### Monday, April 28, 2003

## Math research can be difficult

There have been plenty of arguments on this newsgroup between me and lots of other people, but as far as I'm concerned it can be water under the bridge, which is a saying that it's past and gone.

I say that because math research is extremely unforgiving and can be difficult, while a person can find that they bump up against their own needs and wishes, but the math does not care.

Mathematics is not flexible and it doesn't bend to our whims, so it can be extremely frustrating when you're trying to work things out or pursue an idea, when things don't work out immediately or you find yourself having to go through a lot of mental gyrations to get some kind of understanding.

It also makes it fun and rewarding when you get something. Or it can be fun even if you don't personally discover something but work through to an understanding of other people's discoveries.

To me now it seems that all the past arguing can be boxed up into a nice category of passion. Passions ran high for many reasons and there was a lot of spirited debate that often got nasty.

However, there are comments in the heat of battle and then when you calm down you rethink things you've said. I'd like to address some of the ones I've made.

So, at this time, I'd like to assure you that I am not interested in making sure mathematicians worldwide get fired. I've rethought my desire to go to Congress and try to get funding for mathematicians cut.

And I see no reason to continue saying that mathematics today is a form of white collar welfare.

I do have concerns about "pure math" and the possibility it raises for people to do nothing of value to humanity, while making their work harder to understand than necessary, to hide that fact, but I'm not interested in checking further into the subject.

I do not know if Wiles's work is correct or not, and I do not care.

I do have concerns about the lack of interest in what I've called my prime counting function.

And I'm very concerned with the possibility that despite my irrefutable demonstration of the value of my ideas that people will make false arguments and claim that bogus mathematics is true, depending on a need to be believed.

That is, I'm worried the pointless arguing will continue. Now I like to argue, but I don't like arguing over things I've proven many times before.

I do not consider myself to be a mathematician. I am a problem solver.

It has been my intention to shake things up, have some fun, and hopefully get to the beach soon without worrying about little things.

I have no interest in becoming a mathematician. I don't want to go back to school. I don't want to lecture. I'm not interested in interfering in the affairs of mathematicians any more than is inevitable given my discoveries.

And I don't want to belong.

It is true that my experiences should be worth a considerable sum of money, which I hope to use primarily to help my family, as I could do ok on my own without making spectacular math finds, but I saw this as a route to helping them. It also should help others understand that human imagination and spirit is a formidable force. We all have it. It's rather neat.

And yes, my parents are approaching retirement. I'd like to have all of this handled before they reach it, thank you very much.

I say that because math research is extremely unforgiving and can be difficult, while a person can find that they bump up against their own needs and wishes, but the math does not care.

Mathematics is not flexible and it doesn't bend to our whims, so it can be extremely frustrating when you're trying to work things out or pursue an idea, when things don't work out immediately or you find yourself having to go through a lot of mental gyrations to get some kind of understanding.

It also makes it fun and rewarding when you get something. Or it can be fun even if you don't personally discover something but work through to an understanding of other people's discoveries.

To me now it seems that all the past arguing can be boxed up into a nice category of passion. Passions ran high for many reasons and there was a lot of spirited debate that often got nasty.

However, there are comments in the heat of battle and then when you calm down you rethink things you've said. I'd like to address some of the ones I've made.

So, at this time, I'd like to assure you that I am not interested in making sure mathematicians worldwide get fired. I've rethought my desire to go to Congress and try to get funding for mathematicians cut.

And I see no reason to continue saying that mathematics today is a form of white collar welfare.

I do have concerns about "pure math" and the possibility it raises for people to do nothing of value to humanity, while making their work harder to understand than necessary, to hide that fact, but I'm not interested in checking further into the subject.

I do not know if Wiles's work is correct or not, and I do not care.

I do have concerns about the lack of interest in what I've called my prime counting function.

And I'm very concerned with the possibility that despite my irrefutable demonstration of the value of my ideas that people will make false arguments and claim that bogus mathematics is true, depending on a need to be believed.

That is, I'm worried the pointless arguing will continue. Now I like to argue, but I don't like arguing over things I've proven many times before.

I do not consider myself to be a mathematician. I am a problem solver.

It has been my intention to shake things up, have some fun, and hopefully get to the beach soon without worrying about little things.

I have no interest in becoming a mathematician. I don't want to go back to school. I don't want to lecture. I'm not interested in interfering in the affairs of mathematicians any more than is inevitable given my discoveries.

And I don't want to belong.

