I was concerned that maybe I'd found something TOO SIMPLE and hoped that I'd get some attention from people who could help in case it was too effective.

With no interest from any of the professionals I would work at the problem off and on, where I couldn't seem to get an answer on how effective this could be but recently came up with the simple idea of just assuming

T = p_1*p_2

and working through what that'd mean which is what's now on my Integer Factorization page.

I now believe that page is too long, and I contemplate editing it, now realizing that the simple answer is just so pathetically simple that it is depressing.

I have a key relation that emerges from that page:

-(k+2xp)/p < m < -(k + 2x)/p

where the absolute value of m is a count of non-trivial factorizations given by this technique, and I didn't quite realize the full significance of that for a while.

But it says that if floor(abs((k+x)/p)) is greater than 0, then you MUST have a non-trivial factorization at that point, where surrogate factoring will pull the prime factor p of your target.

So you just need to make sure that the absolute value of k+2x is greater than p, and since x can be positive or negative it makes sense oddly enough to let k=1, which means

x = k*(2)^{-1} mod T = (2)^{-1} mod T

and would have the largest residue modulo T that you can get and should hopefully be larger than a small prime factor p of T.

But you still need to pick n. There are lots of ways to guess but the last big result is that once you get abs(n) over the limit, then any n greater will work.

So that could be a big number, still hard to factor, so why is this a big deal?

Well, you can get hundreds or millions of these surrogates for ANY target T, looking for one that's easy to factor and trying out any factorizations that you can get, where on my page I show an example where a truly trivial factorization of the surrogate works!

What makes this story astounding is my inability to get the math community to pay attention to it, and it took a lot of guts to bug the NSA and try to get their attention as well, and here I am with the simple answer I thought might be there, wondering about a lot of supposedly smart people.

I feel confident in my analysis and it indicates that modern mathematicians were wrong and factoring is NOT a hard problem, and in fact, by the standards of complexity of modern mathematics it has pathetically simple answer, using just a bit of algebra.

In other words, factoring is a trivially EASY problem.

It remains to be seen why mathematicians around the world, including those at the NSA, chose to deliberately ignore a new factoring method over this last year. At this point it is a profound mystery to me how they could.

And last year I finally realize that my difficulties had come from lack of simplification, so last August I sat down and thoroughly thought about what I meant by surrogate factoring and out popped only two equations:

x^2 = y^2 mod T and k^2 = 2xk mod T

where now I've simplified even further and write as

y^2 = x^2 mod T

and

k = 2x mod T

as then it's easier to see that the second one has to exist, if y and x exist, and then it's easier to show multiplying it by k, and adding to the top one:

y^2 + k^2 = x^2 + 2xk mod T, so y^2 + 2k^2 = (x+k)^2 mod T, and reverse to get

(x+k)^2 = y^2 + 2k^2 mod T, and go explicit with

(x+k)^2 = y^2 + 2k^2 + nT

and you get x and y from factoring 2k^2 + nT, and they may be the answers to your congruence of squares

x^2 = y^2 mod T

that factor T.

I had that LAST YEAR in August after that time pondering the question and I began thinking off and on about when this should work while talking about it to mathematicians, and I informed the NSA.

Yup, I was concerned that maybe I'd found something TOO SIMPLE and hoped that I'd get some attention from people who could help in case it was too effective.

With no interest from any of the professionals I would work at the problem off and on, where I couldn't seem to get an answer on how effective this could be but recently came up with the simple idea of just assuming

T = p_1*p_2

and working through what that'd mean which is what's now on my Integer Factorization page.

I now believe that page is too long, and I contemplate editing it, now realizing that the simple answer is just so pathetically simple that it is depressing.

I have a key relation that emerges from that page:

-(k+2xp)/p < m < -(k + 2x)/p

where the absolute value of m is a count of non-trivial factorizations given by this technique, and I didn't quite realize the full significance of that for a while.

But it says that if floor(abs((k+x)/p)) is greater than 0, then you MUST have a non-trivial factorization at that point, where surrogate factoring will pull the prime factor p of your target.

So you just need to make sure that the absolute value of k+2x is greater than p, and since x can be positive or negative it makes sense oddly enough to let k=1, which means

x = k*(2)^{-1} mod T = (2)^{-1} mod T

and would have the largest residue modulo T that you can get and should hopefully be larger than a small prime factor p of T.

But you still need to pick n. There are lots of ways to guess but the last big result is that once you get abs(n) over the limit, then any n greater will work.

So that could be a big number, still hard to factor, so why is this a big deal?

Well, you can get hundreds or millions of these surrogates for ANY target T, looking for one that's easy to factor and trying out any factorizations that you can get, where on my page I show an example where a truly trivial factorization of the surrogate works!

What makes this story astounding is my inability to get the math community to pay attention to it, and it took a lot of guts to bug the NSA and try to get their attention as well, and here I am with the simple answer I thought might be there, wondering about a lot of supposedly smart people.

I feel confident in my analysis and it indicates that modern mathematicians were wrong and factoring is NOT a hard problem, and in fact, by the standards of complexity of modern mathematics it has pathetically simple answer, using just a bit of algebra.

In other words, factoring is a trivially EASY problem.

It remains to be seen why mathematicians around the world, including those at the NSA, chose to deliberately ignore a new factoring method over this last year. At this point it is a profound mystery to me how they could.

## JSH: What if I succeed?

For years I've been talking about using the factoring problem to force acceptance of my other mathematical research as I figured that if I demonstrate my problem solving ability with a practical math problem away from "pure math" then I can get people to pay attention to my other decried research.

My take on responses from this newsgroup at least has been a contemptuous dismissal of any possibility of my success.

But what if I succeed?

The scary part to me is not the world's markets but the possibility that the academic math world IS as dumb as I've feared as if I just trot out a solution that shows that factoring is a trivially easy problem, and do it with ideas I've talked out over a year, how can it be smart?

Especially if it has spent the last five years ignoring my proof of FLT and my prime counting function along with other research?

And, um, can I ethically present a stunningly simple proof of how to make my latest "surrogate factoring" approach work at a time like this when the world's financial markets are already on edge?

But where can I go with the research?

I've already contacted the NSA in the past when I was wrong, and I've bugged cryptology people over the years, so how do I go to anyone now?

But more importantly, how can it be so trivial?

How could R, S and A present something so pathetically easy to solve to the world as a security system, and manage to convince people?

Should I hide math for the good of financial markets to protect a math culture that seems to be stupid on a scale hard to comprehend about even the most basic mathematics?

Or let fly? I think I have no choice but to simply present the research and let the chips fall where they may because the situation is so unprecedented that there is no rulebook.

There simply is no right answer.

My take on responses from this newsgroup at least has been a contemptuous dismissal of any possibility of my success.

But what if I succeed?

The scary part to me is not the world's markets but the possibility that the academic math world IS as dumb as I've feared as if I just trot out a solution that shows that factoring is a trivially easy problem, and do it with ideas I've talked out over a year, how can it be smart?

Especially if it has spent the last five years ignoring my proof of FLT and my prime counting function along with other research?

And, um, can I ethically present a stunningly simple proof of how to make my latest "surrogate factoring" approach work at a time like this when the world's financial markets are already on edge?

But where can I go with the research?

I've already contacted the NSA in the past when I was wrong, and I've bugged cryptology people over the years, so how do I go to anyone now?

But more importantly, how can it be so trivial?

How could R, S and A present something so pathetically easy to solve to the world as a security system, and manage to convince people?

Should I hide math for the good of financial markets to protect a math culture that seems to be stupid on a scale hard to comprehend about even the most basic mathematics?

Or let fly? I think I have no choice but to simply present the research and let the chips fall where they may because the situation is so unprecedented that there is no rulebook.

There simply is no right answer.

### Friday, August 24, 2007

## JSH: My factoring research, update

I finally am getting a handle on a crucial metric for determining how likely you are to factor using what I call surrogate factoring where if your target composite to factor is T, the following equations

x^2 = y^2 mod T

and

k = 2x mod T

can be used to derive

(x+k)^2 = y^2 + 2k^2 + nT

and by factoring 2k^2 + nT, you may factor your target composite.

I call 2k^2 + nT the surrogate, and also use S, for it.

S = 2k^2 + nT

The two big questions in the year since I came up with these simpler equations (note I usually wrote the second congruence as k^2 = 2xk mod T) have been, how effective can this be as a factoring method, and how do you pick k and n?

After trying various ideas over the last year or so I finally focused on assuming that T=p_1*p_2, where the p's are primes, and a solution to the congruence of squares:

x+y = c_1*p_1 and x-y = c_2*p_2

and that was a crucial step allowing me to start answering questions about factoring probability and how to pick k and n.

A key relationship that emerged is

-(k+2xT)/T < m_1 < -(k + 2xp_1)/T

where the absolute value of m_1 is a count of non-trivial factorizations associated with p_2. Its partner is m_2 whose absolute value is a count of non-trivial factorizations associated with p_1.

So if both are non-zero the count of possible factorizations is abs(m_1*m_2). The key range shown tells you that a LARGE k is needed to give a large m_1. Of course you don't know p_1 if you're trying to factor but at least you can see how changing k impacts the range of possible factorizations.

Using the maximum for m_1, with a particular surrogate S, you can approximately find m_2:

m_2 > (S - k^2 - kT(4x + m_1) - (4x^2 + 2x*m_1)T^2)/(kT + T^2 + 2x*m_1*T^2)

Here x is not exactly known, but you have that the residue of x modulo T is (2)^{-1}k mod T. That is, the modular inverse of 2 modulo T times k. So x is just that plus or minus some multiple of T.

Now your choice of surrogate S means you have a choice of n, like an n you're guessing at but want to know the probability approximately before doing any factoring.

But you need one more n to make the range and that is given approximately by

n_min < -k^2/T + 4xk + 4x^2*T

where I call it n_min, though it may be larger than your other n, and you would need to make some educated guesses about x.

Now though finally you can get to an approximate probability with

abs(m_1*m_2/(n-n_min))*100%

where if either of the m's is 0, then you'd substitute 1 for that one.

Where I've assumed that the number of factorizations—given by abs(m_1*m_2)—spread evenly among n's.

Oh yeah, those equations are all for the probability that BOTH primes are factored out, but you can get a non-trivial factorization if only one is, which is like the math thinking T=p_1.

I won't go into detail about how that shifts all the equations but will just how how it shifts the key limit one:

-(k+2xT)/T < m_1 < -(k + 2xp_1)/T

becomes

-(k+2x*p_1)/p_1 < m_1 < -(k + 2x)/p_1

and, of course, you do not know what p_1 is so that equation is useful primarily to tell you how the size of k and x impact the number of possible factorizations, and the lesson is they need to both be BIG, but not as big as to get both primes at the same time.

Notice that in general the worst value for k is k=2, as with x=1 that gives

-2 < m_1 < 0

so you'd have just one possible factorization.

The research is fresh as I've just worked out most of this approach this week. But it is exciting to already have some key answers.

I have a page going into more detail at my Extreme Mathematics group:

http://groups.google.com/group/extrememathematics/web/integer-factorization

x^2 = y^2 mod T

and

k = 2x mod T

can be used to derive

(x+k)^2 = y^2 + 2k^2 + nT

and by factoring 2k^2 + nT, you may factor your target composite.

I call 2k^2 + nT the surrogate, and also use S, for it.

S = 2k^2 + nT

The two big questions in the year since I came up with these simpler equations (note I usually wrote the second congruence as k^2 = 2xk mod T) have been, how effective can this be as a factoring method, and how do you pick k and n?

After trying various ideas over the last year or so I finally focused on assuming that T=p_1*p_2, where the p's are primes, and a solution to the congruence of squares:

x+y = c_1*p_1 and x-y = c_2*p_2

and that was a crucial step allowing me to start answering questions about factoring probability and how to pick k and n.

A key relationship that emerged is

-(k+2xT)/T < m_1 < -(k + 2xp_1)/T

where the absolute value of m_1 is a count of non-trivial factorizations associated with p_2. Its partner is m_2 whose absolute value is a count of non-trivial factorizations associated with p_1.

