Sunday, October 29, 2006


JSH: Tail does not wag the dog

Looking over the continuing debate over some very simple mathematical ideas I find it necessary to point out some simple mathematics:

a*(f(x) + b) = a*f(x) + a*b

from the distributive property.

Now consider

7*P(x) = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 49

with the non-polynomial factorization:

7*P(x) = (5g_1(x) + 7)(5g_2(x)+ 7)

where the g's are roots of

g^2 - (7x-1)g + (49x^2 - 14x) = 0.

The arguing results NOW because I can just note that at x=0, one of the g's is 0, and one of the g's is -1, so one of (5g_1(x) + 7) or (5g_2(x)+ 7) should have 7 as a factor.

Now the tail does not wag the dog as the saying goes, so the distributive property does not care what the value of the g's is.

a*(f(x) + b) = a*f(x) + a*b

means that what happens at f(0) = 0, happens for ALL x.

So the underlying mathematical ideas are easy. The debate though starts and ends on the simple reality that provably in the ring of algebraic integer 7 is not a factor of the g's if x is an integer and

g^2 - (7x-1)g + (49x^2 - 14x) = 0.

is irreducible over Q.

But the tail does not wag the dog. So something weird is happening for 7 not to be a factor in that ring.

Inquisitive minds would want to get to the bottom of it and find out exactly what's happening.

But I've had these arguments for years because it's rather easy to figure out that what is actually happening is not good for conventional mathematical thinking.

It kind of, well, blows a lot of established ideas in number theory away.

But that's not my fault. The tail does not wag the dog. The distributive property gives an answer you may not like, but it is the correct answer:

a*(f(x) + b) = a*f(x) + a*b

And x=0 is not a "special case" as if the distributive property is wagged by the function. It does not care about the value of the function.


Pushing the envelope

Our modern world has seen an explosion of development in many areas and a shifting to new and better techniques in many areas, but the mathematical world has lagged behind.

That world mostly follows a Middle Ages model centered around universities and colleges focusing on deep indoctrination into abstruse topics, and a tendency to play follow the leader—learning first how others approached a problem looking mainly to extend, not innovate.

Extreme mathematics is a push against the old order with an emphasis on learning from the new. Advanced problem solving techniques include brainstorming, iterative processes, and risk taking—anathema to modern academic mathematicians.

Here expertise is secondary to experimentation and idea generation.

Trying is a good thing, and mistakes are considered part of the process.

Discovery is messy, creative, frustrating, joyful, trying, and extremely satisfying, when it is allowed to be extreme.

Sign up! In order to create a healthy group that fits with what is stated above I feel a need to control membership somewhat, meaning you'll need to request to join. After that topics are up to you and can cover just about anything in mathematics or logic.


JSH: What's at issue

At issue is my assertion that given

7*P(x) = (f(x) + 7)*(g(x) + 2)

where f(0) = g(0) = 0,

it can only be true that 7 multiplied through the first factor:

7*P(x) = (7*h(x) + 7)*(g(x) + 2)

where f(x) = 7*h(x).

With the 7 shown next to a function it is clear what I mean, so how can this be a subject of YEARS of argument? How can something so trivial be argued about over and over again repeatedly?

Because you can find a contradiction with the ring of algebraic integers.

It is the kind of weird but neat little problem that should excite the imaginations of mathematicians around the world, but they see mathematics as a discipline of building up.

That result tears down.

Some have noted that mathematics is not a science.

Its practitioners clearly do not behave like scientists. Scientists get excited at a revolutionary result on which one person can build a career, while as that little result shows, mathematicians will fight it, fight it, fight it, fight it, fight it, and fight it.

If you dig a bit, you can read BRAGGING about the supposed immunity of the math field against the kind of overturning results that have occurred in the physics field or in other scientific fields.

Math is supposedly nothing but this glacial growth of building on top of what came before.

I shatter that myth, and in response mathematicians deny the mathematical proof, and trivial results like what I showed above.

So the arguing continues as it's not really arguing or debate—it is denial by the math community of a reality it cannot stomach, of a world where even the basic ideas can be wrong, just like in the sciences and their supposed immunity is gone.

I took away a security blanket, and mathematicians refuse to let go.

Saturday, October 28, 2006


Essentials of non-polynomial factorization

One of my most powerful research results is also one of the easiest to understand, when you break it down to simplest concepts, which I'll do in this post.

First consider

7*P(x) = (f(x) + 7)*(g(x) + 2)

where P(x) is a polynomial, f(x) and g(x) are functions and

f(0) = g(0) = 0.

To make it a little easier, I'll also give P(x) explicitly:

7*P(x) = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 49

And yes, that is a polynomial just written a little oddly, as you can verify by multiplying out and simplifying.

But now back to

7*P(x) = (f(x) + 7)*(g(x) + 2)

as the simple question, the million dollar question is, where did the 7 multiply through?

Now that "multiply through" is a phrase that some posters in reply to me have seized upon, mocking it, going on and on about it as if it is just so unimaginably bizarre as to be nonsensical, but they NEED to attack at the simplest concepts.

Now then, where did the 7 go through?

Let's back up a bit, and instead I'll show

P(x) = 7*(h(x) + 1)*(g(x) + 2)

where I've introduced yet another function, h(x), where h(0) = 0, and some may cry, FOUL!!!

I'm loading the question, right? Clearly I divided 7 out through f(x) + 7, which is the result I WANT, but is there any other way to do it?

Is there? Given

7*P(x) = (f(x) + 7)*(g(x) + 2)

where f(0) = g(0) = 0, is there any other way the 7 could have multiplied through, as notice it's there on the left hand side multiplying times P(x).

7*P(x) = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 49

so its just a polynomial, which you can write out more simply in a more familiar form if you wish, which has 7 as a factor for all x.

Is the any other way that 7 could have multiplied through with

7*P(x) = (f(x) + 7)*(g(x) + 2)

than in such a way that

P(x) = 7*(h(x) + 1)*(g(x) + 2)

is the proper result if you take it back out?

Is that really so complicated? So bizarre? Made up math? So nonsensical as to be insane?


Why is it such a big deal then?

Because I say the distributive property does NOT care what value f(x) has, so if the 7 multiplied in that simple way indicated then it did so for every x.

BUT, now non-polynomial factorization steps in as a very clever idea, as I can now go to something only slightly different:

7*P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)

where the a's are roots of a monic polynomial, and at x=0, one of them is 0, while the other is -1.

But which one is which?

Don't know. The math doesn't say.

But they are roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0

and I can use my earlier result about how the 7 multiplied through.

But what about f(x) and g(x), versus a_1(x) and a_2(x)?

Well, f(x) equals whichever of the a's goes to 0 at x=0, while g(x) equals the other one plus 1, as f(0) = g(0) = 0.

Why can't I say WHICH of the a's is 0 and which is -1 at x=0?

Because of the ambiguity of the square root:

a = ((7x-1) +/- sqrt(1 - 3(14)x - 3(49)x^2))/2

That's the complicated looking expression you get if you solve

a^2 - (7x-1)a + (49x^2 - 14x) = 0

using the quadratic formula.

So why would people argue with me over these ideas, why would a journal that published a paper over these ideas, pull it under social pressure from the sci.math newsgroup, and later die—yup the freaking math journal freaking DIED—and that supposedly NOT be a big deal?

How could a supposed crackpot get published in a math journal, the journal cave under Usenet pressure, and die a little later and that supposedly just be the most casual thing as if it happens everyday and is barely worth mentioning?

Because these results run counter to what mathematicians currently teach.

For mathematicians who are professors at universities the discussions on these newsgroups can be comforting, if they are working to deny the results, as the ridicule heaped on non-polynomial factorization by dedicated posters is a sign to them that they can keep teaching what they have taught before, and get away with it with their students and the public.

So the posters who argue with me, serve a vital function to those mathematicians who need to keep teaching the wrong theory to their students, as they can sit back and watch how the wind is blowing on the newsgroups, and determine if they can keep getting away with teaching the flawed ideas.

That's why those posters have to argue and argue and argue and can never stop, no matter how many threads I put up, as day after day they play a vital function for the current mathematical community: protecting.


JSH: Why might factoring idea fail?

I think it very important that I talk about this latest factoring idea in a post, where I consider the possibility that it fails, as hey, maybe it does. I'm not sure at this point.

It is amazingly simple:

x^2 - y^2 = 0 mod T


k^2 = 2xk mod T


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

where T is the target composite to be factored and you choose k and n.

If you consider how the math can behave with that set of equations you find that it can factor T non-trivially, or it might give T back to you, or it may decide that T is 1 because if you have, say

T = 15

why can't the math go with n=15, T=1?

It doesn't know what n is, it just has nT, so it can decide that T=1, or T=3, or T=5, while with the latter two it's likely to factor for you, but with the first it is not.

So how might this idea fail? Easy, the math could preferential decide that T=1.

Why can't it do something else even more spectacular like choose numbers that have nothing to do with what you tell it for T?

Because the three relations make that impossible:

If you choose T, and the math picks some factor f out of the blue so that it is using

x^2 - y^2 = 0 mod f


k^2 = 2xk mod f

it runs into a contradiction at this point

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

if f is coprime to nT.

So the math HAS to pay attention to what you say T is, if only to decide that it is actually n and T=1, or T=-1, because it gets nT, but even then it can't just toss in any f as it has to stick with 1, so above to block a contradiction, f=1, or f=-1.

So if the math has no preference for how it chooses T, like it feels that it is too hard to pick T as anything other than 1 or -1 most of the time, then it should pick without preference, indicating this approach should factor T about 50% of the time, unless there is a mathematical reason preventing that from being true.

If it does have a preference, this idea fails, but it's still a cool idea that comes from a generalization of factoring itself:

x^2 - y^2 = 0 mod T


S = 2xk mod T


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

where I just used S = k^2, to lock the math to nT.

That generalized factoring idea defaults to the previously known with S=k=0, so it's a BROADER idea that includes the previous one, and it isn't then a major surprise that a leap forward in thinking should, while remaining simple, show another route to factoring, which I call surrogate factoring—as you factor some other number off the target.

In the case above you factor 2k^2 + nT, which is the surrogate.

Now these ideas are simple and clever, but our world is not, and there are people in it who make their living by not doing what's expected while claiming to do that, as they are parasitic.