It is true that my experiences should be worth a considerable sum of money, which I hope to use primarily to help my family, as I could do ok on my own without making spectacular math finds, but I saw this as a route to helping them. It also should help others understand that human imagination and spirit is a formidable force. We all have it. It's rather neat.

And yes, my parents are approaching retirement. I'd like to have all of this handled before they reach it, thank you very much.

### Saturday, April 26, 2003

## Non algebraic integer "integer like" number, demonstrated

Well I use very simple ideas to try and get nifty answers to some old math problems. And I post about them. However, I am not a mathematician so at times I find myself in territory where I can be bullied a bit by mathematicians making bold claims that I'm wrong, and I have had that problem with a key area which has to do with the

coverage of algebraic integers.

Several posters have argued with me for some time about a relatively simple result that I have. The arguments have gone back and forth, and for some odd reason people don't seem to believe me!

However, looking carefully over the information posted by a guy who calls himself "Fred the Wonder Worm" I noticed that it demonstrated that the people arguing with me were wrong, and revealed an algebraic integer unit, which has a multiplicative inverse which is itself not an algebraic integer.

The following is taken from his post

message id: <b82u36$5aq$1@spacebar.ucc.usyd.edu.au>

I'll keep his letters to lessen confusion, and first notice from

x^12 + 6*x^11 - 181*x^10 - 960*x^9 + 7963*x^8 + 37678*x^7 - 117533*x^6 -

488570*x^5 + 686101*x^4 + 2232396*x^3 - 1726811*x^2 - 2931010*x + 987100

that taking a root, using 'a' as he did, and dividing through by 'a' gives

a^11 + 6*a^10 - 181*a^9 - 960*a^8 + 7963*a^7 + 37678*a^6 - 117533*a^5 -

488570*a^4 + 686101*a^3 + 2232396*a^2 - 1726811*a - 2931010 + 987100/a=0

where you can see that if 'a' is a root that has non unit algebraic integer factors of 5, then 987100/a, must STILL have non unit algebraic integer factors of 5, which means that a/5 does not.

For instance, if 'a' has a factors of 5sqrt(5), then because 987100 has a factor of 25 that would leave only sqrt(5), but a full factor of 5 has to still divide across because of 2931010 and all the other terms with a's, so there's be a contradiction.

Now looking at r1, from Fred's post I have

Putting in tr I have

176866667462880 r1 = 9862453210*a^11 + 54243492655*a^10 -

1198409518887700 - 1762144506621*a^9 - 8336476474707*a^8 + 73634871906622*a^7 +

297005313003728*a^6 - 915848551626666*a^5 - 3051857657480243*a^4 +

3027640866778493*a^3 + 7747460630371967*a^2 + 403832110132962*a

and dividing through by 5, as well as collecting a bit gives

35373333492576 r1 = 1972490642*a^11 + 10848698531*a^10 - 239681903777540 -

a/5( 1762144506621*a^8 + 8336476474707*a^7 - 73634871906622*a^6 -

297005313003728*a^5 + 915848551626666*a^4 + 3051857657480243*a^3 -

3027640866778493*a^2 - 7747460630371967*a + 403832110132962)

Let k = 1762144506621*a^8 + 8336476474707*a^7 - 73634871906622*a^6 -

297005313003728*a^5 + 915848551626666*a^4 + 3051857657480243*a^3 -

3027640866778493*a^2 - 7747460630371967*a + 403832110132962

and note that if 'a' has factors of 5, that k does not because

403832110132962

is, of course, NOT divisible by 5.

So I have

35373333492576 r1 = 1972490642*a^11 + 10848698531*a^10 - 239681903777540 - ak/5

Now then, if r1 shares any non unit algebraic integer factors of 5 with 'a', then they can divide across because of

- 239681903777540

which, of course, has a factor of 5.

Then ak/5

But as I've shown above ak/5 cannot have non unit algebraic integer factors of 5, which forces r1 to be a unit, as I quote from Fred:

What makes that fascinating is that if you invert the polynomial using y=1/x, you get

5x^5 + 4176 x^3 - 1488 y^2 + 1

which has now been proven to have a unit root, which is, NOT an algebraic integer, but its inverse IS an algebraic integer.

That settles the argument I've had for several years now.

I think the math is nifty.

Here's the background if you're wondering.

I demonstrated recently a key result using the polynomial-like expression

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

with v=-1 + mf^{2j}, where m=1, j=1, with the factorization

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

that ONLY TWO of the a's could have a factor of f.

To make that simple enough, I showed f=0, which I'll give here quickly as it is

3 x^2 y^2 z^2 + x^3 y^3 = (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy)

and it's easy enough to see that ONLY two of the a's are 0.