So if both are non-zero the count of possible factorizations is abs(m_1*m_2). The key range shown tells you that a LARGE k is needed to give a large m_1. Of course you don't know p_1 if you're trying to factor but at least you can see how changing k impacts the range of possible factorizations.

Using the maximum for m_1, with a particular surrogate S, you can approximately find m_2:

m_2 > (S - k^2 - kT(4x + m_1) - (4x^2 + 2x*m_1)T^2)/(kT + T^2 + 2x*m_1*T^2)

Here x is not exactly known, but you have that the residue of x modulo T is (2)^{-1}k mod T. That is, the modular inverse of 2 modulo T times k. So x is just that plus or minus some multiple of T.

Now your choice of surrogate S means you have a choice of n, like an n you're guessing at but want to know the probability approximately before doing any factoring.

But you need one more n to make the range and that is given approximately by

n_min < -k^2/T + 4xk + 4x^2*T

where I call it n_min, though it may be larger than your other n, and you would need to make some educated guesses about x.

Now though finally you can get to an approximate probability with

abs(m_1*m_2/(n-n_min))*100%

where if either of the m's is 0, then you'd substitute 1 for that one.

Where I've assumed that the number of factorizations—given by abs(m_1*m_2)—spread evenly among n's.

Oh yeah, those equations are all for the probability that BOTH primes are factored out, but you can get a non-trivial factorization if only one is, which is like the math thinking T=p_1.

I won't go into detail about how that shifts all the equations but will just how how it shifts the key limit one:

-(k+2xT)/T < m_1 < -(k + 2xp_1)/T

becomes

-(k+2x*p_1)/p_1 < m_1 < -(k + 2x)/p_1

and, of course, you do not know what p_1 is so that equation is useful primarily to tell you how the size of k and x impact the number of possible factorizations, and the lesson is they need to both be BIG, but not as big as to get both primes at the same time.

Notice that in general the worst value for k is k=2, as with x=1 that gives

-2 < m_1 < 0

so you'd have just one possible factorization.

The research is fresh as I've just worked out most of this approach this week. But it is exciting to already have some key answers.

I have a page going into more detail at my Extreme Mathematics group:

http://groups.google.com/group/extrememathematics/web/integer-factorization

### Wednesday, August 22, 2007

## JSH: Surrogate factoring analysis, done?

After my failure with what I deluded myself into thinking (for a while) was the perfect factoring algorithm, I just bit the bullet and began the detailed and tedious analysis that my latest insight indicated which took me into the calculus which was kind of fun, and a bit surprising, and now it looks to me like if I didn't screw up the analysis is done.

Remember what I call surrogate factoring is an integer factorization technique that involves factoring some other number than your target to in fact factor your target composite, which is a lateral thinking kind of move.

I have a complete page at my Extreme Mathematics group:

http://groups.google.com/group/extrememathematics/web/integer-factorization

Or you can go to my math blog.

For those who'd like a synopsis or a teaser, whatever you might like to call it, the key result is that the method works within certain limits, where the key limits is:

-(k+2xT)/T < m < -(k + 2xp)/T

where p is a prime factor of your target T, and m is less important than the range, as

floor(-(k + 2xp)/T - (-(k+2xT)/T))

is the exact count (assuming I did my analysis correctly!!!) of possible solutions that will non-trivially factor with the smallest absolute value of the surrogate, where with a congruence of squares

x^2 = y^2 mod T

you have that both x+y and x-y share prime factors with T.

If only ONE of them does then the limit shifts to

-(k+2xp)/ < m < -(k + 2x)/p

as the range that gives the total number of non-trivial solutions that will factor out one prime only, and practically that is what you get most of the time, with the smallest prime getting yanked out, so my analysis explains the why of it all.

So now it makes sense when and why the surrogate factoring equations work.

Lots more detail in the complete analysis than I expected so I'd appreciate any comments noting errors in that analysis.

I am rather excited by this analysis as it was back in August of LAST YEAR that I finally boiled down the surrogate factoring concept to some simple key equations and I've puzzled over how and why they work—or do not—since then, off and on.

Now finally I think I have the answers, and know the 'why' of surrogate factoring.

Looking for critiques! Especially anyone noticing mistakes!!!

http://groups.google.com/group/extrememathematics/web/integer-factorization

You can check it all out there. I like to think I'm clever but did I get the calculus right?

I AM an amateur playing at it so this is on topic for alt.math.recreational, and remember the issue has not been about whether or not the surrogate factoring can work, but whether or not it can be a viable factoring technique which required the kind of detailed and in-depth analysis that I think I've completed.

And it took about a year, not that I was working at it full time, but for those looking to do any kind of big research, expect years of effort. YEARS!!!

It will take you years if you have any ideas that might actually pan out.

Remember what I call surrogate factoring is an integer factorization technique that involves factoring some other number than your target to in fact factor your target composite, which is a lateral thinking kind of move.

I have a complete page at my Extreme Mathematics group:

http://groups.google.com/group/extrememathematics/web/integer-factorization

Or you can go to my math blog.

For those who'd like a synopsis or a teaser, whatever you might like to call it, the key result is that the method works within certain limits, where the key limits is:

-(k+2xT)/T < m < -(k + 2xp)/T

where p is a prime factor of your target T, and m is less important than the range, as

floor(-(k + 2xp)/T - (-(k+2xT)/T))

is the exact count (assuming I did my analysis correctly!!!) of possible solutions that will non-trivially factor with the smallest absolute value of the surrogate, where with a congruence of squares

x^2 = y^2 mod T

you have that both x+y and x-y share prime factors with T.

If only ONE of them does then the limit shifts to

-(k+2xp)/ < m < -(k + 2x)/p

as the range that gives the total number of non-trivial solutions that will factor out one prime only, and practically that is what you get most of the time, with the smallest prime getting yanked out, so my analysis explains the why of it all.

So now it makes sense when and why the surrogate factoring equations work.

Lots more detail in the complete analysis than I expected so I'd appreciate any comments noting errors in that analysis.

I am rather excited by this analysis as it was back in August of LAST YEAR that I finally boiled down the surrogate factoring concept to some simple key equations and I've puzzled over how and why they work—or do not—since then, off and on.

Now finally I think I have the answers, and know the 'why' of surrogate factoring.

Looking for critiques! Especially anyone noticing mistakes!!!

http://groups.google.com/group/extrememathematics/web/integer-factorization

You can check it all out there. I like to think I'm clever but did I get the calculus right?

I AM an amateur playing at it so this is on topic for alt.math.recreational, and remember the issue has not been about whether or not the surrogate factoring can work, but whether or not it can be a viable factoring technique which required the kind of detailed and in-depth analysis that I think I've completed.

And it took about a year, not that I was working at it full time, but for those looking to do any kind of big research, expect years of effort. YEARS!!!

It will take you years if you have any ideas that might actually pan out.

## JSH: My recent posting purpose

Some may wonder, why now? Why am I back with sci.physics and talking like a physics student again, as my B.Sc. is in physics as I've mentioned before, versus just arguing with the math people on math newsgroups?

Well, I'm brainstorming connections between "pure math" and physics that have puzzled me, where I have stated positions while I'd like to be sure, so I'm trying to trigger discussions in key areas, as I've linked primes to real world behavior already—my discrete oscillator—and now am considering how "random" in our world may be a prime phenomena.

Specifically with this question of p mod 3, there is the issue of underlying disorder—random residues modulo primes—beneath absolute structure, like in the 1,2,0, repeat of the residues modulo 3 of the counting numbers.

So chaos is connected to order, or is it not?

If p mod 3 is an example of an ordered system then prediction is not just a random thing or a matter of experiment but of LAWS that govern that underlying order.

Physics is about prediction, but prediction by laws, which I think is a distinction few people understand.

Past "philosophers" might notice leaves falling and come up with predictions based on what they saw, and compare to a brick to make some other prediction. But Galileo in finding underlying laws could figure out that a brick and a feather would fall at the same rate without air (that was Galileo, wasn't it?).

Appeals to statistics with p mod 3 are against the scientific method.

Without a reason there is no law, and prediction for its own sake is not science, nor I suggest to you, is it mathematics.

Well, I'm brainstorming connections between "pure math" and physics that have puzzled me, where I have stated positions while I'd like to be sure, so I'm trying to trigger discussions in key areas, as I've linked primes to real world behavior already—my discrete oscillator—and now am considering how "random" in our world may be a prime phenomena.

Specifically with this question of p mod 3, there is the issue of underlying disorder—random residues modulo primes—beneath absolute structure, like in the 1,2,0, repeat of the residues modulo 3 of the counting numbers.

So chaos is connected to order, or is it not?

If p mod 3 is an example of an ordered system then prediction is not just a random thing or a matter of experiment but of LAWS that govern that underlying order.

Physics is about prediction, but prediction by laws, which I think is a distinction few people understand.

Past "philosophers" might notice leaves falling and come up with predictions based on what they saw, and compare to a brick to make some other prediction. But Galileo in finding underlying laws could figure out that a brick and a feather would fall at the same rate without air (that was Galileo, wasn't it?).

Appeals to statistics with p mod 3 are against the scientific method.

Without a reason there is no law, and prediction for its own sake is not science, nor I suggest to you, is it mathematics.

## JSH: But why? Questioning primes.

I think that "but why?" is the question of the true scientist and that's where things get interesting.

But why?

So you can ask, why would primes care what residue they have modulo some other prime?

If they do not, then residues of one prime modulo another behave randomly.

If so, then EVERY random sequence of two elements only can be represented as being somewhere on the line of p mod 3.

So if you flip a coin for a while you can get some computer to do a search and if it's not too far out, you'll find that sequence of flips in the sequence of p mod 3.

Or, there is an underlying pattern where some prime like 37 PREFERS to have a residue of 1 modulo 3, and if you can figure out that underlying pattern you can get some sense of some special rules that govern primes, but, composites are products of primes! So they'd reflect that pattern, right?

I am not the first person to suggest randomness in this area.

I suspect mathematicians have demonized the others who have.

Why?

Because if p mod 3 is random, as in, there is no rhyme nor reason to why any particular prime has 1 or 2 as its residue modulo 3 than it just does, then huge areas of funding in the math field collapse.

Quite simply, math people no longer get paid in those areas, and mathematicians who specialize in those areas would have to find somewhere else to do research!

But what if p mod 3 IS random? Then you would know that ANY random behavior in our real world would have to be represented within the infinite sequence of primes modulo some other prime, and that bit of "pure math" would have real world application in physics!

But why?

So you can ask, why would primes care what residue they have modulo some other prime?

If they do not, then residues of one prime modulo another behave randomly.

If so, then EVERY random sequence of two elements only can be represented as being somewhere on the line of p mod 3.

So if you flip a coin for a while you can get some computer to do a search and if it's not too far out, you'll find that sequence of flips in the sequence of p mod 3.

Or, there is an underlying pattern where some prime like 37 PREFERS to have a residue of 1 modulo 3, and if you can figure out that underlying pattern you can get some sense of some special rules that govern primes, but, composites are products of primes! So they'd reflect that pattern, right?

I am not the first person to suggest randomness in this area.

I suspect mathematicians have demonized the others who have.

Why?

Because if p mod 3 is random, as in, there is no rhyme nor reason to why any particular prime has 1 or 2 as its residue modulo 3 than it just does, then huge areas of funding in the math field collapse.

Quite simply, math people no longer get paid in those areas, and mathematicians who specialize in those areas would have to find somewhere else to do research!

But what if p mod 3 IS random? Then you would know that ANY random behavior in our real world would have to be represented within the infinite sequence of primes modulo some other prime, and that bit of "pure math" would have real world application in physics!

### Tuesday, August 21, 2007

## JSH: So math people fake it

Most mathematicians stay away from applied mathematics because they lie. They fake it.

The real world is the harshest judge so instead they have a world that is all about taking people's word.

