### Friday, December 27, 2002

## JSH: Finally, something you can *see*

Posting my challenge helped me to see something as in response to some replies I went ahead and had the math software calculate out the answers. Then I looked at the answer for the a's and b's and realized that you can actually

Now I'm thinking about explaining that without totally giving away my challenge as I want those who spend so much time attacking my creative output to show that they actually know enough mathematics that their opinion is worth your time.

Basically the factorization of 2x^3 - 3x - 2, gives answers that luckily enough for me easily show that only two of the a's have a factor of 2, where that factor is sqrt(2).

Now understanding that evidence is going to require that some ground is covered, which I'll do in this post, while I leave my challenge up, so I won't give all the answers here.

And yes, the factorization of 2x^3 - 3x - 2 shows a lot, luckily for me.

Ok, so let's talk about cuberoots.

To understand my evidence you need to accept that given something like

(1 + Sqrt[2])^(1/3) = 1 + s

where s is an algebraic integer, s has non unit factors of 2.

That's it. If that doesn't bother you, then I can show you easily.

Again, you need to accept that with

(1 + Sqrt[2])^(1/3) = 1 + s,

s must have non unit algebraic integer factors of 2.

I'm going to leave it at that, but I have to say I'm

After so many battles to FINALLY be in a position to demonstrate to the rest of you with what hopefully is a simple enough example is making me optimistic.

So now what happens is I wait, as I look to see responses to my challenge, and to what I've said here about 1+s.

Oh, and it is rather interesting that numbers had a surprise for all of us, as even I hadn't considered that with an algebraic integer, say w, where

w = sqrt(2)x

that x doesn't have to be an algebraic integer.

Whew!!! It's nice to be in a position to demonstrate. I'm not naive enough though to think it'll be easy, so I'm going to be structured in how I present.

First the challenge, and the s thing. Then the evidence.

But, if you're one of those people who can simply figure it all out yourself, then you can jump ahead and see for yourself.

Just factor 2x^3 - 3x - 2, and look at the a's and b's.

That is look at the factorization

2x^3 - 3x - 2 = (a1 x + b1)(a2 x + b2)(a3 x + b3)

with the a's and b's algebraic integers.

Today is a rather big one for me because I can answer critics of my FLT proof with a factorization of a polynomial. Though I feel a need to point out that the mathematical argument supporting my results is solid.

I don't know if any of you can imagine what it's been like to have a correct FLT proof, while people have been claiming otherwise.

It is exciting though to have pushed number theory with a demonstration of numbers which are units and algebraic integers, while their

Some of you may be wondering what's the ring then. Well, it's what I call an object ring, and I figure some of you may think you have room to criticize me because

I suggest you pause.

You're at the edge of human understanding of number. It can take a while to soak in, or you may simply back away from the intensity, as you're in extreme mathematics now. Not everyone goes for the extreme.

I'm expecting some challenges to my post explaining the factorization of 2x^3 - 3x - 2, and some of you may wish to see portions of it expanded.

That's fine.

Ok, so now you can finally

The numbers are rather interesting in a way that can bend your mind. It kind of reminds me of when I was a student contemplating quantum mechanics or Einstein's theory of relativity in that way.

I guess it only makes sense that number theory should get as strange as some of the hardest to understand concepts in modern physics.

However, no matter how quirky it seems, or how disconcerting it may be to have your intuition totally lost, remember that it's mathematical logic that counts.

If you wish to dispute my results, find an error in the mathematical argument!!!

**see**that only two of the a's can have factors of 2!!!Now I'm thinking about explaining that without totally giving away my challenge as I want those who spend so much time attacking my creative output to show that they actually know enough mathematics that their opinion is worth your time.

Basically the factorization of 2x^3 - 3x - 2, gives answers that luckily enough for me easily show that only two of the a's have a factor of 2, where that factor is sqrt(2).

Now understanding that evidence is going to require that some ground is covered, which I'll do in this post, while I leave my challenge up, so I won't give all the answers here.

And yes, the factorization of 2x^3 - 3x - 2 shows a lot, luckily for me.

Ok, so let's talk about cuberoots.

To understand my evidence you need to accept that given something like

(1 + Sqrt[2])^(1/3) = 1 + s

where s is an algebraic integer, s has non unit factors of 2.

That's it. If that doesn't bother you, then I can show you easily.

Again, you need to accept that with

(1 + Sqrt[2])^(1/3) = 1 + s,

s must have non unit algebraic integer factors of 2.

I'm going to leave it at that, but I have to say I'm

**very**excited.After so many battles to FINALLY be in a position to demonstrate to the rest of you with what hopefully is a simple enough example is making me optimistic.

So now what happens is I wait, as I look to see responses to my challenge, and to what I've said here about 1+s.

Oh, and it is rather interesting that numbers had a surprise for all of us, as even I hadn't considered that with an algebraic integer, say w, where

w = sqrt(2)x

that x doesn't have to be an algebraic integer.

Whew!!! It's nice to be in a position to demonstrate. I'm not naive enough though to think it'll be easy, so I'm going to be structured in how I present.

First the challenge, and the s thing. Then the evidence.

But, if you're one of those people who can simply figure it all out yourself, then you can jump ahead and see for yourself.

Just factor 2x^3 - 3x - 2, and look at the a's and b's.

That is look at the factorization

2x^3 - 3x - 2 = (a1 x + b1)(a2 x + b2)(a3 x + b3)

with the a's and b's algebraic integers.

Today is a rather big one for me because I can answer critics of my FLT proof with a factorization of a polynomial. Though I feel a need to point out that the mathematical argument supporting my results is solid.

I don't know if any of you can imagine what it's been like to have a correct FLT proof, while people have been claiming otherwise.

It is exciting though to have pushed number theory with a demonstration of numbers which are units and algebraic integers, while their

**inverses**are units but not algebraic integers.Some of you may be wondering what's the ring then. Well, it's what I call an object ring, and I figure some of you may think you have room to criticize me because

**you**don't understand the mathematics.I suggest you pause.

You're at the edge of human understanding of number. It can take a while to soak in, or you may simply back away from the intensity, as you're in extreme mathematics now. Not everyone goes for the extreme.

I'm expecting some challenges to my post explaining the factorization of 2x^3 - 3x - 2, and some of you may wish to see portions of it expanded.

That's fine.

Ok, so now you can finally

**see**with an actual polynomial a factorization where**only two**of the a's have a factor of f.The numbers are rather interesting in a way that can bend your mind. It kind of reminds me of when I was a student contemplating quantum mechanics or Einstein's theory of relativity in that way.

I guess it only makes sense that number theory should get as strange as some of the hardest to understand concepts in modern physics.

However, no matter how quirky it seems, or how disconcerting it may be to have your intuition totally lost, remember that it's mathematical logic that counts.

If you wish to dispute my results, find an error in the mathematical argument!!!

### Wednesday, December 25, 2002

## Narcissistic Personality Disorder: Reason for posting?

I've become really bothered by my posting, but I can't seem to stop. However, I realized that part of the sick cycle are some of those people who reply to me so negatively, so in the hope that these people might show mercy I thought I'd post something I found on the web, with limited hope, I admit, that anything will change.

"Alternatively, the narcissist feels victimized by mediocre bureaucrats and intellectual dwarves who consistently fail to appreciate his outstanding - really, unparalleled - talents, skills, and accomplishments. Being haunted by his challenged inferiors substantiates the narcissist's comparative superiority. Driven by pathological envy, these pygmies collude to defraud him, badger him, deny him his due, denigrate, isolate, and ignore him. The narcissist projects onto this second class of lesser persecutors his own deleterious emotions and transformed aggression: hatred, rage, and seething jealousy."