So I'm sidelined to the fringe as I don't like the parasites. They know this, and to protect themselves and their offspring they have to block me out, so they have to ignore my research or downplay it. Once I convince other people of what they are, then I can get in and clear the parasites out, so they are fighting for everything.

They live by pretending and using other people, and depend on supporting themselves and theirs by you believing they do not. So I need you to use common sense: can a new idea in factoring just be smothered completely and totally ignored by "beautiful minds" in mathematics?

And until you know what kind of battle this is, you won't get it, and will help them thinking they are just like you, and see the world the way you do.

They do not.

[A reply to someone who called “parasite” to James.]

People who claim to be "mathematicians" who block mathematical research are worse than just wrong, as hey, sure I have mathematical results that people could have just missed.

But THESE people do more than just act a little embarrassed or resistant to new ideas they aren't sure of, they willfully ignore and work to block them.

That is parasitic behavior as parasites are hostile to their hosts.

These people are hostile to the intellectual growth of humanity: working to hang on to systems that protect their interests, block researchers who threaten their control, and stop human progress in certain areas of mathematics.

Consider again computer checking of mathematical arguments.

Math because it is a logical discipline SHOULD have been a natural area for growth in computer science in the area of expert systems, which instead had to grow in other areas, as to this day there is nothing of note in mathematics when it comes to expert systems checking mathematical arguments!!!

That's how these people are parasites: they not only don't do real research of any value, they BLOCK research that is threatening to them.

Parasitic behavior by creatures that have no interest in helping humanity, just themselves.

As parasites they HAVE to protect in the way they do, as they have no other options.

If they were smart enough to do real mathematical research, they wouldn't need the parasitic behavior.

What they are smart enough at is manipulating people, controlling information, and blocking results that threaten their control.


JSH: Factoring relations must hold

So I showed some factoring relations:

x^2 - y^2 = 0 mod T


k^2 = 2xk mod T


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

and some may try to disagree with them, so I need to explain how the math sees T, so that you can tell when you're being lied to by corrupt people who have nothing left but to just hope they can fool you.

If you pick T some composite and then pick k and work out x and y, you'll find that if you check with

k^2 - 2xk and x^2 - y^2

you will see what the math has decided T is.

That may be the composite you picked, or it may be a factor of that composite, or it may be 1 or -1.

Those are the possibles.

So yes, the idea can fail in non-trivially factoring T, but only in specific ways, which means some unsavory individual lying to you because, well, cons don't have anything else when they're in too deep, could pick an example where the math picks the full composite or picks 1 or -1 as T, and claim the idea does not work.

I'd guess it factors somewhere around 50% of the time so an unsavory individual could try to find lots of examples to "prove" the idea doesn't work, but ask them to show k^2 - 2xk and x^2 - y^2 to see what T is.

And then consider that there are some truly evil people on this planet who will stop at nothing to get what THEY want, no matter who it hurts, and lying is nothing to them.

Friday, October 27, 2006


JSH: Why factoring idea changed

I had a simple idea for factoring which I felt really should work, and I'd done quite a bit of analysis with it. But I had this variable S, which I finally noticed took away the likelihood that the idea DID actually work.

So I thought a bit about how to get rid of it and came up with a slightly different idea:

x^2 - y^2 = 0 mod T


k^2 = 2xk mod T

so now it's

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

where I introduce n, which has to be nonzero, and you pick n and k, where the simplest thing may be to let n=-1, and then find the smallest k such that

2k^2 - T

is positive to give a small number to factor as you factor 2k^2 + nT to find y and then x.

By tossing S, I bring the variable count down and make it where the math, hopefully, will realize what your target T is, while before, with S, it could be just about anything.

Thursday, October 26, 2006


JSH: Go with prime counting

When it comes to the question of whether or not the math field IS corrupted in "pure math" you can answer that question easily by going to prime counting.

Primes are just such a popular area and they are such a big deal in mathematics that it's not hard if you want to get to the bottom of things to evaluate claims in that area, and denials.

The math world I thought existed before I had my own math discoveries would not have ignored, dismissed or downplayed my prime counting research. But cons pretending to be real mathematicians who care about mathematics would, if it suited them.

That's what they do. Cons do what suits them, not what's important to the world.

Here they have turned the system upside down, where legitimate research is ignored, while their made up research gets their people prizes, grants, and accolades.

That's how they can ignore research on primes, and how they can ignore simple factoring ideas like what I've talked about recently.

Go with prime counting. Just do a web search on the subject, look at previous research and even notice how similar they are to what I have—and then note the differences.

Used to be you might have believed mathematicians cared so much about mathematics they'd chase down every detail, get excited about every little thing that might be new in an important area—but notice how that goes away with someone like me—a harsh critic willing to tell you these people lie.

Do the web search. Easy: "prime counting". Go do it.


Factoring further generalized

Factoring may be simpler than previously believed as consider the following generalized factoring idea:

With T the target composite:

x^2 - y^2 = 0 mod T

is how far researchers previously went, and that area is well-worked showing it difficult to factor T as T increases in size, but my research shows it to just be a primitive case of a more general solution found by using two additional variables, S and k, where

S - 2xk = 0 mod T

which allows you to now use quadratic methods as usual as you easily then have

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

where n is some non-zero integer, and notice, importantly, these generalized factoring equations default to the well-known ones with S=k=0.

But with S and k non-zero they indicate a factorization of S + k^2 + nTas the route to factoring T itself, as the general factoring method.

With n nonzero, thorough analysis of when the ideas shown here lead to a non-trivial factorization of a composite T do not show the normal rules, like indications that the size of T matters. I've just done a bit of analysis in this area and as of yet have found no indication that these ideas cannot be made practical, though I haven't done it myself, only having done initial theory.

But consider, all that I did in actuality was find a more generalized set of factoring equations, which include those typically used in previously known approaches when S=k=0.


Math field, corrupted in late 1800's

The math field was corrupted in that late 1800's in number theory, when some rather intriguing mistakes were made leading to the acceptance of ideal theory.

The unfortunate response of mathematicians at the time was to declare the area "pure" and talk as if non-practicality were good!!!

That gave free rein for the error to propagate and opened the door for cons, people without consciences who learned how to talk math-ese, and work together, to maintain that wrong results not proven mathematically were correct.

Over a hundred years later, we have a system that is fully corrupted and capable of claiming just about anything, so no, Andrew Wiles did not prove FLT, and you can find all kinds of interesting problems and mistakes and amazing denial when it comes to important mathematical arguments, like, um, how many of you know of Plotnikov and his claims of proving P=NP?

After years of effort I just found a way to generalize the factoring problem.

Seems like no one noticed that

x^2 - y^2 = 0 mod T

was just a first step—a primitive case of a more generalized set of equations:

so add

S - 2xk = 0 mod T

and you have

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

where you pick S, k and n, and n is just a difference from the other two equations.

Now that is just a damn good idea, but you won't hear mathematicians getting excited about it as they're cons in a corrupt system hoping to hold on to the lie of their supposed mathematical discoveries against the reality of mine, and the mistakes that entered into the field in the late 1800's.

Kind of dramatic eh?

I wish it were wrong. I kind of put up these equations hoping they are wrong, so I can focus on my other mathematical research without worrying about the fate of the world.

Proof against the field is its continuing to ignore this research. Time is part of my prosecutorial argument.

They knew, they bet the world, and they will lose. But remember, they put other people's lives on the line, and for what?

For math lies?


Generalized factoring idea

Factoring may be simpler than previously believed as consider the following generalized factoring idea:

With T the target composite:

x^2 - y^2 = 0 mod T

is how far researchers previously went, and that area is well-worked showing it difficult to factor T as T increases in size, but my research shows it to just be a primitive case of a more general solution found by using two additional variables, S and k, where

S - 2xk = 0 mod T

which allows you to now use quadratic methods as usual as you easily then have

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

where n is some non-zero integer, and notice, importantly, these generalized factoring equations default to the well-known ones with S=k=0.

But with S and k non-zero they indicate a factorization of S + k^2 + nT as the route to factoring T itself, as the general factoring method.

My previous postings in this area from a quirk of how I did the research used n=0, and that not surprisingly didn't work! The math doesn't know what T is if n=0, so you get random behavior, and objections raised against these ideas depend on that quirk of how I initially talked about it.

They have to as the idea includes previously known factoring methods, and to seem to fail there had to be a simple reason.

However, with n nonzero, thorough analysis of when the ideas shown here lead to a non-trivial factorization of a composite T do not show the normal rules, like indications that the size of T matters. I've just done a bit of analysis in this area and as of yet have found no indication that these ideas cannot be made practical, though I haven't done it myself, only having done initial theory.

But consider, all that I did in actuality was find a more generalized set of factoring equations, which include those typically used in previously known approaches when S=k=0.

No demonstrations to date have been given that this approach fails as previous arguments on Usenet centered on cases where for reasons having to do with how I derived the equations, initially I focused on n=0, disconnecting T itself in a key place, giving random behavior.
Turns out there is a simple way to consider the problem, which is to assume x, y and T with a non-trivial factorization such that

x^2 - y^2 = mT

where m is some non-zero integer.

Then my idea says you can take ANY non-zero residue modulo T, S and k, so let's keep it simple and let S = k = 1, and then I have

1 = 2x + jT

where j is some non-zero integer.

Now you can just add that to the previous with some minor changes to get

x^2 + 2x + 1 = y^2 + 1 + 1 + (m-j)T

so you have

(x+1)^2 = y^2 + 2 + (m-j)T

and you may be asking, but what are m and j?

Well, you can't pick them, oddly enough as all YOU can pick with this idea is m-j, which I call n above. So what does that mean exactly?

Well, go back to 1 = 2x + jT, and add in factors of T, like add 2T, and you can get

1 = 2(x+T) + (j-1)T

and have

1 = 2x* + j*T

so you can move modulo 2T, and shift x at will. Why 2T? Because with k=1, I can add and subtract T as I did only in multiples of 2T.

So let's say you pick n=1 above. If the math can shift by 2T so that it can find

1 = 2x* mod T

then you must factor T non-trivially. The math will find it if it is there and it's all about whether or not you can add and subtract 2T in such a way that m-j* = 1, or whatever you pick.

