Tuesday, March 31, 2009


JSH: Why you're more interesting

I've had some of the most powerful mathematical techniques ever discovered for years. But something weird happened—your community rejected the knowledge.

I find you fascinating.

It's almost a weird thrill to read through your posts, listen to your confidence, watch how giddy some of you are at my errors.

Wonder what it'd be like to feel like you do. To have such wonderful ignorance.

You're special, so I keep you safe.

For a little while longer I can keep you in your bubbles before you're taken away from me.

But for now, we can play our little games, and I can wonder: What are you?

I know what you are. But I like to ask you, and see what you THINK you are.


JSH: General alternate Pell's Equation result

I've noted several simple alternates to Pell's Equation, which includes the negative Pell's Equation, which allow solving the main one more easily, even with continued fractions, but that has been when D is a prime number.

Here is the general result which includes composite D:

Given D such that D=f_1*f_2, where the f's are positive integer factors and one of them can be 1, you have the following alternates to Pell's Equation with which you can solve it:

f_1*j^2 - f_2*k^2 = -1, where x = 2f_2*k^2 - 1 = 2f_1*j^2 + 1?


f_1*j^2 - f_2*k^2 = -2, where x = f_2*k^2 - 1 = f_1*j^2 + 1


f_1*j^2 - f_2*k^2 = 2, where x = f_2*k^2 + 1 = f_1*j^2 - 1

and x^2 - Dy^2 = 1.

One of those equations will be valid for each f_1, and f_2 available.

Here's an example with D=21:

3j^2 - 7k^2 = -1

I can multiply both sides by 3, and pull the 3 in with the j, so I have

(3j)^2 - 21k^2 = - 3

and notice that k=4, works if 3j = 9, so j=3, so x = 2*3*9 + 1 = 55,


55^2 - 21*12^2 = 1.

The alternates provide an easier route to solving Pell's Equation in general, as solutions for j tend to be approximately sqrt(x) of the solution for Pell's Equation, and continued fractions and other approaches known to Pell's Equation are available with the alternates.

So it's actually dumb to try and solve Pell's Equation directly!!!

You should use an alternate.


JSH: "Fodder for the Beast"

That's what I call them. The math people whose minds snap: Fodder for the Beast.

Mathematics is a difficult discipline.

I burn through a lot of people worldwide.

I do feel guilty about it, but I don't make these people post, or read my posts.

And their minds would NOT snap, if they'd simply accept what's mathematically correct.

The mental destruction is a result of denial.

And make no mistake, these people are no longer, whole.

Many of them may be permanently damaged.

Monday, March 30, 2009


JSH: Could they be aliens?

Hmmm…people who are so hostile to mathematical knowledge that for years they worked like mad demons to hide every major mathematical result of mine, even managing to destroy a mathematical journal in the process, and NOW they seem certain they can hide a result that shows that the famous Pell's Equation can be more easily solved for prime D by using alternate equations including the famous negative Pell's Equation.

That's just strange.

So are they even people?

I know, sounds crazy, but how much do we really know about our reality? Why couldn't aliens infiltrate to try and make sure that a species like ours never advances its mathematics beyond a certain point?

After all, how could human beings behave in this way?

Aliens may be more exciting, but I think the sad answer is, they are human and it's class war. Or as it's usually described, racism.

There is a comfort for some people in believing they are members of a superior race, and part of that belief that I've often heard is claiming the races are equal in intelligence is just political correctness, as notice, they'll say, no major discoverers (supposedly) from certain races.

I am a major discoverer.

I upset the applecart, blowing away a racist line of argument.

I think that explanation is simpler than aliens. But, of course, we can study these people later and see, um, if they're actually people.

Sunday, March 29, 2009


JSH: Purpose of these posts on this newsgroup

Hey, I made my BIG post today on other newsgroups.

These posts on THIS newsgroup are only to draw the kind of postings I've seen for years, to really drive home my case against this newsgroup and the people here.

Many of you despise mathematics.

THAT is my point.

Your pretending otherwise has just made my life more difficult for years as many of you have waged very successful campaigns against important mathematical research.

Mathematical research that is important for the future of humanity.

The window that opened was a research result that is clear in a way that can't be talked down, as the mathematical literature has to be updated, and the negative Pell's Equation has been a research area.

The purpose of this post is to talk about the purpose of the others and I think, of course, mathematicians around the world will later claim that they shouldn't be held accountable by postings from newsgroup people.

Of course by then the world will have pictures and interviews, I hope, of most of you who have been so obsessive in your replies to me all these years.

You will have an opportunity to explain yourselves.

But my belief is that mathematicians around the world have taken comfort in your behavior and believed they could succeed in ignoring my research indefinitely, to the detriment of the human race.

Mathematics is an important discipline.

People who fought against the discipline in the hopes that they would never face accountability, need to face it.

Email records. Computer records of any kind. Posting history where it's there.

Whatever. Conversations.


IF you fought against mathematical progress then you are not truly a mathematician. And shouldn't be paid to be one.


JSH: Pell's Equation alternates

So remarkably in a well-worked area where supposedly no simple new research results were available I've found that for every odd prime D, there are three alternates to Pell's Equation, which also solve it, for which only one is true for a particular D:

j^2 - Dk^2 = -1

where x = 2Dk^2 - 1 = 2j^2 + 1


j^2 - Dk^2 = -2

where x = Dk^2 - 1 = j^2 + 1


j^2 - Dk^2 = 2

where x = Dk^2 + 1 = j^2 - 1

and x^2 - Dy^2 = 1.

What's remarkable then is that past mathematicians were approaching Pell's Equation BACKWARDS when D is a prime, as the solutions above for j are approximately the square root of x, so they're easier to find!!!

Here's an example, with D=19:

13^2 - 19*(3)^2 = -2

So j=13. With that alternate I have:

x = Dk^2 - 1 = j^2 + 1, so x = 13^2 + 1 = 170

And 170^2 - 19*(39)^2 = 1, as required.

Of interest now then is checking the mathematical literature to see what has been said about these alternates and it appears that only the so-called negative Pell's Equation has gotten attention with research done attempting to find its frequency.

I suggest readers search on: negative Pell's Equation

But an immediate guess from the result shown here is that x^2 - Dy^2 = -1, is available for an odd prime D approximately 1/3 of the time.

So that is an immediate research route to check the distribution for an ODD PRIME D, as I'd guess that researchers were simply looking at D an integer, including composites.

An astounding mystery now though is, why is this result new?

That is, however, not a question for mathematical research but one for mathematical scholars.

Readers looking to find out more about these equations and how they are derived should go to my math blog:


I highly recommend doing searches or going to the library and checking on Pell's Equation first.

I think it's a remarkable result but unfortunately, for me, it's just another major research find, where in the past I've faced resistance from the mathematical community.

I hope that is not the case here, as hey, math is fun! And in this area you can compare your own research findings to that of past mathematicians over 2000 years of human history.

Now when is the last time you could so something like that?

The door is open. There is no way of knowing at this point what major result may lurk around the corner from paths taken with this line of research. But the math community has to be willing to step through…


JSH: Understanding the situation

The alternates to Pell's Equation provide a simpler route to finding solutions when D is a prime number.

That is, given x^2 - Dy^2 = 1, it is EASIER to find solutions by instead using:

j^2 - Dk^2 = -1, j^2 - Dk^2 = -2, or j^2 - Dk^2 = 2

which are required to exist for D a prime, where only one will be valid when D is an odd prime, but one must.

They are easier as the first j that solves is roughly the square root of the first x that solves.

Now those are the mathematical facts.

The social results are more complicated:
  1. Math society needs to acknowledge the result properly.

  2. Math textbooks around the world that talk about Pell's Equation will have to be updated.

  3. Math curricula around the world where Pell's Equation is discussed will have to be updated.
I can see from posts I'm still getting that some of you not only have no clue about the reality of the situation, you don't realize what you're revealing to readers who do.

Your fight to hide knowledge is clear here.

No one who is for the human race, or for mathematical progress, or is intellectually honest or who has good intentions at all, would fight to hide this result.

