### Sunday, April 29, 2007

## JSH: Gotcha

Start with

x^2 - 6x + 35 = 0

and solve using the quadratic formula to get

x = (3 +/- sqrt(-26)

and now comes a simple yet clever trick—multiply the last two coefficients of the original polynomial with one solution for x and I'll use x = 3 + sqrt(-26), so I have

x^2 - 6(3+sqrt(-26))x + 35(3 + sqrt(-26)) = 0

where the purpose here is to make sure that BOTH roots must have 3 + sqrt(-26) as a factor, but notice that one root must have 7 as a factor as well, as the last coefficient has 7 as a factor, so this is just a tricky way of forcing one root to have 7 as a factor without forcing them both.

Now I can group my radicals to one side and everything else on the other to get

sqrt(-26)(-6x + 35) = -x^2 + 18x - 105

and square both sides to get

(-26)(36x^2 -420x + 1225) = x^4 - 36x^3 + 534x^2 - 3780x + 11025

which is

x^4 - 36x^3 +1470x^2 - 14700x + 42875 = 0

where at least one root must have 7 as a factor.

But that polynomial is irreducible over Q.

So you see, I've been right all along and you people have been fighting very valuable and important mathematics.

You've been fighting against the discipline of mathematics itself in a futile attempt to destroy the future of humanity.

You have been fighting history.

x^2 - 6x + 35 = 0

and solve using the quadratic formula to get

x = (3 +/- sqrt(-26)

and now comes a simple yet clever trick—multiply the last two coefficients of the original polynomial with one solution for x and I'll use x = 3 + sqrt(-26), so I have

x^2 - 6(3+sqrt(-26))x + 35(3 + sqrt(-26)) = 0

where the purpose here is to make sure that BOTH roots must have 3 + sqrt(-26) as a factor, but notice that one root must have 7 as a factor as well, as the last coefficient has 7 as a factor, so this is just a tricky way of forcing one root to have 7 as a factor without forcing them both.

Now I can group my radicals to one side and everything else on the other to get

sqrt(-26)(-6x + 35) = -x^2 + 18x - 105

and square both sides to get

(-26)(36x^2 -420x + 1225) = x^4 - 36x^3 + 534x^2 - 3780x + 11025

which is

x^4 - 36x^3 +1470x^2 - 14700x + 42875 = 0

where at least one root must have 7 as a factor.

But that polynomial is irreducible over Q.

So you see, I've been right all along and you people have been fighting very valuable and important mathematics.

You've been fighting against the discipline of mathematics itself in a futile attempt to destroy the future of humanity.

You have been fighting history.

## Simple disproof, Galois Theory impact

A simple exercise with basic quadratics and a quartic shows a problem with some standard views in modern number theory.

Start with

x^2 - 6x + 35 = 0

and solve using the quadratic formula to get

x = (3 +/- sqrt(-26)

and now comes a simple yet clever trick—multiply the last two coefficients of the original polynomial with one solution for x and I'll use x = 3 + sqrt(-26), so I have

x^2 - 6(3+sqrt(-26))x + 35(3 + sqrt(-26)) = 0

where the purpose here is to make sure that BOTH roots must have 3 + sqrt(-26) as a factor, but notice that one root must have 7 as a factor as well, as the last coefficient has 7 as a factor, so this is just a tricky way of forcing one root to have 7 as a factor without forcing them both.

Now I can group my radicals to one side and everything else on the other to get

sqrt(-26)(-6x + 35) = -x^2 + 18x - 105

and square both sides to get

(-26)(36x^2 -420x + 1225) = x^4 - 36x^3 + 534x^2 - 3780x + 11025

which is

x^4 - 36x^3 +1470x^2 - 14700x + 42875 = 0

where at least one root must have 7 as a factor, as long as I got the algebra right. Easier to do this sort of thing with math software.

Assuming that is right now it's just a matter of checking it for reducibility over Q. I only went that far for those of you who might wonder how you'd get to a monic with integer coefficients.

Remember that trivially you have that one of the roots must have 7 itself as a factor because I forced that with

x^2 - 6(3+sqrt(-26))x + 35(3 + sqrt(-26)) = 0

but if that polynomial is irreducible you can just use x = 7y, substitute and find you have a non-monic polynomial.

Start with

x^2 - 6x + 35 = 0

and solve using the quadratic formula to get

x = (3 +/- sqrt(-26)

and now comes a simple yet clever trick—multiply the last two coefficients of the original polynomial with one solution for x and I'll use x = 3 + sqrt(-26), so I have

x^2 - 6(3+sqrt(-26))x + 35(3 + sqrt(-26)) = 0

where the purpose here is to make sure that BOTH roots must have 3 + sqrt(-26) as a factor, but notice that one root must have 7 as a factor as well, as the last coefficient has 7 as a factor, so this is just a tricky way of forcing one root to have 7 as a factor without forcing them both.

Now I can group my radicals to one side and everything else on the other to get

sqrt(-26)(-6x + 35) = -x^2 + 18x - 105

and square both sides to get

(-26)(36x^2 -420x + 1225) = x^4 - 36x^3 + 534x^2 - 3780x + 11025

which is

x^4 - 36x^3 +1470x^2 - 14700x + 42875 = 0

where at least one root must have 7 as a factor, as long as I got the algebra right. Easier to do this sort of thing with math software.

Assuming that is right now it's just a matter of checking it for reducibility over Q. I only went that far for those of you who might wonder how you'd get to a monic with integer coefficients.

Remember that trivially you have that one of the roots must have 7 itself as a factor because I forced that with

x^2 - 6(3+sqrt(-26))x + 35(3 + sqrt(-26)) = 0

but if that polynomial is irreducible you can just use x = 7y, substitute and find you have a non-monic polynomial.

### Sunday, April 22, 2007

## Nothing sacred to mathematicians

Carefully considering everything that has happened over the last few years I have concluded that mathematicians routinely lie about results that they know are not actually correct, which they know will be defended if challenged—which is a rarity anyway—by mathematicians as a group claiming the results are correct.

To some extent I think it is a view that sees modern mathematics as no longer connected to the real world in any way, where progress in the sciences is so unrelated that it no longer matters if a mathematician's claim is actually true as long as the argument looks a certain way and is convincing to other mathematicians.

See: "When is a proof?"

If you read that article or have read it before, and wonder what's the big deal, I say, what's wrong with you? How can a mathematician say that what is true is what is believed to be true?

People have believed that the earth rode on the back of elephants. They have believed that everything on the planet was made out of earth, fire, water and air.

Group belief is not proof, and isn't there a better way than just having some people SAY what is true?

Sure, either real world checks, like those of mathematics used in science, or have computers that check, but that's where the current situation gets mind boggling as to the gullibility of a world that has decided to forget that people cheat, as it allows mathematicians to shy away from computer checking of mathematical arguments.

To how many of you does it even occur that it might be possible?

Mathematicians lie on this subject more boldly than in other areas as they have to convince against the grain, to say that a supposedly logical and rigid discipline with a special formal language meant to make it hard for ambiguity cannot be checked by computer systems that supposedly love such things.

After all, programming languages like Java are special languages meant to limit ambiguity.