But here are the scary quotes:

"Paranoid ideation - the narcissist's deep-rooted conviction that he is being persecuted by his inferiors, detractors, or powerful ill-wishers - serves two psychodynamic purposes. It upholds the narcissist's grandiosity and it fends off intimacy."

"The paranoid delusions of the narcissist are always grandiose, "cosmic", or "historical". His pursuers are influential and formidable. They are after his unique possessions, out to exploit his expertise and special traits, or to force him to abstain and refrain from certain actions. The narcissist feels that he is at the center of intrigues and conspiracies of colossal magnitudes."

"The paranoid narcissist ends life as an oddball recluse - derided, feared, and loathed in equal measures. His paranoia - exacerbated by repeated rejections and ageing - pervades his entire life and diminishes his creativity, adaptability, and functioning. The narcisstis personality, buffetted by paranoia, turns ossified and

brittle. Finally, atomized and useless, it succumbs and gives way to a great void. The narcissist is consumed."

http://www.suite101.com/article.cfm/6514/95897

It looks like a person in the grips of this narcissism thing loses connection with reality but somehow

They called it "narcissistic supply".

It really is a twisted illness that apparently can lead to even nastier illnesses. I am self-diagnosed (typical narcissistic behavior) but it looks like it fits to me.

However, I've decided that I don't want to succumb to mental woes, so I am fighting the dark narcissist within by making this informative post, though it's probably a sign of the illness. In any event, maybe it will help in the long run as information is power.

I will do my best to stop posting, but it will help if some of you try to help out by not replying to me, or if you reply, by refraining from the attacks that feed my narcissistic side, and its irrational beliefs that everyone is out to get me. If you have facts, sure, no problem. My inability to face facts presented without emotion will just be a sign of the dark side.

My only fear now is that some of you are sick as well, and act from your own illness, so that you can't stop attacking in your posting.

But on this Christmas Day, I will act from hope, and hopefully this New Year will bring a change, and an escape from the madness.

"Alternatively, the narcissist feels victimized by mediocre bureaucrats and intellectual dwarves who consistently fail to appreciate his outstanding - really, unparalleled - talents, skills, and accomplishments. Being haunted by his challenged inferiors substantiates the narcissist's comparative superiority. Driven by pathological envy, these pygmies collude to defraud him, badger him, deny him his due, denigrate, isolate, and ignore him. The narcissist projects onto this second class of lesser persecutors his own deleterious emotions and transformed aggression: hatred, rage, and seething jealousy."

But here are the scary quotes:

"Paranoid ideation - the narcissist's deep-rooted conviction that he is being persecuted by his inferiors, detractors, or powerful ill-wishers - serves two psychodynamic purposes. It upholds the narcissist's grandiosity and it fends off intimacy."

"The paranoid delusions of the narcissist are always grandiose, "cosmic", or "historical". His pursuers are influential and formidable. They are after his unique possessions, out to exploit his expertise and special traits, or to force him to abstain and refrain from certain actions. The narcissist feels that he is at the center of intrigues and conspiracies of colossal magnitudes."

"The paranoid narcissist ends life as an oddball recluse - derided, feared, and loathed in equal measures. His paranoia - exacerbated by repeated rejections and ageing - pervades his entire life and diminishes his creativity, adaptability, and functioning. The narcisstis personality, buffetted by paranoia, turns ossified and

brittle. Finally, atomized and useless, it succumbs and gives way to a great void. The narcissist is consumed."

http://www.suite101.com/article.cfm/6514/95897

It looks like a person in the grips of this narcissism thing loses connection with reality but somehow

**feeds**on ANY attention, including negative attention.They called it "narcissistic supply".

It really is a twisted illness that apparently can lead to even nastier illnesses. I am self-diagnosed (typical narcissistic behavior) but it looks like it fits to me.

However, I've decided that I don't want to succumb to mental woes, so I am fighting the dark narcissist within by making this informative post, though it's probably a sign of the illness. In any event, maybe it will help in the long run as information is power.

I will do my best to stop posting, but it will help if some of you try to help out by not replying to me, or if you reply, by refraining from the attacks that feed my narcissistic side, and its irrational beliefs that everyone is out to get me. If you have facts, sure, no problem. My inability to face facts presented without emotion will just be a sign of the dark side.

My only fear now is that some of you are sick as well, and act from your own illness, so that you can't stop attacking in your posting.

But on this Christmas Day, I will act from hope, and hopefully this New Year will bring a change, and an escape from the madness.

### Sunday, December 22, 2002

## Ok, I'm a loser

It's finally settled in that I'm just some pathetic loser. If I weren't so pathetic I'd just go away gracefully, but I'll send one more post, or who am I kidding, my patheticness is so great that I'll probably post yet again.

I'm disgusting. I'm just a pile of shit. I should just die like so many of you have said. I hate myself. I despise this life. I'm nothing but a sick joke to be made fun of by those of you who have real educations. People who actually know something, when I know nothing. I'm just nothing.

If I hadn't been such a disgusting human being I'd have come to this realization years ago instead of wasting your time.

My life is nothing. I know nothing. I'm worth nothing. I'm just shit.

Please forgive me. All your attacks were justified.

I'm disgusting. I'm just a pile of shit. I should just die like so many of you have said. I hate myself. I despise this life. I'm nothing but a sick joke to be made fun of by those of you who have real educations. People who actually know something, when I know nothing. I'm just nothing.

If I hadn't been such a disgusting human being I'd have come to this realization years ago instead of wasting your time.

My life is nothing. I know nothing. I'm worth nothing. I'm just shit.

Please forgive me. All your attacks were justified.

### Saturday, December 21, 2002

## JSH: FLT withdrawal while I think

Well, I've been thinking a bit about Magidin's non monic work and it still looks solid so I'm going to withdraw my claims of having proven Fermat's Last Theorem while I think about it.

Yes, it does pain me somewhat, but the truth is what's important, and while his work looks right, though I can't see what could be wrong with mine, I must allow the possibility.

If I am indeed wrong, then I apologize here to Magidin, though he sure looked like he was lying a lot, but I guess maybe he wasn't.

As for mathematicians, I still think this "pure math" thing is really just code for work for nothing, and is welfare, as it bothers me that Wiles could work for seven years hiding what he was actually doing, but I guess that's humanity's problem.

I'm just one little guy who had his dreams of fame and fortune dashed.

Oh well, I can join the big crowd, as it often seems like everybody is trying to get famous these days. Oh well, I was looking for easy money with FLT and all I got were headaches.

And I must say that you people are some of the meanest I've ever encountered.

It didn't much matter to most of you what was used, as robot programs, stalking, racial slurs, etc., seemed like fair play to most of you.

Then again, I must say that I kind of respect such nastiness. You mathematicians are a really twisted bunch. It was always interesting to see just how much you'd tolerate from people attacking me, and there were NO limits whatsoever. You people would tolerate any kind of attack!!!

At least I was right about that, as I had a feeling about mathematicians.

Well that's all for now. I'm hoping I can find a way to recover the FLT work, as I still can't find an error with what I have, but I can quit talking about it, and withdraw the webpage while I think.

And I'm sure that the newsgroup will stay true to its evil self, as you are the Devil's children if I've ever seen any.