If you could work as a professor without ever accomplishing anything that is correct but keep your job because none of the other professors or researchers in your field had either, and you just published papers that LOOKED a certain way, which your group would proclaim as being important, why wouldn't you?

Money for nothing.

Easy money. Just learn what is to you gobbledy-gook that gets you paid, and drive on.

It's a cruel world, right? If that "pure" math were important people would catch you, right?

Technology keeps progressing, so why does it matter if number theorists lie about primes, and lie about claims of proving this or that?

Who does it really hurt, right?

Put these people back in history though and let them destroy the underpinnings of algebra or calculus, or geometry.

And they'd take away this present.

I suggest to you that by them blocking the real mathematical research that we do not know where all of it could lead, they destroy the future.

Publish or perish?

How about publish and perish the human race?

The real world is the harshest judge so instead they have a world that is all about taking people's word.

If you could work as a professor without ever accomplishing anything that is correct but keep your job because none of the other professors or researchers in your field had either, and you just published papers that LOOKED a certain way, which your group would proclaim as being important, why wouldn't you?

Money for nothing.

Easy money. Just learn what is to you gobbledy-gook that gets you paid, and drive on.

It's a cruel world, right? If that "pure" math were important people would catch you, right?

Technology keeps progressing, so why does it matter if number theorists lie about primes, and lie about claims of proving this or that?

Who does it really hurt, right?

Put these people back in history though and let them destroy the underpinnings of algebra or calculus, or geometry.

And they'd take away this present.

I suggest to you that by them blocking the real mathematical research that we do not know where all of it could lead, they destroy the future.

Publish or perish?

How about publish and perish the human race?

### Monday, August 20, 2007

## JSH: Critical mass

My last post is about showing how mathematicians directly impact the physics field with what I think are rather bold lies.

But how do they succeed?

Well, why would any physicists check p mod 3, where p is a prime other than 3 to see if you get a random distribution?

That's a mathematician kind of question.

And for them that is a dangerous question as what I'm sure is the truth takes away federal funds for research.

And what cushy funds those are as there is no possible answer if you deny randomness, so they can work indefinitely.

Talk is one thing but statistical analysis is another.

Is there a pattern in p mod 3?

If you can prove it is random, then you can take out the mathematicians, and wreck entire careers of people who got away with taking money to muddle around in an area where the simplest answer, is random.

I think there are a critical mass of mathematicians willing to lie about math for money which is how they can do this and go to sleep at night, never worrying about getting caught because, well, because people TRUST them.

But how do they succeed?

Well, why would any physicists check p mod 3, where p is a prime other than 3 to see if you get a random distribution?

That's a mathematician kind of question.

And for them that is a dangerous question as what I'm sure is the truth takes away federal funds for research.

And what cushy funds those are as there is no possible answer if you deny randomness, so they can work indefinitely.

Talk is one thing but statistical analysis is another.

Is there a pattern in p mod 3?

If you can prove it is random, then you can take out the mathematicians, and wreck entire careers of people who got away with taking money to muddle around in an area where the simplest answer, is random.

I think there are a critical mass of mathematicians willing to lie about math for money which is how they can do this and go to sleep at night, never worrying about getting caught because, well, because people TRUST them.

## About random, primes and statistics

Knowing that statistics is an area of mathematics that physics students get rather familiar with I thought it'd be of interest to explain a simple area where mathematicians routinely lie—I think in order to keep research grants.

First you need to learn a bit of number theory, as 7 mod 3 = 1, where 1 is the residue modulo 3, so the bit of math is that shown x mod y, you take x-ky where k is the largest positive integer that will fit, and use what's left over, which is the residue.

I picked 3 because there is a fascinatingly boring thing about numbers modulo 3—perfect regularity:

Starting at 1 and counting up you have

1, 2, 3 followed by 4, 5, 6 followed by 7, 8, 9 on out to infinity

which modulo 3 gives

1, 2, 0 followed by 1, 2, 0 followed by 1, 2, 0 repeated on out to infinity.

(It's like a perfect waltz to infinity!!!)

That is an absolute and I'd say a trivial one at that but it will challenge everything you think you know about mathematicians as decent researchers as you have primes and you have composites and composites are products of primes—kind of like their children!—so if primes tended to pick a particular residue modulo 3, then composites would follow along!

But they don't. They split evenly between 0, 1 and 2.

Therefore, I strongly suggest to you, primes split evenly between having a residue of 1 and 2 modulo 3.

If that is true then the residue of a prime other than 3 modulo 3 will be random.

Here is what you get with the first 23 primes greater than 3:

5 mod 3 = 2, 7 mod 3 = 1, 11 mod 3 = 2, 13 mod 3 = 1, 17 mod 3 = 2, 19 mod 3 = 1, 23 mod 3 = 2, 29 mod 3 = 2, 31 mod 3 = 1, 37 mod 3 = 1, 41 mod 3 = 2, 43 mod 3 = 1, 47 mod 3 = 2, 53 mod 3 = 2, 59 mod 3 = 2, 61 mod 3 = 1, 67 mod 3 = 1, 71 mod 3 = 2, 73 mod 3 = 1, 79 mod 3 = 1, 83 mod 3 = 2, 89 mod 3 = 2, 97 mod 3 = 1

So the sequence is

2, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1

and I suggest to you it is random, like flipping a coin, but better as it is absolutely random.

Now then what if it were NOT random so that there was some pattern in there?

Well, composites are products of primes, so where primes go, so do the composites, so if primes tended to have say, a residue of 1 modulo 3, then so would composites, like 7*13 = 91 and 91 mod 3 = 1.

If there were some more hidden pattern, for instance, when I've brought this subject up before sci.math'ers would claim that if 1 tends to be followed by 2 and 2 by 1 and they produced statistical tests claiming those proved that, then the sequence is not random!

But if 1 tends to be followed by 2 and 2 by 1, how would they push the composites? Into what pattern?

Now then, let's leap forward and say you accept that the rigidity of the ordering of counting numbers, where you have

1, 2, 3 followed by 4, 5, 6 followed by 7, 8, 9 on out to infinity

which modulo 3 gives

1, 2, 0 followed by 1, 2, 0 followed by 1, 2, 0 repeated on out to infinity

convinces you that primes other than 3 modulo 3 give a random distribution, why should anyone care?

Well, random is useful in physics but more intriguingly, if you accept that then you end a lot of research paths in modern mathematics!!!

Because if you push the argument to p_1 mod p_2, where the p's are differing primes, then it turns out some supposedly big questions in modern math, like the Twin Primes Conjecture are easily answered!

But there are mathematicians getting federal funds for research in those areas, if random is the call then those funds cease.

BUT if no one notices, those mathematicians can work endlessly in an area where they can never get an answer, at least not a correct one.

That is a highlight of how you can find controversy in the math field in a simple area where you can run your own statistical analysis to see.

Oh yeah, and remember a while back the math awards where one guy famously refused to accept his, while the others did accept?

Well, if primes are random then one of those people won a prize for supposedly innovative research into an area where no pattern actually lies.

Maybe you believe I'm wrong and there is some hidden pattern in primes modulo 3, or more technically

p mod 3

with p not equal to 3, as you go out to positive infinity.

I've given the beginning of the sequence:

2, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1

where the start was the number 2 which has itself as a residue modulo 3.

In answering my bringing this issue up before posters on sci.math claimed that this area has been checked and that statistical analysis proved non-randomness but there is no why here given.

Why would primes care? Especially when their products, the composites can't swing one way with the primes swinging another?

I suggest to you that math people lie in this area, where you can do the statistical analysis yourself to see that they lie, or I'm wrong. I've been wrong before, but I don't think I'm wrong here.

I know people lie. And I know some of you in the physics field who like to play with statistics can crank up some machinery and play with primes to see for yourself.

Remember, BIG research grants are wrapped up in this where some mathematicians have worked in this area for decades, so their entire livelihoods are about it NOT being true that p mod 3 is random.

Their entire LIVES are about that not being true, so they are totally invested.

First you need to learn a bit of number theory, as 7 mod 3 = 1, where 1 is the residue modulo 3, so the bit of math is that shown x mod y, you take x-ky where k is the largest positive integer that will fit, and use what's left over, which is the residue.

I picked 3 because there is a fascinatingly boring thing about numbers modulo 3—perfect regularity:

Starting at 1 and counting up you have

1, 2, 3 followed by 4, 5, 6 followed by 7, 8, 9 on out to infinity

which modulo 3 gives

1, 2, 0 followed by 1, 2, 0 followed by 1, 2, 0 repeated on out to infinity.

(It's like a perfect waltz to infinity!!!)

That is an absolute and I'd say a trivial one at that but it will challenge everything you think you know about mathematicians as decent researchers as you have primes and you have composites and composites are products of primes—kind of like their children!—so if primes tended to pick a particular residue modulo 3, then composites would follow along!

But they don't. They split evenly between 0, 1 and 2.

Therefore, I strongly suggest to you, primes split evenly between having a residue of 1 and 2 modulo 3.

If that is true then the residue of a prime other than 3 modulo 3 will be random.

Here is what you get with the first 23 primes greater than 3:

5 mod 3 = 2, 7 mod 3 = 1, 11 mod 3 = 2, 13 mod 3 = 1, 17 mod 3 = 2, 19 mod 3 = 1, 23 mod 3 = 2, 29 mod 3 = 2, 31 mod 3 = 1, 37 mod 3 = 1, 41 mod 3 = 2, 43 mod 3 = 1, 47 mod 3 = 2, 53 mod 3 = 2, 59 mod 3 = 2, 61 mod 3 = 1, 67 mod 3 = 1, 71 mod 3 = 2, 73 mod 3 = 1, 79 mod 3 = 1, 83 mod 3 = 2, 89 mod 3 = 2, 97 mod 3 = 1

So the sequence is

2, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1

and I suggest to you it is random, like flipping a coin, but better as it is absolutely random.

Now then what if it were NOT random so that there was some pattern in there?

Well, composites are products of primes, so where primes go, so do the composites, so if primes tended to have say, a residue of 1 modulo 3, then so would composites, like 7*13 = 91 and 91 mod 3 = 1.

If there were some more hidden pattern, for instance, when I've brought this subject up before sci.math'ers would claim that if 1 tends to be followed by 2 and 2 by 1 and they produced statistical tests claiming those proved that, then the sequence is not random!

But if 1 tends to be followed by 2 and 2 by 1, how would they push the composites? Into what pattern?

Now then, let's leap forward and say you accept that the rigidity of the ordering of counting numbers, where you have

1, 2, 3 followed by 4, 5, 6 followed by 7, 8, 9 on out to infinity

which modulo 3 gives

1, 2, 0 followed by 1, 2, 0 followed by 1, 2, 0 repeated on out to infinity

convinces you that primes other than 3 modulo 3 give a random distribution, why should anyone care?

Well, random is useful in physics but more intriguingly, if you accept that then you end a lot of research paths in modern mathematics!!!

Because if you push the argument to p_1 mod p_2, where the p's are differing primes, then it turns out some supposedly big questions in modern math, like the Twin Primes Conjecture are easily answered!

But there are mathematicians getting federal funds for research in those areas, if random is the call then those funds cease.

BUT if no one notices, those mathematicians can work endlessly in an area where they can never get an answer, at least not a correct one.

That is a highlight of how you can find controversy in the math field in a simple area where you can run your own statistical analysis to see.

Oh yeah, and remember a while back the math awards where one guy famously refused to accept his, while the others did accept?

Well, if primes are random then one of those people won a prize for supposedly innovative research into an area where no pattern actually lies.

Maybe you believe I'm wrong and there is some hidden pattern in primes modulo 3, or more technically

p mod 3

with p not equal to 3, as you go out to positive infinity.

I've given the beginning of the sequence:

2, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1

where the start was the number 2 which has itself as a residue modulo 3.

In answering my bringing this issue up before posters on sci.math claimed that this area has been checked and that statistical analysis proved non-randomness but there is no why here given.