And computer programmers manage the level of correctness necessary for computers to run.

Checking a mathematical proof is in a sense, like running a computer program through a compiler.

I suggest to you that mathematicians do NOT want to be checked because they are often wrong, and not only are they often wrong, where it is extremely difficult for you or any one else to figure out that they are wrong, but if you dare ask them or challenge them, they will attack you, including using verbal abuse.

How dumb are you? How dare you trust when the New York Times or some other media just tells you these people are brilliant when no one actually checks and they have managed to fool a world in the most dramatic area we have as supposedly now computers are too stupid for their research?

Do they do mathematics or math-magic? I think to most of you what mathematicians do is some mystical, magical thing that cannot be encapsulated by our meager toys like computers so of course they cannot be checked by computers!

You don't see them as human beings but as magicians, so they get away with actually being con artists.

And if you have your own research outside of their system, as do I, at best they will ignore you, no matter what it is because nothing is sacred to them. Yet our world today might not exist if people like them had taken over when algebra was being discovered as who knew then?

Who knew at that time how important those beginnings would be to our future?

Today people in power can rob the world of those beginnings, for no reason at all, except to maintain their own power—their ability to claim correctness without being meaningfully checked.

Realize, these people gain nothing but the power and prestige of their position and the income related to it, to say things that are not true, and try to block future progress out of short-sightedness, meager real intellect, and parasitism.

But it is the world that lets them.

And all you have to do instead is ask for computers to check. Really ask, not politely and run away as soon as the mathematicians start abusing you (when some of you will truly realize they ARE cons), but really ask and push the issue.

If I am wrong about them lying, what's the loss? We'd have computers that can check as I assure you of one thing, logical arguments under a system with a special formal language CAN be checked by computers as that's all programs are. Sure there can be other problems like with programs having bugs that escape compilers, but we do use spell checks without expecting computers to write the entire book, now don't we?

Don't let simple dodges from con artists fool you. Computers do NOTHING today in checking most of these people. Anything on top of that would be a vast improvement, or do you really just want to trust like you do now?

To some extent I think it is a view that sees modern mathematics as no longer connected to the real world in any way, where progress in the sciences is so unrelated that it no longer matters if a mathematician's claim is actually true as long as the argument looks a certain way and is convincing to other mathematicians.

See: "When is a proof?"

If you read that article or have read it before, and wonder what's the big deal, I say, what's wrong with you? How can a mathematician say that what is true is what is believed to be true?

People have believed that the earth rode on the back of elephants. They have believed that everything on the planet was made out of earth, fire, water and air.

Group belief is not proof, and isn't there a better way than just having some people SAY what is true?

Sure, either real world checks, like those of mathematics used in science, or have computers that check, but that's where the current situation gets mind boggling as to the gullibility of a world that has decided to forget that people cheat, as it allows mathematicians to shy away from computer checking of mathematical arguments.

To how many of you does it even occur that it might be possible?

Mathematicians lie on this subject more boldly than in other areas as they have to convince against the grain, to say that a supposedly logical and rigid discipline with a special formal language meant to make it hard for ambiguity cannot be checked by computer systems that supposedly love such things.

After all, programming languages like Java are special languages meant to limit ambiguity.

And computer programmers manage the level of correctness necessary for computers to run.

Checking a mathematical proof is in a sense, like running a computer program through a compiler.

I suggest to you that mathematicians do NOT want to be checked because they are often wrong, and not only are they often wrong, where it is extremely difficult for you or any one else to figure out that they are wrong, but if you dare ask them or challenge them, they will attack you, including using verbal abuse.

How dumb are you? How dare you trust when the New York Times or some other media just tells you these people are brilliant when no one actually checks and they have managed to fool a world in the most dramatic area we have as supposedly now computers are too stupid for their research?

Do they do mathematics or math-magic? I think to most of you what mathematicians do is some mystical, magical thing that cannot be encapsulated by our meager toys like computers so of course they cannot be checked by computers!

You don't see them as human beings but as magicians, so they get away with actually being con artists.

And if you have your own research outside of their system, as do I, at best they will ignore you, no matter what it is because nothing is sacred to them. Yet our world today might not exist if people like them had taken over when algebra was being discovered as who knew then?

Who knew at that time how important those beginnings would be to our future?

Today people in power can rob the world of those beginnings, for no reason at all, except to maintain their own power—their ability to claim correctness without being meaningfully checked.

Realize, these people gain nothing but the power and prestige of their position and the income related to it, to say things that are not true, and try to block future progress out of short-sightedness, meager real intellect, and parasitism.

But it is the world that lets them.

And all you have to do instead is ask for computers to check. Really ask, not politely and run away as soon as the mathematicians start abusing you (when some of you will truly realize they ARE cons), but really ask and push the issue.

If I am wrong about them lying, what's the loss? We'd have computers that can check as I assure you of one thing, logical arguments under a system with a special formal language CAN be checked by computers as that's all programs are. Sure there can be other problems like with programs having bugs that escape compilers, but we do use spell checks without expecting computers to write the entire book, now don't we?

Don't let simple dodges from con artists fool you. Computers do NOTHING today in checking most of these people. Anything on top of that would be a vast improvement, or do you really just want to trust like you do now?

### Thursday, April 19, 2007

## JSH: Hard success

I started with

2975x^2 - 15x + 2 = 2(f(x) + 1)*(g(x) + 1)

and stepped through a straightforward bit of algebra where I say I relied on the principle that

p(a+b)*(c+d) = (p*a + p*b)*(c+d)

without regard to the value of 'a' and 'c' but I think some of you think, wait a minute, what if 'a' and 'c' are algebraic numbers—as I haven't mentioned a ring—and they divide off factors in common with p?

I have various proofs that address that issue but none of them have resonated so I figured out another which was different from what I thought would work, where I've said that before these last couple of days, and been wrong.

The problem with my previous approach yesterday was that I figured out an angle that forced a function I call g(x) to be an integer function with integer x, given the conditions I was using, which I realized today didn't help me.

So I was stuck, again, trying to figure out some other path, some other way, hoping against hope that I wasn't actually wrong…but maybe I was?

So I finally faced reality, and noted that my argument if true with the two derivations I am now using would require that at least one solution match between two different quadratics at least to unit factors, and I came up with another way, and the answer seems so hard to explain.

It's a difference between squares of factors with 289 or 49 versus 17 and 7 and I wonder if anything can convince you people and does it really matter?

I just came up with it today and maybe tomorrow I'll find a mistake, but it's such a brilliant idea.

Succinctly, my argument required that except for unit factors, I could match ONE root of two different quadratics to each other—if I was right.

Pursuing that path I found that I appear to have a direct proof that I am right—if I didn't make some stupid mistake.

But that proof requires that you understand the difference between squares of factors, in this case, squares of factors of 289 or 49, versus factors of 17 and 7.

Such an odd thing and what does it really mean?

What if I DID make a mistake, yet again? What if I just screwed up, one more time?

But I didn't.

I don't know where that answer came from, and that kind of scares me. How did I ever figure out to take that path?

Mu.