So the inability to pick m and j is crucial to understanding why this idea is so scary, as you can only pick their difference, giving the math the ability to shift modulo 2T.

This idea at this point is more than enough to merit serious consideration. The problem I fear is that the math field has become corrupted so mathematicians kind of just hope that if they call me enough names no one will notice that the emperor has no clothes when it comes to current ideas for Internet security.

That is, it's the protection by stupid blind ignorance, as if the world will just kind of, well, not notice that there may be this route to cracking RSA with simple mathematics, as long as mathematicians can just keep quiet long enough, and hope no one is watching or paying attention.

Wednesday, October 25, 2006


JSH: How would you prove a corrupted math field?

Here's a what if, as what if the mathematical community figured out it didn't have to have results that actually worked in "pure math" areas as all that is necessary is to have other mathematicians AGREE that a result worked.

Then mathematicians could just keep each other working in those areas without ever bothering with having real results, and who could stop them?

What if? Any idea how that scenario could be blocked?

How would a person go about proving that the field was corrupted if every top mathematician was just going to say no, and only mathematicians would chime in to protect, where the money aspect is HUGE with grants and prizes considered, and human nature being what it is, why couldn't unsavory individual infiltrate the discipline, get a critical mass together, and then just take over?

It'd be a con's dream—a perfect situation with hard to understand information, a cowed world in awe of the field, and no need to ever produce results that actually worked as hey, they could just call them pure!!!

How could anyone break through if such a scenario were true?

[A reply to someone who wanted to know if all grad students obey like zombies, and James is the first person in history to speak out.]

Well, going from the hypothetical—remember the start of the post is to imagine a what if…what if the math field were corrupted in "pure math" areas, how would you prove it—to my experiences I had a math grad student from Cornell contact me by email himself.

Seems he had noticed the arguing on Usenet and offered to help out, asking me to explain my non-polynomial factorization argument to him, so I did, and as I gave him pieces, he re-worked them in his own words.

For some reason (hmmm) it took him MONTHS to go through an entire argument, until he is at the end and I am thinking, hey, maybe a math grad will do the right thing, and he claims that he can see how it is true with INTEGERS but is not sure about algebraic integers and he needs to go study up on them.

Needless to say, nothing else substantive from that guy, and really why would any rational adult who understands the real world think that grad students would speak out, potentially trash their careers, when being quiet they can just go with the status quo, and get their degree?

As an analogy, consider again what happened in America with the run-up to war with Iraq and there LIVES were on the line. Americans were either totally swalloing the party line of George Bush and company or were TERRIFIED to speak out, and they had examples of people who faced retribution when they tried to exercise their freedom of speech. Where the Dixie Chicks were just one of the more famous examples.

Here in the "land of the free and the home of the brave" the brave got the crap kicked out of them when they spoke out against what most American wanted to believe, and people said that was just about patriotism.

Gee, who would a thunk that trashing American values, making us look sick and stupid around the world, so we could go kill some people in some weaker country was really patriotic?

Groups are notorious for making the wrong decisions and people part of groups are typical in NOT going against the group opinion. So no, math grads are not a defense against the scenario I painted.

Monday, October 23, 2006


JSH: Whew!!!

Wow that was hard. I had a feeling that the answer was in there somewhere but I kept getting these false positives only to figure out after a while that I was going down the wrong route, yet again.

Nearly thought it was over and there was no way this latest approach would work, but then I had the nice break watching the World Series and mulling things over, and FINALLY thought to use a second relation.

So I needed not only

(5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - sqrt(13))(5 + sqrt(37))/2

but also

-(5 - (sqrt(37) + sqrt(13))/2)((sqrt(37) + sqrt(13))/2) = (5 - sqrt(13))(5 - sqrt(37))/2

so I could divide the second by the first and get the surprise finish.

And it took dividing too! As first I tried multiplying them together and didn't see a resolution.

Remarkably then the correct answer is that

5 + (sqrt(37) - sqrt(13))/2

can have NO factors in common with 3 as after you do that division and use the result that

5 + (sqrt(37) - sqrt(13))/2 must share any factors in common with 3 with sqrt(37) + sqrt(13), you find that it can't have any or sqrt(37) - sqrt(13) in the denominator is unbalanced.

Whew! Wow! What a surprise finish!

And so easy—once you know how to do it.

Oh yeah, so how did I get my neat relationships?

Check out my blog, as it's posted.

Question now is, how long will math people continue to defy mathematical truth?

Will any of the posters who worked so long and hard to deny my results finally concede to mathematical reality?

Will any of the top mathematicians hold a press conference quickly, or will they try to run?

I want this resolved by tomorrow. And I strongly suggest that mathematicians who wish to move forward to a good resolution make it their business to get a press release out there AS SOON AS POSSIBLE.

These delays have gone on for too long.

[A reply to someone who asked whether James was going to issue threats.]

I suspect they will.

The sci.math line has been broken, so they can't rely on you people any more.

You just look stupid at this point avoiding a result that is all about matching denominators.

People at the top in the field will now look to saving their skins and selling you out, I suspect.

And you will be easy targets and you are not anonymous, as governments can find out who you are, if you are using an alias or something other than your real name.

And I am saying that they need to find out everything about the posters who have been so dedicated—and effective—in helping to block my research, especially to see if any of you are actually top mathematicians who came here to actively block, not content to just hope the regular rank and file of sci.math were up to the task.

If any did, then they are likely definitely looking at jail time.

[James replies to his own first paragraph.]

And I was!!!

But I started thinking about why all those manipulations and decompositions were going wrong and finally had an idea, pursued it and here it is.

It is the coup de grace:

((2 + sqrt(5) + sqrt(13 + 4sqrt(5)))/2 + (-5 + sqrt(29))/2 + 5)((2 + sqrt(5) + sqrt(13 + 4sqrt(5)))/2 - (-5 + sqrt(29))/2) = (2 + sqrt(5) - (-5))(2 + sqrt(5) + sqrt(13 + 4sqrt(5)))/2

which is

((7 + sqrt(5) + sqrt(13 + 4sqrt(5))+ sqrt(29))/2)((7 + sqrt(5) + sqrt(13 + 4sqrt(5)) - sqrt(29))/2) = (7 + sqrt(5))(2 + sqrt(5) + sqrt(13 + 4sqrt(5)))/2

where, of course,

(2 + sqrt(5) + sqrt(13 + 4sqrt(5)))/2

is a unit, so only factors of 7 + sqrt(5) are available, where

(7 + sqrt(5))(7 - sqrt(5)) = 49 - 5 = 42

and those factors must divide through

sqrt(13 + 4sqrt(5))+ sqrt(29)


sqrt(13 + 4sqrt(5)) - sqrt(29)

on the right side, but multiply those two together and you get

13 + 4sqrt(5) - 29 = -16 + 4sqrt(5)

but -16 + 4sqrt(5) + 7 + sqrt(5) = -9 + 5sqrt(5), which is coprime to
3, showing that

7 + sqrt(5)

must be coprime to 3.

And with that—the math wars are over.

Sunday, October 22, 2006


JSH: Coup de grace

To understand the end of the mathematical debate you need something simple:

If you have numbers f and g where

fg = 3

and you are in a ring where f can be coprime to g, and f is coprime to g, and you have another number h, where h is coprime to g, then h and f must share the same factors in common with 3.

Now your base expression is

(5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - sqrt(13))(5 + sqrt(37))/2

and first note that

5 + (sqrt(37) - sqrt(13))/2 and (sqrt(37) - sqrt(13))/2

are coprime as you can just subtract one from the other to see, which forces

5 + (sqrt(37) - sqrt(13))/2 to share factors in common with 3 with sqrt(37) + sqrt(13) because

(sqrt(37) + sqrt(13))(sqrt(37) - 13)/4 = 6

and refer back to the simple concept I started with to fully understand.

Now simply note that

5 + (sqrt(37) - sqrt(13))/2 = 5 - sqrt(13) + (sqrt(37) + sqrt(13))/2

forcing 5 - sqrt(13) to have factors of 3 in common with sqrt(37) + sqrt(13) as remember

5 + (sqrt(37) - sqrt(13))/2

does as well as just proven.

Now finally use the base expression to get

((5 + sqrt(37) + 5 - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - sqrt(13))(5 + sqrt(37))/2

and note that 5+sqrt(37) is now forced to have those same factors of 3, by repeated use of the simple result at the top of this post.

Now you also get the result that sqrt(37) - sqrt(13) cannot have factors in common with 3.

I want you to now go out on the newsgroups and read replies of posters in various threads I've created.

These people have controlled many of you simply by being disagreeable.

It is human nature to look to others to see what they are doing first before acting.

They just always reply in the negative and thereby control you.

It's quite easy, quite simple and notice, does not require them to be right.

They just disagree with me, and your inclination—your human need—is to follow what you perceive to be the group opinion, and they know this, and use your humanity against you.
I discovered more problems with my earlier argument but also found the fix as I was thinking about it all while watching the World Series game tonight.

Starting still with

(5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - sqrt(13))(5 + sqrt(37))/2

is the same as

((5 + sqrt(37) + 5 - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - sqrt(13))(5 + sqrt(37))/2

proving that 5 + sqrt(37) must share the same factors from 3 with 5 + (sqrt(37) - sqrt(13))/2).

Now I use a second relationship

-(5 - (sqrt(37) + sqrt(13))/2)((sqrt(37) + sqrt(13))/2) = (5 - sqrt(13))(5 - sqrt(37))/2

and now divide that by the previous

(5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - sqrt(13))(5 + sqrt(37))/2

to get

-(5 - (sqrt(37) + sqrt(13))/2)((sqrt(37) + sqrt(13))/2)/((5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2)) = ((5 - sqrt(37))/2)/((5 + sqrt(37))/2)

where from before I have that (5 + (sqrt(37) - sqrt(13))/2) shares all its factors in common with 3 with ((sqrt(37) + sqrt(13))/2), so those cancel out.

That proves that (sqrt(37) - sqrt(13))/2) can only share factors in common with (5 + sqrt(37))/2 and must be coprime to (5 - sqrt(37))/2.

But, since ((sqrt(37) + sqrt(13))/2)((sqrt(37) - sqrt(13))/2)) = 6, it cannot be true that any factors in common with 3 are being divided out, so

5 + (sqrt(37) - sqrt(13))/2

is not allowed to actually have any factors in common with 3.