Yes, it does pain me somewhat, but the truth is what's important, and while his work looks right, though I can't see what could be wrong with mine, I must allow the possibility.

If I am indeed wrong, then I apologize here to Magidin, though he sure looked like he was lying a lot, but I guess maybe he wasn't.

As for mathematicians, I still think this "pure math" thing is really just code for work for nothing, and is welfare, as it bothers me that Wiles could work for seven years hiding what he was actually doing, but I guess that's humanity's problem.

I'm just one little guy who had his dreams of fame and fortune dashed.

Oh well, I can join the big crowd, as it often seems like everybody is trying to get famous these days. Oh well, I was looking for easy money with FLT and all I got were headaches.

And I must say that you people are some of the meanest I've ever encountered.

It didn't much matter to most of you what was used, as robot programs, stalking, racial slurs, etc., seemed like fair play to most of you.

Then again, I must say that I kind of respect such nastiness. You mathematicians are a really twisted bunch. It was always interesting to see just how much you'd tolerate from people attacking me, and there were NO limits whatsoever. You people would tolerate any kind of attack!!!

At least I was right about that, as I had a feeling about mathematicians.

Well that's all for now. I'm hoping I can find a way to recover the FLT work, as I still can't find an error with what I have, but I can quit talking about it, and withdraw the webpage while I think.

And I'm sure that the newsgroup will stay true to its evil self, as you are the Devil's children if I've ever seen any.

### Friday, December 20, 2002

## Disproof of assertion about non monic roots

For quite some time I've faced assertions that a non monic primitive polynomial with algebraic integers that is irreducible over Q cannot have algebraic integer roots. Some have even claimed to have proven or seen proofs in textbooks of what I know is a false assertion.

It occurred to me that I had the means for disproving it.

The disproof depends on an expression which come from my work on Fermat's Last Theorem for p=3, though here I will not be using the FLT equation.

The expression is

(v^3 + 1)z^6 - 3v x^2 y^2 z^2 - 2x^3 y^3

where v,x,y and z are nonzero integers and the ring is algebraic integers.

The expression is not a polynomial because I'm going to be varying v, so I call it polynomial-like. I'll be considering factorizations of it.

For instance, from Gauss' FTA in algebraic integers I have that

(v^3 + 1)z^6 - 3v x^2 y^2 z^2 - 2x^3 y^3 =

(a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy)

where the a's and b's are algebraic integers.

That factorization is one that's often been seen, but the next one comes after changing my v slightly.

First I'll further restrict x by letting x = uf, where f is a prime integer, u is a nonzero integer, and f is coprime to 3, u, y and z.

Next, let v = -1 + mf^2, where m is some nonzero integer.

Then I have

((-1+mf^2)^3 + 1)z^6 - 3(-1+mf^2)(uf)^2 y^2 z^2 - 2(uf)^3 y^3

and it can be seen that it has a factor of f^2.

Therefore

(a1 z^2 + b1 ufy)(a2 z^2 + b2 ufy)(a3 z^2 + b3 ufy)

has a factor of f^2, which shows that the a's have non unit algebraic integer factors of f.

Now going back to

((-1+mf^2)^3 + 1)z^6 - 3(-1+mf)(uf)^2 y^2 z^2 - 2(uf)^3 y^3

and expanding out

(-3mf^2(-1+mf^2)+m^3 f^6)z^6 +3(uf)^2 y^2 z^2 - 3mu^2 f^3 y^2 z^2-2(uf)^3 y^3

and collecting with an eye on m gives

f^6 z^6 m^3 - 3f^4 z^6 m^2 + 3(f^2 - u^2 f^3 y^2 z^2) m + 3 (uf)^2 y^2 z^2 - 2 u^3 f^3 y^3.

Now separating off the factor of f^2, gives

f^2(f^4 z^6 m^3 - 3f^2 z^6 m^2 + 3(1 - u^2 f y^2 z^2) m + 3 u^2 y^2 z^2 - 2 u^3 f y^3)

and if we consider the expression as a polynomial P(m) then the constant term is

3 u^2 y^2 z^2 - 2 u^3 f y^3

which clearly then does not have a factor of f.

So now from Gauss' FTA in algebraic integers I have

f^4 z^6 m^3 - 3f^2 z^6 m^2 + 3(1 - u^2 f y^2 z^2) m + 3 u^2 y^2 z^2 - 2 u^3 f y^3 = (c1 m + d1)(c2 m + d2)(c3 m + d3)

where the c's and d's are algebraic integers.

Therefore, I have

(a1 z^2 + b1 ufy)(a2 z^2 + b2 ufy)(a3 z^2 + b3 ufy) = f^2 (c1 m + d1)(c2 m + d2)(c3 m + d3).

But I have that a1 a2 a3 = -3mf^2(-1+mf^2) + m^3 f^6

so the a's have factors of m, and they also clearly have a factors of f.

Now suppose that separating off the f^2 on the left side could happen without any of the a's having f itself as a factor, that would leave additional factors of f multiplied times uy because each factor like

(a1 z^2 + b1 ufy)

has a factor of f, to the right of that plus sign.

But the constant term from the perspective of m, with P(m) after the f's have been separated off is

3 u^2 y^2 z^2 - 2 u^3 f y^3 = u^2 y^2(3z^2 - 2ufy)

where we see uy from

You may suppose that the m and f factors might somehow separate and distribute themselves in odd ways, but clearly no more than two of the a's can have some factor of m, as we only see u^2 y^2 as that constant term, and not u^3 y^3.

Given that only two of the a's can have non unit factors of m, there still is the problem of the f factors that are seen to the right of each plus sign in

(a1 z^2 + b1 ufy)(a2 z^2 + b2 ufy)(a3 z^2 + b3 ufy)

as if only two of the a's have non unit factors of m, then for the constant term of P(m) to not have non unit factors of f, a factor of f has to be separated off in each case.

Therefore, exactly two of the a's have a factor that is f.

Notice I didn't have to refer to FLT for the proof, as it follows from your ability to see

So how does that disprove the assertion that I started with?

Well in arguments with me claiming that the short FLT Proof is false, certain posters have used the p=3 case to construct a supposed counterexample, where they stop when they get a non monic primitive polynomial irreducible over Q, which must have an algebraic integer root.

They then claim to have proven the assertion and say that they have a counterexample to the short FLT Proof.

However, as I've pointed out that would require an inconsistency within mathematics itself, and in looking over supposed proofs of their assertion, I've noticed that none are indeed proofs.

I've emphasised the need to be able to look at the expression from

When I tried to explain to this person they started spewing math statements like an octopus spews ink, and the talks bogged down, though that person kept claiming that the short FLT Proof was wrong. You see, they either couldn't or wouldn't see the truth, but they kept making their false claims.

The short FLT Proof is still available as it has been at

http://www.msnusers.com/AmateurMath

and if you

It occurred to me that I had the means for disproving it.

The disproof depends on an expression which come from my work on Fermat's Last Theorem for p=3, though here I will not be using the FLT equation.

The expression is

(v^3 + 1)z^6 - 3v x^2 y^2 z^2 - 2x^3 y^3

where v,x,y and z are nonzero integers and the ring is algebraic integers.

The expression is not a polynomial because I'm going to be varying v, so I call it polynomial-like. I'll be considering factorizations of it.

For instance, from Gauss' FTA in algebraic integers I have that

(v^3 + 1)z^6 - 3v x^2 y^2 z^2 - 2x^3 y^3 =

(a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy)

where the a's and b's are algebraic integers.