It's knowledge. It's interesting knowledge, and as a "pure math" result it's fascinating for many reasons.

But as a social instrument it is incredible that it has already revealed that some of you are as far out there against mathematics as is possible, yet you have been operating on newsgroups freely for years, as something else.

The world of mathematics has changed because of one discovery.

For me, it is just one of many. But for a lot of you, it is the one you can't just run away from, or be distracted from, or trust that some way there must be more that you don't know so that you can let it stay hidden.

I am the world's latest major mathematical discoverer.

The history of my predecessors is very rich, and their discoveries did much to advance our world.

My hope is to live up to the very high bar of those who preceded me.

And more importantly, to not disgrace the role, for those I truly hope will follow.

As if I am the last major discoverer, it is the end of the human race.

I am increasingly confident, now with this result, that thankfully, I am not the last.

The future should be brighter than the past.

History will be the judge. But make no mistake, you are now a part of history in the making.


JSH: Finally! They're caught.

Now readers can finally see clearly what I've had to face for years: some of those among you are anti-knowledge.

It has been so frustrating watching these people get away with it for years. And now the result that proves my point by how they react to it.

Notice, they have all the warnings too!!! Yet they still are fighting it. That is telling.

It is such a beautiful mathematical result too. A family of three equations connected to Pell's Equation, where primes come in as well with D being prime, where they are EASIER to solve.

That creates a story where none of the old lies and manipulating techniques they use can hold.

But some of them are trying anyway which is an expression of contempt for you.

I don't think they see most of you as being anything but their tools.

They've ruled you for so long that they have no respect for your intelligence.

Yup. They've RULED you. That's why they fight so hard.

You have been slaves in mind, if not in body.

You have been servants to a class of people who have nothing but contempt for you.


JSH: Why they lurk

There are people on the newsgroups who often reply to me with replies that betray a woeful lack of even basic mathematics.

So why are they here?

Mathematics does not require a math Ph.D, and does not require that person be a member of a particular social order, or follow social rules.

Mathematics does not care if you are beneath a certain level in society.

However, mathematical discovery can elevate a person in society.

So it is a gaping hole in walls built by certain people looking for security.

A person from a "second class" world can make a major discovery and elevate themselves in the status quo.

So they lurk to enforce a social order that is trying to get only established people acknowledged for mathematical discovery.

They intimidate and insult people not part of that order to try and stop them from working on mathematics.

And they fight discoveries made to try and hide them from the world so that their social classes will not be disturbed.

I like my recent Pell's Equation results as you, probably a disbelieving person who can't imagine such behavior, can see them in action.

The threads are already there. Read through them, and see the watchers who have monitored all along.

Saturday, March 28, 2009


JSH: Quadratic residues and large composite factoring

I've been brainstorming a factoring idea for less than 24 hours which involves using quadratic residues to factor, relying on the fact that non-square residues should be shared by two large prime factors for a particular quadratic residue only roughly 50% of the time.

The idea is bizarrely simple, given a target composite T, where T=p_1*p_2, where the p's are large primes and p_1>p_2,

with m = [sqrt(T)] + 1, find residues r(n) = (n*m)^2 mod T, from n=1, to some relatively small number, like 1000 or 10000, and take the mean of r(n).

For approximately 50% of the cases the non-square quadratic residues for p_1 and p_2 will match, and for those case r(n) will tend to be LESS than p_2.

For the other approximately 50% of the cases the non-square quadratic residues for p_1 and p_2 will not match—or "bump" as I like to say—and for those cases r(n) will tend to be GREATER than p_2.

So the mean should approximately give p_2 itself!!!

What a remarkably simple idea for finding approximately what the value of p_2 is for large prime factors!!!

Anyone heard of this before? I've been brainstorming it out for, what, I think about, 16 hours now.

I'm curious about whether or not it's new.

Anyone wonder why it would be useful for factoring if you approximately new p_2? After all, you need to know it exactly, right?

Ok, short answer is that then you can just use Newton-Raphson's method with an appropriate multiplier.


JSH: The 'why' of the behavior

With yet another "pure math" result that tests the boundaries of incredulity when people insist I'm just an insane crackpot with no research of value, I like the demonstration from the reactions to tell those of you who are honest what is really going on: certain people decided that no matter what I discovered they would reject my research and its value.

It's a classed reaction. It is all about class rules and a real "glass ceiling" as they say.

To understand classed behavior you need to go back to where class ruled: feudal Europe.

Once nobility was founded, people in the noble classes often were LESS than any idea of great in their behaviors and skillsets. You can say they were walking contradictions—nobles that were far less than noble.

So social rules were put in place where class was just by birth, and not about ability.

It was a social institution and wasn't expected to say anything really about the actual worth or abilities of a person in some abstract sense, but only about the value that SOCIETY put on that person.

In a classed society, the king was the king. You might be smarter than the king. But the king owned the land. You might be faster than the king, better looking than the king, and all around a much more likeable person than the king.

But the king was the king, and the king ruled.

So in modern classed societies like the academic world, the professor is the professor.

You may be smarter than the professor. Your research may be better. The professor's may be wrong!!!

But the professor is the professor as our modern academic system is a anachronistic throwback to medieval Europe where it was created.

So yeah, I have great research. I've even proven major errors with the work done by the pseudo-nobility of the professorial mathematical class, but their class is socially supported, and like classes in feudal Europe they maintain a glass ceiling.

They are incompetent, but they rule—because society gives them the power to do so.


JSH: Probing composites with quadratic residues idea

I've been thinking still about factoring as my other "pure math" research is as useless as ever in breaking the impasse—mathematicians can choose to ignore any "pure math" result—so I'm thinking more about quadratic residues and how primes will "bump" each other.

Given T a target composite that is the product of two odd primes p_1 and p_2, so T=p_1*p_2, I'm considering

m^2 - T = r

where m^2>T, with m near sqrt(T), but I'm interested in the smaller prime, as there is only roughly a 50% chance that r will be a non-square quadratic residue for BOTH primes, so the math has an out to which I will return.

So consider about 1000 checks, so I'll let m be a function:

m(n)^2 - T = r(n)

where I want it linear, so you'd say, take floor(sqrt(T)) + 1, for m(0), and m(1) would be m(0) + 1, and m(2) = m(1) + 1 and so forth.

For approximately 50% of the cases, p_1 and p_2 will have r itself as a quadratic residue.

In EACH of those cases r < p_1 and r < p_2.

For approximately the other half of cases p_1 will have r itself as a quadratic residue, but for p_2, r mod p_2 will be a perfect square, as the math takes the out.

And in EACH of those cases r < p_1 and r > p_2.

So with your 1000 checks roughly 1/2 of your values for r(n) should be above p_2 and roughly 1/2 should be below.

So you find p_2 in the middle. That is, p_2 is given approximately by the mean, that is r(n)/1000.

With 10000 checks you should get a sharper view of p_2, in what is kind of like a pixelated approach to factoring.

Just an idea as I grasp for straws here. I'm beginning to hate "pure math" as it's so much easier for me to do, while for a practical result, the only thing big enough is factoring. And I'm looking for my third solution to the factoring problem as the other two are just too hard for people to grasp for some reason.

But this one should be easy.


JSH: Too simple factoring idea, quadratic residues

I've been playing around with Pell's Equation still but off and on still thinking about factoring after my MASSIVE failure with my general solution to Pell's Equation which didn't factor like I really, really, wanted it to, and I realized something simple:

Given a target composite T, and an integer m, such that m^2 > T:

with m^2 - T = r,

the residue r must be a quadratic residue modulo EACH prime factor of T, which is just so trivial that I know this quick factoring idea must be known, but here goes anyway.

Well each prime has (p-1)/2 quadratic residues, so the probability that two primes have exactly the same non-square r as a quadratic residue is roughly, 50%. So there is roughly a 50% chance that they do not. I say they bump each other.

So then what? Well the math has no choice but to give them a shared perfect square residue—as both DO have that one—and then you can factor.

For example, T = 35 = 5(7). Note that 36 - 35 = 1, and 6+1 = 7, and 6 - 1 = 5.