Why would primes care? Especially when their products, the composites can't swing one way with the primes swinging another?

I suggest to you that math people lie in this area, where you can do the statistical analysis yourself to see that they lie, or I'm wrong. I've been wrong before, but I don't think I'm wrong here.

I know people lie. And I know some of you in the physics field who like to play with statistics can crank up some machinery and play with primes to see for yourself.

Remember, BIG research grants are wrapped up in this where some mathematicians have worked in this area for decades, so their entire livelihoods are about it NOT being true that p mod 3 is random.

Their entire LIVES are about that not being true, so they are totally invested.

## JSH: Our lying world

Reality is that people just do lie.

And they lie across the board so why wouldn't they lie about math?

Mathematicians check mathematicians so if you get paid by presenting stuff in a mathematical way that looks complicated and abstruse so you can just berate anyone who questions you on it—claim they just don't understand—why can't you just lie?

Posters reply like that's silly as if mathematicians don't get anything for what they do, and act as if to be worth it a con needs to make someone rich.

But if you can make $50,000 per year doing nothing but babbling math-ese that isn't even right, why wouldn't you?

And if there is a critical mass of people like you who can laugh and joke about the stupidity of the world trusting you, why couldn't you?

The world was so shocked when the tapes of Enron people laughing about loss of power in California came out, as if, as if in today's age we are really so gullible.

And Catholic priests molesting children? How dare you even suggest such a thing, right? But it was true.

People lie about everything.

There are no real checks for most mathematicians except some other mathematician.

I say, they lie. They lie repeatedly. And they lie knowing full well why they lie, and they rely on complex stuff that few people can even follow so that they can say things that are not true, so that if you could follow the argument you'd find it lead nowhere.

We have a lying world.

People start wars over lies. Keep wars going over lies. Burnt bodies and mutilated corpses over lies.

Why would any sane human being think that mathematicians are somehow above being human?

And they lie across the board so why wouldn't they lie about math?

Mathematicians check mathematicians so if you get paid by presenting stuff in a mathematical way that looks complicated and abstruse so you can just berate anyone who questions you on it—claim they just don't understand—why can't you just lie?

Posters reply like that's silly as if mathematicians don't get anything for what they do, and act as if to be worth it a con needs to make someone rich.

But if you can make $50,000 per year doing nothing but babbling math-ese that isn't even right, why wouldn't you?

And if there is a critical mass of people like you who can laugh and joke about the stupidity of the world trusting you, why couldn't you?

The world was so shocked when the tapes of Enron people laughing about loss of power in California came out, as if, as if in today's age we are really so gullible.

And Catholic priests molesting children? How dare you even suggest such a thing, right? But it was true.

People lie about everything.

There are no real checks for most mathematicians except some other mathematician.

I say, they lie. They lie repeatedly. And they lie knowing full well why they lie, and they rely on complex stuff that few people can even follow so that they can say things that are not true, so that if you could follow the argument you'd find it lead nowhere.

We have a lying world.

People start wars over lies. Keep wars going over lies. Burnt bodies and mutilated corpses over lies.

Why would any sane human being think that mathematicians are somehow above being human?

### Sunday, August 19, 2007

## JSH: Musing on failure

Yesterday I started with a long and detailed post talking about the latest insight with my factoring research, but after that post I had several more claiming that I had now solved the factoring problem which all were flawed.

So there was the one detailed and correct post and then a leap to believing I was done, instead of in a position where a lot more work might be ahead, so it was just a fantasy reaching thing.

I WISH I could just get a stupendously simple solution that could end all the debate and finally corner mathematicians and show them for what I know them to be, but its mathematics. You have to prove.

Oh, so does it really worry me about what people might think?

No, because if I DO figure out the factoring problem none of this will matter and everything will be over in a blink, as in overnight.

Immediately everything would change and my past failures would mean nothing.

That is the burden on the math world, hope I fail but know that if I do succeed you lose everything—overnight.

So feel secure with that if you want, as I'm still thinking about this research and did get a key insight recently.

It boggles my mind that I have to find something practical to break the math world because it has left mathematical proof behind to live in its own world of pretend. And factoring is it.

I have my problems with wishing but I admit them. You people do NOTHING right in huge areas but hold on to the fantasy that you are correct for the social and ego benefits.

I can admit failure, even huge failure.

You are not strong enough to live in the real world.

So I have to find a mathematical solution that would make obsolete the system that underpins the world's Internet security to prove beyond anyone's doubt that the current math world is full of cons and liars.

Gauss and Newton in comparison had it easy. But they did live in a simpler world.

So there was the one detailed and correct post and then a leap to believing I was done, instead of in a position where a lot more work might be ahead, so it was just a fantasy reaching thing.

I WISH I could just get a stupendously simple solution that could end all the debate and finally corner mathematicians and show them for what I know them to be, but its mathematics. You have to prove.

Oh, so does it really worry me about what people might think?

No, because if I DO figure out the factoring problem none of this will matter and everything will be over in a blink, as in overnight.

Immediately everything would change and my past failures would mean nothing.

That is the burden on the math world, hope I fail but know that if I do succeed you lose everything—overnight.

So feel secure with that if you want, as I'm still thinking about this research and did get a key insight recently.

It boggles my mind that I have to find something practical to break the math world because it has left mathematical proof behind to live in its own world of pretend. And factoring is it.

I have my problems with wishing but I admit them. You people do NOTHING right in huge areas but hold on to the fantasy that you are correct for the social and ego benefits.

I can admit failure, even huge failure.

You are not strong enough to live in the real world.

So I have to find a mathematical solution that would make obsolete the system that underpins the world's Internet security to prove beyond anyone's doubt that the current math world is full of cons and liars.

Gauss and Newton in comparison had it easy. But they did live in a simpler world.

### Saturday, August 18, 2007

## JSH: Looking back, wondering

Wow, I didn't know the solution to the factoring problem would turn out to be that simple.

It seems strange to think back over the years since I first started looking at factoring one number by using another and consider all the failed approaches and all the other, with the actual solution now in hand.

But I got this feeling before with the proof of Fermat's Last Theorem and with the prime counting function, though I must admit this answer is the most succinct, the most definitive and probably the simplest result I have yet.

So why didn't some other person figure it out before?

That question is one that comes up time and time again, and growing up I would get amazed when I'd learn something like trigonometry or basic calculus and realized how easy it was and you wonder about why it took so long for people to figure it out.

Or ranging further, the Romans were playing with steam over two thousand years ago and even built simple devices that were novelties demonstrating steam power, but didn't figure out a steam engine.

If they had history would have changed. We might be in interstellar spacecraft today.

They had a mechanical computer as well and Archimedes had quite a bit of calculus which was mostly lost to be re-discovered much later by Leibniz and Newton.

With several major discoveries now under my belt I find myself thinking about history in different ways and I have a different perspective now on what it takes to make a major discovery and what it takes for it to be accepted where of course the big story here has been the successful fight by mainstream mathematicians to ignore my other discoveries.

There is that amazing difference between wanting to be something and actually being it, where I see most modern mathematicians as playing pretend, which is why they picked factoring anyway because they lacked the mathematical intuition that said a simple solution probably did exist.

Oh yeah, so it is amazing all the machinery and effort done around factoring, when people just didn't know how to do it.

Students who learn about the factoring problem in the future with the simple answer presented, say, in some textbook will not have the perspective many of you have had with BELIEVING it was a hard problem (many of you probably still believe it at this point with the solution only hours old) so it will be remarkable to see how they try to understand how it could have been such a big deal.

With so many mathematical questions now answered from FLT, to what I've learned with primes which cover the Twin Primes Conjecture and the Goldbach Conjecture, where I guess I haven't really bothered with going in detail with the Riemann Hypothesis because it doesn't interest me (I think it's false), there is an odd kind of closing of the door on the discipline of mathematics as it was.

Given that I have also presented the definition of mathematical proof and united mathematics and logic in doing so, as well as discovered the object ring, it is clear that a door to mathematical discoveries that were impossible before has been opened and with that a door to science that could never have been learned is now open as well.

I have joked about mathematicians ignoring my research maybe being aliens trying to destroy the future of humanity, but there is something to looking at it from that perspective as without the mathematics there is no way to get the science and no way to push technology into areas that seem like fantasy today.

After all to even Newton flying was a thing of fantasy. And going to the moon could only be imagined in dreams.

Our world today may be as primitive to a near future now possible because of crucial mathematical advances as Newton's was to us and could Newton have ever dreamed of our modern world?

It seems strange to think back over the years since I first started looking at factoring one number by using another and consider all the failed approaches and all the other, with the actual solution now in hand.

But I got this feeling before with the proof of Fermat's Last Theorem and with the prime counting function, though I must admit this answer is the most succinct, the most definitive and probably the simplest result I have yet.

So why didn't some other person figure it out before?

That question is one that comes up time and time again, and growing up I would get amazed when I'd learn something like trigonometry or basic calculus and realized how easy it was and you wonder about why it took so long for people to figure it out.

Or ranging further, the Romans were playing with steam over two thousand years ago and even built simple devices that were novelties demonstrating steam power, but didn't figure out a steam engine.

If they had history would have changed. We might be in interstellar spacecraft today.

They had a mechanical computer as well and Archimedes had quite a bit of calculus which was mostly lost to be re-discovered much later by Leibniz and Newton.

With several major discoveries now under my belt I find myself thinking about history in different ways and I have a different perspective now on what it takes to make a major discovery and what it takes for it to be accepted where of course the big story here has been the successful fight by mainstream mathematicians to ignore my other discoveries.

There is that amazing difference between wanting to be something and actually being it, where I see most modern mathematicians as playing pretend, which is why they picked factoring anyway because they lacked the mathematical intuition that said a simple solution probably did exist.

Oh yeah, so it is amazing all the machinery and effort done around factoring, when people just didn't know how to do it.

Students who learn about the factoring problem in the future with the simple answer presented, say, in some textbook will not have the perspective many of you have had with BELIEVING it was a hard problem (many of you probably still believe it at this point with the solution only hours old) so it will be remarkable to see how they try to understand how it could have been such a big deal.

With so many mathematical questions now answered from FLT, to what I've learned with primes which cover the Twin Primes Conjecture and the Goldbach Conjecture, where I guess I haven't really bothered with going in detail with the Riemann Hypothesis because it doesn't interest me (I think it's false), there is an odd kind of closing of the door on the discipline of mathematics as it was.

Given that I have also presented the definition of mathematical proof and united mathematics and logic in doing so, as well as discovered the object ring, it is clear that a door to mathematical discoveries that were impossible before has been opened and with that a door to science that could never have been learned is now open as well.

I have joked about mathematicians ignoring my research maybe being aliens trying to destroy the future of humanity, but there is something to looking at it from that perspective as without the mathematics there is no way to get the science and no way to push technology into areas that seem like fantasy today.

After all to even Newton flying was a thing of fantasy. And going to the moon could only be imagined in dreams.

Our world today may be as primitive to a near future now possible because of crucial mathematical advances as Newton's was to us and could Newton have ever dreamed of our modern world?

## JSH: Perfect factoring algorithm

Algorithm for guaranteeing the factorization of a target composite T of any size.

Use (x+k)^2 = y^2 + 2k^2 + n*T.

Use (x+k)^2 = y^2 + 2k^2 + n*T.

- Pick k=2 and an n that gives you an absolute value of 2k^2 + n*T above the minimums, where you can choose

abs(2k^2 + n*T)> 2x_res*T where

x_res = k*(2)^{-1} mod T

and solve for x and y by factoring 2k^2 + n*T, iterating through integer combinations to check with all possible values for integer x and y. - If your first n does not factor non-trivially increment its absolute value by 1 and try again, and do this for a maximum of 3 increments and you are guaranteed to factor T without regard to the size of T.

## JSH: Solving the factoring problem

After the explanation of my current factoring research that I just posted I realized that wrapped up in this latest analysis is solution to the factoring problem.