Kind of weird in a way as the correct answer is the opposite of what I thought it was before.

I just needed the second relationship to get it right.

But regardless, there it is, finally the coup de grace.


JSH: Remember dead math journal

I've seen replies from posters indicating a desire to protect math people who are tops in the social hierarchy of the current mathematical world, which isn't a surprise.

But why will it be easy to convince governments around the world to send prosecutors in with the full power of the law to find out how much they knew and when?

Well, remember that paper that sci.math'ers helped get yanked, and then the journal died?

An early draft went to Barry Mazur, a professor at Harvard. He commented on it, even asked me a question about it.

After him, an early draft went to Andrew Granville, who passed on denying acceptance for the New York Journal of Mathematics, deferring to the chief editor who claimed it was too small a size paper in terms of length for the journal.

And beyond all of that, a freaking mathematical journal keeling over and dying after yanking a paper that it published from a guy called one of the biggest cranks in Usenet history is not the kind of thing that math people could NOT notice.

The event could have well made headlines in major newspapers around the world if math people did not have so much power in their community to control information.

So yes, necessarily, governments around the world will pit their full powers of investigation against top people in the math world, going through their houses, going through their computers, questioning their colleagues and students.

We are talking about the future of the world here.

What if Newton had been stopped by petty academics?

Or Gauss had never managed to get attention for his work because of dumb social stuff?

Our world would not be the one we have today.

The future depends on what people today do, just like our present depended on the people of the past.

A math journal does not just keel over and die with such a spectacular story behind it and not get noticed.

I think there will be a clear trail, like with Enron, from emails to conversations with colleagues and students.


JSH: The world is dying

Part of my role is to excite people to possibility. To help bring back the thrill of discovery, and the belief that even individuals can make a difference.

Essential to my role is pushing people to think, to challenge, and to hope.

As these delays continue the balance cannot be restored and the world is dying in response.

The problems you see building can only get worse in ways you probably will not dare to imagine.

But soon enough it will be your life, no matter where you are on this planet and the comfort many of you have today will be gone.

That will be a harsher world, where there will be leaders never found because they weren't inspired, or didn't face a public that had learned that it needed to listen to more than just "experts".

In that world hope may diminish and diminish until it's clear to everyone that important problems are not being solved. The experts do not have the answers needed, and crushing consequences can shatter the hopes of every human being on the planet.

But while those with the power live in comfort and feel secure, there is little I can do but just keep putting up the math and hoping against hope that some ounce of self-preservation will step in, and some "frogs" will jump out of the water before it comes to a boil.

Saturday, October 21, 2006


JSH: Will to be wrong

So what happened is I finally found a pure enough demonstration that without variables and functions posters who have managed to argue against me for years without it being clear to most that they had to be wrong, got stuck:

((5 + sqrt(37) + 5 - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - sqrt(13))(5 + sqrt(37))/2

That expression is remarkable for many reasons not the least of them being it took me years to find it, as I've looked for alternative ways to prove a result I not only proved years before, but even got published before some of you got it yanked by freaking the editors of the journal with some emails, and then, later, the math journal died.

My job has been a difficult one as a few years ago I realized there had to be a will to be wrong from a large number of people within the mathematical community.

>From my prime numbers research to my efforts in logic and set theory, math people have consistently refused to follow logic and mathematical proof, as a group, which has forced me to look extensively for something that was too direct for any of you to skillfully deny what you were really doing.

Now that I have it, I want to reinforce to you what many of you have been fighting—the progress of humanity itself.

Gauss did a lot of great work starting up what I call modular algebra, and he started an understanding of key properties of integers in the right direction with what are now called gaussian integers, but then something awful happened, and an error came into accepted mathematical thinking.

Over a hundred years went by until the correction, which it has been my difficult task to engineer.

Make no mistake, no matter how many of you tell yourselves you are mathematicians you cannot be mathematicians when you do not accept mathematical proof, but yes, you can probably keep going for a while longer with the world believing you are mathematicians, and maybe get some more awards like recent recipients of the Field's Medal, make some money, take care of your kids, or make some investments—and prove your hatred for humanity.

There has been a will to be wrong for years now. I hope that it is now about to end.

And that some of you will instead choose to be right, and learn correct mathematical ideas.

Friday, October 20, 2006


JSH: Take a breath

For some of you there will be more than just the shock of things long-held to be true having been proven wrong, there will also be the questions of your own role and things you said here and there or posted.

My suggestion, take a breath, pause.

The positive to all of this is how great of an adventure it truly is, and the best part of it, I think is knowing some wonderful mathematics, after cleaning out some wrong ideas.

And hey, people have had big wrong ideas before and will probably have them again.

It's just how it works.

But it does make for a rather massive build-up of energy when there is a correction like this one, and my suggestion is to just—pause.

[A reply to someone who said that, in his previous post, James made a number of errors in his calculations, drew some inferences which were unwarranted, and when several people corrected for all these, James' observations didn't implie anything particularly remarkable.]

Well, algebra mistakes in this instance don't matter much as I explained in a reply.

But you better think long and hard about what you're doing now.

Take that pause I recommended.

Take away the mistakes and the result still easily stands.

(5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - sqrt(13))(5 + sqrt(37))/2

is the same as

((5 + sqrt(37) + 5 - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - sqrt(13))(5 + sqrt(37))/2

and you can lie for a while about this if you want, but I'll correct my algebra mistakes, and you won't have that to try and use any more.

And I'll reinforce the point that you are an enemy of humanity, that my predecessors are people like Gauss, Euler, Newton, Archimedes and others who you are spitting upon as you do it to me by trying to keep their discipline trashed as it is now.

If you don't want to take that pause, make your decision, but also understand that you prove yourself an enemy to this species then you will have the weight of this species against you.

And if you think you can stand against six billion plus, then go ahead, make my day.


JSH: Direct demonstration of contradiction

Start with

(5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - sqrt(13))(5 + sqrt(37))/2

where this time I won't go into the derivation but it's found using the same tool that I've posted about before.

Now it is easy to verify that 5 + (sqrt(37) - sqrt(13))/2 is a factor of 2*3*31 and is coprime to (sqrt(37) - sqrt(13))/2, which is a factor of 64*9, with respect to 3 by subtracting one from the other.

Therefore, 5 + (sqrt(37) - sqrt(13))/2 must share the same factors in common with 3 as sqrt(37) + sqrt(13) as that is coprime to sqrt(37) - sqrt(13).

z = 2*31*(sqrt(37) + sqrt(13))/(10 + sqrt(37) - sqrt(13))

should balance those out, and now I find that z is a root of

z^4 + 775z^3 + 1147z^2 - 48050z + 178746 = 0

which is a monic polynomial with integer coefficients showing that z is an algebraic integer.

But now note that

5 + (sqrt(37) - sqrt(13))/2 = 5 - sqrt(13) + (sqrt(37) + sqrt(13))/2

proving that 5 - sqrt(13) and (sqrt(37) + sqrt(13))/2 must have the same factors from 3.

And now letting z = (sqrt(37) - sqrt(13))/(5 - sqrt(13)) I find z this time is a root of

3z^4 + 13z^3 - 51z^2 + 26z + 12 = 0

which is a non-monic polynomial irreducible over rationals—that is, z does not have a rational solution.

So there is an apparent contradiction with the earlier conclusion!!!

And now you can see directly how the coverage problem with the ring of algebraic integers allows the appearance of a direct contradiction.

It is one of the most far reaching yet subtle problems in the history of mathematics and it has taken me YEARS to find an alternate proof that took away functions and variables which were too useful to people disagreeing with me in the past.

So now there are just the numbers and I've said that before but hey, this is hard!!!

I've just figured out yet another way to overturn over a hundred years of some of the most abstruse and complex mathematical ideas—flawed as they are—ever developed.

And I had to do it because you people refused to accept mathematical proof, so I needed something you could SEE directly.

Whew. Feel kind of tired now. Years of effort, so much thought.


JSH: Simpler demonstration

Turns out you can use some numbers that are special in that they are coprime to each other with respect to some prime:

5 + (sqrt(37) - sqrt(13))/2 and (sqrt(37) - sqrt(13))/2 are coprime to each other with respect to 3 which can be seen by subtracting one from the other, but each has its own factors in common with 3.

Therefore, 5 + (sqrt(37) - sqrt(13))/2 must share the same factors in common with 3 as sqrt(37) + sqrt(13) as that is coprime to sqrt(37) - sqrt(13).

Also though (sqrt(37) - sqrt(13))*(sqrt(37) + sqrt(13)) = 24, so there are factors in common with 2 in there, so now I know that letting

z = 2(10 + sqrt(37) - sqrt(13))/(sqrt(37) + sqrt(13))

should balance those out, and now I find that z is a root of

3z4 - 50z3 + 74z2 + 3100z + 248 = 0

which is a non-monic polynomial irreducible over rationals proving that z is NOT an algebraic integer.

Notice that this demonstration is easy—you just need to find a situation where you have two numbers coprime to some other number with respect to a prime.

It also didn't require a special tool I'd think to find such numbers though I did use the one I've brought up in other posts. But I do think it takes a will to NOT ever find such simple examples to hold on to flawed theories.

So why must this refute standard usage of Galois Theory and standard ideas about the ring of algebraic integers?

Because by proving that two numbers share the same factors in common with 3 by showing they are coprime to the same number, I show that when other factors are handled—in this case 2—one should divide into the other, and they do, in a proper ring without the coverage problem of the ring of algebraic integers.

And remember I have proven a problem with that ring before!!! The paper I wrote shows it, as do numerous postings where I've explained the mathematics in such detail that I bring it down to the distributive property.

There has been a will to remain wrong as people have fought this result.