The two primes bump because 5 has no non-square quadratic residues in common with 7, so the math is forced to use a perfect square.

Here's an example where two factors do share one: T = 7(17) = 119, as they both have 2 as a non-square quadratic residue, and yup, 121 - 119 = 2, as the math uses that shared one and there isn't any prime fighting going on.

But hey, what are the odds of THREE primes all in agreement? Let's try tossing in 3, and I have:

19^2 - 3(5)(7) = 4

so yeah, they bumped into each other!

And notice (19 - 2) = 17, and (19+2) = 21, and I pulled out a non-trivial factorization by playing probabilities.

(Ok, I know that one is weird as 3 doesn't have ANY non-square residues!!! So what gives?)

Playing around with this idea a bit I noticed you seem to need a prime smaller than the prime factors of the target, or maybe this example is just a random case, as, um, I've only done one.

But it is kind of cool, and very obvious: non-square quadratic residues dominate, especially with bigger primes, and the probability of two primes having the same one is about 50%.

I DO assume that the RSA people check for this possibility?

If not you can factor an RSA public key with it, about 50% of the time.

Friday, March 27, 2009


JSH: Wacky quadratic residues result

Ok, so finding variants that will solve Pell's Equation, gave me a result I think is kind of wacky, which is that EVERY prime number must have 2, -1, or -2 as a quadratic residue.

Is that previously known? If so, what is the prior proof?

I've noted it's true from the result that for every solution to x^2 - Dy^2 = 1, where D is a prime number there must also exist a solution to exactly one variant as follows:

j^2 - Dk^2 = 2 where k = sqrt((x-1)/D) and j = sqrt(x+1)


j^2 - Dk^2 = -2 where k = sqrt((x+1)/D) and j = sqrt(x-1)


j^2 - Dk^2 = - 1 where k = sqrt((x+1)/2D) and j = sqrt((x-1)/2)

And that means that for at least one of those cases when D is a prime, D must have 2, -2 or -1 as a quadratic residue.

Interestingly, if D is a composite and NONE of those variants are true, then a solution to Pell's Equation will non-trivially factor D.

I'm curious if anyone can find old research results which cover any of these areas.

Of special curiosity would be the method used to prove—if they exist in the literature—where my curiosity is mostly sparked by the 2, -2, or -1, quadratic residue result for primes.

Wednesday, March 25, 2009


JSH: Why you're fun

Years ago I likened mathematicians letting me run wild on newsgroups without the world realizing really what my discoveries actually are, is like letting Shaquille O'Neal play pick-up ball with neighborhood kids. My problem is, if you understand the truth, then no one wants to play with me.

And you don't understand because you've never had a major discoverer around. You just read about them.

So every one of you I'm sure has some idea in mind of what someone who is a major discoverer would do, but you're all wrong, because none of you are one, and all you know of previous ones is what you read in history books.

I like to delete posts as I've often wondered what we really know about people in those history books, like, what if they deliberately screw with things?

So I delete original papers, toss away things, and make things up, as I'm curious later when I read what historians say, how much will they get wrong? And then maybe I'll have an idea of how much was wrong that I thought was true about, say, Archimedes, or Einstein.

What can you do if the major historical figure is screwing up the records on purpose?

History will reflect on me.

It is something I ponder a lot as I think about each of you, especially the ones who seems so happy at times ripping on me, as if it matters. How happy they are at the mistakes. How serious they seem in believing they're accomplishing something by arguing with me.

I'm the guy no one would play with if I didn't let them win, sometimes.

This latest Pell's Equation result fascinates me, not because of the math, which IS beautiful, but because I wonder about the reaction. I eagerly await your rationalizations, and ponder how some of you will keep going, keep trying, keep thinking you're doing real mathematics, as you cling ever more tightly to the status quo.

I sent a paper with my blog post on the subject to Lenstra. Do you know why?

Because Pell's Equation was one of his pet areas. I cc'd Mazur and Ribet. No replies.

Of course. No replies.

I am by myself.

If you allow yourselves to understand how history must see me, then you will not play with me.

To history, I am the only mathematician of this age, even though I say I am not a mathematician.

To history, I am an impossible figure. A person that is unbelievable.

But today I am a crackpot mouthing off on a newsgroup to people who disrespect all that I've accomplished, and that is more than enough, as then at least, I don't have to sit quietly and be nice, and wonder, and look at people who can't really see me, and wonder, about what they think, as we all wonder, how can this be.

I know I'm here, but I don't know how I'm here, or why I'm so far beyond anyone else.

I just know that history will ponder, and be all the angrier because I know I'll fight it. I destroy information. I shred papers. Lie about details. Make things up, and in general force them to think…what do we really know about him?

I don't even know any more about me.


JSH: Understanding the 'why' of Pell's Equation

With my factoring hopes using Pell's Equation having taken a huge battering as I finally simplified my general solution, I'm now looking for a silver lining, and feel like I've found it with the 'why' of Pell's Equation revealed from the simplified equations.

The simplified general solution to Pell's Equation:

In rationals, x^2 - Dy^2 = 1

you can solve for x and y explicitly in terms of the integer factors of D-1, and a rational independent variable v, with:

y = 2[f_2*v - 1]/[D - (f_2*v - 1)^2]


x = [D + (f_2*v - 1)^2]/[D - (f_2*v - 1)^2]

where f_1*f_2 = D-1 where v is non-zero.

The v's and factor of D-1 may seem redundant, but now it's possible to consider when Pell's Equation is an integer, and since v can be a fraction, let v = n/m, then:

n = (sqrt[D(B-1)/(B+1)]+1)m/f_2

Where B is some integer. Intriguingly it turns out that x = B, for a solution to Pell's Equation.

And that relation tells the 'why' of Pell's Equation, as notice from some examples:

For D=7, I have x=8, y=3, is a solution, as 8^2 - 7(9) = 1, so B=8, f_1*f_2 = 6.

n = (sqrt[7(7)/(9)]+1)m/f_2 = (7/3 + 1)m/f_2 = 10m/(3f_2).

Want something bigger?

For D = 29, x = 9801, and y = 1820 are solutions, so f_1*f_2 = 28, and B = 9801, so

n = (sqrt[29(9800)/(9802)]+1)m/f_2


n = (sqrt[29(8)(35^2)/((2)(29)(169)]+1)m/f_2

which is

n = (70/13+1)m/f_2 = 83m/(13f_2).

So you can see that for D=29, the math had to find a number that achieved two base conditions, for some integers j and k:

x = 2Dk^2 - 1, and x = 2j^2 + 1

Another case with primes you will see is:

x = Dk^2 - 1, and x = j^2 + 1

And intriguingly, those two cases mean that:

j^2 - Dk^2 = -1

for the latter case, while for the former:

j^2 - Dk^2 = - 2.

So now you know why the so-called negative Pell's Equation is not always true. If it's not true then the second equation with -2, is true.

And now you know trivially how to find a solution to the negative Pell's Equation when it exists, from a solution to Pell's Equation.

For over a thousand years people have not known why a particular solution worked, even though they knew how to find them.

These equations don't necessarily help you find solutions to Pell's Equation, but they tell you why a particular solution works.

So, for the first time in thousands of years of human history working with the so-called Pell's Equation, people can finally know why a particular solution is required.


Pell's Equation simplified

I've proven that given:

x^2 - Dy^2 = 1

you can solve for x and y explicitly in terms of the integer factors of D-1, and a rational independent variable v, with:

y = 2[f_2*v - 1]/[D - (f_2*v - 1)^2]


x = [D + (f_2*v - 1)^2]/[D - (f_2*v - 1)^2]

where f_1*f_2 = D-1

where v is non-zero.

The equations may seem to some like just a curiosity but you can do figure out why Pell's Equation behaves as it does using them.

One result I just got from them is that if p+2 is a square, then if D = p, Pell's Equation has x = p+1 as a solution.

For instance, for D=7, x=8, works, as 8^2 - 7(9) = 1, and for D=23, x=24, works as 24^2 - 23^2(25) = 1.

The behavior of Pell's Equation is about factoring. Everything has to do with how D itself factors.

The equations are remarkably simple given the history around Pell's Equation.