That factorization is one that's often been seen, but the next one comes after changing my v slightly.

First I'll further restrict x by letting x = uf, where f is a prime integer, u is a nonzero integer, and f is coprime to 3, u, y and z.

Next, let v = -1 + mf^2, where m is some nonzero integer.

Then I have

((-1+mf^2)^3 + 1)z^6 - 3(-1+mf^2)(uf)^2 y^2 z^2 - 2(uf)^3 y^3

and it can be seen that it has a factor of f^2.

Therefore

(a1 z^2 + b1 ufy)(a2 z^2 + b2 ufy)(a3 z^2 + b3 ufy)

has a factor of f^2, which shows that the a's have non unit algebraic integer factors of f.

Now going back to

((-1+mf^2)^3 + 1)z^6 - 3(-1+mf)(uf)^2 y^2 z^2 - 2(uf)^3 y^3

and expanding out

(-3mf^2(-1+mf^2)+m^3 f^6)z^6 +3(uf)^2 y^2 z^2 - 3mu^2 f^3 y^2 z^2-2(uf)^3 y^3

and collecting with an eye on m gives

f^6 z^6 m^3 - 3f^4 z^6 m^2 + 3(f^2 - u^2 f^3 y^2 z^2) m + 3 (uf)^2 y^2 z^2 - 2 u^3 f^3 y^3.

Now separating off the factor of f^2, gives

f^2(f^4 z^6 m^3 - 3f^2 z^6 m^2 + 3(1 - u^2 f y^2 z^2) m + 3 u^2 y^2 z^2 - 2 u^3 f y^3)

and if we consider the expression as a polynomial P(m) then the constant term is

3 u^2 y^2 z^2 - 2 u^3 f y^3

which clearly then does not have a factor of f.

So now from Gauss' FTA in algebraic integers I have

f^4 z^6 m^3 - 3f^2 z^6 m^2 + 3(1 - u^2 f y^2 z^2) m + 3 u^2 y^2 z^2 - 2 u^3 f y^3 = (c1 m + d1)(c2 m + d2)(c3 m + d3)

where the c's and d's are algebraic integers.

Therefore, I have

(a1 z^2 + b1 ufy)(a2 z^2 + b2 ufy)(a3 z^2 + b3 ufy) = f^2 (c1 m + d1)(c2 m + d2)(c3 m + d3).

But I have that a1 a2 a3 = -3mf^2(-1+mf^2) + m^3 f^6

so the a's have factors of m, and they also clearly have a factors of f.

Now suppose that separating off the f^2 on the left side could happen without any of the a's having f itself as a factor, that would leave additional factors of f multiplied times uy because each factor like

(a1 z^2 + b1 ufy)

has a factor of f, to the right of that plus sign.

But the constant term from the perspective of m, with P(m) after the f's have been separated off is

3 u^2 y^2 z^2 - 2 u^3 f y^3 = u^2 y^2(3z^2 - 2ufy)

where we see uy from

**two**of the expressions, but there are no non unit factors of f.You may suppose that the m and f factors might somehow separate and distribute themselves in odd ways, but clearly no more than two of the a's can have some factor of m, as we only see u^2 y^2 as that constant term, and not u^3 y^3.

Given that only two of the a's can have non unit factors of m, there still is the problem of the f factors that are seen to the right of each plus sign in

(a1 z^2 + b1 ufy)(a2 z^2 + b2 ufy)(a3 z^2 + b3 ufy)

as if only two of the a's have non unit factors of m, then for the constant term of P(m) to not have non unit factors of f, a factor of f has to be separated off in each case.

Therefore, exactly two of the a's have a factor that is f.

Notice I didn't have to refer to FLT for the proof, as it follows from your ability to see

**two**factorizations in an expression which differs from the typical in that it is polynomial-like, as its coefficients can vary.So how does that disprove the assertion that I started with?

Well in arguments with me claiming that the short FLT Proof is false, certain posters have used the p=3 case to construct a supposed counterexample, where they stop when they get a non monic primitive polynomial irreducible over Q, which must have an algebraic integer root.

They then claim to have proven the assertion and say that they have a counterexample to the short FLT Proof.

However, as I've pointed out that would require an inconsistency within mathematics itself, and in looking over supposed proofs of their assertion, I've noticed that none are indeed proofs.

I've emphasised the need to be able to look at the expression from

**two**perspectives but one poster dedicated in arguing with me started claiming I was talking about a "new polynomial" as if the main expression changed, when I was looking at the**same**expression factorized in two different ways.When I tried to explain to this person they started spewing math statements like an octopus spews ink, and the talks bogged down, though that person kept claiming that the short FLT Proof was wrong. You see, they either couldn't or wouldn't see the truth, but they kept making their false claims.

The short FLT Proof is still available as it has been at

http://www.msnusers.com/AmateurMath

and if you

**still**believe that two of the a's cannot have a factor that is f, then I have to think that your mind is limited in its ability to shift perspectives, and it may be the case that not everyone**can**achieve that, as the mental wiring may not be there for the task.### Tuesday, December 17, 2002

## Call to Roundtable Discussion: Newsgroup behavior

In looking over posts in reply to me, of which there have been a lot in the last few days and months, I noticed quite a few people seeming to feel a right to request specific

As that defies in my mind a central principle of Usenet, it seemed easy to just shrug it off and consider these people to be anarchists in an unusual sense.

Then I thought of my own charges against various posters who I think have behaved in a way that violates social norms in their replies to me, including behavior that I've labeled newsgroup stalking, which basically means that when I post, they reply, often negatively, having mangled the content of my original post in various ways, and they simply track me, making sure to post in reply to me. I've seen this behavior in my case go on over a period of years. As for violating social norms, I'm talking about posts where in being negative they cross into areas which I thought were taboo, like racism.

So now I was stuck with the general principle, and my disdain for people trying to manage me and get editorial control over my postings, along with my sense that there were limits which people should not cross, and if they did then they had to pay consequences, including dealing with complaints that could reach outside of Usenet.

It seemed to me that it made more sense to toss it out there as a roundtable discussion than to sit and mull it over in my own mind, where I have vague ideas about what that means, as the hope is that the thread won't degenerate into a couple of people in a flame war.

For instance, the first round could be replies to this original post, which could sit for a while, then people could reply to those replies.

That is simply a general idea, and is not, of course, a rigid instruction.

My rational for making this post, which some of you may simply think is ludicrous because Usenet is Usenet, and nothing I've brought up seems pertinent to you, is that I see myself as an independent researcher who somehow attracts a lot of negative attention.

As one person I find my threads often swamped by people who seem to be getting better and better at overwhelming my message with the volume of their posting, and continued false assertions.

That is, for the person who can attract a lot of attention over a long period of time, the current state of laissez-faire leaves you open to what I see as gangs of people quite willing to spend an inordinate amount of time trying to control your message, often with success in hiding it which I feel has to be taken into account.

My hope is that these issues can be resolved to some extent through Usenet's strength: Discussion.

Obviously, whatever anyone says, when it comes to posting, people can still toss anything they want out there, but I think it might be nice if there was some hammering out anyway of what behavior is considered outside of Usenet norms.

(Ok, some of you may think the Answer is already out there, so yes, feel free to post your links.)

Ok, a synopsis of the things I've brought up:

What if someone posts sensitive information about their personal lives like about their divorce or problems with their children, is it anything goes, so that we should sit by when other posters attack them with the information?