2975x^2 - 15x + 2 = 2(f(x) + 1)*(g(x) + 1)

and stepped through a straightforward bit of algebra where I say I relied on the principle that

p(a+b)*(c+d) = (p*a + p*b)*(c+d)

without regard to the value of 'a' and 'c' but I think some of you think, wait a minute, what if 'a' and 'c' are algebraic numbers—as I haven't mentioned a ring—and they divide off factors in common with p?

I have various proofs that address that issue but none of them have resonated so I figured out another which was different from what I thought would work, where I've said that before these last couple of days, and been wrong.

The problem with my previous approach yesterday was that I figured out an angle that forced a function I call g(x) to be an integer function with integer x, given the conditions I was using, which I realized today didn't help me.

So I was stuck, again, trying to figure out some other path, some other way, hoping against hope that I wasn't actually wrong…but maybe I was?

So I finally faced reality, and noted that my argument if true with the two derivations I am now using would require that at least one solution match between two different quadratics at least to unit factors, and I came up with another way, and the answer seems so hard to explain.

It's a difference between squares of factors with 289 or 49 versus 17 and 7 and I wonder if anything can convince you people and does it really matter?

I just came up with it today and maybe tomorrow I'll find a mistake, but it's such a brilliant idea.

Succinctly, my argument required that except for unit factors, I could match ONE root of two different quadratics to each other—if I was right.

Pursuing that path I found that I appear to have a direct proof that I am right—if I didn't make some stupid mistake.

But that proof requires that you understand the difference between squares of factors, in this case, squares of factors of 289 or 49, versus factors of 17 and 7.

Such an odd thing and what does it really mean?

What if I DID make a mistake, yet again? What if I just screwed up, one more time?

But I didn't.

I don't know where that answer came from, and that kind of scares me. How did I ever figure out to take that path?

Mu.

### Tuesday, April 17, 2007

## JSH: Awesome proof

So finally I have a proof that blows away any attempts at lying about it, as with this latest approach I get two quadratics—one defines the a's and another defines the b's.

And I figured out that I can set the solutions for the b's equal the NEGATIVES of the a's, which was actually forced as first I tried to set them equal.

So the functions Q(x) and R(x) by mapping every possible f(x) and g(x), do allow me to find integer solutions that prove my case dramatically.

That means that numerically you can use integer x's and see that abs(g(x)) is the same for both, which blows away any attempts at lying about it—with real mathematicians.

So then you have g(x)=5a_2(x)/7 and g'(x) = 5b_2(x)/17, where g(x) = -g'(x).

So finally I can use numerical methods and directly demonstrate that, yeah, the distributive property DOES hold, even with functions.

The principal underlying the proof is that

p(a+b)*(c+d) = (p*a + p*b)*(c+d)

without regard to the value of 'a' and 'c' so even if they are functions, which they are in my research, it DOES NOT MATTER, so you can just pick a convenience value and the distributive property still holds.

BUT lots of mathematicians inherited a system where mistakes were made before any of you were born, and rather than love mathematical proof they have forced me to find a demonstration that can convince others that they are lying.

Such weakness is what drives the contempt in my postings.

I admit I hate their weakness. Their need for approval.

Their need for you to trust them and believe them even when they are wrong.

If you were Andrew Wiles or Ribet or Taylor, would you tell the truth if you found out that works you were called brilliant for turned out to be crap?

Could you live with people knowing? Or would you break?

Can any of you be real mathematicians?

That's why social crap is so dangerous. Real mathematicians don't need it, so they can walk way from it when the mathematics says so.

To the extent that Wiles, Ribet, Taylor and others cannot, they simply prove they not only are not real mathematicians anyway, regardless of their mistakes, but they have not an inkling of what a real mathematician is.

They need social approval. They need people believing in them.

They need you.

And I figured out that I can set the solutions for the b's equal the NEGATIVES of the a's, which was actually forced as first I tried to set them equal.

So the functions Q(x) and R(x) by mapping every possible f(x) and g(x), do allow me to find integer solutions that prove my case dramatically.

That means that numerically you can use integer x's and see that abs(g(x)) is the same for both, which blows away any attempts at lying about it—with real mathematicians.

So then you have g(x)=5a_2(x)/7 and g'(x) = 5b_2(x)/17, where g(x) = -g'(x).

So finally I can use numerical methods and directly demonstrate that, yeah, the distributive property DOES hold, even with functions.

The principal underlying the proof is that

p(a+b)*(c+d) = (p*a + p*b)*(c+d)

without regard to the value of 'a' and 'c' so even if they are functions, which they are in my research, it DOES NOT MATTER, so you can just pick a convenience value and the distributive property still holds.

BUT lots of mathematicians inherited a system where mistakes were made before any of you were born, and rather than love mathematical proof they have forced me to find a demonstration that can convince others that they are lying.

Such weakness is what drives the contempt in my postings.

I admit I hate their weakness. Their need for approval.

Their need for you to trust them and believe them even when they are wrong.

If you were Andrew Wiles or Ribet or Taylor, would you tell the truth if you found out that works you were called brilliant for turned out to be crap?

Could you live with people knowing? Or would you break?

Can any of you be real mathematicians?

That's why social crap is so dangerous. Real mathematicians don't need it, so they can walk way from it when the mathematics says so.

To the extent that Wiles, Ribet, Taylor and others cannot, they simply prove they not only are not real mathematicians anyway, regardless of their mistakes, but they have not an inkling of what a real mathematician is.

They need social approval. They need people believing in them.

They need you.

### Monday, April 16, 2007

## Relief and wait and see

I was brought up learning modern problem solving techniques, and just took it for granted that using them was understood until I mostly as a lark started working on math problems that have been for a long time considered hard, and ran into a crap storm from math people.

So I made a lot of mistakes along the way, and found people who capitalized on those mistakes and they flamed me across the web as a crackpot, but then I naively thought that once I actually had real, solid results that would end.

And it didn't so I learned that I needed to work a political process and began looking for a mathematical proof that was hard to lie about and yesterday I put the finishing touches on one, and now, I'm in a wait-and-see mode, though I do feel relieved as well.

Where do solutions come from really? Why me and not someone else? Why this way?

But the saddest part of all this is that there are people who'd rather be wrong and congratulate each other as if they were right, than actually learn mathematics, and understand it, and be right.

The will to be wrong is something I think about as our world moves closer to what may be the end of humanity because of people who would rather be wrong and be convincing than actually solve problems and I have to say to myself that if our species goes extinct as a result, we deserved it.

We deserve it if we die out as a species because we let people get away with being wrong, and using group processes and being convincing to not solve problems until the problems can't be solved, and if everything before we were born was then meaningless and all the efforts of human beings before us were meaningless because of THIS generation that kills it all for nothing, then that's just the way it is.

Ultimately it is about backbone.

Maybe ultimately the human species is too weak to handle the truth.

And for that it will deservedly die. But, um, I hope not. I keep trying because I think there is something valuable to us. And I think our species can handle at least a little more of harsh reality.

But first I have to get past the people who would rather kill us all to live empty lies.

So I made a lot of mistakes along the way, and found people who capitalized on those mistakes and they flamed me across the web as a crackpot, but then I naively thought that once I actually had real, solid results that would end.