Sunday, March 22, 2009


Updated general solution to Pell's Equation

A previous posting of mine giving a general solution to Pell's Equation had a sign error. Here is a corrected, and much simpler general solution.

In rationals, given

x^2 - Dy^2 = 1

I have proven:

y = 2[f_2*v - 1]/[(D-1) + 2f_2*v - f_2^2*v^2]


x = [f_2*v^2 - 2f_2*v + (D+1)]/[(D-1) + 2f_2*v - f_2^2*v^2]

where f_1*f_2 = D-1, and the f's are non-zero integer factors, while v is nonzero but is otherwise a free variable.

As Pell's Equation is normally considered in integers as a Diophantine equation note that you find rational v such that x and y are integers, which gives the 'why' of Pell's Equation. For instance, for D=2, f_1*f_2 = 1, so I have:

y = 2(v - 1)/(1 + 2v - v^2)


x = (v^2 - 2v + 3)/(1 + 2v - v^2)

and v = 2, gives y = 2, x = 3, and, of course,

9 - 2(4) = 1.

My apologies for the prior sign error.

Thursday, March 19, 2009


JSH: Now you can understand the parasites

The story of the AIG bonuses can give some of you perspective on the parasitic who have lived among you for so long, as before, they could have gotten away with it. I've talked before about the specifics of the way the human brain works which allowed them to control so many of you, so well, for so long so I won't go into that now. Best way to put it is, many of you were simply being loyal, to creatures who see that as weakness.

Their aim was to divide humanity into two groups: the rulers and the ruled.

They use various psychological tools and amass wealth with which they dominate further, while they will claim to any value you wish them to, but they only hold a parasitic view.

They do not want to earn a living. They believe they should be given everything: your life, your blood, your hopes, your dreams.

Their weakness is what you saw with AIG: they depend on you giving in to them. If you don't, then they fall apart easily enough.

The worst thing humanity ever said to itself was, we are free.

You are only free to the extent that you keep your eyes open. Many of you have been enslaved in mind for years now.

The world you are beginning to see around you is a brand new one, but don't think it gets easier.

Your world does have monsters. The problem is, the monsters have human faces.

Learn to look beneath the surface, and learn to be truly free.

Wednesday, March 18, 2009


JSH: These people are stupid

Hey it's way past time for niceness as I solved the factoring problem. I have a mathematical proof using rather basic algebra that shows that clearly, and I've put up test code showing a basic demonstration for a specialized case.

Yet math people put up headlines around the world for quantum computing when it factored 15.

They follow their own wacky rules!

And one major rule they have is to protect their system, block out outsiders, and maintain that major mathematical proof is only the province of people with math Ph.D's.

To them, it's simple: protect their turf.

To you, your actions on the Internet may be transparent. Your precious public key may be hacked in milliseconds by hostiles using the mathematics that the established mathematicians are trying to ignore.

Civilizations are destroyed over stupidity like is happening now.

The mathematicians do not wish to accept merit. They have a class system. In that class system they do not recognize people like me, so they refuse to recognize major research results from me.

ALL they can see is later having to clap or applaud or pretend to like my research. They have nightmares about the ignominy of seeing me get a math prize and they will not have it.

They are in their own little world.

You are in this one.

Why can't Iran use my mathematical research? Or North Korea? Why can't the US government for that matter? Or Russia? Or China?

What happens to you when everything you do on the Internet—EVERYTHING—is open and transparent because no one will accept that the system is broken so the people with the power are those who are simply using the mathematics anyway?

Mathematicians are changing your life here and now.

They refuse to follow rules. Re-think the evidence now. Remember I was published in a peer reviewed mathematical journal, with other very important research that just so happens to not have the potential of crashing the world economy.

They PULLED my paper after publication. The journal SHUT-DOWN. The hosting university, Cameron University, SCRUBBED ALL MENTION of the journal from its website though it had been around for ten years. These people are not who you think. And right now your life depends on getting a clue.

Google: SWJPAM

for the journal story.

The mathematicians are acting with one goal in mind: stop me from getting rewards for my research or getting elevation in their society.

That's it. They've made up their minds on social class issues.

You need to make up your mind, on the facts.

Tuesday, March 17, 2009


Factoring problem solution, proof of concept code

Ok, here's a posting of some code which provides the proof of concept for my mathematical proof which solves the factoring problem. The code factors numbers of the form 2^n + 1 because it factors D by factoring D-1, so if D = 2^n + 1, then it's trivial to factor D-1 = 2^n, and also trivial to loop through those factors. Further it is proven that ANY number of that form can be non-trivially factored in at most n+1 iterations! Which I think brings Fermat numbers into reach.

I was going to only post the code to comp.theory as it seems more appropriate but I've been making the case to the physics community that I have actually solved the factoring problem, so I'm including them as well, as with physics people, yeah, a demonstration can be very effective and I'm sure plenty of them can run Java.

The factoring problem IS solved. This code isn't meant to allow you to go factor anything but to demonstrate a very beautiful proof that relies on dual factorizations through Pell's Equation.

James Harris

import java.lang.Math;
import java.math.BigInteger;
import java.util.ArrayList;

//Code utilizes my surrogate factoring algorithm where the underlying equations rely on
//Pell's Equation and a dual factorization. Go to mymath.blogspot.com for more info.

//To use input a natural number less than 32, like 5, and the code will, for instance, factor 2^5+1

* The FermatNumber class implements an application that
* attempts to factor a number of the form 2^n + 1.
public class FermatNumber {

static int n;

public static void main(String[] args) {


n = Integer.parseInt(args[0]);
if (n>29){
System.out.println("Number too large. Input has to be a positive
integer less than 30.");

catch (NumberFormatException e){
System.out.println("Not a number.");

FermatNumber Factorer= new FermatNumber();



private long D, f_1, h, mult;
private ArrayList<Long> g_2 = new ArrayList<Long>();

private BigInteger BigD;

private double v, max_range, min_range;
private double temp;

private int max_iter;
private int trivial_factors=0;

public FermatNumber(){

private void process(){

long temp_long;

//D is of the form 2^n + 1
D = 1<<n;

if (checkForSquare(D)){
System.out.println("D is a perfect square.");

BigD = BigInteger.valueOf(D);

f_1 = 1;
mult = 1;

max_iter = n;

for (int i=0; i<max_iter; i++){

f_1 = mult;

//code for the integer v case
h = 1;
for (int j=0; j<i; j++){

temp_long = setIntRad(f_1, h);

if (temp_long>0){

setRange(f_1, temp_long);


h = h<<1;


//negative case
h = 1;
for (int j=0; j<i; j++){

temp_long = setIntRad(f_1, h);

if (temp_long>0){

setRange(f_1, temp_long);


h = h<<1;


//need to make f_1 positive again, may make neater later
f_1 = -f_1;

//Searching for the fractional v case
temp_long = setFracRad(f_1);
if (temp_long<0) break;

setRange(f_1, temp_long);

//now negative values for f_1

f_1 = -f_1;

setRange(f_1, temp_long);

mult = mult<<1;



if (!g_2.isEmpty()){


Object[] factors = g_2.toArray();

if (trivial_factors>=factors.length) System.out.println(D+" is
for (int i=0; i<factors.length; i++){

System.out.println("g_1="+(D/(Long)factors[i])+", "+"g_2="+(Long)


else System.out.println(D+" is prime.");
if (trivial_factors!=0) System.out.println("Trivial factorizations
found: "+trivial_factors);


private long r_c(double v_in){

temp = (D+1)*v*v - 2*f_1*v + f_1*f_1;

return (long)(2*temp);


private void setRange(long f_in, long other_in){

max_range = (f_in + Math.sqrt(other_in))/(D+1);

min_range = (f_in - Math.sqrt(other_in))/(D+1);


private boolean checkGCD(long in_value){

long check_value = BigD.gcd(BigInteger.valueOf(in_value)).longValue

if (check_value!=1){

Long add_value = new Long(check_value);

if (!g_2.contains(add_value)){



if (check_value==D) trivial_factors++;

return true;

return false;


private boolean checkForSquare(long D){

long temp_long = (long)Math.sqrt(D);
if (temp_long*temp_long == D) return true;
else return false;


private long setFracRad(long in_f_1){

return (f_1*f_1 - (f_1*f_1 - D/2)*(D+1));


private long setIntRad(long in_f_1, long in_h){

return (f_1*f_1 - (f_1*f_1 - in_h*in_h*D)*(D+1));


private void searchFrac_v(){

v = (double)((int)(2*min_range))/2;

if (v<max_range){



v = v+ 0.5;



private void searchInt_v(int in_count){

int c = 1;

for (int k=0; k<in_count; k++){

temp = min_range/c;

if (temp>-1 && temp<1) continue;

v = ((int)temp)*c;

if (v<max_range){


v = v + c;


c = c<<1;




[A reply to someone who wrote that James' algorithm didn't strike him as faster than trial division.]