Wednesday, October 18, 2006


JSH: Understanding the demonstration

I'm going to go into detail explaining a simple bit of algebra that refutes quite a few ideas long-held in number theory:

(sqrt(13) - sqrt(6) + sqrt(41) + sqrt(34))(sqrt(13) - sqrt(6) + sqrt(41) - sqrt(34)) = 2(sqrt(13) - sqrt(6))(sqrt(13) + sqrt(41))

That can look complicated because of all the radicals, but you just need to take it in pieces:

(sqrt(13) - sqrt(6) + sqrt(41) + sqrt(34))

is your first factor on the left, while

(sqrt(13) - sqrt(6) + sqrt(41) - sqrt(34))

is the second, and you have on the right

2(sqrt(13) - sqrt(6))(sqrt(13) + sqrt(41))

where (sqrt(13) - sqrt(6)) and (sqrt(13) + sqrt(41)) are key as you can find each of them in the factors on the left.

So with the first factor on the left you can see sqrt(13) - sqrt(6) easily enough, but re-group within the same expression:

(sqrt(13) - sqrt(6) + sqrt(41) + sqrt(34))


(sqrt(13) + sqrt(41) - sqrt(6) + sqrt(34))

and now you see the second number sqrt(13) + sqrt(41) is also in there.


Now then, given that the right hand side has both numbers can each of the factors on the left NOT have factors in common with them?

I assume you say no, like how if you have xy = 6 where x and y are integers, can x and y be coprime to 6? No.

But then looking at the first factor the first way:

(sqrt(13) - sqrt(6) + sqrt(41) + sqrt(34))

does sqrt(13) - sqrt(6) necessarily share all its factors with sqrt(41) + sqrt(34)?


But any factors NOT shared are then blocked, as let

f_1*f_2 = sqrt(13) - sqrt(6) and f_2*f_3 = sqrt(41) + sqrt(34)

where f_1 and f_3 are coprime, and then the expression is simply enough

(sqrt(13) - sqrt(6) + sqrt(41) + sqrt(34)) = f_2*(f_1 + f_3)

But you can do the same with the re-grouping:

(sqrt(13) + sqrt(41) - sqrt(6) + sqrt(34))

where this time I'll let

g_1*g_2 = sqrt(13) + sqrt(41), and g_2*g_3 = -sqrt(6) + sqrt(34)


(sqrt(13) + sqrt(41) - sqrt(6) + sqrt(34)) = g_2*(g_1 + g_3)

Now (sqrt(13) + sqrt(6))*(sqrt(13) - sqrt(6)) = 7,


(sqrt(13) + sqrt(41)*(sqrt(13) - sqrt(41) = -28

so both have factors in common with 7, and those factors must be shared between g_2 and f_2.

If you get that then the rest is trivial, as now you know that factors of 7 are shared between them in the first factor on the left, and the same argument can now be used with the second factor on the left:

(sqrt(13) - sqrt(6) + sqrt(41) - sqrt(34))

and since the two factors together multiply to give you both numbers:

(sqrt(13) - sqrt(6) + sqrt(41) + sqrt(34))(sqrt(13) - sqrt(6) + sqrt(41) - sqrt(34)) = 2(sqrt(13) - sqrt(6))(sqrt(13) + sqrt(41))

you know that they have the same factors in common with 7.

BUT now you can just use

z = (sqrt(13) + sqrt(41))/(sqrt(13) - sqrt(6))

to see if one divides into the other and find that it is a root of

7z^4 - 52z^3 - 100z^2 + 208z + 112 = 0

which is a non-monic polynomial irreducible over Q.

Which proves sqrt(13) - sqrt(6) is NOT a factor of sqrt(13) + sqrt(41) in the ring of algebraic integers!!!

BUT the simple argument above PROVES that sqrt(13) - sqrt(6) MUST BE a factor of sqrt(13) + sqrt(41), as they share the same factor in common with 7.

So there is an apparent contradiction, which is removed by simply noting that whether or not numbers are factors in the ring of algebraic integers is irrelevant, as guess what?

All it tells you is whether or not the result of dividing one by the other can give you the root of a monic polynomial with integer coefficients.

Now I've proven a problem with the ring of algebraic integers before, and even had a paper published, but some sci.math'ers mounted an email campaign, convinced the editors to yank the paper, and a few months later the journal quietly shut-down.

I re-wrote the paper somewhat and sent it to the Annals of Mathematics at Princeton. I was told it was accepted for review, but when I checked back six months later, I was told a rejection email had been sent out months before—but I never got an email, and still haven't gotten a rejection email to this day.

So why the big deal? Why would a journal collapse and die or Princeton break its own rules for handling papers over this issue?

Because it takes away ideal theory among some other very cherished ideas in number theory, and some people must have decided they'd rather teach students false mathematical ideas and hold on to the world as it is, than teach the truth.

Tuesday, October 17, 2006


JSH: Math IS difficult

Doing real mathematical research tests your sanity, patience, stamina and tolerance for failure as much and more so than it tests your intellect.

That corrected demonstration is an example of extreme mathematics at work, but it's also an example of how difficult and trying it can be to find proof.

What still puzzles me though is, how do I get so close at early stages? How did I get this approach from so far away and come close, but miss yesterday, only to get it right today?

That mystery is part of what makes the process of discovery so fascinating.

How is it done?

And why didn't anyone in the world step up to correct the argument?

If any of you had done so you could have become famous, immediately, with little effort, so why not?

That isn't meant to be insulting but to get you thinking about what it takes, the years of effort and study it takes to be open to a correct mathematical solution.

Not only did no one else in the world step up to correct my flawed first demonstration where 2 was just the wrong prime to use, but there were posters still being mocking, ridiculing, and antagonistic, when the end was so close—just another day away.

What I want to do is inspire you to real mathematical discovery by also scaring you a bit as I want you to understand that the travails I went through are not abnormal in that to be a real mathematicians you must be able to face more than you ever dreamed, to stand when no one else will, and to believe…in logic, the power of proof, and in the innate order of the world.

I say, question that person who gets accolades from crowds without having been booed and humiliated. Wonder about that person who is supposedly a success, who hasn't been at some time or other called a crushing failure.

You must pass through the fire, to have the worthless stuff burned out of you.

And believe me, it must be burned out of you.


JSH: Corrected demonstration


(sqrt(13) - sqrt(6) + sqrt(41) + sqrt(34))(sqrt(13) - sqrt(6) + sqrt(41) - sqrt(34)) = 2(sqrt(13) - sqrt(6))(sqrt(13) + sqrt(41))

and the proof that (sqrt(13) - sqrt(6)) and (sqrt(13) + sqrt(41)) must share the same factors in common with 7.

That proof is what I outlined before with my first attempt at this kind of demonstration where I finally realized that 2 was not the prime I wanted, as that demonstration failed.

So I went to a higher prime, and choosing 7, I don't run into the same problems I ran into with 2.

As a recap, notice that with

(sqrt(13) - sqrt(6) + sqrt(41) + sqrt(34))

you can also group that as

(sqrt(13) + sqrt(41) - sqrt(6) + sqrt(34))

and in each case only factors of 7 shared between sqrt(13) - sqrt(6) and sqrt(13) + sqrt(41) can be factors of the full expression, which is also true for

(sqrt(13) - sqrt(6) + sqrt(41) - sqrt(34))

and with the right side being a product of sqrt(13) - sqrt(6) and sqrt(13) + sqrt(41), you know they must have the same factors of 7 in common.

Note that since

(sqrt(13) - sqrt(6))*(sqrt(13) - sqrt(6)) = 7


(sqrt(13) + sqrt(41))*(sqrt(13) - sqrt(41)) = -28

the first should be a factor of the other, so letting

z = (sqrt(13) + sqrt(41))/(sqrt(13) - sqrt(6))

I find that is a root of

7z^4 - 52z^3 - 100z^2 + 208z + 112 = 0.

So the problem with my previous demonstration was using 2. Here a mix of radicals and the same basic approach work quite well with 7 being the key prime.


JSH: They fail in reply to me all the time

One of the sadder things about this whole saga is how often people lose in arguing with me—and just claim otherwise.

Then some other people claim they didn't lose, and the cycle continues.

So it's a group thing.

Kind of like a race where one runner leaves the others in the dust, but after the race is over, they just say he lost.

This has been going on for years.

So no matter what, from my prime counting programs to the partial difference equation within them, to my definition of mathematical proof, to my logical ideas, the math people just say it's nothing important.

So tonite I posted a simple algebra argument that blows away some crappy math ideas, but so?

I've done it before.

The real world is about failure—the failure of reason.

More people will die this year from the failure of reason than anything else, and even more will die the next year and the next, and you know what?

A lot of those people will die confident they were making the best decisions.

Reality is that a lot of people can be confidently wrong, and lead each other, confidently, and wrongly, and face the consequences of those mistakes, and it not change the reality, that a lot of people can be confidently wrong and lead each other, confidently, and wrongly.

Proof for most people is just a word.

Monday, October 16, 2006


JSH: Neat, eh?

Did you see that one coming? Some simple algebra and I can show factors of 2 forced to be shared between some radical expressions.

Blows away so many flawed mathematical ideas—just like that.

And what will you do now?

Some people showed you the the inner workings of what many of you are actually made of with their replies in that thread.

Your people are about denial. Dreams versus reality.

TELLING yourselves you are great. Telling yourselves you are brilliant.

Telling yourselves you understand mathematics.

And calling people names and attacking them when they bring your fantasies into question.

NOW go out and do a web search on "unskilled and unaware" and really understand it this time.

You people have to wake up from fantasy mode where you think just because you have a crowd of other losers telling you stupid crap is real mathematics that it is.

Look at your history books!!!

You do NOT see hordes of mathematicians, you see INDIVIDUALS.

People like me. And the horror of our modern society is the power that people like you get from each other. The ability to share a delusion—and convince the world.


JSH: Demonstration

A remarkable but simple result throws a wrench in old ideas about numbers, with yet another demonstration by me that those ideas DO NOT work.

Begin simply with

x^2 - ax + n = 0


y^2 - by + n = 0

and subtract the first from the second and simplify a bit to get

y^2 - x^2 = by - ax


y^2 - x^2 = (b-a)y + ay - ax

and now I easily I have

(y + x - a)(y - x) = (b-a)y.