And it didn't so I learned that I needed to work a political process and began looking for a mathematical proof that was hard to lie about and yesterday I put the finishing touches on one, and now, I'm in a wait-and-see mode, though I do feel relieved as well.

Where do solutions come from really? Why me and not someone else? Why this way?

But the saddest part of all this is that there are people who'd rather be wrong and congratulate each other as if they were right, than actually learn mathematics, and understand it, and be right.

The will to be wrong is something I think about as our world moves closer to what may be the end of humanity because of people who would rather be wrong and be convincing than actually solve problems and I have to say to myself that if our species goes extinct as a result, we deserved it.

We deserve it if we die out as a species because we let people get away with being wrong, and using group processes and being convincing to not solve problems until the problems can't be solved, and if everything before we were born was then meaningless and all the efforts of human beings before us were meaningless because of THIS generation that kills it all for nothing, then that's just the way it is.

Ultimately it is about backbone.

Maybe ultimately the human species is too weak to handle the truth.

And for that it will deservedly die. But, um, I hope not. I keep trying because I think there is something valuable to us. And I think our species can handle at least a little more of harsh reality.

But first I have to get past the people who would rather kill us all to live empty lies.

### Sunday, April 15, 2007

## JSH: Galois Theory, so what's wrong?

I am sure there will be a lot of confusion about the significance of the result that I have showing a problem with use of the ring of algebraic integers.

It is such a huge problem in number theory that it's hard to grasp the full impact, but I can maybe help at least with Galois Theory.

The result shows that Galois Theory tells you nothing more about non-rationals than it does about rationals.

That is the succinct way to explain the impact there, and why I say it does not say Galois Theory is wrong, exactly, but it greatly limits its usefulness to number theorists as to taking it away for the most part as a meaningful tool.

And I want to emphasize that mathematically it just never was.

People just can make mistakes, and as time goes on those mistakes can be found and the truth learned. It is a process that has gone on for as long as there have been people.

We live. We learn.

It is such a huge problem in number theory that it's hard to grasp the full impact, but I can maybe help at least with Galois Theory.

The result shows that Galois Theory tells you nothing more about non-rationals than it does about rationals.

That is the succinct way to explain the impact there, and why I say it does not say Galois Theory is wrong, exactly, but it greatly limits its usefulness to number theorists as to taking it away for the most part as a meaningful tool.

And I want to emphasize that mathematically it just never was.

People just can make mistakes, and as time goes on those mistakes can be found and the truth learned. It is a process that has gone on for as long as there have been people.

We live. We learn.

## JSH: Strange will to be wrong

I think one of the biggest puzzles for me as I wait and hope that the latest very simple explanation of a problem in number theory might finally get some traction is this strange will to be wrong that many of you clearly have.

One would think that some of you actually would like to use correct mathematics, and not waste time and effort with mathematical ideas proven to be wrong, and that at least some of you would actually care about learning powerful mathematical techniques that DO WORK versus wasting time on flawed ones that have been proven not to work.

But I guess most of you tell yourselves that's not what you're doing so you can do it, which is how this situation has continued for so many years.

And then you must feel pride at learning useless crap that doesn't work, but maybe impresses people because so many people are a part of the error?

So this time I updated a technique that I first thought up late last year as now I start with the factorization

2975x^2 - 15x + 2 = 2(f(x) + 1)*(g(x) + 1)

where the change was to leverage two prime numbers so that I can step through the derivation with 7 and 17 versus just 7 as that didn't work with you as I got some of the same people on sci.math who would come over to other newsgroups mind you as well, tossing up specious objections which sounded intelligent, I guess, and that crap worked like it has worked so many times before.

By using two prime numbers though I remove the ability of those people to make seemingly intelligent objections but people remember I have not only explained and explained as I've come up with different ways to try and get mathematicians to behave like mathematicians and accept mathematical proof, I've even had the result published in a peer reviewed mathematical journal!!!

Usenet crushed the process as sci.math'ers managed to convince the editors to pull my paper, and later the entire journal died.

A an entire mathematical journal died here yet still the error has stayed in place along with the denial in this will to be wrong.

That takes a will to be wrong for that to happen and years to go by while I labor to try and find some way to explain that doesn't allow you to tell yourselves lies about the math so you can keep doing wrong mathematics.

Wrong mathematics is EASIER.

There is a thrill in feeling you are correct because you are using flawed ideas that allow you to convince yourself that you have proven something you haven't.

But now we have a situation where a lot of people around the world are choosing easy, doing wrong mathematics, on a huge scale. So yes, I can understand if say Andrew Wiles wouldn't want to accept this result as it takes away his research, so no, he did not prove Fermat's Last Theorem. And Ribet might want to hide from this because it takes away his, and I'm not saying either of them are doing so, but I'm giving them as dramatic examples to explain the why of the very human behavior here.

But wrong is wrong. You people can keep doing wrong math to "prove" things all you want and you are not doing anything of value no matter how many of you get together to live in the error.

Wrong mathematical ideas are easier because they let you "prove" anything, like the classic examples that boil down to divide by zero errors.

So yeah, do the wrong math, and use the ring of algebraic integers wrong, without understanding its quirks and real mathematical properties, and you can think you proved Fermat's Last Theorem when you didn't.

Wrong math is easier, but it's still wrong.

My hope is that some of you start appreciating mathematics. Because of the last few years Usenet has been doing the opposite, using group processes to make this situation much harder in a fight to be wrong.

And in a fight to keep bringing other people into the error as pity the poor students, who could have been learning the truth years ago, but instead are currently wasting time and mental energy on bogus mathematical ideas that are appealing because they are wrong, and in mathematics, wrong is easier.

It is so, so much harder to be right in mathematics.

One would think that some of you actually would like to use correct mathematics, and not waste time and effort with mathematical ideas proven to be wrong, and that at least some of you would actually care about learning powerful mathematical techniques that DO WORK versus wasting time on flawed ones that have been proven not to work.

But I guess most of you tell yourselves that's not what you're doing so you can do it, which is how this situation has continued for so many years.

And then you must feel pride at learning useless crap that doesn't work, but maybe impresses people because so many people are a part of the error?

So this time I updated a technique that I first thought up late last year as now I start with the factorization

2975x^2 - 15x + 2 = 2(f(x) + 1)*(g(x) + 1)

where the change was to leverage two prime numbers so that I can step through the derivation with 7 and 17 versus just 7 as that didn't work with you as I got some of the same people on sci.math who would come over to other newsgroups mind you as well, tossing up specious objections which sounded intelligent, I guess, and that crap worked like it has worked so many times before.

By using two prime numbers though I remove the ability of those people to make seemingly intelligent objections but people remember I have not only explained and explained as I've come up with different ways to try and get mathematicians to behave like mathematicians and accept mathematical proof, I've even had the result published in a peer reviewed mathematical journal!!!

Usenet crushed the process as sci.math'ers managed to convince the editors to pull my paper, and later the entire journal died.

A an entire mathematical journal died here yet still the error has stayed in place along with the denial in this will to be wrong.

That takes a will to be wrong for that to happen and years to go by while I labor to try and find some way to explain that doesn't allow you to tell yourselves lies about the math so you can keep doing wrong mathematics.

Wrong mathematics is EASIER.