You're just repeating other posters who say that over and over again as a "talking point".

The test code says what I've been saying: you CAN factor one number by instead factoring another.

That test code demonstrates factoring D=2^n+1, by looping through factors of 2^n, which is just frakking easy.

Now many of you pretend that you are about knowledge and learning while there are these "crackpots" who are never right, who just annoy you so much, but I say you are nasty, sloppy people who just hate so you spew on other people because you're nasty human beings who like doing nothing better.

For others, what does this result really mean?

Quite simply: you can factor a target composite D, by instead factoring D-1.

RSA encryption relies on having a public key that is supposed to be hard to factor because it is the product of two large primes, so D = p_1*p_2, with big primes.

My approach though allows you to take an alternate approach to factoring D, where rather than go at it head-on, which is what most modern approaches, like say the Number Field Sieve, do, it goes at it at an angle, so now you can factor D, by instead factoring D-1, so you factor p_1*p_2 - 1, and find p_1 and p_2 that way, so your public key can go down by someone factoring some other number.

It is an awesome concept, which I've researched for about 5 years now, and as a basic research concept it produced a hunt like no other, and yes, nasty human beings loved ripping on me about my failures.

But that's not the worst part: it's the fight against success that wraps us and the entire world up into this saga.

Real life isn't so much about high drama and grand things you can see from a mile away. It's not about being able to predict what the next big thing is, or knowing what to do or not to do, but it does punish for inconsistency.

IF the mathematical community were consistent even in its love of "pure math" then a solution using the world famous Pell's Equation, connecting a brilliant idea of finding an alternate approach to factoring would have rang bells, but the math community has NOT been consistent, so now, no one knows what will happen.

But what you can know for certain is: your life will change because of this result.

And that's true no matter who you are, no matter where you are on this planet.

Real life is what happens. It's not a movie, or some book. It's not what you think can happen as reality doesn't give a damn what you think.

If it did, then the world would really be frakking boring, now wouldn't it?

Sunday, March 15, 2009


JSH: So what happened?

Some of you may have noticed a very big discussion over the last few weeks where I've claimed to have solved the factoring problem and various posters have said I haven't and put up issues with my claims.

So here's a quick synopsis to get you up to speed on what happened:

1. I noticed that you can use Pell's Equation to do dual factorizations to attack factoring a composite target D by factoring D-1 instead:

In rationals, x^2 - Dy^2 = 1

requires that

(D-1)j^2 + (j+/-1)^2 = (x+y)^2

where j = ((x+Dy) -/+1)/D.

Ok, so I know you've seen that but the issue was how to actually get a key variable called v, where j=uv, and I had an explicit solution for x that looked like:

x = +/-(f_1 + f_2*v^2)/(f_1 - f_2*v^2 - 2v) - [+/-2Dv/(f_1 - f_2*v^2 - 2v) +/- 1 -/+(f_1 + f_2*v^2)/(f_1 - f_2*v^2 - 2v)]/(D-1)

where f_1*f_2 = D-1, and the f's are integers, while v is a free variable, which is a mess I didn't want to deal with, so I didn't try to simplify it, and just hoped others with math software would do so, but I talked about factoring D non-trivially with its numerator and denominator, as with functions I call r(v) and t(v), and


it is trivial that if they are integers and abs(r(v) - t(v)) < D, and abs(r(v) + t(v)) < D, then you non-trivially factor D.

And I'd proven the existence of integer solutions so I thought I was done.

But posters started claiming they couldn't work, and some put up:

r(v) + t(v) = 2Dv^2

which to me implied that v had to have a non-trivial factor in common with D before you could use it, but hey, that couldn't work as v is squared, so you'd always have a fraction! So I figured they were lying, and said so repeatedly as I didn't want to bother with that monster expression I had for x, and kept thinking that all those plus or minuses meant a lot of variations.

But I was wrong. I finally simplified things further and found a simple equation for x:

x = [(D+1)v^2 - 2f_1*v + f_1^2]/[f1(f_1 - f_2*v^2 - 2v)]

And lo and behold if adding the numerator to the denominator didn't give what I'd claimed had to be wrong!!!

Turned out the math was more subtle than I'd imagined and was slapping a freaking non-trivial factor of D on the numerator and denominator. That divides off easily enough—if you know what it is—but when you add them, you get 2Dv^2.

But NOT a big deal as instead of taking the gcd of r(v) + t(v) with D as I'd originally said, you take it with r(v). No worries, I thought.

But then I noticed that I could NOT get

r(v) = (D+1)v^2 - 2f_1*v + f_1^2


t(v) = f_1*(f_1 - f_2*v^2 - 2v)

to both be integers with a rational v. It was a what the f-moment.

Turns out the algebra had one more trick up its sleeve as both have been divided by 2.

They have to both be integers to force a non-trivial solution with abs(r(v)) < D—oh yeah, I found that as a primary condition—so I ended up needing to multiply both by 2, and that's it.

However, a month of me working through these issues, especially weeks of me refusing to simplify x left the door open to posters who spread confusion about the result! One funny part though was when they specifically claimed it couldn't factor 15. But I had an example of a factorization of 15, so I knew they were wrong.

And that's the story, of how there can have been so much debate about a solution to the factoring problem for weeks.

Synopsis: I didn't feel like simplifying a key equation. The algebra had some subtleties that opened the door to confusion if you DID simplify the damn thing. And even after that point you needed to figure out one more wacky thing as in the functions had been divided by 2, and then everything works just fine.

Factoring problem is solved. No more room for confusion about the result. All issues are resolved.


Coding opportunity with factoring problem solution with Fermat numbers

I've solved the factoring problem, which took a lot of effort. Now the new method that solves the factoring problem is rather easy in a lot of ways, but also it is almost perfect for factoring Fermat numbers, and can, if one exists, find the next Fermat prime.

That opens the door to a remarkable, earthshaking result for anyone willing to do a little code work.

The possibility here is in finding the next Fermat prime.

Here is the factoring algorithm:

Given an odd composite D, to be factored, you factor it by taking its gcd with a function I call r_c(v), where:

r_c(v) = 2((D+1)v^2 - 2f_1*v + f_1^2)

and f_1 is a factor of D-1, so with Fermat numbers, it is always just 1 or 2 raised to some natural number power, where you find rational v, such that v is in the following range:

(f_1 - sqrt(f_1^2 - [f_1^2 - D/2](D+1)))/(D+1) < v < (f_1 + sqrt(f_1^2 - [f_1^2 - D/2](D+1)))/(D+1)

and r_c(v) is an integer, where if that range gives an integer v, then you're done, as you just plug that into r_c(v), and you're guaranteed have a factorization of D, from the gcd with r_c(v).

Notice for rational range f_1 <= sqrt(D/2).

If no integers are available for v, not to worry! Turns out a rational solution MUST exist if there is rational range at all, and its denominator is, 2.

Yup. 2. So like 1/2 or -1/2 will probably pop up a lot.

There is a mathematical proof that the method I just gave you WILL give each non-trivial factorization of D, if any exist, so if none pop up, D is prime.

It is screamingly fast because notice, for D = 2^2048 + 1, you have 2049 iterations to give each possible f_1.

So this method is most optimal for Fermat numbers, and may be the only hope for finding another Fermat prime if one exists.