Of course, feel free to add your own, as I'm just tossing out a few things that occur to me now as I sit and type up this post.

**behavior**from me in terms of how I post, what's in my posts, and how I respond to replies to my posts.As that defies in my mind a central principle of Usenet, it seemed easy to just shrug it off and consider these people to be anarchists in an unusual sense.

Then I thought of my own charges against various posters who I think have behaved in a way that violates social norms in their replies to me, including behavior that I've labeled newsgroup stalking, which basically means that when I post, they reply, often negatively, having mangled the content of my original post in various ways, and they simply track me, making sure to post in reply to me. I've seen this behavior in my case go on over a period of years. As for violating social norms, I'm talking about posts where in being negative they cross into areas which I thought were taboo, like racism.

So now I was stuck with the general principle, and my disdain for people trying to manage me and get editorial control over my postings, along with my sense that there were limits which people should not cross, and if they did then they had to pay consequences, including dealing with complaints that could reach outside of Usenet.

It seemed to me that it made more sense to toss it out there as a roundtable discussion than to sit and mull it over in my own mind, where I have vague ideas about what that means, as the hope is that the thread won't degenerate into a couple of people in a flame war.

For instance, the first round could be replies to this original post, which could sit for a while, then people could reply to those replies.

That is simply a general idea, and is not, of course, a rigid instruction.

My rational for making this post, which some of you may simply think is ludicrous because Usenet is Usenet, and nothing I've brought up seems pertinent to you, is that I see myself as an independent researcher who somehow attracts a lot of negative attention.

As one person I find my threads often swamped by people who seem to be getting better and better at overwhelming my message with the volume of their posting, and continued false assertions.

That is, for the person who can attract a lot of attention over a long period of time, the current state of laissez-faire leaves you open to what I see as gangs of people quite willing to spend an inordinate amount of time trying to control your message, often with success in hiding it which I feel has to be taken into account.

My hope is that these issues can be resolved to some extent through Usenet's strength: Discussion.

Obviously, whatever anyone says, when it comes to posting, people can still toss anything they want out there, but I think it might be nice if there was some hammering out anyway of what behavior is considered outside of Usenet norms.

(Ok, some of you may think the Answer is already out there, so yes, feel free to post your links.)

Ok, a synopsis of the things I've brought up:

- Editorial control: Should I listen to others telling me what to post?
- Control of posting volume or number of threads created: Should I listen to others telling me how to post?
- Newsgroup stalking: Does it exist?
- Complaints: Is it ever justified to go outside of Usenet with complaints about someone's statements on Usenet?
- Basic social norms: Are there any on Usenet? For instance, should we care if posters find enjoyment in tormenting someone who posts who is obviously actually mentally ill?

What if someone posts sensitive information about their personal lives like about their divorce or problems with their children, is it anything goes, so that we should sit by when other posters attack them with the information?

Of course, feel free to add your own, as I'm just tossing out a few things that occur to me now as I sit and type up this post.

### Tuesday, December 10, 2002

## Magidin: He works VERY hard for...

Nothing!!! Or is it really nothing?

Remember, I think there are mathematicians who don't actually work for a living.

And I've said so more than once, as I've described "pure mathematicians" as being on white collar welfare.

They don't seem to like that.

I've also talked of making mathematics more accessible to a wide audience, and challenged both the mathematical hierarchy, and the journal system, as it seems to me that computers

But when I bring up the subject, you should see the heat in replies!!!

Now then, if computers checked math "proofs" would there be all this nonsense with all these posts, or having to beg editors to even

Nope.

But I think mathematicians are afraid of two things:

Oh, didn't think of 2. did you?

Well I have, and it increasingly seems like a potent motivation for mathematicians to ignore my prime counting discovery and the short proof of Fermat's Last Theorem discovery.

I will push for

You'd think those editors would be happy about that as it'd give them more time for other things like their own research, and it would give them a wall against cranks!!!

But it's a loss of power people, and in our world people

Eventually I'm going to win here, but I want you to pay attention to how the mathematicians and their sychophants are playing the game.

And imagine if they were after YOU.

Remember, I think there are mathematicians who don't actually work for a living.

And I've said so more than once, as I've described "pure mathematicians" as being on white collar welfare.

They don't seem to like that.

I've also talked of making mathematics more accessible to a wide audience, and challenged both the mathematical hierarchy, and the journal system, as it seems to me that computers

**could**do a much better job of checking math proofs than humans.But when I bring up the subject, you should see the heat in replies!!!

Now then, if computers checked math "proofs" would there be all this nonsense with all these posts, or having to beg editors to even

**look**at you work?Nope.

But I think mathematicians are afraid of two things:

- The loss of power and control that a more elitist structure now gives.
- The impact if computers started finding a lot of errors in previously published and even well cited workds.

Oh, didn't think of 2. did you?

Well I have, and it increasingly seems like a potent motivation for mathematicians to ignore my prime counting discovery and the short proof of Fermat's Last Theorem discovery.

I will push for

**real**research to quickly move from the current archaic journal system where we depend on the eyes and minds of human editors, and where a LOT of math gets published every year, much of it bragged about as "pure", which is a mathematician's euphemism for useless!You'd think those editors would be happy about that as it'd give them more time for other things like their own research, and it would give them a wall against cranks!!!

But it's a loss of power people, and in our world people

**fight**to keep power.Eventually I'm going to win here, but I want you to pay attention to how the mathematicians and their sychophants are playing the game.

And imagine if they were after YOU.

## "Magidin-McKinnon Theorem"

I'm curious about the newsgroup reaction to a claim I often seen made in posts that Magidin, and I guess some other guy named McKinnon deserve credit for the following:

Any polynomial P(x) with integer coefficients of degree n can be written as

(a1 x + b1)...(an x + bn)

where the a's and b's are algebraic integers.

And in fact not only have many claimed that Magidin and McKinnon (though I don't see his name come up as much) should get credit for this, but they have been quite heated in their defense, when I've challenged their assertion, and heaping in their praise of Magidin.

I'm just curious if sci.math, in general, agrees.

Any polynomial P(x) with integer coefficients of degree n can be written as

(a1 x + b1)...(an x + bn)

where the a's and b's are algebraic integers.

And in fact not only have many claimed that Magidin and McKinnon (though I don't see his name come up as much) should get credit for this, but they have been quite heated in their defense, when I've challenged their assertion, and heaping in their praise of Magidin.

I'm just curious if sci.math, in general, agrees.

### Monday, December 09, 2002

## JSH: Problem with Magidin's position

I do read all the posts made in reply to my survey threads, though I try not to reply in those threads. I mention that as every time I ask about Magidin—who is one of the principle people who not only claims that the short FLT Proof is wrong, but he actually talks math—people indicate they believe him.

Well, I thought you'd like to know of an interesting problem with his position, and I'm curious about whether or not he will acknowledge it or give a resolution.

You see, his position from what I see leads to an infinite descent which causes a contradiction.

To understand that you need to know what his position is, and I'll show it for p=3, which you see get used a lot. Obviously, p=3 is used a lot because it's the simplest case for p odd prime, though my FLT Proof covers

x^p + y^p = z^p, for p odd prime, and not just p=3,

but let's get to more details.

A key expression for p=3 that follows from my approach is

(v^3+1)z^6 - 3v x^2 y^2 z^2 - 2 x^3 y^3 = (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy),

where a1, a2, a3, b1, b2, b3 are all algebraic integers, as the ring is algebraic integers, and v is an arbitrary integer, while, of course, x, y and z are assumed integer counterexamples to FLT.