But now let's go to work,

let a = sqrt(5), b=sqrt(13), and n=-3, then I have as a solution

y = (sqrt(13) + sqrt(13 + 12))/2 = (sqrt(13) + 5)/2


x = (sqrt(5) + sqrt(5 + 12))/2 = (sqrt(5) + sqrt(17))/2


((sqrt(13) + 5)/2 + (sqrt(5) + sqrt(17))/2 - sqrt(5))((sqrt(13) + 5)/2 - (sqrt(5) + sqrt(17))/2) = (sqrt(13) - sqrt(5))((sqrt(13) + 5)/2)

which is

(sqrt(13) - sqrt(5) + 5 + sqrt(17))(sqrt(13) - sqrt(5) + 5 - sqrt(17))/2 = (sqrt(13) - sqrt(5))((sqrt(13) + 5)

which proves that sqrt(13) - sqrt(5) and sqrt(13) + 5 must share the same factors in common with 2.

And easily enough key ideas in the standard teaching of Galois Theory go out the window.

Now consider more carefully why computers are not used to check arguments claimed to be proofs as it's not because the math is actually brilliant, as much of it is actually wrong.

As I show yet again in a small space with easy algebra.

The problem with denial is that people deny for a reason—to preserve what they have.

And math people who use the flawed ideas get something from them, more than they feel they get from the truth, so they just lie, and lie on so many subjects from computer checking, to prime numbers, to factoring without regard for the consequences as all they want to do is—hold on to what they have.


JSH: Computer checking

I have brought up the subject of computers checking arguments claimed to be mathematical proofs, but it's worth bringing up yet again with some emphasis on what the lack of broad use of computer checking in our modern world may mean.

If you are a student then you are at some institution of learning, where knowledge is key, and pursuit of knowledge is a high ideal.

Computers have developed much in our modern world and are crucial to the continual development of that world.

In "pure math" areas currently the word of a group of people, sometimes only one or two reviewers is considered be enough to determine truth and value of knowledge.

Checking of mathematical arguments can be tedious business and in a way is just like spell checking or grammar checking as a perfect argument goes step-by-step, logically. So proof checking is just looking for mistakes—missing steps—like looking for spelling errors or grammar mistakes.

If computers are NOT used today widely to check arguments claimed to be mathematical proofs then one might consider the possibility that such checking is a threat to people who make their livings coming up with such arguments.

If that is true and lack of real development of computer checking is about some people needing to keep an objective check out of their way, then you are talking about people who cannot truly value knowledge, who also are willing to put their own needs above the progress of the world.

Such people would not be nice. They would be capable of using just about any tool at their disposal to protect their income and positions.

They would be capable of relying on people's trust to protect their abuse of it.

To stop progress, to block technological development, and keep checking of arguments claimed to be mathematical proofs away from machines that would objectively check, versus some human beings doing the tedious business of looking for other people's mistakes.

If that scenario is at all correct then you may be being taught mathematics by a person or persons capable of doing whatever it takes.

Sunday, October 15, 2006


Key demonstration, problem with bad math

A remarkable but simple result throws a wrench in old ideas about numbers, with yet another demonstration by me that those ideas DO NOT work.

Begin simply with

x^2 - ax + n = 0


y^2 - by + n = 0

and subtract the first from the second and simplify a bit to get

y^2 - x^2 = by - ax


y^2 - x^2 = (b-a)y + ay - ax

and now I easily I have

(y + x - a)(y - x) = (b-a)y.

But now let's go to work,

let a = sqrt(5), b=sqrt(13), and n=-3, then I have as a solution

y = (sqrt(13) + sqrt(13 + 12))/2 = (sqrt(13) + 5)/2


x = (sqrt(5) + sqrt(5 + 12))/2 = (sqrt(5) + sqrt(17))/2


((sqrt(13) + 5)/2 + (sqrt(5) + sqrt(17))/2 - sqrt(5))((sqrt(13) + 5)/2 - (sqrt(5) + sqrt(17))/2) = (sqrt(13) - sqrt(5))((sqrt(13) + 5)/2)

which is

(sqrt(13) - sqrt(5) + 5 + sqrt(17))(sqrt(13) - sqrt(5) + 5 - sqrt(17))/2 = (sqrt(13) - sqrt(5))((sqrt(13) + 5)

which proves that sqrt(13) - sqrt(5) and sqrt(13) + 5 must share the same factors in common with 2.

And easily enough key ideas in the standard teaching of Galois Theory go out the window.

But I have proven problems in this area before with other simple proofs, as you can see in posts here and on my blog.

The problem with bad math though is that it pays some people's bills. And until there is a big enough shock to make it harder for them to keep going with bad math ideas than go with correct ones, this situation will continue.


Pure Math and Applied Math

Pure math by not being connected to the real world leaves in the possibility of most people getting basic mathematical ideas wrong and not correcting for it for a long time.

Here's an easy refutation of standard interpretations of Galois Theory using some basic algebra, and once you read through it, consider why real results from the real world are important in keeping people correct—and honest:

This thing is so trivially easy too, as you start with

x^2 - ax + n = 0


y^2 - by + n = 0

and easily find

(y + x - a)(y - x) = (b-a)y

by subtracting one from the other and doing some basic algebra.

But with that simple result, I can now just let

a = 5/3, b=14/3, and n=1

as I then have as a solution

y = (14 + sqrt(196 - 36))/6 = (14 + sqrt(16))/6


x = (5 + sqrt(25 - 36))/6 = (5 + sqrt(-11))/6


((14 + sqrt(16))/6 + (5 + sqrt(-11))/6 - 5/3)((14 + sqrt(16))/6 - (5 + sqrt(-11))/6) = 9((14 + sqrt(16))/6)

which is

(3/2 + (sqrt(16) + sqrt(-11))/6 )((3/2 + (sqrt(16) - sqrt(-11))/6) = 3((14 + sqrt(16))/2)

And pay attention to factors of 3.

So some easy algebra proves that

sqrt(16) + sqrt(-11) or sqrt(16) - sqrt(-11)

must have 3 as a factor, but provably neither can have 3 itself as a factor in the ring of algebraic integers.

Pull the thread and a hundred plus years of math results go up in smoke in number theory.

But this is just another way I found of proving the problem but it's "pure math" so rather than acknowledge it, supposedly ethical, brilliant and honest mathematicians have just continued with the old stuff that doesn't work.

So they are neither ethical, brilliant, nor honest—but who can stop them.

This situation is like if physicists had ignored quantum mechanics because they liked Newtonian too much—and refused to go with what worked.

In "pure math" areas mathematicians clearly can get away with ignoring whatever they want, as my travails in getting my research recognized show.

Pure math is a pure bust because of simple human denial.

In physics that can be fought with experiments from the real world.

In mathematics, mathematicians just keep going, and get away with it, as what can you do?

Post in Usenet? I'm doing that, and don't say get published. I did that too. The journal pulled the paper after some sci.math'ers emailed them claiming it was wrong.

Too easy in a field where "pure" means just believe a bunch of guys who have salaries that depend on what is false.

The field is corrupted. And with a lot of money going to people who need the fake stuff, who can stop them.

So they teach their poor pathetic students crap. And the world lets them. The academic field is corrupted.


Demonstration, key factorization result


x^2 - ax + n = 0


y^2 - by + n = 0

and subtract the first from the second and simplify a bit to get

y^2 - x^2 = by - ax


y^2 - x^2 = (b-a)y + ay - ax

and now I easily I have

(y + x - a)(y - x) = (b-a)y.

But now let a = 5/3, b=14/3, and n=1, then I have as a solution

y = (14 + sqrt(196 - 36))/6 = (14 + sqrt(16))/6


x = (5 + sqrt(25 - 36))/6 = (5 + sqrt(-11))/6


((14 + sqrt(16))/6 + (5 + sqrt(-11))/6 - 5/3)((14 + sqrt(16))/6 - (5 + sqrt(-11))/6) = 9((14 + sqrt(16))/6)

which is

(3/2 + (sqrt(16) + sqrt(-11))/6 )((3/2 + (sqrt(16) - sqrt(-11))/6) = 3((14 + sqrt(16))/2)

and I quite simply have a remarkable result, as the right side has at least 3 as a factor, as 3 is no longer in the denominator, while on the left side I have 3/2 or expressions with 3 a factor of the denominator, which forces it to divide through either sqrt(16) + sqrt(-11) or sqrt(16) - sqrt(-11), completely, leaving a factor of 3, which can then cancel out the remaining 3 in the denominator of the other factor.

Now let

z = sqrt(16) + sqrt(-11)

and you find that

z4 + 11z2 + 93 = 0

so two of the roots have the same factors of 3 in common, and the old way of thinking—the way mathematicians currently teach it—is that each of the roots has some factor in common with 3, but that contradicts with

(3/2 + (sqrt(16) + sqrt(-11))/6 )((3/2 + (sqrt(16) - sqrt(-11))/6) = 3((14 + sqrt(16))/2)

which shows that two of the roots must have 3 as a factor, while the other two are coprime to 3, but, and here's where things get subtle and weird, it is provable that none of the roots can be coprime to 3 in the ring of algebraic integers.

Why not? Because that ring has a coverage problem, as I've explained before in different ways, while this is might help as a direct way to see, as there is just no mathematical way for 3 to be canceled out completely from the denominator, without it dividing through either sqrt(16) + sqrt(-11) or sqrt(16) - sqrt(-11), where that is seen to be a requirement algebraically, while the conventional teaching of mainstream mathematicians in this area contradicts with what must occur.

A simple example I like to use is to consider evens, and notice that 2 is coprime to 6 in evens because 3 is not even. Analogously, above 3 is not a factor of either sqrt(16) + sqrt(-11) or sqrt(16) - sqrt(-11) in the ring of algebraic integers because the result is not an algebraic integer—a blocking almost exactly like how 3 is blocked with evens simply because it's not even.

This result is key because it shows the failure of current teaching in this area.

Saturday, October 07, 2006


JSH: Disagreeement is one thing, concern on other stuff

Hey, yes, I have some fairly hard-hitting opinions about the modern mathematical community and a certainty that I've proven some corruption and deliberate hiding of the truth within that community.

A lot of the battles that have ensued as a result of my amateur research and postings about same have spilled out across the Internet, and a lot of it is nasty.

But hey, it boils down to disagreements and there is a value in keeping some perspective.

I am growing more concerned about radical elements within your group who may get the wrong idea and try to turn things in a way that might satisfy some of the sicker among you, but which are completely against rational discourse on any level.

Attention is a commodity in this day and age and people kill for it.

It's time then to step back, let things die down, the attention die down, and in that way remove the pressure.