There is a thrill in feeling you are correct because you are using flawed ideas that allow you to convince yourself that you have proven something you haven't.

But now we have a situation where a lot of people around the world are choosing easy, doing wrong mathematics, on a huge scale. So yes, I can understand if say Andrew Wiles wouldn't want to accept this result as it takes away his research, so no, he did not prove Fermat's Last Theorem. And Ribet might want to hide from this because it takes away his, and I'm not saying either of them are doing so, but I'm giving them as dramatic examples to explain the why of the very human behavior here.

But wrong is wrong. You people can keep doing wrong math to "prove" things all you want and you are not doing anything of value no matter how many of you get together to live in the error.

Wrong mathematical ideas are easier because they let you "prove" anything, like the classic examples that boil down to divide by zero errors.

So yeah, do the wrong math, and use the ring of algebraic integers wrong, without understanding its quirks and real mathematical properties, and you can think you proved Fermat's Last Theorem when you didn't.

Wrong math is easier, but it's still wrong.

My hope is that some of you start appreciating mathematics. Because of the last few years Usenet has been doing the opposite, using group processes to make this situation much harder in a fight to be wrong.

And in a fight to keep bringing other people into the error as pity the poor students, who could have been learning the truth years ago, but instead are currently wasting time and mental energy on bogus mathematical ideas that are appealing because they are wrong, and in mathematics, wrong is easier.

It is so, so much harder to be right in mathematics.

## JSH: Newsgroups mistakes, updated explanation and proof

One of the more horrible additions to mathematical history will be the role of the sci.math newsgroup in acting to block the world's knowledge of important new mathematical research, where I post now to point out a simplified mathematical argument that proves my point about a problem with the ring of algebraic integers in a way that removes the ability of others to make seemingly intelligent objections.

I start with

2975x^2 - 15x + 2 = 2(f(x) + 1)*(g(x) + 1)

where x is a non-zero algebraic integer and f(x) and g(x) are functions of x determined by the following derivations. I've done something like this before in posts, so it's important to note upfront that these demonstrations are different in that I do two versus one.

Multiplying through the first factor by 2 and multiplying both sides by 7, picked because it is a prime factor of 2975, as 2975 = 7(17)(25) I have

7(2975x^2 - 15x + 2) = 7*(2f(x) + 2)*(g(x) + 1)

now re-order in a special way on the left side and now pick one way to multiply by 7 on the right:

7(7(17)(5^2)x^2 - (3)(5)x + 2) = (2f(x) + 2)*(7g(x) + 7)

Now let

2f(x) = 5a_1(x) + 5

and

7g(x) = 5a_2(x)

so I have:

(7(7(17)(5^2)x^2 - (3)(5)x + 2) = (5a_1(x) + 7)*(5a_2(x) + 7)

Now multiply through by 7 on the left, add and subtract 35 and introduce the polynomial Q(x), where I add and subtract it so everything still balances, and now I have

(49(17)x^2+7(17)Q(x))5^2 - (3x+1+5(17)Q(x))(5)(7) + 7^2 = (5a_1(x)+7)*(5a_2(x)+7)

where now a general solution can be seen by multiplying out on the right side as that gives

a_1(x)*a_2(x) = 49(17)x^2 + 7(17)Q(x)

and

a_1(x) + a_2(x) = -(3x + 1 + 5(17)Q(x))

which determines the following of which the a's must be roots:

a^2 + (3x + 1 + 5(17)Q(x))a + (17)(49x^2 + 7Q(x)) = 0

Then with Q(x) = -2x, the a's are roots of

a^2 - (168x-1)a + (17)(49x^2 - 14x) = 0

And letting x=1 gives as a solution:

a^2 - 167a + 595 = 0

where I say the simple argument above shows that ONLY ONE ROOT can have 7 as a factor, where of course in the ring of algebraic integers neither has 7 itself as a factor, and objectors have pointed at solving

7g(x) = 5a_2(x)

to get

g(x) = 5a_2(x)/7

and claim that the 7 does not divide through a_2(x), so now let's consider the second derivation.

Starting yet again with

2975x^2 - 15x + 2 = 2(f(x) + 1)*(g(x) + 1)

Still multiplying through the first factor by 2, but this time multiplying both sides by 17:

17(2975x^2 - 15x + 2) = 17*(2f(x) + 2)*(g(x) + 1)

and again re-order in a special way on the left side and now pick one way to multiply by 17 on the right:

17(7(17)(5^2)x^2 - (3)(5)x + 2) = (2f(x) + 2)*(17g(x) + 17)

Now let

2f(x) = 5b_1(x) + 15

and

17g(x) = 5b_2(x)

so I have:

(17(7(17)(5^2)x^2 - (3)(5)x + 2) = (5b_1(x) + 17)*(5b_2(x) + 17)

Now multiply through by 17 on the left, add and subtract 255 and introduce the polynomial R(x) so I have

(289(7)x^2+119R(x))5^2-(3x+3+5(7)R(x))(5)(17)+17^2= (5b_1(x)+17)*(5b_2(x)+17)

where now a general solution can be seen by multiplying out on the right side as that gives

b_1(x)*b_2(x) = (7)(289x^2 + 17R(x)

and

b_1(x) + b_2(x) = -(3x + 3 + 5(7)R(x))

which determines the following of which the b's must be roots:

b^2 + (3x + 3 + 5(7)R(x))b + ((7)(289x^2 + 17R(x)) = 0

but now notice I have 17g(x) = 5b_2(x), so

g(x) = 5b_2(x)/17

and the prior objection MUST go away as 17 was not in the denominator with the first derivation as instead you had 7.

Where of course I started EACH DERIVATION with

2975x^2 - 15x + 2 = 2(f(x) + 1)*(g(x) + 1)

so, you have the SAME g(x) in BOTH DERIVATIONS.

Readers should note that you can solve for a_2(x) and b_2(x), as the first is a root of

a^2 + (3x + 1 + 5(17)Q(x))a + ((17)(49x^2 + 7Q(x)) = 0

while the second is a root of

b^2 + (3x + 3 + 5(7)R(x))b + ((7)(289x^2 + 17R(x)) = 0

and solving the first using the quadratic formula gives

a_2(x) = (-(3x + 1 + 5(17)Q(x)) +/- sqrt((3x + 1 + 5(17)Q(x))^2 - 4((17)(49x^2 + 7Q(x)) )/2

so g(x) = 5a_2(x)/7 means that

g(x) = 5(-(3x + 1 + 5(17)Q(x)) +/- sqrt((3x + 1 + 5(17)Q(x))^2 - 4((17)(49x^2 + 7Q(x)) )/14

but from the SECOND solution, you have g(x) = 5b_2(x)/17, so

g(x) = 5(-(3x + 3 + 5(7)R(x))) +/- sqrt((3x + 3 + 5(7)R(x)))^2 - 4((7)(289x^2 + 17R(x)) )/34

so with one case you have 7 as a factor in the denominator and with the other you have 17, so 7 must divide though with the first and 17 must divide through with the second for at least one solution.