Even if one does not, you can get a record for the largest Fermat number factored, easily enough.


JSH: Factoring solution test run, Fermat primes

In case you missed it, I've found what I claim is a solution to the factoring problem involving factor D-1 in order to factor a target composite D, where I used Pell's Equation.

It was brought to my attention by Penny Hassett in a comment on my math blog that Fermat primes might be a good area to bring this new solution to bear, as factoring D-1 is then trivial.

The possibility here is in finding the next Fermat prime. Here is the factoring algorithm:

Given an odd composite D, to be factored, you factor it by taking its gcd with a function I call r_c(v), where:

r_c(v) = 2((D+1)v^2 - 2f_1*v + f_1^2)

and f_1 is a factor of D-1, so with Fermat numbers, it is always just 2 raised to some natural number power, where you find rational v, such that v is in the following range:

(f_1 - sqrt(f_1^2 - [f_1^2 - D/2](D+1)))/(D+1) < v < (f_1 + sqrt(f_1^2 - [f_1^2 - D/2](D+1)))/(D+1)

and r_c(v) is an integer, where if that range gives an integer v, then you're done, as you just plug that into r_c(v), and you're guaranteed have a factorization of D, from the gcd.

If no integers are available for v, not to worry! Turns out a rational solution MUST exist if there is rational range at all, and it's denominator is, 2.

Yup. 2. So like 1/2 or -1/2 will probably pop up a lot.

Ok, so one of you code jockeys can script that up and give it a whirl against Fermat numbers.

When you see that it works, break some records.

It will be screamingly fast. Like nothing you've ever seen. The records should be easily within reach, within minutes.


JSH: Remember the goal

My stated aim for years has been to end the academic system as it currently exists worldwide, and I need your help in making my case. And your failures in the face of mathematical proof represent that help.

I think our academic system is a medieval anachronism that is all about feudalism, and it fails all the time now, so it has to be revamped in a major way.

And what I'm learning more and more is that money makes the difference, so the simplest way to help the system is to remove the money.

Once academics know that being a professor provides no guarantees as tenure is destroyed, and learn that failing to be a researcher means you stop getting paid to do research, the non-functional ones among you will leave.

It's that simple.

Before this story ends, quite a few tenured academics will find they are no longer professors, as tenure is destroyed and I'll worry about the legal battles later. But don't assume if you are tenured in any academic field that you will remain in your position without results.

The white collar welfare system of our modern academic world is about to end.

THAT is one of my primary goals.

Saturday, March 14, 2009


JSH: Closing moments on first part of saga

The battle over the solution to the factoring problem is drawing to a close as I finally have the full algorithm and correct method, but I find it interesting that certain posters who have warred for so long against my research are still fighting the fight.

But the war is over on the example of D=15, where I see one attempt already at moving to obscure the result, which I answered.

The math wars have been about the very future of humanity.

Some would send us down a dark path based on a sense of the futility of life, and a belief that the weak must always be dominated by the strong.

Their attempt at a neo-feudal reality was defeated recently in the United States with the end of the "reign" of George W. Bush.

There darkest efforts, however, have been an attempt to arrest the development of the entire human race.

By keeping science and technology at a controllable level, they sought to place themselves as a permanent aristocracy which would rule over the human race.

Their defeat is having repercussions around our world, but for those of you who missed the war or its deeper implications, just remember, your future depended on the efforts of so many people you will never fully know, as the parasitic clan that sought to enslave our world didn't understand that if they succeeded, the human race would have gone extinct, soon thereafter.

The rules of this reality are more complicated than their primitive intellects allowed.

Our species must progress, or die.

Friday, March 13, 2009


JSH: Critical year

The fate of the human species revolves around this year. Sir Isaac Newton actually calculated that the end of the world was in 2010, while I know now popularly 2012 is supposedly the one. But what do you people know?

Consider that Sir Isaac was someone like me.

You people still have no clue what reality is, or how it works.

For some of you that will mean, if you are not very careful that THIS year is your last.

I still suggest for some of you that savings are unnecessary. You might wish to consider partying while there is still time left in your life.

Figure out what you enjoy most, and start appreciating what time you have left.

The world as you know it is about to come to an end.


Factoring problem solved, remaining issues resolved

For weeks I've been talking about having a proof that solves the factoring problem by using Pell's Equation and posters have argued with me. Well it turns out they were arguing over details of implementation, and ignoring the proof.

I've carefully figured out the answer to the implementation detail which they criticized:

r(v) = [(D+1)*v^2 - 2f_1*v + f_1^2]/2^k


t(v) = [f_1*(f_1 - f_2*v^2 - 2v)]/2^k

where D is the target composite to be factored, f_1 and f_2 are integer factors of D-1, where f_1*f_2 = D-1, and k is a natural number, so it cannot be non-zero, which is the critical element that was missing, which is necessary for implementation of the proof.

The proof notes that with r(v) and t(v) coprime to each other, and abs(r(v))<D, if you have integer r(v) for a rational v, then r(v) + t(v) or r(v) - t(v) MUST factor D non-trivially, and in fact, if D is prime, those conditions cannot be met, but the proof shows that if it is not, they can ALWAYS be met. Which is why the proof solves the factoring problem.

Now others may disagree saying that the proof should tell you how get r(v) and t(v), but it doesn't.

But I say that's like Einstein's theories don't tell you how to build a nuclear bomb.

Implementation is separate from theory.

The proof DOES give you the route to finding r(v) and t(v) by giving you the conditions they must meet, and as I've been enamored with the proof I've ignored appeals to give r(v) and t(v), and then gave them wrong repeatedly as I didn't bother to figure them out, but after enough arguing I finally decided to settle down and figure them out.

I know that a Patricia Shanahan keeps asking for pseudo-code, but in this case the answer is easy—once you have r(v) and t(v).

Finding them involves using whatever rules you know to get k above, which is just about making the expressions coprime with respect to 2, with the caveat that if there is a way to do that with k=0 that will NOT work, so k has to be a natural number, so it is nonzero.

What makes this story kind of sad is that now, yes, the factoring problem is solved, and yes, I had a mathematical proof all along.

If posters had focused on the proof then they'd have realized it HAD to be correct, and that arguments were over an implementation detail.

So if intelligence services listened to them, or their own analysts looked the argument over and agreed with their objection, then they failed the most basic test of mathematical understanding: believe the proof first.

If you believed in mathematical proof, then you knew there had to be an answer and could find it, as I did.

I did write the definition of mathematical proof. Just Google: define mathematical proof

The factoring problem is solved. Posters ignoring mathematical proof focused on one issue looking for me to be wrong, but they were focused on an implementation detail. Now with that detail removed ANYONE who wishes can factor, including factoring RSA public keys.

RSA encryption is, as I've said for a while now, obsolete.

[A reply to his own definitions of r and t.]

That's freaking STUPID. Of course if I just divide by 2^k, r(v) + t(v) will still have the same problem!!!

Ok. I realized that. Cursed myself a few times. Pondered my own stupidity and came up with the correct answer.

r(v) and t(v) share a non-trivial factor with D

So you can just take the gcd of r(v) with D, if you find abs(r(v)) < D, with the equation:

r(v) = [(D+1)*v^2 - 2f_1*v + f_1^2]

with a rational v that gives integer r(v).

The proof assumes coprime r(v) and t(v), but those are not necessary to reach.

So that's it. I really reached desperately at first which was embarrassing. The correct answer is easier, but why didn't I think of it first?

Wednesday, March 11, 2009


Factoring problem solution simplified

After weeks of debate on the mathematical proof showing I've solved the factoring problem using Pell's Equation, I've finally gone ahead and worked out the simple equations that result from it, which solve the factoring problem.

A critical equation that posters have bugged me about has only two variants, where one is:

r(v) = (D+1)*f_2*v^2 - 2(D-1)v + (D-1)*f_1

If D is the target to be factored, and f_1 and f_2 are integers where f_1*f_2 = D-1, I have proven that if rational v can be found such that r(v) is an integer and abs(r(v)) < D, then D MUST be non-trivially factored. The non-trivial factorization is then forced, as in a mathematical absolute!