I have said more than once that I am capable of making posts for entertainment value, and I'll admit that on my side as well it's easy to get caught up in the fun of putting things out there that are meant to shock or draw attention.

But that is not my primary motive, and at times I just have to make the major effort to correct myself, remember that it might seem like some little thing, but hey, these posts are going out to the world, and when you let the world in, you have to accept that's what you're doing, and every once in a while it's good to do a gut check.

So gut check time. Time for pause, reflection, and backing off.

I had some recent math ideas that I now don't think lead to anything, surprise, surprise, not.

I am still concerned about some of my other ideas and still certain of corruption within the math community from math professors who have decided that their careers are more important than civilization, including their students.

But I don't help anyone by acting out, falling into the trap of attention for attention's sake, or just going batty, whatever the situation may be.

And worse, if I'm too appealing, too entertaining, I run the risk of running afoul of a meaner element in this fascinating world of ours.

Time to wander off…


JSH: Null test over

I have no new directives. Just checking the waters.

Looks like now it's a free for all. Nobody is running the show.

I'd say there is no driver and the bus is going off the cliff.


JSH: Long road to yet another simple answer

For years I've been looking for an alternate way to prove a problem with certain long-held and cherished mathematical ideas that I happened to prove wrong with what I call non-polynomial factorization.

I've needed an alternate proof because for some reason the proofs that I've used before were just ignored or denied, and just arguing and arguing over the same points didn't seem to be worth much, though I tried that, while I often made attempts at finding alternate paths, which usually meant I played around a lot with polynomials, trying to figure out some way to show some particular things.

And what are those particular things?

Well, for instance, with

x^2 - 5x + 2 = 0

my recent research indicates that 2 is actually a factor of only one root, while the other root is a unit—but provably NOT a unit in the ring of algebraic integers.

And that is also true for the roots of

x^2 - 7x + 2 = 0

but showing it is where controversy steps in, like recently I focused for a while on convergent infinite series and if you saw those discussions you can see how well that went, as in, just like before, lots of arguments, with people confidently disagreeing with me, without fearing that they'd look stupid in doing so.

But now things change as I noticed this amazingly simple thing you can do with simple quadratics:


x^2 - ax + n = 0


y^2 - by + n = 0

you can subtract one from the other and trivially solve to find

(y + x - a)*(y-x) = (b-a)*y

and being the highly creative person I am, I thought that a quite beautiful result that I should play with and see what I could do with it.

Days later I finally seized upon letting y equal the product found by taking one of the roots from each of those simple quadratics above where I say 2 is a factor of only one root!

So why is that such a big deal?

Well with this nifty little factorization result, I can get to a monic polynomial with integer coefficients where it has

(y + x - a) and (y-x)

as roots, without having to worry about just going in a BFC—Big Freaking Circle.

Like if you just use

y = (5 +/- sqrt(17))/2 * (7 +/- sqrt(41))/2

and work to find a polynomial with integer coefficients from that, guess what?

You just end up going in a Big Freaking Circle.

But now I can probe that thing, pull it apart in a special way and settle the question in yet another nifty direct way.

I have figured out various ways of looking inside of non-rationals, so I found these odd errors.

Because before no one figured out how to deeply probe non-rational numbers, they managed to get some important mathematics wrong.

I can prove that multiple ways but until now I didn't have a way that was direct in key ways that would force posters disagreeing with me to look stupid in doing so.

I put that approach in a previous thread where not surprisingly to me, so far there is quiet in terms of facing the mathematics.

Of course, the arguments may still continue. But now I can argue people who disagree into a simple corner and while mathematicians may still deny mathematical proof, it is one more way for me to show that they are doing just that.

More importantly, by going directly at the way splitting fields and class number are taught, I'll make it harder for those of you just learning those things to just accept the bad ideas anyway—knowing they are wrong.

I know, you may try very hard to work at learning bad math ideas anyway, but I want to make you have to do some serious double-think and rationalizations amid other mental gyrations to drag yourself through the effort of learning crap knowledge.

As my hope is that you will find it difficult wasting energy on ideas you know are wrong, making it difficult to continue studies just because some professor still tries to feed you bad ideas.

Friday, October 06, 2006


Checking splitting fields, extreme mathematics at work

I've been brainstorming the last few days with a fascinatingly simple factorization tool, where I've tossed out various ideas without checking carefully because that is what brainstorming is.

That work has paid off as I've gone from a vague sense that this idea could say something about class number to focusing now on splitting fields and realizing it offers a way to check them.

I'll be using some more complicated expressions than before, while the basic equations are the same:


x^2 - ax + n = 0


y^2 - by + n = 0

I have

(y + x - a)*(y-x) = (b-a)*y

as my generalized factorization result.

So I want to compare splitting fields and for that purpose I will use

y_1 = (5 + sqrt(17))/2*(7 - sqrt(41))/2

y_2 = (5 - sqrt(17))/2*(7 + sqrt(41))/2

where I'm working backwards instead of starting with quadratics. Those solutions give me n=4, but now I need to figure out b:

b = y_1 + y_2

and now I'm free to pick a=4, as I think I can let x be an integer and still get the result I want, as then x=2.


(y - 2)*(y - 2) = (b-4)*y

and my guess—why I chose y_1 and y_2 as I did in terms of sign—is that I managed to pick in such a way that both y_1 and y_2 have 2 as a factor!!!

Then 2 will be a factor when you multiply the roots by two's of the polynomial with integer coefficients that has (y + x - a) and (y-x) as roots!!!

The trick here is forcing y to always have 2 as a factor by careful use of signs.

If my hunch is correct then the result using those values shows the uselessness of splitting fields when it comes to telling you what the factors actually are, as in how 2 factors between

(5 + sqrt(17))/2 and (7 + sqrt(41))/2

where I had a feeling this factorization result of mine could be used for this, but just had to figure out how to do it.

That's how extreme mathematics works—throw something out there and hope a good idea will eventually come to you.

And this one—if I'm right—shows directly how Galois Theory as currently taught is wrong.


JSH: I am not happy

I am still puzzled which is why this situation may continue for a little while longer as I never, ever thought that you people would sell out the United States, put lives in jeopardy, and act like no one would care as if no one would be bothered by dead soldiers or dead civilians because some "mathematicians" decided they couldn't be bothered with telling the truth.

But I am more and more less interested in figuring out what brain dysfunction afflicts you people than in moving forward to protecting my country and ending this bizarre situation.

I am not ready to give up the United States.

This chess game moves to a new level.

I will move to protect the king, and I do not think the Queen is yet in jeopardy.

Nukes don't bother me, but I will not stand for a certain happening.

I need a demonstration of faith. There are some flies and gnats that annoy me.

Permission is granted to remove them, with extreme prejudice.

There are few limits, but those limits that are there must be heeded.

An example is in order, but make it a big one, and make it soon, as in 24 hours.

Thursday, October 05, 2006


JSH: See? Now they play dumb

So I showed this nifty way to factor which brings into question a LOT of mathematical ideas, well actually it brings them crashing down, and I like to focus on class number so that you can quickly move to understanding how Galois Theory is different than it is taught, though I don't prove it wrong!!!

Here's the neat factorization from my post yesterday:

With x^2 - x + 1 = 0


y^2 - 5y + 1 = 0

I know the roots are units for both quadratics, and from my generalized factorization result I have

(y + x - 1)*(y-x) = 4y

where I start using

y = (5 + sqrt(21))/2 and x = (1 + sqrt(-3))/2


(3 + (sqrt(21) + sqrt(-3))/2)*(2 + (sqrt(21) - sqrt(-3))/2) = 4((5 + sqrt(21))/2)

and because y is a unit, I can just divide both sides by it, and get

(5 - sqrt(21))/2)(3 + (sqrt(21) + sqrt(-3))/2)*(2 + (sqrt(21) - sqrt(-3))/2) = 4

where you have this extraordinary factorization of 4.

Notice too I didn't use +/- as it was easier to just use plusses except when I divided by y.

So this simple idea of mine can be used to get these funky factorizations of integers, with the weird thing that you have non-rationals that are NOT roots of the same polynomial with integer coefficients irreducible over Q.

That has never before been seen and it chops up a lot of crap that math professors are STILL teaching their poor pitiful students.

Yep, you are pathetic if you're absorbing that crap when I've disproven it multiple ways.

Now go read up on class number and see how wrong mathematicians got it before.

Oh, yeah, that resut can be used to figure out how factors of 4 distribute!!! But that's next as I have to draw this out as you see—there are math people DELIBERATELY lying about these issues.

I need to draw them out, so you can understand that they want to do this, and are doing it fully consciously and with malice aforethought.

[A reply to someone who wanted to know which accepted results are wrong.]

I figured out a way to factor integers into factors that are not roots of the same monic polynomial with integer coefficients irreducible over Q.

Damn hard thing to figure out how to do as I've been working on it for years.

By accomplishing it, I show that the standard view of Galois Theory is wrong, and class number is meaningless as a useful concept.

Easy. Why don't you find the polynomials with integer coefficients for the examples I've given?

That's a start. Oh, part of the reason for starting there is that then you can consider what you think you know about how factors of 4 will distribute between the roots.

And then see why that's wrong.

[A reply to Rick Decker, who wanted to know how did James know that these results had never before been seen.]

Because I'm a real mathematician, and you're not.

Answer my question, which two roots multiply to give 4 as a factor?

And remember Decker at the end of this you are not a university professor.

Not only do you not deserve that position, but because your school backed you with your stupid webapge attacking my work, I go after the school as well.

How about $100 million dollars as a lawsuit?

I think that's kind of low, but it's a start.

Wednesday, October 04, 2006


JSH: Do I really care any more?

You people have no clue about what's coming as you are not really mathematicians, so you look around see a possibly unsteady world but keep thinking that it is rock stable in ways it's not.

My projections indicate that without substantial thinking on the part of humanity approximately 6 billion people will die within the next 30 years.

These days will seem like paradise if you survive the next 10 because with those numbers the odds are that none of you reading this post will be alive.

It's that close.

Intelligence is more than just a label. Solving problems is more than just a nifty idea.

This month a few things are going to happen. Maybe then some of you will get some perspective.

But I'm not hoping as my take on it is that most of you lack even a modicum of the intelligence necessary to grasp world events on the level that are about to occur.