(Which means I'm asking "What if…?" and looking to see if that doesn't create a problem, which it does, which proves FLT, which is the point of contention as people like Magidin claim that I haven't found that problem.)

It so happens that I've focused on v+1, which is a factor of v^3+1 as

v^3 + 1 = (v+1)(v^2 - v + 1),

and I've given v a value (it's arbitrary remember) that gives it a prime factor of f, which just means that f could be 5, or 7, or 11 or some other prime number other than 3, as, in general, p itself is excluded, to handle a complication that would arise if x were divisible by p.

And my proof depends on only two of the a's, say a1 and a2, having a factor of f, where things are a little more complicated because the a's are themselves not integers, though they are algebraic integers.

Magidin has disputed that only two of the a's have a factor of f, and has instead claimed that ALL the a's have non unit algebraic integer factors of f, and he has based his claim, among other things on Galois Theory.

However, people seem to forget that my choice of v+1 was from a math perspective

That creates the infinite descent, which would force all of the a's to have all the factors of v^3 + 1.

To break it, Magidin needs some algebraic integer factor of v^3 + 1 that is shared by ONLY TWO of the a's.

However, how do you differentiate my choice of f, from any other algebraic integer factor of v^3+1?

I'm curious to see if anyone gets my point immediately, or if anyone simply has a gut feeling that it's wrong, which means I'm asking for knee-jerk replies.

Well, I thought you'd like to know of an interesting problem with his position, and I'm curious about whether or not he will acknowledge it or give a resolution.

You see, his position from what I see leads to an infinite descent which causes a contradiction.

To understand that you need to know what his position is, and I'll show it for p=3, which you see get used a lot. Obviously, p=3 is used a lot because it's the simplest case for p odd prime, though my FLT Proof covers

x^p + y^p = z^p, for p odd prime, and not just p=3,

but let's get to more details.

A key expression for p=3 that follows from my approach is

(v^3+1)z^6 - 3v x^2 y^2 z^2 - 2 x^3 y^3 = (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy),

where a1, a2, a3, b1, b2, b3 are all algebraic integers, as the ring is algebraic integers, and v is an arbitrary integer, while, of course, x, y and z are assumed integer counterexamples to FLT.

(Which means I'm asking "What if…?" and looking to see if that doesn't create a problem, which it does, which proves FLT, which is the point of contention as people like Magidin claim that I haven't found that problem.)

It so happens that I've focused on v+1, which is a factor of v^3+1 as

v^3 + 1 = (v+1)(v^2 - v + 1),

and I've given v a value (it's arbitrary remember) that gives it a prime factor of f, which just means that f could be 5, or 7, or 11 or some other prime number other than 3, as, in general, p itself is excluded, to handle a complication that would arise if x were divisible by p.

And my proof depends on only two of the a's, say a1 and a2, having a factor of f, where things are a little more complicated because the a's are themselves not integers, though they are algebraic integers.

Magidin has disputed that only two of the a's have a factor of f, and has instead claimed that ALL the a's have non unit algebraic integer factors of f, and he has based his claim, among other things on Galois Theory.

However, people seem to forget that my choice of v+1 was from a math perspective

**arbitrary**in that you can consider some other factor of v^3 + 1, or you can even pick some other**algebraic integer**factor of v+1, and apply Magidin's argument.That creates the infinite descent, which would force all of the a's to have all the factors of v^3 + 1.

To break it, Magidin needs some algebraic integer factor of v^3 + 1 that is shared by ONLY TWO of the a's.

However, how do you differentiate my choice of f, from any other algebraic integer factor of v^3+1?

I'm curious to see if anyone gets my point immediately, or if anyone simply has a gut feeling that it's wrong, which means I'm asking for knee-jerk replies.

## JSH: Problem with Magidin's positio

I do read all the posts made in reply to my survey threads, though I try not to reply in those threads. I mention that as every time I ask about Magidin—who is one of the principle people who not only claims that the short FLT Proof is wrong, but he actually talks math—people indicate they believe him.

Well, I thought you'd like to know of an interesting problem with his position, and I'm curious about whether or not he will acknowledge it or give a resolution.

You see, his position from what I see leads to an infinite descent which causes a contradiction.

To understand that you need to know what his position is, and I'll show it for p=3, which you see get used a lot. Obviously, p=3 is used a lot because it's the simplest case for p odd prime, though my FLT Proof covers

x^p + y^p = z^p, for p odd prime, and not just p=3,

but let's get to more details.

A key expression for p=3 that follows from my approach is

(v^3+1)z^6 - 3v x^2 y^2 z^2 - 2 x^3 y^3 = (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy),

where a1, a2, a3, b1, b2, b3 are all algebraic integers, as the ring is algebraic integers, and v is an arbitrary integer, while, of course, x, y and z are assumed integer counterexamples to FLT.

(Which means I'm asking "What if…?" and looking to see if that doesn't create a problem, which it does, which proves FLT, which is the point of contention as people like Magidin claim that I haven't found that problem.)

It so happens that I've focused on v+1, which is a factor of v^3+1 as

v^3 + 1 = (v+1)(v^2 - v + 1),

and I've given v a value (it's arbitrary remember) that gives it a prime factor of f, which just means that f could be 5, or 7, or 11 or some other prime number other than 3, as, in general, p itself is excluded, to handle a complication that would arise if x were divisible by p.

And my proof depends on only two of the a's, say a1 and a2, having a factor of f, where things are a little more complicated because the a's are themselves not integers, though they are algebraic integers.

Magidin has disputed that only two of the a's have a factor of f, and has instead claimed that ALL the a's have non unit algebraic integer factors of f, and he has based his claim, among other things on Galois Theory.

However, people seem to forget that my choice of v+1 was from a math perspective

That creates the infinite descent, which would force all of the a's to have all the factors of v^3 + 1.

To break it, Magidin needs some algebraic integer factor of v^3 + 1 that is shared by ONLY TWO of the a's.

However, how do you differentiate my choice of f, from any other algebraic integer factor of v^3+1?

I'm curious to see if anyone gets my point immediately, or if anyone simply has a gut feeling that it's wrong, which means I'm asking for knee-jerk replies.

Well, I thought you'd like to know of an interesting problem with his position, and I'm curious about whether or not he will acknowledge it or give a resolution.

You see, his position from what I see leads to an infinite descent which causes a contradiction.

To understand that you need to know what his position is, and I'll show it for p=3, which you see get used a lot. Obviously, p=3 is used a lot because it's the simplest case for p odd prime, though my FLT Proof covers

x^p + y^p = z^p, for p odd prime, and not just p=3,

but let's get to more details.

A key expression for p=3 that follows from my approach is

(v^3+1)z^6 - 3v x^2 y^2 z^2 - 2 x^3 y^3 = (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy),

where a1, a2, a3, b1, b2, b3 are all algebraic integers, as the ring is algebraic integers, and v is an arbitrary integer, while, of course, x, y and z are assumed integer counterexamples to FLT.

(Which means I'm asking "What if…?" and looking to see if that doesn't create a problem, which it does, which proves FLT, which is the point of contention as people like Magidin claim that I haven't found that problem.)

It so happens that I've focused on v+1, which is a factor of v^3+1 as

v^3 + 1 = (v+1)(v^2 - v + 1),

and I've given v a value (it's arbitrary remember) that gives it a prime factor of f, which just means that f could be 5, or 7, or 11 or some other prime number other than 3, as, in general, p itself is excluded, to handle a complication that would arise if x were divisible by p.