It's fairly easy but as usual I'll post what I think it is to see if I made any mistakes:

Starting as usual with the key congruences:

y^2 = x^2 mod T

and

k = 2x mod T

I note that a solution must exist for those equations for any non-zero x coprime to T, as I worked out in detail in my prior post.

Using them I easily get

(x+k)^2 = y^2 + 2k^2 + n*T

where n is some non-zero integer, which I've also explained and now I can address the value of n--and solve the factoring problem.

As we know a solution to the difference of square that non-trivially factors T must exist, so assume you have that x, and some arbitrary n, can it be reached?

The answer is yes if it is even, because

k = 2x mod T

explicitly is k = 2x + m*T, where m is some non-zero integer, so I can always use

(k+/-a*T) = 2x + ( m+/-a)*T

and letting k* = k+/-a*T, I then have

(x+k*)^2 = y^2 + 2k*^2 + n*T

so if k=2, I can add or subtract as needed to reach ANY even n, so n needs to be even.

Notice that knowing how much is being added or subtracted from k to get k* is unimportant as you're looking at everything mod T, so with even n, and k=2, the size limit discussed in my prior post is the only block and over that size limit you are 100% guaranteed to factor T non-trivially, which solves the factoring problem.

Note: You are moving in multiples of k*T, as remember k = 2x mod T is first multiplied by k and then added back to the congruence of squares.

The factoring problem is done.

It's fairly easy but as usual I'll post what I think it is to see if I made any mistakes:

Starting as usual with the key congruences:

y^2 = x^2 mod T

and

k = 2x mod T

I note that a solution must exist for those equations for any non-zero x coprime to T, as I worked out in detail in my prior post.

Using them I easily get

(x+k)^2 = y^2 + 2k^2 + n*T

where n is some non-zero integer, which I've also explained and now I can address the value of n--and solve the factoring problem.

As we know a solution to the difference of square that non-trivially factors T must exist, so assume you have that x, and some arbitrary n, can it be reached?

The answer is yes if it is even, because

k = 2x mod T

explicitly is k = 2x + m*T, where m is some non-zero integer, so I can always use

(k+/-a*T) = 2x + ( m+/-a)*T

and letting k* = k+/-a*T, I then have

(x+k*)^2 = y^2 + 2k*^2 + n*T

so if k=2, I can add or subtract as needed to reach ANY even n, so n needs to be even.

Notice that knowing how much is being added or subtracted from k to get k* is unimportant as you're looking at everything mod T, so with even n, and k=2, the size limit discussed in my prior post is the only block and over that size limit you are 100% guaranteed to factor T non-trivially, which solves the factoring problem.

Note: You are moving in multiples of k*T, as remember k = 2x mod T is first multiplied by k and then added back to the congruence of squares.

The factoring problem is done.

## JSH: Explaining my latest factoring research

Here's a post to go over the latest with my factoring research and explain a key breakthrough, as well as address the issue of demonstration.

First some background, since August of last year I have focused on very simple congruences in an attempt to develop an integer factorization method, where now I present as

y^2 = x^2 mod T

which is of course the difference of squares with T the target prime, and the one addition I make is introducing a variable I call k, where

k = 2x mod T

which is the second congruence. That equation I multiply by k and then add to the first, so that I can complete the square and since I'm moving to actually using it, I go to an explicit equation with

(x+k)^2 = y^2 + 2k^2 + n*T

where n is some non-zero integer, and I have had the gist of that in various form as I've puzzled over it and tried to get it to work well for about a year now, but I'm talking a lot about this now so what's new?

The answer is that my various attempts at analysis have not explained exactly when and why this system works when it does, and why it might NOT work, like, how do you pick k and n which are the variables you must choose and the only variables that you can fiddle with as you try to factor the target composite T?

So two key variables where if you pick them correctly you factor the target T non-trivially, but how do you choose?

The breakthrough came when I focused on a hypothetical T that is made up of two primes:

T = p_1*p_2

where the p's are differing primes, and wondered about picking x, and in that way choose k, and I found that with that hypothetical choice there were two more key equations that stepped in:

x+y = c_1*p_1 and x-y = c_2*p_2

solving the difference of squares where the c's are non-zero integers, so I can just solve for x and y to get

2x = c_1*p_1 + c_2*p_2

from which I have c_1 = (2x)*(p_1)^{-1} mod p_2, so you can pick ANY non-zero x coprime to p_1 and p_2,

and get a solution as then you have

y = (c_1*p_1 - c_2*p_2)/2

and importantly since c_2*p_2 is subtracted in one case and added in another it must be true that either x or y is at least the size of c_1*p_1, so how big can c_1 get?

Well c_1 = (2x)*(p_1)^{-1} mod p_2, tells us that its maximum is the largest residue available mod p_2 which is

p_2 - 1

so it must be true that abs(c_1)<abs(2x*p_2), so

abs(c_1*p_1)<abs(2x*p_2*_1) < abs(2x*T)

which is a crucial result, showing that it is the choice of x that is key.

So now, of course, it pays to look at the solution for x and y from

(x+k)^2 = y^2 + 2k^2 + n*T

more carefully, so introducing f_1 and f_2 where

4f_1*f_2 = 2k^2 + n*T

I have that y = f_1 - f_2 and x = f_1 + f_2 - k.

So I have a way to roughly estimate the size necessary for 2k^2 + n*T, as it has got to be rather large if either of the prime factors are large since x or y must be as large as c_1*p_1.

I think that it wouldn't be hard to generalize to a composite with an arbitrary number of primes, but for now getting an explanation is key so that can left for clean-up later.

In the past year, I have worked with various algorithms programmed in Java as I've tried to get the equations to work, and puzzled over their behavior, but now have an explanation for that behavior as I used to do things like let k=floor(T/30) which would have made x move about in some random way.

But how is this result relevant to that? Well if I pick a k and n that makes 2k^2 + n*T too small then the primary congruences:

y^2 = x^2 mod T and k = 2x mod T

are NOT satisfied which is what I call decoupling.

The best guess now for picking k is to pick k=2, corresponding to x = 1 mod T.

Savvy readers may note that the mathematics CAN also give you a non-trivial factorization if with

T = p*C

where p is a prime and C is a composite:

y^2 = x^2 mod p and k = 2x mod p

where the size limit would still come into play but would be 2x*p, and I have an example of such a factorization that I did before when I was still puzzling over how to pick k:

T = 732367903, k=floor(T/30) = 24412263, n= -2

2k^2 + n*T = 1191915704826532 = ( 2^2 )( 7 )( 73 )( 583129014103 )

f_1 = 7/2 and f_2 = 85136836059038

y=-170273672118069/2 and x=170273623293557/2

so, x+y=-24412256, which has 223 as a factor.

T = 732367903 = (223)(3284161).

Notice that you get the target factored with a trivial factorization of 2k^2 + n*T, which remarkably enough is just a bit larger than 2x*T.

So why does this matter?

Well, if you are over the minimums then there could be a very high probability of a non-trivial factorization while if you are BELOW the minimums there is a 0% probability of factoring—the equations decouple—so it's an absolute cut-off.

Above the minimums the mathematics goes looping through possible values for

2x = c_1*p_1 + c_2*p_2

and

y = (c_1*p_1 - c_2*p_2)/2

where it has infinity, so with your choice of k, you force possible x's and those force the math to go looking for c's that will give you a y for the size of abs(2k^2 + n*T) that you have.

If none exists, then you are decoupled.

But of course you can increase that size by increasing abs(n) as you can ONLY move in increments or decrements of T itself.

If a solution does exist then you have a non-trivial factorization.

Oh yeah, so if it's so great why can't I factor some huge number?

Well, I just got an explanation of mathematical behavior which I've puzzled over for a year and JUST got that explanation a couple of days ago.

So I'm musing over it. To me mathematics is a joy and these kind of searches are fun and I like to talk about them, but yes factoring is serious business as well so understandably some people would like me to produce some spectacular factorization now early in the research phase.

But that is what people who don't understand basic research often ask of researchers.

It's human nature to want something definite as early as possible, but the reality of real research that answers of practical importance, especially for business reasons, often take some time to reach.

First some background, since August of last year I have focused on very simple congruences in an attempt to develop an integer factorization method, where now I present as

y^2 = x^2 mod T

which is of course the difference of squares with T the target prime, and the one addition I make is introducing a variable I call k, where

k = 2x mod T

which is the second congruence. That equation I multiply by k and then add to the first, so that I can complete the square and since I'm moving to actually using it, I go to an explicit equation with

(x+k)^2 = y^2 + 2k^2 + n*T

where n is some non-zero integer, and I have had the gist of that in various form as I've puzzled over it and tried to get it to work well for about a year now, but I'm talking a lot about this now so what's new?

The answer is that my various attempts at analysis have not explained exactly when and why this system works when it does, and why it might NOT work, like, how do you pick k and n which are the variables you must choose and the only variables that you can fiddle with as you try to factor the target composite T?

So two key variables where if you pick them correctly you factor the target T non-trivially, but how do you choose?

The breakthrough came when I focused on a hypothetical T that is made up of two primes:

T = p_1*p_2

where the p's are differing primes, and wondered about picking x, and in that way choose k, and I found that with that hypothetical choice there were two more key equations that stepped in:

x+y = c_1*p_1 and x-y = c_2*p_2

solving the difference of squares where the c's are non-zero integers, so I can just solve for x and y to get

2x = c_1*p_1 + c_2*p_2

from which I have c_1 = (2x)*(p_1)^{-1} mod p_2, so you can pick ANY non-zero x coprime to p_1 and p_2,

and get a solution as then you have

y = (c_1*p_1 - c_2*p_2)/2

and importantly since c_2*p_2 is subtracted in one case and added in another it must be true that either x or y is at least the size of c_1*p_1, so how big can c_1 get?

Well c_1 = (2x)*(p_1)^{-1} mod p_2, tells us that its maximum is the largest residue available mod p_2 which is

p_2 - 1

so it must be true that abs(c_1)<abs(2x*p_2), so

abs(c_1*p_1)<abs(2x*p_2*_1) < abs(2x*T)

which is a crucial result, showing that it is the choice of x that is key.

So now, of course, it pays to look at the solution for x and y from

(x+k)^2 = y^2 + 2k^2 + n*T

more carefully, so introducing f_1 and f_2 where

4f_1*f_2 = 2k^2 + n*T

I have that y = f_1 - f_2 and x = f_1 + f_2 - k.

So I have a way to roughly estimate the size necessary for 2k^2 + n*T, as it has got to be rather large if either of the prime factors are large since x or y must be as large as c_1*p_1.

I think that it wouldn't be hard to generalize to a composite with an arbitrary number of primes, but for now getting an explanation is key so that can left for clean-up later.

In the past year, I have worked with various algorithms programmed in Java as I've tried to get the equations to work, and puzzled over their behavior, but now have an explanation for that behavior as I used to do things like let k=floor(T/30) which would have made x move about in some random way.

But how is this result relevant to that? Well if I pick a k and n that makes 2k^2 + n*T too small then the primary congruences:

y^2 = x^2 mod T and k = 2x mod T

are NOT satisfied which is what I call decoupling.

The best guess now for picking k is to pick k=2, corresponding to x = 1 mod T.

Savvy readers may note that the mathematics CAN also give you a non-trivial factorization if with

T = p*C

where p is a prime and C is a composite:

y^2 = x^2 mod p and k = 2x mod p

where the size limit would still come into play but would be 2x*p, and I have an example of such a factorization that I did before when I was still puzzling over how to pick k:

T = 732367903, k=floor(T/30) = 24412263, n= -2

2k^2 + n*T = 1191915704826532 = ( 2^2 )( 7 )( 73 )( 583129014103 )

f_1 = 7/2 and f_2 = 85136836059038

y=-170273672118069/2 and x=170273623293557/2

so, x+y=-24412256, which has 223 as a factor.

T = 732367903 = (223)(3284161).

Notice that you get the target factored with a trivial factorization of 2k^2 + n*T, which remarkably enough is just a bit larger than 2x*T.

So why does this matter?