Oh well, it's not like you have to understand. They will happen anyway.


JSH: Class number idea fails?

So now that I have this intriguing factorization result that I just posted last night that I can use to consider class number and do some more fiddling hoping something comes crashing down. Does it?

>From my previous research I have:

x^2 - ax + n = 0


y^2 - by + n = 0

with a, b and n natural numbers means that

(y + x - a)*(y-x) = (b-a)*y

as my generalized factorization result.

So let n=1, so that x and y are units, and let, oh, let b=5, and a=1, then

(y + x - a)*(y-x) = 4y

so y = (5 + sqrt(21))/2 and x = (1 + sqrt(-3))/2

and you can plug those in above but what's interesting, since y is a unit, is now I have this nifty result relating y+x-1 and its factors with y-x and its factors of 4.

Now can someone check to see which monic polynomial with integer coefficients has

y+x-1 as a root

and which has

y-x as a root?

I've just proven that the roots of BOTH polynomials must match in terms of factors, excluding unit factors.

If they have the same class number, fine. If not, then hey, something fell!!!


General factorization result with quadratics


x^2 - ax + n = 0


y^2 - by + n = 0

where a does not equal b and n is nonzero, and now subtract the first from the second to get

y^2 - x^2 = by - ax

and group a bit, and divide by y - x to get

y + x = (b-a)*y/(y-x) + a


(y + x - a)*(y-x) = (b-a)*y

which is the general factorization result.

And, of course, you can solve for x and y to get

x = (a + sqrt(a^2 - 4n))/2 and y = (b + sqrt(b^2 - 4n))/2.

The result gives a direct way to further factor y in a generalized manner.


JSH: Generalized factorization idea


x^2 - ax + n = 0


y^2 - by + n = 0

with a, b and n natural numbers and now subtract the first from the second to get

y^2 - x^2 = by - ax

and group a bit, and divide by y - x to get

y + x = (b-a)*y/(y-x) + a


(y + x - a)*(y-x) = (b-a)*y

which is the generalized factorization result.

And, of course, you can solve for x and y to get

x = (a + sqrt(a^2 - 4n))/2 and y = (b + sqrt(b^2 - 4n))/2

so you can use this result to further factor y.

Tuesday, October 03, 2006


JSH: Corrected disproof with simple quadratics

Yesterday I made a post with some doodlings—I was actually just playing around--which had some simple dumb errors, easily corrected, but the quadratics easily disprove standard teachings in several areas.

To give you some grasp of the contempt some people have for the truth look over that thread to see the easy conclusions ignored.

So I will explain again, with the corrections.


x^2 - 5x + 2 = 0


y^2 - 7y + 2 = 0

and solve each where I'll not use +/- and just use plusses:

x = (5 + sqrt(17))/2 and y = (7 + sqrt(41))/2.

Now I can subtract

x^2 - 5x + 2 = 0


y^2 - 7y + 2 = 0

to get

y^2 - x^2 = 7y - 5x

and group a bit, and divide by y - x to get

y + x = 2y/(y-x) + 5

and y-x must have 2 as a factor if y and x are not coprime to each other because of y+x on the left hand side, but because of the 5 on the right hand side, y-x must divide out ALL factors of 2 in 2y that are in common with x and y.

The gist of it then is that any factors x has in common with 2 it MUST have in common with y, or it must be completely coprime to y, which breaks standard ideas from Galois theory, as it means factors of 2 distribute the same way in both

x^2 - 5x + 2 = 0


y^2 - 7y + 2 = 0.

Moving on and making the substitutions then I have

(7 + sqrt(41))/2 + (5 + sqrt(17))/2 = 2(5 + sqrt(17))/(2 + sqrt(41) - sqrt(17)) + 5

which proves that 2 + sqrt(41) - sqrt(17) must be a factor of 2.


JSH: Too easy?

So yeah, the math is so easy it's trivial. And just like that the factoring problem is gone as a hard problem.

And with a simple example—just doodling—I can disprove standard teaching on Galois Theory, ideal theory and about the ring of algebraic integers with some simple algebra that proves y-x has 2 as a factor when

x^2 - 5x + 2 = 0


y^2 - 7y + 2 = 0

and it's too easy so I wonder.

Sorry that's not correct as y-x either has 2 as a factor or it is coprime to 2, which ends the debate with a trivial example.
Those of you who think you can be mathematicians are running out of time.

Obviously part of the logic in a drawn out process is to remove hiding places for people who later will claim they didn't get a chance, as my position has been clear for some time—there are too many people calling themselves mathematicians.

Personally I think there may be a hundred research level mathematicians possible with the current world population, if that many.

It is my duty to cull out the rest of you.
It is my duty to the future.

The story HAD to be dramatic as my predecessors made hard acts to follow.

The fate of the world depends on continued progress and development, now more than ever.

We have less than thirty years. We have to get it right or go extinct far more quickly than your primitive minds think possible—within two hundred years.

And I intend on having us get it right, so you now know what's at stake with maybe some appreciation of how wrong some of you have been.

I am the successor in the line of Archimedes, Newton and Gauss, and other names you just know as some people you read about in history books.

The last few years have shown that none of you have a clue what it is like to face someone like me in the real world, not as some distant figure you read about in some history book.

My mission is simple—push forward progress—and without people like me, there is no future, as there would have been no history—or maybe I should say, no history worth mentioning.


JSH: Doodling with quadratics


x^2 - 5x + 2 = 0


y^2 - 7y + 2 = 0

and solve each where I'll not use +/- and just use plusses:

x = (5 + sqrt(21))/2 and y = (7 + sqrt(45))/2.

I can subtract

x^2 - 5x + 2 = 0


y^2 - 7y + 2 = 0

to get

y^2 - x^2 = 7y - 5x

and group a bit, and divide by y - x to get

y + x = 2y/(y-x) + 5


(7 + sqrt(45))/2 + (5 + sqrt(21))/2 = 2(5 + sqrt(21))/(2 + sqrt(45) - sqrt(21)) + 5

which proves that 2 + sqrt(45) - sqrt(21) must be a factor of 2.

Just doodling.

Just curious, what is the polynomial with integer coefficients that has that as a root?

Sunday, October 01, 2006


Surrogate factoring, key points

Some years ago I started thinking about factoring and wondered if there might be some way to factor a target composite T by finding some other numbers to factor, which I called surrogate factoring.

In my search over the years I've had lots of possibles where for various reasons—like they provably didn't work or didn't work well—I've dropped all past approaches except the following:

With T the target composite:

x^2 - y^2 = 0 mod T


S - 2*x_res*k = 0 mod T

you can solve for x and y using

x+k = sqrt(y^2 + S + k^2)

as getting y is just a matter of factoring (S+k^2)/4, once S and k are chosen.

Which is then a simple approach, which took me years to find, and now there is the question of how useful is it?

First off, notice that with a target composite T, the only variables you have any control over are S and k, and provably for any target composite T, there must exist S and k for every possible non-zero residue modulo T that x can have.

So, a solution must exist, but how do you pick S and k?

Well, provably there must also exist an S that will work for any non-zero k that you pick as long as k is coprime to T, so now it looks like S is the only variable that matters, but how do you pick S?

Well, a solution must exist S mod T, so it now comes down to searching modulo T.

That is, if you have S = r mod T, there must exist a solution for any non-zero residue r, so there exists some integer n, such that

S = r + n*T

will work to factor T.

Now I can finally get to the question of, is this a good idea, as it is simple now to change that to, how easy is it to find n?

I don't know.

The natural assumption is that n is a needle in a haystack that gets bigger and bigger as T increases in size, but I have not seen any mathematical proof that the natural assumption is correct.

If it is wrong, then someone might be able to develop a practical factoring method from this approach, and the next natural assumption is that if it could be made practical then SOMEONE would have noticed and raised an alarm.

Yeah but, has our world impressed you with being that brilliant? If so, fine. Rest easy and don't worry about it.

Maybe guardian angels are protecting us or something, or hey, the idea just can't be made practical, as I don't know, I refuse to check further out of sheer fear.

But I do know that I haven't found a mathematical reason for n to be a needle in a haystack, nor has anyone replied to me in posts—as I've talked about this idea before—with even a basic understanding of how this thing works with a cogent response showing that n would be ever harder to find as T increases in size.

Posters instead have avoided the proper mathematical analysis to look for ways to claim it doesn't work without showing it doesn't work!!!

I suspect the same will happen in reply to this post, as remember there is only one variable where a search is needed as you're looking for a solution to

S = r + n*T

where r can be any residue modulo T, so only n matters, and there must exist some integer n, such that

x^2 - y^2 = 0 mod T

where you solve for x and y using

x+k = sqrt(y^2 + S + k^2)

and it's surrogate factoring because you get y by factoring S + k^2.

Does it work?

Yes. It's easy to prove solutions must exist.

But can it be made practical?

Don't know.

But consider, you can't find this idea in math textbooks, or in the mathematical mainstream, so I just thought about a problem for a couple of years and figured out a new way to factor, previously unknown.

[A reply to someone who said that he had programmed James' methods, and that some have failed entirely, but most of them do find factors.]

Unfortunately this idea is stupid simple—as in so damn easy it takes effort to get it wrong.

I think you made that effort.

Now I refuse to implement because I am too terrified both of the idea and of a world that lets something like this just sit out there because a few people are, well, pathological.

I fear this idea is what turned the Hezbollah and Israel conflict.

I fear that this idea is now allowing terrorist organizations to get funding by simply scooping money out of the world network.

And I fear that terrorists have not yet fully reacted because they are in shock that they can do this and keep doing it, and even have evidence that they are doing it, which is being ignored, so they can keep doing it, when supposedly countries like the United States and Israel are so brilliant.

I fear they are not quite reacting yet because they can't quite comprehend this either.

It makes no sense. This situation should be impossible. I find I can't wrap a logical argument around it.

It is puzzling to an extreme. Denial on this level shouldn't be possible.

But countries in the know are maybe increasingly bold as they know they have knowledge of an Achilles heel, giving them the power to do, well, just about anything they want.

So like now North Korea wants to test a nuke. What can stop them? What if they can break military codes? What kind of bloodbath in Iraq might we expect? Where will this end?

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