Finding v is as simple as solving for r(v) < D, with the quadratic formula. That gives you v:

v < ((D-1) +/- sqrt((D-1)^2 - [(D-1)*f_1 - D][(D+1)*f_2]))/(D+1)

Using calculus it's easy to see the lowest bound for v is:

v >= (D-1)/[(D+1)*f_2]

To get the non-trivial factorization you just need one more equation called t(v), where:

t(v) = (D-1)*(f_1 - f_2*v^2 - 2v)

as non-trivial factors of D MUST be available from r(v) + t(v), or r(v) - t(v), by the gcd.

ALL of the arguing over this research is needless, pointless and dumb, as if I'm correct, I just gave you enough algebra to factor an RSA public key. Just make it D.

I really didn't want to just give the damn thing, but I hoped by now that the NSA would have contacted me, as the algebra is easy.

They have not. No one in the US Government or any other friendly or hostile or any government has contacted me.

All has been scarily and depressingly silent. Our intelligence services all failed.

I don't quite know why posters argued about this result as if it were so complicated, why others on the newsgroups let them, nor why no one just used the easy equations to factor.

The mathematical proof says they do work, though I have one more hedge as there is one more equation for r(v) as there are two variants, just in case the one above doesn't work.

Tuesday, March 10, 2009


JSH: r(v), s(v) and t(v)

I've proven that Pell's Equation can be generally solved in rationals. So given

x^2 - Dy^2 = 1

I have x and y as functions of an independent variable v, where further x = r(v)/t(v), and y = s(v)/t(v)

where with integer f_1 and f_2 where f_1*f_2 = D-1, I have two sets of equations:

r(v) = [((D-1)f_1 + (D-1)f_2*v^2) - [2Dv + (D-1)(f_1 - f_2*v^2 - 2v) - (f_1 + f_2*v^2)]]

s(v) = [2Dv + (D-1)(f_1 - f_2*v^2 - 2v) -(f_1 + f_2*v^2)]


t(v) = [(f_1 - f_2*v^2 - 2v)(D-1)].

And the second set:

r(v) = [-((D-1)f_1 + (D-1)f_2*v^2) - [-2Dv - (D-1)(f_1 - f_2*v^2 - 2v) + (f_1 + f_2*v^2)]]

s(v) = [-2Dv - (D-1)(f_1 - f_2*v^2 - 2v) +(f_1 + f_2*v^2)]


t(v) = [(f_1 - f_2*v^2 - 2v)(D-1)].

I have proven that if abs(r(v)) < D, and a rational v gives an integer r(v), then a non-trivial factorization of D must be found from

r(v) + t(v), or r(v) - t(v).

For several weeks a group of posters have mounted a rather effective campaign to hide the truth about these equations.

That truth is, they solve the factoring problem.

The posters may be part of a gang of international hackers, looking to lengthen the time until the truth is known, so they can infiltrate computer systems around the world, or, I could be completely wrong.

I don't know. The situation is beyond bizarre. But if it's the worst case, then the Internet has been thoroughly compromised by highly intelligent individuals who can break RSA encryption at will.


JSH: Competency problem

I've started looking at x and y more closely, including giving r(v) and t(v), where x = r(v)/t(v) on my math blog, and the more I look at them and consider the mathematical proof that underpins this research, my take on the postings I've seen is there is a massive competency problem from posters, as in lack of competency.

These equations CAN be kind of complicated to handle, to understand what to do, but for instance, there are two sets. The plus or minus is all in combination as in if you have + with one of them, then it's plus down the line, and the -/+ if it's negative it's negative for all of them.

THAT is not hard. Or you can simply derive the two sets from the underlying equations, no biggee.

Looking at r(v) + t(v) and r(v) - t(v), I'm not seeing anything that says these quadratics can't deliver,

Like here's an example of what I mean:

r(v) = [+/-((D-1)f_1 + (D-1)f^2*v^2) - [+/-2Dv +/- (D-1)(f_1 - f_2*v^2 - 2v) -/+(f_1 + f_2*v^2)]]

where f_1*f_2 = D-1.

All you need is abs(r(v)) < D, and you're guaranteed a non-trivial factorization if D is an odd composite.

Now the way the plus or minuses work you actually have just two equations:

r(v) = [((D-1)f_1 + (D-1)f^2*v^2) - [2Dv + (D-1)(f_1 - f_2*v^2 - 2v) - (f_1 + f_2*v^2)]]


r(v) = [-((D-1)f_1 + (D-1)f^2*v^2) - [-2Dv - (D-1)(f_1 - f_2*v^2 - 2v) + (f_1 + f_2*v^2)]]

and you need r(v) < D, so you can solve the quadratic for that, or if r(v) tends to be negative r(v) > - D.

That's it.

Now those are simpler so I can ask you yahoos if your position is that given r(v) above, is v always non-rational? Or is there never an integer v such that r(v) < D, or r(v) > -D, with those equations?

That's an easier question than the bigger one of whether or not I solved the factoring problem.

Now in my mind, many of you are just incompetent at mathematics, but you talk a lot on newsgroups.

So that's what you are: people who cannot really do math, but people who will despite that, talk a lot on newsgroups.

You're incompetent.

Saturday, March 07, 2009


JSH: So what's your explanation?

With a simple proof of a solution to the factoring problem there is just no more room for denial about the reality of people lying about my research, while one remaining question is, why?

x^2 - Dy^2 = 1

requires that

(D-1)j^2 + (j+/-1)^2 = (x+y)^2

where j = ((x+Dy) -/+1)/D.

And it's easy algebra to go from there and factor D with factors of D-1, but I've watched posters lie for days and posters CONTINUE lying about the simple reality here.

I used mathematical objects I call tautological spaces against Pell's Equation just to see what they might do, through what I call my Quadratic Diophantine Theorem.

Similar mathematical objects to what I used to prove Fermat's Last Theorem. But in this case I generally solved binary quadratic Diophantine equations, and part of the bonus of that accomplishment was a solution to the factoring problem.

Easy algebra.

Mathematical society to date has to my knowledge ignored the general solution to binary quadratic Diophantine equations and is currently trying to ignore the solution to the factoring problem.

I imagine researchers around the world still trying to squeeze out improvements with things like the Number Field Sieve—and probably relying on public funds in many cases—despite the problem being solved.

There is a willfulness in that I believe. A desire to be wrong as long as they maintain their flawed view of the mathematics.

At this time it's not clear what the full impact will be, but it is increasingly clear that mathematics as a discipline took a wrong turn over a hundred years ago. Current practitioners are quite simply, possibly, fakes.

The fate of the human race lies in correct mathematics. We're running on fumes mathematically now mostly using mathematics that is over one hundred years old, with mathematicians doing bogus research.

It seems impossible for greater advancement of science to occur without the correct mathematics. It is unclear how bad the impact is on things we think we know, but it may be huge.

Our reality itself may be very different than people imagine because there has been this massive error corrupting things.

Our physics might be a hundred years further than it is now, without the error. Question is, will we recover from this massive blow?

What will be the full impact on the survival potential of the human species?

Thursday, March 05, 2009


JSH: So how can they lie?

If any of you have looked over my solution to the factoring problem, you may find yourselves very conflicted as it's easy algebra with a rather basic concept. I just found a way to factor one number by connecting it to a factorization of that number minus one.

Once you see the equations, it's obvious that has been achieved.

Easy algebra.

But how can math people lie about it?

Because they HAVE lied about my research for years. They are simply following the same strategy as before, except this time I've found a problem where that strategy can have real world consequences.

So, shouldn't I just implement it?

Even if I had the skill and the time, and could get something together fast enough—before some critical breach of some security system—I'd have a major problem after, as math society would simply then accept the result, and claim they don't lie about anything else!!!

They'd lie about lying.

I have SEEN that behavior with my prime counting function, as after all, it does count primes correctly! They lie about lying and shift as needed when facts force their hand.

The situation here is nearly impossible. A major field got corrupted. Given the weight we give to major fields and experts in those fields it is nearly impossible to prove such a thing.