Some might wonder about Q(x) and R(x), but those are polynomials which you select, as remember above I selected Q(x) = -2x to show that only one root of

a^2 - 167a + 595 = 0

has 7 as a factor. And importantly note that if x is an algebraic integer and Q(x) is as well then

a_2(x) = (-(3x + 1 + 5(17)Q(x)) +/- sqrt((3x + 1 + 5(17)Q(x))^2 - 4((17)(49x^2 + 7Q(x)) )/2

shows you that the a's are algebraic integers as that is the solution for both a's while I put a_2(x) as I was focusing on it.

So if you try to fudge in factors of 17 in the denominator with g(x), you run into the wall of

g(x) = 5a_2(x)/7

as there is just the 7, and also you can't say that factors of 7 are in the denominator because of

g(x) = 5b_2(x)/17

so the simply brilliant idea I had to remove even the semblance of an intelligent objection to this approach was to use TWO PRIME NUMBERS in concert with each other.

I will note though that mathematicians have steadfastly refused to accept what is mathematically true in this area for years now, creating the necessity for clever explanations to get around emphatic denials of what is mathematically correct.

The reason for those denials is that now you know that a lot of Galois Theory is not quite right, and you know that the ring of algebraic integers is used in a flawed way, as in that ring NEITHER of the roots of

a^2 - 167a + 595 = 0

can have 7 as a factor!!!

This error has been explained by me many times, and the refusal of the mathematical community to acknowledge it is best explained by the importance of the wrong math for research done by quite a few mathematicians who care more about their careers and belief that they are right than on what is mathematically true.

And they are willing to teach error to trusting students coming to them for a mathematical education.

That must stop. These students are no longer to be victims.

Remember at this point the error is willful. And it betrays a hatred of mathematical truth.

Mathematicians who tolerate this error and teach it are like arsonists who become firefighters, not because they actually hate error, but because they love it, and hate mathematical proof.

[A reply to someone who explained James that there is more that one pair of functions f(x) and g(x) that satisfy his equation.]

You are correct as I don't necessarily have the same g(x) in both cases.

Excellent reply.

Thinking it over I realized that Q(x) and R(x) are the functions that handle the full space of possible solutions for f(x) and g(x), and what I want is for Q(x) and R(x) to give the SAME quadratic defining the a's and b's, with an integer x, where both are integer functions in that they themselves give an integer value.

That is, I want the solution for the a's and the b's to be the same, with all integers.

Turns out that because I have two equations with three degrees of freedom since I have x, Q(x) and R(x), I can solve out and find congruence relationships that should define values that will work!!!

The nice thing about that is that then you can also directly see it work by plugging in actual values, noting that you have the same solutions for the a's and b's and therefore for g(x), which completes the demonstration I need.

I am going now to update the paper, but any of you who are serious mathematicians can step through the exercise on your own as it should be truly sublime mathematics.

It is just a remarkable proof, and it leave no shadow of doubt.

Have fun. I'll be checking in later to see if you any of you bothered.

[A reply to someone who told James not to worry because this time he would be right.]

William Hughes gave a cogent objection that was RIGHT ON TARGET.

And in doing so he showed at least some mathematical sophistication.

I gave a direct reply acknowledging the issue and giving what I thought was a resolution.

So what's your problem?

As an update I went ahead to try that approach and found it didn't work!!!

Turns out that I can't force the a's and b's to be equal, but I CAN force them to be equal by absolute value, so it's the same difference.

To me hanger's on like you are just such a waste of the time of people who actually care about doing some mathematics.

I may be loud and obnoxious often, and get things wrong a lot, but I am working, and it takes me years at times to figure things out, but I keep at it until I do.

It's done people. The direct demonstration now allows one to use numerical methods to prove that with the defined congruences g(x) is the SAME by absolute value for both the a's and b's though they are opposite in sign, which is why I talk about the absolute value.

That is, I figured out a route where you get g(x) and g'(x) where g(x) = -g'(x), and one is given by the a's and the other by the b's and the sign doesn't matter, but I had to do that as the other way didn't work.

If you figure it out you will be floored. I mean freaking floored as it is just so beautiful.

Worth all the crap. Definitely worth all of it.

One of the most beautiful proofs I've ever seen or imagined.

Maybe life does have meaning after all. Such a beautiful, beautiful, beautiful proof.

I am simply floored.

[A reply to someone who explained James that he is repeatedly rude, obnoxious, abusive, and that he believes that others have to accept that.]

It's not about abuse.

Figuring out mathematics can just simply be hard beyond belief.

It stresses people out. Pushes them outside of normal limits.

And most importantly, it makes you realize just how small you are.

Just to keep your head above water, yes, you may come over the top, and sound very arrogant and full of yourself, but that's mostly bravado.

Mathematics is so much bigger than any of us.

It is overwhelming and if you ever really get a good feel for just how big it is, and accept that, then you just might stop, and never try to figure anything out.

I do not fully understand why I do what I do.

But this latest proof gives me a greater sense of so much that is so far beyond me.

I don't know what people like you think is so important about other people liking you, or telling you that you are whatever it is that you think they should tell you.

But what people tell me means little to me.

What I can prove means so much more to me.

And what I can prove takes me so far beyond what you can get from people liking you, or telling you whatever it is that is so important for you to hear.

Mathematical proof is certainty. And it's knowing the truth without wondering about motivations, or wondering what's in it for some other person, or wondering why this or that person likes you or doesn't like you.

Mathematical proof is truth.

You will die someday, as will I. And when you die, will you be running through your mind all the good things people said about you? Or worrying about all the nasty things?

Either way, will it matter? When you die, you'll still be dead.

You will die and when you die what will any of this matter?

You will still be dead and everyone you know now will die, and everything you build will be torn down.

I know you will die. I know that no matter what you say now billions of years will pass and none of that will be around to matter any more.

Nothing any of you say will still be around billions of years from now.

But this proof that I discovered will still be true.

And as I comprehend absolutes, I understand certainty, and appreciate that which will still be here long after we are all dead.

Take your social nothings with you to your grave which will come soon enough.

You will die. Everyone you know now will die. Everything you build will be torn down.

What do you think is so permanent about your silly social crap?

I take mathematical proof.

As with mathematical proof I hear what God hears, and understand just a little bit, as much as I can, and that is the best that I can do.

[A reply to someone who wrote James usually starts his arguments with statements that are true.]

I have a record of actually trying versus acting like I am trying.

One of the more profound things I realized some years ago was that sci.math'ers were deliberately acting to confuse versus caring about what was mathematically true and that happened when I wrote a very detailed rebuttal to "Nora Baron" only to have that poster delete out everything and just repeat flawed claims.

It's easier to claim you are honest than to actually be honest.

I screw up. I make lots of mistakes in my search for truth and quite a few times over the years I have been rude, very wrong, and regretted things I've posted.

But I at least admit my humanity.

And in the world I thought I knew when I was growing up, human beings do those things and then some, because they work harder at being right, than just trying to look right.

If mathematicians proved me wrong I'd accept that because I know fantasy, and I prefer knowing truth.

But to the extent that they claim they prefer truth and rip on me as if the truth were the opposite, when I admit to being a very flawed human being just trying to get some things straight while these monsters pretend to be perfect—or very damn close to perfect—while they lie, and lie, and lie then I say we have a serious problem people.