And my proof depends on only two of the a's, say a1 and a2, having a factor of f, where things are a little more complicated because the a's are themselves not integers, though they are algebraic integers.

Magidin has disputed that only two of the a's have a factor of f, and has instead claimed that ALL the a's have non unit algebraic integer factors of f, and he has based his claim, among other things on Galois Theory.

However, people seem to forget that my choice of v+1 was from a math perspective

**arbitrary**in that you can consider some other factor of v^3 + 1, or you can even pick some other**algebraic integer**factor of v+1, and apply Magidin's argument.That creates the infinite descent, which would force all of the a's to have all the factors of v^3 + 1.

To break it, Magidin needs some algebraic integer factor of v^3 + 1 that is shared by ONLY TWO of the a's.

However, how do you differentiate my choice of f, from any other algebraic integer factor of v^3+1?

I'm curious to see if anyone gets my point immediately, or if anyone simply has a gut feeling that it's wrong, which means I'm asking for knee-jerk replies.

### Sunday, December 08, 2002

## JSH: IN CASE YOU MISSED IT

Some of you may not realize that some people make posts in reply possibly to keep you from noticing that various positions have been successfully refuted in my original post by relying on a tendency of posters to read the last post in a thread first. That is important to two key threads: "JSH: Assessing the truth", where a G. Frege has made a cleverly edited post, and "Short FLT Proof: Attacks refuted" where a Virgil has lied in his reply.

## JSH: Newsgroup survey

- Given the amount of attention given to my work on the newsgroup, how did you explain to yourself that attention if you thought it invalid?
- Do you still believe posters such as Magidin who claim that among other things Galois Theory supports a refutation of a key result in what I claim is a short proof of FLT?
- Do you believe that a counter example to what I claim is a proof exists?
- Why do you suppose people argue with me?
- Do you think that if I'm correct that it will have a negative or positive impact on mathematicians in general?
- Do you accept that my prime counting work is in fact trivial? If so, why?
- Do you believe the Riemann Hypothesis?

## Short FLT Proof: Attacks refuted

There have been several recent claims of counter examples and/or counter arguments to the short proof of Fermat's Last Theorem.

I noted that those claims contradicted an independent verification of a key piece of my work, which I called Area One.

Some questioned that, and continued to make said claims.

Here's a refutation, which focuses on expressions that follow from

x^3 + y^3 = z^3

for simplicity.

It assumes some familiarity with the proof.

Consider

(v^3+1)z^6 - 3v x^2 y^2 z^2 - x^3 y^3 =

w1 w2 w3(s1 z^2 + t1 xy)(s2 z^2 + t2 xy)(s3 z^2 + t3 xy)

where s1 = s2 = sqrt(v+1), and

w1 s1 = a1, w2 s2 = a2, w3 s3 = a3, w1 t1 = b1,

w2 t2 = b2, w3 t3 = b3,

so

w1 w2 w3 (s1 z^2 + t1 xy)(s2 z^2 + t2 xy)(s3 z^2 + t3 xy) = (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy),

where a1, a2, a3, b1, b2 and b3 are algebraic integers, and the ring is formed by adjoining the square roots of the integers, t1, t2 and t3 to the ring of integers.

Now there is also the factor f, which is a prime integer factor of sqrt(v+1) as described in the short FLT Proof.

(Reference http://www.msnusers.com/AmateurMath/fltproof.msnw )

Then the Area One result within the specified ring can be written as

(s1 z^2 + t1 xy)(s2 z^2 + t2 xy) = 0(mod f^{n+2j}),

which is

(sqrt(m)f^j z^2 + t1 xy)(sqrt(m)f^j z^2 + t2 xy) = 0(mod f^{n+2j}),

so

(sqrt(m) z^2 + t1 uy)(sqrt(m) z^2 + t2 uy) = 0(mod f^n),

This result has been independently corroborated.

I moved from a ring like the one here described to the ring of algebraic integers as I thought it gave a more beautiful proof, and was more in line with what I'd originally envisioned.

Within the ring of algebraic integers the Area One result is

(a1 z^2 + b1 uy)( a2 z^2 + b2 uy) = 0(mod f^n),

and posters have claimed that it is false.

But consider that

w1 w2 w3(s1 z^2 + t1 xy)(s2 z^2 + t2 xy)(s3 z^2 + t3 xy) = (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy),

so their claim that a1 does not have a factor of f within the ring of algebraic integers requires that w1 s1 not have a factor of f either.

But, s1 = sqrt(v+1) = m f^j, so that position requires that f must be a unit in the ring I described above.

However, the correctness of the Area One result depends on f NOT being a unit in the ring, as if it is then

(sqrt(m) z^2 + t1 uy)(sqrt(m) z^2 + t2 uy) = 0(mod f^n),

is meaningless, and in fact, it has been independently shown to not be a unit in that ring.

In considering the arguments above the important point is to realize that the shift in rings does not change the *value* of the factors which is

(v^3+1)z^6 - 3v x^2 y^2 z^2 - x^3 y^3

so results in different rings must be linked.

What I think is of interest here is that f must be a unit in the ring for counter claims to be valid, which is a position that was taken, heatedly argued, and refuted. The refutation of the claim that f was a unit was accepted. Yet now a position that requires that f be a unit has been from what I've seen generally accepted.

In surveys on the newsgroup that I've conducted, posters seem to rely on the notion that there is a counterexample or counterarguments, which is still puzzling to me. I'm curious as to whether or not anyone can explain.

So, succinctly, certain posters made claims that required that f be a unit within the ring, those claims were refuted by someone other than myself, and the refutation accepted on the newsgroup. Then posters came back later, made arguments which required that f be a unit, as I've shown in this post, and these claims were accepted on the newsgroup.

However, though human beings can be inconsistent in this way, mathematics cannot, so their claims are false.

I noted that those claims contradicted an independent verification of a key piece of my work, which I called Area One.

Some questioned that, and continued to make said claims.

Here's a refutation, which focuses on expressions that follow from

x^3 + y^3 = z^3

for simplicity.

It assumes some familiarity with the proof.

Consider

(v^3+1)z^6 - 3v x^2 y^2 z^2 - x^3 y^3 =

w1 w2 w3(s1 z^2 + t1 xy)(s2 z^2 + t2 xy)(s3 z^2 + t3 xy)

where s1 = s2 = sqrt(v+1), and

w1 s1 = a1, w2 s2 = a2, w3 s3 = a3, w1 t1 = b1,

w2 t2 = b2, w3 t3 = b3,

so

w1 w2 w3 (s1 z^2 + t1 xy)(s2 z^2 + t2 xy)(s3 z^2 + t3 xy) = (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy),

where a1, a2, a3, b1, b2 and b3 are algebraic integers, and the ring is formed by adjoining the square roots of the integers, t1, t2 and t3 to the ring of integers.

Now there is also the factor f, which is a prime integer factor of sqrt(v+1) as described in the short FLT Proof.

(Reference http://www.msnusers.com/AmateurMath/fltproof.msnw )

Then the Area One result within the specified ring can be written as

(s1 z^2 + t1 xy)(s2 z^2 + t2 xy) = 0(mod f^{n+2j}),

which is

(sqrt(m)f^j z^2 + t1 xy)(sqrt(m)f^j z^2 + t2 xy) = 0(mod f^{n+2j}),

so

(sqrt(m) z^2 + t1 uy)(sqrt(m) z^2 + t2 uy) = 0(mod f^n),

This result has been independently corroborated.