But you can look at a very clever and very correct solution to the factoring problem, and read through replies to me where math people are lying. And guess what? I found someone at RSA with an email address public and emailed her yesterday.

I AM working to try and prove this result exists!!!

The problem is the math people are fighting me.

My fear is that the impasse may be broken by something terrible happening.

And no, she hasn't emailed me back yet. I may have not even got past her spam filter. It is far more difficult than you imagine to inform people about even such a result as solving the factoring problem.

I've been trying now for over three weeks.

Wednesday, March 04, 2009


JSH: Weird feeling

For me it is wacky how this story has worked out and such a weird feeling to be commenting after my post of a fairly straightforward and to the point stepping through of a solution to the factoring problem.

First off, why was it necessary?

Well I've had major mathematical discoveries for years now and people in math society have lied about them. So I threatened to solve the factoring problem to find something they couldn't lie about, and got lucky and somehow did it after somewhere around 5 years of searching.

The lying I've seen has been so weird, and problematic.

It left me few options in considering how I should proceed in this case because I have in the past made efforts to work through say, the NSA. I have made contacts with or tried to make contacts with people in the intelligence community both here and abroad.

But scarily, I've also faced the fear that hey, maybe mathematicians had no limits, and like, the NSA is FULL of mathematicians.

What might they do if cornered?

STILL I've been talking about this solution in various ways as I worked through it for three weeks now, and I find it incredible that no intelligence services around the world picked up on it, despite my efforts.

Like using words and phrasings that I'd think would be picked up. I was so desperate I recently started naming Britain specifically and talking about disaster.

But I also hoped someone would step up, some other person out there in the world, who would understand the importance here, and make the contacts, and I've been waiting now for weeks for someone from some government agency to contact me.

None have.

With a situation completely beyond anything I expected, the best solution was to go forward. Later inquiries can find out why intelligence services around the world failed so badly. And inquiries can determine why people lied about this result. And deeper inquiries can determine why they lied about my prior research.

For me the weirdest thing about this situation has been that lying as it seems to me that no one who expected to ever get caught would put themselves in the position these people have—so they never expected to get caught.

And THAT is scary, as my mathematical finds are, well, as time goes forward that may be best left to others to fully explain, but at a minimum they are worth the world knowing, and properly acknowledging.

Time is of the essence people. Your delusions about what can happen are no match for reality.

Your imaginations are not big enough. I assure you that waiting and hoping no one will notice is not a strategy.

It is suicidal.


Dual factorization equations and factoring

The following post steps through the mathematical underpinnings of a remarkably simple factoring method which relies on dual factorizations. It steps off from the familiar Pell's Equation—with rationals—and through basic algebra proves a viable factoring technique.

In rationals starting with Pell's equation:

x^2 - Dy^2 = 1

I have proven that

(D-1)j^2 + (j+/-1)^2 = (x+y)^2

where j = ((x+Dy) -/+1)/D.

Here the +/- indicates that one variation will work so it is an OR and not an AND. Either plus OR minus will give a valid j.

It suffices to substitute out j, and simplify, which will give Pell's Equation again, showing the relations are valid.

Dual factorizations:

The utility of the expressions comes from dual factorizations:

(x-1)(x+1) = Dy^2

gives an opportunity to factor D, which I will declare to be an odd composite.


(D-1)j^2 = (x+y - (j+/-1))(x+y + (j+/-1))

gives the opportunity factor D-1.

Focusing on the second factorization, note to generalize I need additional variables.

Introducing u and v, where j = uv, I further need factors f_1 and f_2 where f_1*f_2 = D-1.

Then I have for generality

(x+y - (uv+/-1)) = u*f_1


(x+y + (uv+/-1)) = u*v^2*f_2

Note that f_1 and f_2 can be declared to be integers, and further that x and y uniquely determine u and v, as consider:

(x+y + (uv+/-1)) = (uv)*v*f_2

where I remind that j = ((x+Dy) -/+1)/D = uv, so given any x and y in rationals, I can simply separate off all non-unit non-zero integer factors in common with D-1, for f_1 in the first, and for f_2 in the second, leaving u and v as rationals, and trivially solved and proven to exist as rationals.

Therefore, it is proven that for every x and y that fulfill Pell's Equation there is a u and v pair.

Given that there must exist x and y, such that (x-1)(x+1) = Dy^2, non-trivially factors D, it is proven there exists a u and v pair unique to any such cases.

That completes the underlying mathematical proof showing the dual factorization and proving the existence of rational v, for a non-trivial factorization of a composite D.

It is now possible to move further into the utilities of these expressions and reveal a direct factoring method.

There are then three equation available:

(x+y - (uv+/-1) = u*f_1

(x+y + (uv+/-1) = u*v^2*f_2

((x+Dy) -/+1)/D = uv

Considering x and y to be unknowns it follows then that x and y can be determined as functions of v. Doing so gives:

y = [+/-2Dv/(f_1 - f_2*v^2 - 2v) +/- 1 -/+(f_1 + f_2*v^2)/(f_1 - f_2*v^2 - 2v)]/(D-1)


x = +/-(f_1 + f_2*v^2)/(f_1 - f_2*v^2 - 2v) - [+/-2Dv/(f_1 - f_2*v^2 - 2v) +/- 1 -/+(f_1 + f_2*v^2)/(f_1 - f_2*v^2 - 2v)]/(D-1)

Given that y and x are now expressed as functions of v, f_1 and f_2, it is now possible to elucidate a fairly simple factoring method. Said method requires the addition of new functions, so let x = r(v)/t(v), and y = s(v)/t(v), where r(v) is the numerator of the x function, s(v) is the numerator of the y function, and t(v) is the denominator.

Note that t(v) is the easiest given as:

t(v) = f_1 - f_2*v^2 - 2v

Then making the substitutions into the factorization of Pell's Equation: (x-1)(x+1)=Dy^2, and simplifying slightly, I now have

(r(v) - t(v))(r(v) + t(v)) = D(s(v))^2

Note that it suffices if r(v) and t(v) are integers for

abs(r(v) - t(v)) < D and abs(r(v) + t(v)) < D

to guarantee a non-trivial factorization. The absolute values may at first seem problematic, but note that if D is positive then the right side must be positive, forcing the two expressions to match by sign. So it is possible to simply add them, giving the result:

abs(2r(v)) < D

So remarkably, it is proven that rational v must be found such that r (v) is an integer and abs(r(v)) < D/2, and you are guaranteed to non-trivially factor D.

That completes the proof. The factoring problem is then, solved.

Comments or questions are appreciated.

Thank you for your time and attention.


JSH: Britain's choice

For what it's worth, I like Britain.

I am giving that region another option. Survival.

MI-6 needs to get a clue. Forget the NSA.

I hate being so blunt or so open but you have DAYS to react to this threat.

The NSA is not your friend here. It is worse than an enemy.

You have comfortable people who don't want to believe that their codes are broken.

Remember Hitler during World War II. Hubris has no boundaries.

The factoring problem is solved. Act accordingly.

Sunday, March 01, 2009


JSH: Of course they're lying

Posters have lied about my research for years. The question isn't whether or not they do so, but how do they get away with it? Why do any of you believe them?

And where is your sense of self-preservation?

The reason for me picking the factoring problem is that the lies cannot hold.

Currently the posting cabal that works against me is cornered on the issue of factoring 15, as they say these methods cannot do it, when of course they can.

But they're stuck. Decker, one of the leaders, put forward equations that cannot factor anything, claiming they result from this approach. ALL of the other posters have come in line with his original claims.

But I can then just show how it DOES factor 15, showing his position is false, and refute them all in one blow.

However, I know they will simply turn to some new lie, so I pause to put forward how remarkable this situation is!!!

Your group claims to care about mathematics, but a simple solution to the factoring problem has been found.

It's so trivially easy that a cabal dedicated to lying about my mathematics is now trapped in what should be a checkmated position over the issue of factoring 15.

However I know that they will simply lie when that is fully revealed and return to their previous behavior, so I'm pausing to put up this post to highlight that your group in fact displays an amazing contempt for mathematics.

You hate mathematics, no matter what you SAY.

The only thing you hate more is the truth.

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