Well, if you are over the minimums then there could be a very high probability of a non-trivial factorization while if you are BELOW the minimums there is a 0% probability of factoring—the equations decouple—so it's an absolute cut-off.

Above the minimums the mathematics goes looping through possible values for

2x = c_1*p_1 + c_2*p_2

and

y = (c_1*p_1 - c_2*p_2)/2

where it has infinity, so with your choice of k, you force possible x's and those force the math to go looking for c's that will give you a y for the size of abs(2k^2 + n*T) that you have.

If none exists, then you are decoupled.

But of course you can increase that size by increasing abs(n) as you can ONLY move in increments or decrements of T itself.

If a solution does exist then you have a non-trivial factorization.

Oh yeah, so if it's so great why can't I factor some huge number?

Well, I just got an explanation of mathematical behavior which I've puzzled over for a year and JUST got that explanation a couple of days ago.

So I'm musing over it. To me mathematics is a joy and these kind of searches are fun and I like to talk about them, but yes factoring is serious business as well so understandably some people would like me to produce some spectacular factorization now early in the research phase.

But that is what people who don't understand basic research often ask of researchers.

It's human nature to want something definite as early as possible, but the reality of real research that answers of practical importance, especially for business reasons, often take some time to reach.

### Wednesday, August 15, 2007

## JSH: May have it!

I've been working on the factoring problem with a simple approach I call surrogate factoring where I had a breakthrough yesterday by considering T = p_1*p_2 directly and wondering what would happen if I just picked x, with the governing equations being

x^2 = y^2 mod T

and k = 2x mod T, where T is the target composite to be factored.

I realized that the surrogate I use to factor needed to be roughly equal to T or greater than it, but when you solve everything out as I have I actually put in the freaking thing that x*T is what you need to get near!

What I've usually done is just kind of randomly pick k and worry about the k/T ratio, but you can pick x to force k, with

k = 2x

just using the explicit, and pick a very small x, like x=1, which would then indicate that the surrogate needs to be really close to T, OR if you DO pick k, you need to solve for x, using

x = k*(2)^{-1} mod T

and pick the smallest value to multiply times T to see the minimum size of your surrogate, if I'm not still missing something but that may be the last clue with which 100% factoring may be achieved.!!!

Could it be nearly over? Could I finally have it?

If so, then math people need to get off the namecalling thing and start paying attention to my research!

As I may finally have the factoring problem in hand.

So I guess it's the Hammer.

I've talked a lot about the Hammer over the years and the reference is to Thor's Hammer, as I've always been into mythology and like Norse and Celtic mythology and Thor is a fascinating character.

Note that Thor is HUGE and has red hair and a fiery red beard, not the Goldilocks of modern American comic books.

He was a mythic hero to me as a child. You need heroes in this day and age to keep going, when there seems to be no hope.

That's what heroes are for, to keep going, even when there seems to be no hope.

Thank you Thor and all the gods of the various mythologies…

The Hammer I think is here.

There will be more checking, but I feel greater confidence so I will state more.

I think Mjolnir has arrived…and the earth shakes yet again in recognition.

The gods cheer. Even the giants roar.

And the earth shakes.

[A reply to someone who wrote that James might now try to win the RSA challenge.]

They didn't even drop the money until after I came up with the first part of this approach.

I did that LAST August back in 2006.

Mathematicians had an ENTIRE YEAR to accept that this approach might work and help the world to adjust, but you people seem more intent on running a political campaign against my research.

If this works then I would think that some of you can be charged with treason by your various countries for that and it will not take me demonstrating as if this works then someone will use it.

So yes, the wager now is your life.

Not mine, as I've done everything I can in an impossible situation, but now at this end if you people play the same stupid games as you fight against the progress of the entire human species if it is your life that is the price then you gave it.

YOU gave it.

You cannot make demands of me.

To me you are already traitors—of the human race.

[A reply to someone who, concerning the number 17413590682008709732516359924590332789077936369050, asked James which

That is not a bad question. Answer is k=2. That matches with x = 1 mod T, and should give you the minimum surrogate to factor.

I've been thinking about this idea all day and now realize how crucial it is to know what x is, since the surrogate will roughly be about the size of x*T.

Before I focused on the k/T ratio which gave random behavior and decreasing factoring odds as T increased in size.

Trivial math throughout the proof, but simple as it is, you have to figure it out first to be able to look back and see how simple it is.

Everything is mod T, so you quickly get some serious limits on how hard it can be.

x^2 = y^2 mod T

and k = 2x mod T, where T is the target composite to be factored.

I realized that the surrogate I use to factor needed to be roughly equal to T or greater than it, but when you solve everything out as I have I actually put in the freaking thing that x*T is what you need to get near!

What I've usually done is just kind of randomly pick k and worry about the k/T ratio, but you can pick x to force k, with

k = 2x

just using the explicit, and pick a very small x, like x=1, which would then indicate that the surrogate needs to be really close to T, OR if you DO pick k, you need to solve for x, using

x = k*(2)^{-1} mod T

and pick the smallest value to multiply times T to see the minimum size of your surrogate, if I'm not still missing something but that may be the last clue with which 100% factoring may be achieved.!!!

Could it be nearly over? Could I finally have it?

If so, then math people need to get off the namecalling thing and start paying attention to my research!

As I may finally have the factoring problem in hand.

So I guess it's the Hammer.

I've talked a lot about the Hammer over the years and the reference is to Thor's Hammer, as I've always been into mythology and like Norse and Celtic mythology and Thor is a fascinating character.

Note that Thor is HUGE and has red hair and a fiery red beard, not the Goldilocks of modern American comic books.

He was a mythic hero to me as a child. You need heroes in this day and age to keep going, when there seems to be no hope.

That's what heroes are for, to keep going, even when there seems to be no hope.

Thank you Thor and all the gods of the various mythologies…

The Hammer I think is here.

There will be more checking, but I feel greater confidence so I will state more.

I think Mjolnir has arrived…and the earth shakes yet again in recognition.

The gods cheer. Even the giants roar.

And the earth shakes.

[A reply to someone who wrote that James might now try to win the RSA challenge.]

They didn't even drop the money until after I came up with the first part of this approach.

I did that LAST August back in 2006.

Mathematicians had an ENTIRE YEAR to accept that this approach might work and help the world to adjust, but you people seem more intent on running a political campaign against my research.

If this works then I would think that some of you can be charged with treason by your various countries for that and it will not take me demonstrating as if this works then someone will use it.

So yes, the wager now is your life.

Not mine, as I've done everything I can in an impossible situation, but now at this end if you people play the same stupid games as you fight against the progress of the entire human species if it is your life that is the price then you gave it.

YOU gave it.

You cannot make demands of me.

To me you are already traitors—of the human race.

[A reply to someone who, concerning the number 17413590682008709732516359924590332789077936369050, asked James which

*k*would he pick and does it factor.]That is not a bad question. Answer is k=2. That matches with x = 1 mod T, and should give you the minimum surrogate to factor.

I've been thinking about this idea all day and now realize how crucial it is to know what x is, since the surrogate will roughly be about the size of x*T.

Before I focused on the k/T ratio which gave random behavior and decreasing factoring odds as T increased in size.

Trivial math throughout the proof, but simple as it is, you have to figure it out first to be able to look back and see how simple it is.

Everything is mod T, so you quickly get some serious limits on how hard it can be.

## JSH: Shattering power

Ok, so they're stupid. But we're talking about people who are really actors playing at being mathematicians in a world that hasn't seen a major mathematical discoverer in a long time.

Ancient mathematicians are just those people they read about in books.

No one takes it seriously, right? Not really, right?

But Mathematics built human civilization.

It is the most powerful discipline on the planet.

Without it you do not have music, you do not have guns, you do not have planes, you do not have refrigerators, you do not have nuclear weapons.

There is no way that you stop a major mathematical discoverer. It has never been done. It cannot be done indefinitely.

It is simply suicidal to try.

Ancient mathematicians are just those people they read about in books.

No one takes it seriously, right? Not really, right?

But Mathematics built human civilization.

It is the most powerful discipline on the planet.

Without it you do not have music, you do not have guns, you do not have planes, you do not have refrigerators, you do not have nuclear weapons.

There is no way that you stop a major mathematical discoverer. It has never been done. It cannot be done indefinitely.

It is simply suicidal to try.

### Tuesday, August 14, 2007

## JSH: Factoring integers, more analysis

ON August 9th I stepped through integer factorization equations in my post "JSH: On integer factorization" and earlier today I thought some more about when that approach might work to give a non-trivial factorization.

Those equations were

y^2 = x^2 mod T

and

k = 2x mod T

where T is the target composite to be factored. You can easily find by multiplying the second by k, adding to the first and completing the square and then going to an explicit equation that

(x+k)^2 = y^2 + 2k^2 + n*T

where n is some non-zero integer, so that now you may factor T by instead factor 2k^2 + nT, but how do you pick k and n?

I have approached the problem by wondering what would happen if

T = p_1*p_2

where the p's are differing primes, if you picked x.

Well that gives you k, but can any x work? The answer comes from the first congruence as you need to solve for x using two linear equations:

x+y = c_1*p_1

and

x-y = c_2*p_2

where the c's non-zero integers and that is trivial and gives you

2x = c_1*p_1 + c_2*p_2

so a solution can always be found with

c_1 = (2x)*(p_1)^{-1} mod p_2

now assume p_1>p_2, then the maximum size of the modular inverse of p_1 with respect to p_2 is p_2 - 1, so the maximum size for c_1 is

c_1<= (2x)*(p_2-1)

so you have roughly c_1*p_1 < 2x*T.

Now you can also solve for y with those two linear equations to get

y = (c_1*p_1 - c_2*p_2)/2

so then y<x*T, so if that is in the size range of 2k^2 + n*T then you factor non-trivially.

As from before I have that

(x+k)^2 = y^2 + 2k^2 + n*T

so if I have rationals f_1 and f_2 where

4f_1*f_2 = 2k^2 + n*T

then

y = f_1 - f_2,

so the issue is managing the size of 2k^2 + n*T, so first assume that you find the smallest absolute size given by

n = -floor(2k^2/T)

and it may work, but if it does NOT work, then subtract 1 from n and you push up the absolute size, but I'm debating how much that increases the likelihood of a non-trivial factorization.

If p_1 is very much larger than p_2 then it seems reasonable to suppose that y will roughly be the size of p_1, and it doesn't seem likely that it can be much smaller. So 2k^2 + n*T has to be above a minimum absolute size, and above that minimum I'd think factoring should occur with a high probability.

[A reply to someone who asked James if is at the stage where he can write some code that will factor efficiently.]

I like theory. Theoretically the size of 2k2 + nT is the key variable, as if it is too small, making it easier to factor, then y cannot be roughly the same size as the largest prime factor, if you have two prime factors where one is much larger than the other.

Now that is math.

Questioning the math, well that is politics.

It occurs to me that some sci.math'ers must believe that talking down math can be enough to convince people not to use it, or you may think that a viable factoring approach requires that I implement in some stunning algorithm that shuts down all debate.

And you believe that because the world has not noticed that I DID prove Fermat's Last Theorem, it fails to care when I show that Andrew Wiles didn't—even when I show a stupid logical flaw in his work—and can even ignore my prime number research.

And people thought mathematicians were in love with primes!

So people like you hope that you can talk down any mathematics, even if, if you fail, the world pays the price and people who trusted and saved lose those savings, and everything they built for decades because mathematicians learned to lie about math and thought they'd never get caught.

Oh yeah, didn't Jim Ferry claim that he programmed an efficient algorithm into Mathematica?

Was he lying? Hey, Jim! You out there? Were you lying?

If so, then you better say so, as I am increasingly confident in this approach, as I'm finally understanding when and how it should work, assuming I didn't make any major mistakes in my analysis.

But now comes the scary part, what to do? I KNOW mathematicians routinely lie about math, probably because they never took it seriously, and couldn't appreciate the consequences.

They make even stupid mistakes and play pretend smart for their niche audience that loves to believe they're smart no matter what.