I moved from a ring like the one here described to the ring of algebraic integers as I thought it gave a more beautiful proof, and was more in line with what I'd originally envisioned.

Within the ring of algebraic integers the Area One result is

(a1 z^2 + b1 uy)( a2 z^2 + b2 uy) = 0(mod f^n),

and posters have claimed that it is false.

But consider that

w1 w2 w3(s1 z^2 + t1 xy)(s2 z^2 + t2 xy)(s3 z^2 + t3 xy) = (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy),

so their claim that a1 does not have a factor of f within the ring of algebraic integers requires that w1 s1 not have a factor of f either.

But, s1 = sqrt(v+1) = m f^j, so that position requires that f must be a unit in the ring I described above.

However, the correctness of the Area One result depends on f NOT being a unit in the ring, as if it is then

(sqrt(m) z^2 + t1 uy)(sqrt(m) z^2 + t2 uy) = 0(mod f^n),

is meaningless, and in fact, it has been independently shown to not be a unit in that ring.

In considering the arguments above the important point is to realize that the shift in rings does not change the *value* of the factors which is

(v^3+1)z^6 - 3v x^2 y^2 z^2 - x^3 y^3

so results in different rings must be linked.

What I think is of interest here is that f must be a unit in the ring for counter claims to be valid, which is a position that was taken, heatedly argued, and refuted. The refutation of the claim that f was a unit was accepted. Yet now a position that requires that f be a unit has been from what I've seen generally accepted.

In surveys on the newsgroup that I've conducted, posters seem to rely on the notion that there is a counterexample or counterarguments, which is still puzzling to me. I'm curious as to whether or not anyone can explain.

So, succinctly, certain posters made claims that required that f be a unit within the ring, those claims were refuted by someone other than myself, and the refutation accepted on the newsgroup. Then posters came back later, made arguments which required that f be a unit, as I've shown in this post, and these claims were accepted on the newsgroup.

However, though human beings can be inconsistent in this way, mathematics cannot, so their claims are false.

### Saturday, December 07, 2002

## JSH: Assessing the truth

Some of you may honestly be confused about whether or not I've presented a short proof of Fermat's Last Theorem, whether or not there's a counter example or counter argument to my proof, and whether or not people have been caught in actual lies about the proof.

Unfortunately, I'm quite certain that mathematicians have strong motivations not to tell you the truth. While I admit it's quite possible that many mathematicians have not heard of the short FLT Proof, I'm also very suspicious to the extent that I point out that they may simply believe that not saying anything is not only a potent weapon against anyone knowing the truth, but it might seem like the best defense for the future.

So, if you're someone who prides themselves on knowing the actual truth versus political or social "truths" then you may wonder what hope you have here, especially if you don't know a lot of number theory.

I suggest your common sense is a good start, along with a consideration of the facts NOT in dispute.

You may wonder about relevance of some of these facts, but I've chosen them and their order quite deliberately.

Unfortunately, I'm quite certain that mathematicians have strong motivations not to tell you the truth. While I admit it's quite possible that many mathematicians have not heard of the short FLT Proof, I'm also very suspicious to the extent that I point out that they may simply believe that not saying anything is not only a potent weapon against anyone knowing the truth, but it might seem like the best defense for the future.

So, if you're someone who prides themselves on knowing the actual truth versus political or social "truths" then you may wonder what hope you have here, especially if you don't know a lot of number theory.

I suggest your common sense is a good start, along with a consideration of the facts NOT in dispute.

You may wonder about relevance of some of these facts, but I've chosen them and their order quite deliberately.

- I did find a way to count prime numbers that is not in any established reference, which is easily verifiable. It is also easy to check anyone claiming that is not true by asking them to provide the reference, and then simply looking.

The prime counting function I found is

dS(x,y) = (pi((x/y), y-1) -pi(y-1, sqrt(y-1))[pi(y, sqrt(y)) - pi(y-1, sqrt(y-1))],

S(x,1) = 0.

And pi(x, y) = floor(x) - S(x, y) - 1, and you get S by summing dS from y = 2 to sqrt(x).

though

dS(x,y) = (pi((x/y), (sqrt(x/y)) - pi(y-1, sqrt(y-1))[ pi(y, sqrt(y)) - pi(y-1, sqrt(y-1))]

when sqrt(x/y) < y-1, makes for faster calculation.

Here pi(x,sqrt(x)) gives the prime count for x.

For instance pi(10,sqrt(10)) will give you 4, corresponding to 2, 3, 5 and 7.

Now it should be easy enough to see if that's in any reference, and please note there is NOT dispute about whether or not it works. - Most of the short FLT proof I've presented has been verified as being correct, including points that
**had**been controversial. Here the important thing is to pay attention to those claiming some error has been found.

For me that's an important point because I was a central figure in arguments that went on for MONTHS, and even when I was vindicated time and time again, these same people would keep acting as if they'd never been wrong!!!

Verifying that is something that could take a bit of effort, so it's not as easy as the prime counting function.

However, I did make it somewhat easy for you by separating out a key argument as "Area One" so you can just go to groups.google.com and do a search where I am the author on the sci.math newsgroup for the first mention of the phrase "Area One" and you will see me talking quite a while back about what I was going to do, and what it would show.

Then you can see how long the arguing went on, and most importantly**who**I argued with.

Of course, you may know that a "proof" can be mostly correct and still be wrong, but what's fascinating here is that certain people have clearly been caught arguing against results before they were independently shown to be proven, pausing, and then arguing against those**same**results later!

If you don't believe it, be sure to read what I said when I first started talking about "Area One" in posts, and pay attention to what actually happened over a period of months. - Despite my prime counting function not being in any reference, the same people who are telling you that the short FLT Proof is wrong, are usually claiming that the prime counting work is unimportant, or in even more bizarre cases, claiming that I haven't presented anything new!

Now I've contacted leading mathematicians by email, and for important reasons I'm keeping their names and the complete contents of those emails private, and what I've been told by them is not that it's not new, but that they don't think it's important.

Also, they keep talking about fast prime counting, and what I have is not the fastest around, though I still believe that algorithms based on it can be.

What's important here is for you to ask yourselves, how unimportant could a method for counting PRIME NUMBERS truly be such that mathematicians don't seem bothered with it NOT being published?

Here the proper assessment, I strongly suggest, is that it is VERY important, which is why mathematicians, to my knowledge, aren't talking about it.

Here what you believe probably depends on whether or not you're one of those "conspiracy people" who suppose that it is possible aliens landed at Roswell and the government covered it up, and those who think it's complete hogwash.

Personally, since I mentioned it, I don't think aliens crash landed at Roswell, but I'm hoping some of you will smell the possibility of a conspiracy of silence here.

Now I hope those three points are a start.

Some of you may labor under the misapprehension that what is happening is strange historically, but it's not.

You may have seen the life of Evariste Galois mentioned or read my post linking John Nash's schizophrenia to lack of proper recognition, based on an established psychological theory called the double bind theory, which you can easily enough look up on the web to see that I didn't make it up.

Do you think that intelligent people—people intelligent enough to get doctorates in mathematics—really don't know the profound effect that simply**ignoring**a person's work can have on that person?

Maybe those who repeatedly post in reply to me at least acknowledging the existence of my work are better than that shadowy majority of mathematicians who by now may know much about it, but realize that their most potent and vicious weapon—is silence.