The monsters have taken the high ground, and made it illegal to be human.

They spit on mistakes made in the search for truth.

And they piss on failures made in the pursuit of truth, to tell you lies, when they do not actually try to know what is mathematically correct—preferring only to rely on what they can convince you is correct.

I fail. And I cheer my ability to not only fail in trying, but to admit that I fail, and I fail, and I fail.

And through failing honestly, I find success, honestly.

While the demons tell you lies.

I start with

2975x^2 - 15x + 2 = 2(f(x) + 1)*(g(x) + 1)

where x is a non-zero algebraic integer and f(x) and g(x) are functions of x determined by the following derivations. I've done something like this before in posts, so it's important to note upfront that these demonstrations are different in that I do two versus one.

Multiplying through the first factor by 2 and multiplying both sides by 7, picked because it is a prime factor of 2975, as 2975 = 7(17)(25) I have

7(2975x^2 - 15x + 2) = 7*(2f(x) + 2)*(g(x) + 1)

now re-order in a special way on the left side and now pick one way to multiply by 7 on the right:

7(7(17)(5^2)x^2 - (3)(5)x + 2) = (2f(x) + 2)*(7g(x) + 7)

Now let

2f(x) = 5a_1(x) + 5

and

7g(x) = 5a_2(x)

so I have:

(7(7(17)(5^2)x^2 - (3)(5)x + 2) = (5a_1(x) + 7)*(5a_2(x) + 7)

Now multiply through by 7 on the left, add and subtract 35 and introduce the polynomial Q(x), where I add and subtract it so everything still balances, and now I have

(49(17)x^2+7(17)Q(x))5^2 - (3x+1+5(17)Q(x))(5)(7) + 7^2 = (5a_1(x)+7)*(5a_2(x)+7)

where now a general solution can be seen by multiplying out on the right side as that gives

a_1(x)*a_2(x) = 49(17)x^2 + 7(17)Q(x)

and

a_1(x) + a_2(x) = -(3x + 1 + 5(17)Q(x))

which determines the following of which the a's must be roots:

a^2 + (3x + 1 + 5(17)Q(x))a + (17)(49x^2 + 7Q(x)) = 0

Then with Q(x) = -2x, the a's are roots of

a^2 - (168x-1)a + (17)(49x^2 - 14x) = 0

And letting x=1 gives as a solution:

a^2 - 167a + 595 = 0

where I say the simple argument above shows that ONLY ONE ROOT can have 7 as a factor, where of course in the ring of algebraic integers neither has 7 itself as a factor, and objectors have pointed at solving

7g(x) = 5a_2(x)

to get

g(x) = 5a_2(x)/7

and claim that the 7 does not divide through a_2(x), so now let's consider the second derivation.

Starting yet again with

2975x^2 - 15x + 2 = 2(f(x) + 1)*(g(x) + 1)

Still multiplying through the first factor by 2, but this time multiplying both sides by 17:

17(2975x^2 - 15x + 2) = 17*(2f(x) + 2)*(g(x) + 1)

and again re-order in a special way on the left side and now pick one way to multiply by 17 on the right:

17(7(17)(5^2)x^2 - (3)(5)x + 2) = (2f(x) + 2)*(17g(x) + 17)

Now let

2f(x) = 5b_1(x) + 15

and

17g(x) = 5b_2(x)

so I have:

(17(7(17)(5^2)x^2 - (3)(5)x + 2) = (5b_1(x) + 17)*(5b_2(x) + 17)

Now multiply through by 17 on the left, add and subtract 255 and introduce the polynomial R(x) so I have

(289(7)x^2+119R(x))5^2-(3x+3+5(7)R(x))(5)(17)+17^2= (5b_1(x)+17)*(5b_2(x)+17)

where now a general solution can be seen by multiplying out on the right side as that gives

b_1(x)*b_2(x) = (7)(289x^2 + 17R(x)

and

b_1(x) + b_2(x) = -(3x + 3 + 5(7)R(x))

which determines the following of which the b's must be roots:

b^2 + (3x + 3 + 5(7)R(x))b + ((7)(289x^2 + 17R(x)) = 0

but now notice I have 17g(x) = 5b_2(x), so

g(x) = 5b_2(x)/17

and the prior objection MUST go away as 17 was not in the denominator with the first derivation as instead you had 7.

Where of course I started EACH DERIVATION with

2975x^2 - 15x + 2 = 2(f(x) + 1)*(g(x) + 1)

so, you have the SAME g(x) in BOTH DERIVATIONS.

Readers should note that you can solve for a_2(x) and b_2(x), as the first is a root of

a^2 + (3x + 1 + 5(17)Q(x))a + ((17)(49x^2 + 7Q(x)) = 0

while the second is a root of

b^2 + (3x + 3 + 5(7)R(x))b + ((7)(289x^2 + 17R(x)) = 0

and solving the first using the quadratic formula gives

a_2(x) = (-(3x + 1 + 5(17)Q(x)) +/- sqrt((3x + 1 + 5(17)Q(x))^2 - 4((17)(49x^2 + 7Q(x)) )/2

so g(x) = 5a_2(x)/7 means that

g(x) = 5(-(3x + 1 + 5(17)Q(x)) +/- sqrt((3x + 1 + 5(17)Q(x))^2 - 4((17)(49x^2 + 7Q(x)) )/14

but from the SECOND solution, you have g(x) = 5b_2(x)/17, so

g(x) = 5(-(3x + 3 + 5(7)R(x))) +/- sqrt((3x + 3 + 5(7)R(x)))^2 - 4((7)(289x^2 + 17R(x)) )/34

so with one case you have 7 as a factor in the denominator and with the other you have 17, so 7 must divide though with the first and 17 must divide through with the second for at least one solution.

Some might wonder about Q(x) and R(x), but those are polynomials which you select, as remember above I selected Q(x) = -2x to show that only one root of

a^2 - 167a + 595 = 0

has 7 as a factor. And importantly note that if x is an algebraic integer and Q(x) is as well then

a_2(x) = (-(3x + 1 + 5(17)Q(x)) +/- sqrt((3x + 1 + 5(17)Q(x))^2 - 4((17)(49x^2 + 7Q(x)) )/2

shows you that the a's are algebraic integers as that is the solution for both a's while I put a_2(x) as I was focusing on it.

So if you try to fudge in factors of 17 in the denominator with g(x), you run into the wall of

g(x) = 5a_2(x)/7

as there is just the 7, and also you can't say that factors of 7 are in the denominator because of

g(x) = 5b_2(x)/17

so the simply brilliant idea I had to remove even the semblance of an intelligent objection to this approach was to use TWO PRIME NUMBERS in concert with each other.

I will note though that mathematicians have steadfastly refused to accept what is mathematically true in this area for years now, creating the necessity for clever explanations to get around emphatic denials of what is mathematically correct.

The reason for those denials is that now you know that a lot of Galois Theory is not quite right, and you know that the ring of algebraic integers is used in a flawed way, as in that ring NEITHER of the roots of

a^2 - 167a + 595 = 0

can have 7 as a factor!!!