They were just cons playing a simple game with stuff most people don't care about, but now there is a world on financial edge and it is deadly serious.

And mathematicians are just cons who played with fire, so why would they have any clue what to do now?

I don't know. For all I know, tomorrow could be a very bad day for the world's financial markets, but who really is to blame?

Those equations were

y^2 = x^2 mod T

and

k = 2x mod T

where T is the target composite to be factored. You can easily find by multiplying the second by k, adding to the first and completing the square and then going to an explicit equation that

(x+k)^2 = y^2 + 2k^2 + n*T

where n is some non-zero integer, so that now you may factor T by instead factor 2k^2 + nT, but how do you pick k and n?

I have approached the problem by wondering what would happen if

T = p_1*p_2

where the p's are differing primes, if you picked x.

Well that gives you k, but can any x work? The answer comes from the first congruence as you need to solve for x using two linear equations:

x+y = c_1*p_1

and

x-y = c_2*p_2

where the c's non-zero integers and that is trivial and gives you

2x = c_1*p_1 + c_2*p_2

so a solution can always be found with

c_1 = (2x)*(p_1)^{-1} mod p_2

now assume p_1>p_2, then the maximum size of the modular inverse of p_1 with respect to p_2 is p_2 - 1, so the maximum size for c_1 is

c_1<= (2x)*(p_2-1)

so you have roughly c_1*p_1 < 2x*T.

Now you can also solve for y with those two linear equations to get

y = (c_1*p_1 - c_2*p_2)/2

so then y<x*T, so if that is in the size range of 2k^2 + n*T then you factor non-trivially.

As from before I have that

(x+k)^2 = y^2 + 2k^2 + n*T

so if I have rationals f_1 and f_2 where

4f_1*f_2 = 2k^2 + n*T

then

y = f_1 - f_2,

so the issue is managing the size of 2k^2 + n*T, so first assume that you find the smallest absolute size given by

n = -floor(2k^2/T)

and it may work, but if it does NOT work, then subtract 1 from n and you push up the absolute size, but I'm debating how much that increases the likelihood of a non-trivial factorization.

If p_1 is very much larger than p_2 then it seems reasonable to suppose that y will roughly be the size of p_1, and it doesn't seem likely that it can be much smaller. So 2k^2 + n*T has to be above a minimum absolute size, and above that minimum I'd think factoring should occur with a high probability.

[A reply to someone who asked James if is at the stage where he can write some code that will factor efficiently.]

I like theory. Theoretically the size of 2k2 + nT is the key variable, as if it is too small, making it easier to factor, then y cannot be roughly the same size as the largest prime factor, if you have two prime factors where one is much larger than the other.

Now that is math.

Questioning the math, well that is politics.

It occurs to me that some sci.math'ers must believe that talking down math can be enough to convince people not to use it, or you may think that a viable factoring approach requires that I implement in some stunning algorithm that shuts down all debate.

And you believe that because the world has not noticed that I DID prove Fermat's Last Theorem, it fails to care when I show that Andrew Wiles didn't—even when I show a stupid logical flaw in his work—and can even ignore my prime number research.

And people thought mathematicians were in love with primes!

So people like you hope that you can talk down any mathematics, even if, if you fail, the world pays the price and people who trusted and saved lose those savings, and everything they built for decades because mathematicians learned to lie about math and thought they'd never get caught.

Oh yeah, didn't Jim Ferry claim that he programmed an efficient algorithm into Mathematica?

Was he lying? Hey, Jim! You out there? Were you lying?

If so, then you better say so, as I am increasingly confident in this approach, as I'm finally understanding when and how it should work, assuming I didn't make any major mistakes in my analysis.

But now comes the scary part, what to do? I KNOW mathematicians routinely lie about math, probably because they never took it seriously, and couldn't appreciate the consequences.

They make even stupid mistakes and play pretend smart for their niche audience that loves to believe they're smart no matter what.

They were just cons playing a simple game with stuff most people don't care about, but now there is a world on financial edge and it is deadly serious.

And mathematicians are just cons who played with fire, so why would they have any clue what to do now?

I don't know. For all I know, tomorrow could be a very bad day for the world's financial markets, but who really is to blame?

### Saturday, August 11, 2007

## Problem solving as an art

For me as a kid growing up it was extremely exciting to learn about modern problem solving techniques, where for various reasons I do not these days go into details about the methodology behind modern problem solving as I am more focused these days on DOING something with what I learned as a child, so many decades ago.

I consider myself a problem solver as that's so generic and encompasses so much leaving me the ability to dive into many areas as I choose and having spent years working at mathematics I can easily shift slightly into other areas while the mathematical discoveries already made, of course, stay there.

And getting to mathematics, I think it important to remind that mathematics has an importance for the human race from its ability in helping to solve problems. It is an ACTION discipline which is about solving problems.

I think some today think of it as some weird and very complicated activity that slightly strange people who are very "smart" do for some reason or other that is supposedly important.

So in answer against that too narrow view, I have DMESE, a way that end users can make personal copies of DVD's they buy without content providers needing to worry about losing a lot of money to piracy.

As time goes forward I am working more and more on problems clearly relevant to the mainstream, and in areas where hundreds of millions or even hundreds of billions of dollars are at stake because our civilization is that advanced that those truly adept in the field of problem solving, who are actually at the limits of human power and understanding necessarily have an impact commensurate with that level, where I emphasize the positive.

Problem solving is an art, and those who are the adepts at it are artists.

The goal being to become a master and follow in the footsteps not only of Einstein, Newton, Gauss and Archimedes, but Picasso, Van Gogh, Michelangelo, and of course, Da Vinci.

IN all its forms, the search for truth is the greatest endeavor of them all.

I consider myself a problem solver as that's so generic and encompasses so much leaving me the ability to dive into many areas as I choose and having spent years working at mathematics I can easily shift slightly into other areas while the mathematical discoveries already made, of course, stay there.

And getting to mathematics, I think it important to remind that mathematics has an importance for the human race from its ability in helping to solve problems. It is an ACTION discipline which is about solving problems.

I think some today think of it as some weird and very complicated activity that slightly strange people who are very "smart" do for some reason or other that is supposedly important.

So in answer against that too narrow view, I have DMESE, a way that end users can make personal copies of DVD's they buy without content providers needing to worry about losing a lot of money to piracy.

As time goes forward I am working more and more on problems clearly relevant to the mainstream, and in areas where hundreds of millions or even hundreds of billions of dollars are at stake because our civilization is that advanced that those truly adept in the field of problem solving, who are actually at the limits of human power and understanding necessarily have an impact commensurate with that level, where I emphasize the positive.

Problem solving is an art, and those who are the adepts at it are artists.

The goal being to become a master and follow in the footsteps not only of Einstein, Newton, Gauss and Archimedes, but Picasso, Van Gogh, Michelangelo, and of course, Da Vinci.

IN all its forms, the search for truth is the greatest endeavor of them all.

### Thursday, August 09, 2007

## JSH: On integer factorization

Begin with a difference of squares congruence:

y^2 = x^2 mod T

where T is the target composite to be factored.

Now introduce k, where

k = 2x mod T

and multiply that equation by k so that you have

k^2 = 2xk mod T, which you now add to the first equation to get

k^2 + y^2 = x^2 + 2xk mod T

so you can add k^2 to both sides to complete the square so that you have

2k^2 + y^2 = (x+k)^2 mod T

which, of course is the same as

(x+k)^2 = y^2 + 2k^2 mod T.

So you now have a NEW difference of squares, as going to an explicit equation with the introduction of a non-zero integer n, you have

(x+k)^2 = y^2 + 2k^2 + nT.

I'd guess that with a given composite T, you need T+sqrt(T) < abs(k) < 2T, and can then select n with

n = -floor(2k^2 /T).

Um, it is possible that then looping through the integer factorizations of 2k^2 + nT to get x and y that you will factor your target T with a high probability of success as that gives you a k/T ratio in what I guess is the best area.

I figured ut this approach August of last year and since that time the RSA corporation that profits from the encryption system used to protect the internet has withdrawn monetary prizes for its factoring challenge.

I'd often post ideas in this area on the newsgroup sci.crypt but if you check as I just did with Google Groups you can see that newsgroup has been shut down by spam postings.

One would think coincidentally, as I've noted before, major postings by me in this area have been followed the next day by sharp downward corrections in the stock markets of the world.

My fear is that ego would keep mathematicians from telling the entire world the truth because to them the mathematics I've outlined above is too simple for their tastes, but someone seems to know the truth, or there just happen to be a lot of coincidences and a LOT of naively trusting people, like yourself.

y^2 = x^2 mod T

where T is the target composite to be factored.

Now introduce k, where

k = 2x mod T

and multiply that equation by k so that you have

k^2 = 2xk mod T, which you now add to the first equation to get

k^2 + y^2 = x^2 + 2xk mod T

so you can add k^2 to both sides to complete the square so that you have

2k^2 + y^2 = (x+k)^2 mod T

which, of course is the same as

(x+k)^2 = y^2 + 2k^2 mod T.

So you now have a NEW difference of squares, as going to an explicit equation with the introduction of a non-zero integer n, you have

(x+k)^2 = y^2 + 2k^2 + nT.

I'd guess that with a given composite T, you need T+sqrt(T) < abs(k) < 2T, and can then select n with

n = -floor(2k^2 /T).

Um, it is possible that then looping through the integer factorizations of 2k^2 + nT to get x and y that you will factor your target T with a high probability of success as that gives you a k/T ratio in what I guess is the best area.

I figured ut this approach August of last year and since that time the RSA corporation that profits from the encryption system used to protect the internet has withdrawn monetary prizes for its factoring challenge.

I'd often post ideas in this area on the newsgroup sci.crypt but if you check as I just did with Google Groups you can see that newsgroup has been shut down by spam postings.

One would think coincidentally, as I've noted before, major postings by me in this area have been followed the next day by sharp downward corrections in the stock markets of the world.

My fear is that ego would keep mathematicians from telling the entire world the truth because to them the mathematics I've outlined above is too simple for their tastes, but someone seems to know the truth, or there just happen to be a lot of coincidences and a LOT of naively trusting people, like yourself.

### Tuesday, August 07, 2007

## JSH: So now you can see

I like the simple example using quadratics not because I think it will stop sci.math'ers from lying about my research but because you can see exactly how the ring of algebraic integers fails.

It is stunning mathematics in terms of impact as well as telling you how a simple mistake can undo over a hundred years of human effort.

Now the good news is that I found the short proof of Fermat's Last Theorem which is how I found out that there is this serious issue with the ring of algebraic integers, and even had a key part of that proof published in a peer reviewed mathematical journal, now defunct.

The reaction of the sci.math newsgroup which managed to destroy that journal along the way tells you how big it can be from a human perspective to lose over a hundred years of beliefs.

If you were Andrew Wiles and someone woke you up to show you my simple example, what would you do?

If he gives a press conference he announces that he didn't prove FLT. And he announces that he probably has no major mathematical results at all.

>From the top of the heap, to nothing in the field, in an instant.

Could you do it?

We are coming to the end of summer. If mathematicians fight even this simple example they will soon have more young minds entrusted to them to learn mathematics, and they will teach them false ideas, yet again, for another year.

And Andrew Wiles can be a hero, yet again for another year, if you let them.

I included alt.math.recreational for a reason.

Some of you do not have livelihoods that depend on the lie, and maybe, just maybe you do care more about what is mathematically correct.

I have years of seeing mathematicians fail and reject mathematical proof with many of you sitting by and letting them.

If the fate of the human rests on any one thing then it is in the ability or lack of ability to face the truth.

We have the ability to grasp so much, and the weakness to lose everything, for nothing at all.

It is stunning mathematics in terms of impact as well as telling you how a simple mistake can undo over a hundred years of human effort.

Now the good news is that I found the short proof of Fermat's Last Theorem which is how I found out that there is this serious issue with the ring of algebraic integers, and even had a key part of that proof published in a peer reviewed mathematical journal, now defunct.