This error has been explained by me many times, and the refusal of the mathematical community to acknowledge it is best explained by the importance of the wrong math for research done by quite a few mathematicians who care more about their careers and belief that they are right than on what is mathematically true.

And they are willing to teach error to trusting students coming to them for a mathematical education.

That must stop. These students are no longer to be victims.

Remember at this point the error is willful. And it betrays a hatred of mathematical truth.

Mathematicians who tolerate this error and teach it are like arsonists who become firefighters, not because they actually hate error, but because they love it, and hate mathematical proof.

[A reply to someone who explained James that there is more that one pair of functions f(x) and g(x) that satisfy his equation.]

You are correct as I don't necessarily have the same g(x) in both cases.

Excellent reply.

Thinking it over I realized that Q(x) and R(x) are the functions that handle the full space of possible solutions for f(x) and g(x), and what I want is for Q(x) and R(x) to give the SAME quadratic defining the a's and b's, with an integer x, where both are integer functions in that they themselves give an integer value.

That is, I want the solution for the a's and the b's to be the same, with all integers.

Turns out that because I have two equations with three degrees of freedom since I have x, Q(x) and R(x), I can solve out and find congruence relationships that should define values that will work!!!

The nice thing about that is that then you can also directly see it work by plugging in actual values, noting that you have the same solutions for the a's and b's and therefore for g(x), which completes the demonstration I need.

I am going now to update the paper, but any of you who are serious mathematicians can step through the exercise on your own as it should be truly sublime mathematics.

It is just a remarkable proof, and it leave no shadow of doubt.

Have fun. I'll be checking in later to see if you any of you bothered.

[A reply to someone who told James not to worry because this time he would be right.]

William Hughes gave a cogent objection that was RIGHT ON TARGET.

And in doing so he showed at least some mathematical sophistication.

I gave a direct reply acknowledging the issue and giving what I thought was a resolution.

So what's your problem?

As an update I went ahead to try that approach and found it didn't work!!!

Turns out that I can't force the a's and b's to be equal, but I CAN force them to be equal by absolute value, so it's the same difference.

To me hanger's on like you are just such a waste of the time of people who actually care about doing some mathematics.

I may be loud and obnoxious often, and get things wrong a lot, but I am working, and it takes me years at times to figure things out, but I keep at it until I do.

It's done people. The direct demonstration now allows one to use numerical methods to prove that with the defined congruences g(x) is the SAME by absolute value for both the a's and b's though they are opposite in sign, which is why I talk about the absolute value.

That is, I figured out a route where you get g(x) and g'(x) where g(x) = -g'(x), and one is given by the a's and the other by the b's and the sign doesn't matter, but I had to do that as the other way didn't work.

If you figure it out you will be floored. I mean freaking floored as it is just so beautiful.

Worth all the crap. Definitely worth all of it.

One of the most beautiful proofs I've ever seen or imagined.

Maybe life does have meaning after all. Such a beautiful, beautiful, beautiful proof.

I am simply floored.

[A reply to someone who explained James that he is repeatedly rude, obnoxious, abusive, and that he believes that others have to accept that.]

It's not about abuse.

Figuring out mathematics can just simply be hard beyond belief.

It stresses people out. Pushes them outside of normal limits.

And most importantly, it makes you realize just how small you are.

Just to keep your head above water, yes, you may come over the top, and sound very arrogant and full of yourself, but that's mostly bravado.

Mathematics is so much bigger than any of us.

It is overwhelming and if you ever really get a good feel for just how big it is, and accept that, then you just might stop, and never try to figure anything out.

I do not fully understand why I do what I do.

But this latest proof gives me a greater sense of so much that is so far beyond me.

I don't know what people like you think is so important about other people liking you, or telling you that you are whatever it is that you think they should tell you.

But what people tell me means little to me.

What I can prove means so much more to me.

And what I can prove takes me so far beyond what you can get from people liking you, or telling you whatever it is that is so important for you to hear.

Mathematical proof is certainty. And it's knowing the truth without wondering about motivations, or wondering what's in it for some other person, or wondering why this or that person likes you or doesn't like you.

Mathematical proof is truth.

You will die someday, as will I. And when you die, will you be running through your mind all the good things people said about you? Or worrying about all the nasty things?

Either way, will it matter? When you die, you'll still be dead.

You will die and when you die what will any of this matter?

You will still be dead and everyone you know now will die, and everything you build will be torn down.

I know you will die. I know that no matter what you say now billions of years will pass and none of that will be around to matter any more.

Nothing any of you say will still be around billions of years from now.

But this proof that I discovered will still be true.

And as I comprehend absolutes, I understand certainty, and appreciate that which will still be here long after we are all dead.

Take your social nothings with you to your grave which will come soon enough.

You will die. Everyone you know now will die. Everything you build will be torn down.

What do you think is so permanent about your silly social crap?

I take mathematical proof.

As with mathematical proof I hear what God hears, and understand just a little bit, as much as I can, and that is the best that I can do.

[A reply to someone who wrote James usually starts his arguments with statements that are true.]

I have a record of actually trying versus acting like I am trying.

One of the more profound things I realized some years ago was that sci.math'ers were deliberately acting to confuse versus caring about what was mathematically true and that happened when I wrote a very detailed rebuttal to "Nora Baron" only to have that poster delete out everything and just repeat flawed claims.

It's easier to claim you are honest than to actually be honest.

I screw up. I make lots of mistakes in my search for truth and quite a few times over the years I have been rude, very wrong, and regretted things I've posted.

But I at least admit my humanity.

And in the world I thought I knew when I was growing up, human beings do those things and then some, because they work harder at being right, than just trying to look right.

If mathematicians proved me wrong I'd accept that because I know fantasy, and I prefer knowing truth.

But to the extent that they claim they prefer truth and rip on me as if the truth were the opposite, when I admit to being a very flawed human being just trying to get some things straight while these monsters pretend to be perfect—or very damn close to perfect—while they lie, and lie, and lie then I say we have a serious problem people.

The monsters have taken the high ground, and made it illegal to be human.

They spit on mistakes made in the search for truth.

And they piss on failures made in the pursuit of truth, to tell you lies, when they do not actually try to know what is mathematically correct—preferring only to rely on what they can convince you is correct.

I fail. And I cheer my ability to not only fail in trying, but to admit that I fail, and I fail, and I fail.

And through failing honestly, I find success, honestly.

While the demons tell you lies.

### Saturday, April 14, 2007

## Discussion on non-polynomial-factorization-paper

Updated paper! I wrote this new paper a while back to give an alternate approach to proving an important result with non-polynomial factorization, but felt that the ending was kind of muddled.

Earlier today I realized a very simple way to fix it! I now use the polynomial

a^2 - 167a + 595 = 0

where the constant term now has 7 and 17 as factors as I can step through TWO derivations to prove that only one root can have 7 as a factor, oh yeah, and only one root can have 17 as a factor as well.

It is a really cool little proof.

Earlier today I realized a very simple way to fix it! I now use the polynomial

a^2 - 167a + 595 = 0

where the constant term now has 7 and 17 as factors as I can step through TWO derivations to prove that only one root can have 7 as a factor, oh yeah, and only one root can have 17 as a factor as well.

It is a really cool little proof.