The reaction of the sci.math newsgroup which managed to destroy that journal along the way tells you how big it can be from a human perspective to lose over a hundred years of beliefs.

If you were Andrew Wiles and someone woke you up to show you my simple example, what would you do?

If he gives a press conference he announces that he didn't prove FLT. And he announces that he probably has no major mathematical results at all.

>From the top of the heap, to nothing in the field, in an instant.

Could you do it?

We are coming to the end of summer. If mathematicians fight even this simple example they will soon have more young minds entrusted to them to learn mathematics, and they will teach them false ideas, yet again, for another year.

And Andrew Wiles can be a hero, yet again for another year, if you let them.

I included alt.math.recreational for a reason.

Some of you do not have livelihoods that depend on the lie, and maybe, just maybe you do care more about what is mathematically correct.

I have years of seeing mathematicians fail and reject mathematical proof with many of you sitting by and letting them.

If the fate of the human rests on any one thing then it is in the ability or lack of ability to face the truth.

We have the ability to grasp so much, and the weakness to lose everything, for nothing at all.

### Monday, August 06, 2007

## JSH: It's over with an easy example

If you can follow adding fractions together then you know it's over. No more dumb debates about whether or not the ring of algebraic integers is crap.

Galois Theory as usually taught falls away.

Over a hundred years of number theory texts with errors, Thousands of papers that are crap.

Mathematics is a hard discipline which is not for the faint of heart.

Just one error compounded over a hundred plus years can mean so much heartache on the other end.

And hey! You people wasted some years for me, and have blocked the world's acceptance of a proof of Fermat's Last Theorem, THE prime counting function which I am very confident can help resolve the Riemann Hypothesis, and other tidbits here and there, like oh yeah, Goldbach's Conjecture is false, while the Twin Primes Conjecture is true. And I gave the definition of mathematical proof, and a few other things here and there.

Galois Theory as usually taught falls away.

Over a hundred years of number theory texts with errors, Thousands of papers that are crap.

Mathematics is a hard discipline which is not for the faint of heart.

Just one error compounded over a hundred plus years can mean so much heartache on the other end.

And hey! You people wasted some years for me, and have blocked the world's acceptance of a proof of Fermat's Last Theorem, THE prime counting function which I am very confident can help resolve the Riemann Hypothesis, and other tidbits here and there, like oh yeah, Goldbach's Conjecture is false, while the Twin Primes Conjecture is true. And I gave the definition of mathematical proof, and a few other things here and there.

## JSH: Simple construct

Let

7x^2 + 5x - 8 = 0

and

7y^2 + 5y - 1 = 0

so subtracting the second from the first I have

7(x^2 - y^2) + 5(x-y) - 7 = 0

so

7(x+y) + 5 = 7/(x-y).

Therefore 1/y has 7 as a factor, right?

Wrong. For one root of 7y^2 + 5y - 1 = 0, 1/y has 7 as a factor in a proper ring, while for the other root it is coprime to 7.

However, in the ring of algebraic integers for neither root does 1/y have 7 as a factor.

What's nice about this example is that it is trivially easy for you to understand if you mastered adding fractions together.

If you didn't then yeah, you may still have problems.

Odd that such a simple way to show the ring of algebraic integers is worthless for analysis was always there, eh? I wonder why I just thought of it now.

[A reply to someone who explained James why he was wrong.]

Sigh. So you have problems adding fractions?

Look again at

7(x+y) + 5 = 7/(x-y)

and notice that x+y must divide off all factors of 7 on the left side so that you can add with 5 which is coprime to 7, to get a result that has factors in common with 7.

On the right side you are just multiplying 7 by more factors.

It is a fascinatingly simple example that you should see quickly if you can handle fractions in your head, or if you have trouble, solve out for x and y and try again.

7x^2 + 5x - 8 = 0

and

7y^2 + 5y - 1 = 0

so subtracting the second from the first I have

7(x^2 - y^2) + 5(x-y) - 7 = 0

so

7(x+y) + 5 = 7/(x-y).

Therefore 1/y has 7 as a factor, right?

Wrong. For one root of 7y^2 + 5y - 1 = 0, 1/y has 7 as a factor in a proper ring, while for the other root it is coprime to 7.

However, in the ring of algebraic integers for neither root does 1/y have 7 as a factor.

What's nice about this example is that it is trivially easy for you to understand if you mastered adding fractions together.

If you didn't then yeah, you may still have problems.

Odd that such a simple way to show the ring of algebraic integers is worthless for analysis was always there, eh? I wonder why I just thought of it now.

[A reply to someone who explained James why he was wrong.]

Sigh. So you have problems adding fractions?

Look again at

7(x+y) + 5 = 7/(x-y)

and notice that x+y must divide off all factors of 7 on the left side so that you can add with 5 which is coprime to 7, to get a result that has factors in common with 7.

On the right side you are just multiplying 7 by more factors.

It is a fascinatingly simple example that you should see quickly if you can handle fractions in your head, or if you have trouble, solve out for x and y and try again.

## JSH: Now the fun part

Assume I'm correct in that from the ring of objects you can prove that units are being used in the ring of algebraic integers where they are not units to wrap up constants like 7 so that you can divide them off and remain in that ring.

Then you find a consistent argument as I've stated.

For those who don't know I have the following definition for the ring of objects:

The object ring is defined by two conditions, and includes all numbers such that these conditions are true:

The ring of algebraic integers is included in the object ring while numbers like 1/2, 1/3, or 1/sqrt(2) are excluded.

>From the ring of objects my Wrapper theorem (shown at my math blog) shows how you can appear to divide off integers within the ring of algebraic integers in what I call non-polynomial factorizations by multiplying with units in the ring of objects that are NOT units in the ring of algebraic integers. I call those numbers wrappers.

There is no mathematical argument blocking that scenario.

Then you find a consistent argument as I've stated.

For those who don't know I have the following definition for the ring of objects:

The object ring is defined by two conditions, and includes all numbers such that these conditions are true:

- 1 and -1 are the only rationals that are units in the ring.
- Given a member m of the ring there must exist a non-zero member n such that mn is an integer, and if mn is not a factor of m, then n cannot be a unit in the ring.

The ring of algebraic integers is included in the object ring while numbers like 1/2, 1/3, or 1/sqrt(2) are excluded.

>From the ring of objects my Wrapper theorem (shown at my math blog) shows how you can appear to divide off integers within the ring of algebraic integers in what I call non-polynomial factorizations by multiplying with units in the ring of objects that are NOT units in the ring of algebraic integers. I call those numbers wrappers.

There is no mathematical argument blocking that scenario.

## JSH: Like parallel postulate, when both sides are right

So I've been thinking about what I call the Wrapper theorem, and considering ways to finally prove how you're forced out of the ring of algebraic integers, when it finally dawns on me that with my own theorem I'm explaining how to STAY in the ring of algebraic integers!

And now I'm thinking about the years of arguing that went on in the past about the Parallel Postulate, and it seems that I need to shift focus from trying to prove others wrong on details where instead I can prove them right!

The Wrapper theorem explains exactly how to remain in the ring of algebraic integers with the kind of non-polynomial factorizations I use, and does so from the perspective of factorizations valid in my ring of objects.

So yeah, you can divide constants off as functions in the ring of algebraic integers in those special cases created by my non-polynomial factorizations.

So there was never any way that I could prove that you cannot, as you can.

And now I'm thinking about the years of arguing that went on in the past about the Parallel Postulate, and it seems that I need to shift focus from trying to prove others wrong on details where instead I can prove them right!

The Wrapper theorem explains exactly how to remain in the ring of algebraic integers with the kind of non-polynomial factorizations I use, and does so from the perspective of factorizations valid in my ring of objects.

So yeah, you can divide constants off as functions in the ring of algebraic integers in those special cases created by my non-polynomial factorizations.

So there was never any way that I could prove that you cannot, as you can.

### Thursday, August 02, 2007

## JSH: Impossible factorization in ring of algebraic integers

In the ring of algebraic integers the following factorization is not possible:

P(x) = (g_1(x) + 2)*(g_2(x) + 1)

whern P(x) is a non-monic polynomial with integer coefficients and f_1(0) = f_2(0) = 0, if the f's are non-rational with rational x.

Took a while to get to all the conditions but that is water under the bridge.

Note then that given

d_1*d_2*P(x) = (f_1(x) + d_1)*(f_2(x) + d_2)

if f_1(0) = f_2(0) = 0 and the f's have non-rational values with rational x, then it's not possible to divide of d_1 and d_2 in the ring of algebraic integers in general as then you would end up with the impossible factorization in that ring.

Why?

Well, from that factorization you can get to

d_1*d_2*P(x) = (h_1(x) + d_1*d_2)*(h_2(x) + d_1*d_2)

and if you assume functions w_1(x) and w_2(x) where w_1(x)*w_2(x) = d_1*d_2, then you can divide off to get

P(x) = (h_1(x)/w_1(x) + w_2(x))*(h_2(x)/w_2(x) + w_1(x))

and then use

h_1(x)/w_1(x) + w_2(x) = j_1(x) + 2

and

h_2(x)/w_2(x) + w_1(x) =j_2(x) + 2

to get

P(x) = (j_1(x) + 2)*(j_2(x) + 1)

and the impossible factorization proving that you can't use functions to divide off the constants, contradicting the objection that Dik Winter originally produced.

Notice that prior posts on sci.math only failed in terms of conditions as if P(x) is monic—not surprising considering how algebraic integers are defined—then you CAN find the factorization as demonstrated by William Hughes.

To me the difference between me and most of you is that I care about what is true, so I just kept going to get to the final answer while too many of you only care to preserve what you believe without regard to mathematical truth, so you only look for an excuse to hold on.

That makes me a discoverer.

Kind of a neat exercise though any of you who bothered to read my last paper on non-polynomial factorization could have seen what the conditions had to be from the example in it, as you have a non-monic P(x).

But to see that you'd have to care about what is mathematical truth.

You'd have to be a real mathematician.

P(x) = (g_1(x) + 2)*(g_2(x) + 1)

whern P(x) is a non-monic polynomial with integer coefficients and f_1(0) = f_2(0) = 0, if the f's are non-rational with rational x.

Took a while to get to all the conditions but that is water under the bridge.

Note then that given

d_1*d_2*P(x) = (f_1(x) + d_1)*(f_2(x) + d_2)

if f_1(0) = f_2(0) = 0 and the f's have non-rational values with rational x, then it's not possible to divide of d_1 and d_2 in the ring of algebraic integers in general as then you would end up with the impossible factorization in that ring.

Why?

Well, from that factorization you can get to

d_1*d_2*P(x) = (h_1(x) + d_1*d_2)*(h_2(x) + d_1*d_2)

and if you assume functions w_1(x) and w_2(x) where w_1(x)*w_2(x) = d_1*d_2, then you can divide off to get

P(x) = (h_1(x)/w_1(x) + w_2(x))*(h_2(x)/w_2(x) + w_1(x))

and then use

h_1(x)/w_1(x) + w_2(x) = j_1(x) + 2

and

h_2(x)/w_2(x) + w_1(x) =j_2(x) + 2

to get

P(x) = (j_1(x) + 2)*(j_2(x) + 1)

and the impossible factorization proving that you can't use functions to divide off the constants, contradicting the objection that Dik Winter originally produced.

Notice that prior posts on sci.math only failed in terms of conditions as if P(x) is monic—not surprising considering how algebraic integers are defined—then you CAN find the factorization as demonstrated by William Hughes.

To me the difference between me and most of you is that I care about what is true, so I just kept going to get to the final answer while too many of you only care to preserve what you believe without regard to mathematical truth, so you only look for an excuse to hold on.

That makes me a discoverer.

Kind of a neat exercise though any of you who bothered to read my last paper on non-polynomial factorization could have seen what the conditions had to be from the example in it, as you have a non-monic P(x).

But to see that you'd have to care about what is mathematical truth.

You'd have to be a real mathematician.