Then I went to f=sqrt(2), which gives the polynomial

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

where it's easy to demonstrate that ONLY TWO of the a's have a factor of sqrt(2), but then I went to f=sqrt(5), and ran into a problem.

The problem is in demonstrating the factorization, but I stumbled upon a route that worked by accident.

First off the polynomial for f=sqrt(5) is

65 z^6 - 12 x^2 y^2 z^2 + x^3 y^3 = (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy)

but because it's irreducible there's a twist because there have been posters arguing with me for over a YEAR that because it's irreducible, now ALL the a's would have non unit algebraic integer factors of 5. (Don't get confused on the usage here as it's like 3 is a factor of 12, so if I talk about factors of 12, I don't have to mean 12 itself.)

Well I have a VERY simple argument that proves these people wrong, but the arguing continued.

Now one of them had demonstrated a technique to isolate factors of 5, by solving for the a's first, which because b1 b2 b3=1 is easy here as I'd have

u^3 - 12 u^2 + 65 = 0

and the negative of the roots gives the a's, and using the substitution sqrt(5)w, I have

5sqrt(5) w^3 - 12(5) w^2 + 65 = 0, and dividing off 5, I get

sqrt(5) w^3 - 12 w^2 + 13 = 0, so

sqrt(5) w^3 = 12 w^2 - 13

5w^6 = 144 w^4 - 312 w^2 + 169, so

5w^6 - 144 w^4 + 312 w^2 - 169 =0

and the assertion by these people that argue with me is that because that polynomial is non monic primitive and irreducible over Q, none of the I know that the a's from before must all have factors of 5.

Now I erroneously believed that if they were right the factorization

5w^6 - 144 w^4 + 312 w^2 - 169 = (c1 w + d1)…(c6 w + d6)

where the c's and d's are algebraic integers could not be found.

Well two people responded by posting the polynomials defining the c's and d's where they were clearly algebraic integers, and I did a cursory check but haven't seen anything wrong with their factorization.

coverage of algebraic integers.

Several posters have argued with me for some time about a relatively simple result that I have. The arguments have gone back and forth, and for some odd reason people don't seem to believe me!

However, looking carefully over the information posted by a guy who calls himself "Fred the Wonder Worm" I noticed that it demonstrated that the people arguing with me were wrong, and revealed an algebraic integer unit, which has a multiplicative inverse which is itself not an algebraic integer.

The following is taken from his post

message id: <b82u36$5aq$1@spacebar.ucc.usyd.edu.au>

Let a be a root of this polynomial:

x^12 + 6*x^11 - 181*x^10 - 960*x^9 + 7963*x^8 + 37678*x^7 - 117533*x^6 -

488570*x^5 + 686101*x^4 + 2232396*x^3 - 1726811*x^2 - 2931010*x + 987100

Define tr, r1, r2, r3 to be the following values:

tr = 176866667462880

r1 = (9862453210*a^11 + 54243492655*a^10 - 1762144506621*a^9 -

8336476474707*a^8 + 73634871906622*a^7 + 297005313003728*a^6 -

915848551626666*a^5 - 3051857657480243*a^4 + 3027640866778493*a^3 +

7747460630371967*a^2 + 403832110132962*a - 1198409518887700) / tr

r2 = (6573801085*a^11 + 29503072348*a^10 - 1208403339336*a^9 -

4366240041747*a^8 + 54222010610962*a^7 + 148750611762872*a^6 -

793269747747321*a^5 - 1487188701233408*a^4 + 3985921497989948*a^3 +

4839390026896655*a^2 - 5550606915303438*a - 5671414313897740) / tr

r3 = (6573801085*a^11 + 42808739587*a^10 - 1141875003141*a^9 -

6752351088912*a^8 + 44278396405132*a^7 + 248393754086744*a^6 -

459258252043281*a^5 - 2663297151231683*a^4 + 1065284831907473*a^3 +

7093998365144384*a^2 - 1778334433861968*a - 130135870547080) / tr

Note that r1, r2, r3 are all algebraic integers, as they are roots of

the (irreducible) monic polynomial x^6 - 1488*x^4 + 4176*x^2 - 5.

I'll keep his letters to lessen confusion, and first notice from

x^12 + 6*x^11 - 181*x^10 - 960*x^9 + 7963*x^8 + 37678*x^7 - 117533*x^6 -

488570*x^5 + 686101*x^4 + 2232396*x^3 - 1726811*x^2 - 2931010*x + 987100

that taking a root, using 'a' as he did, and dividing through by 'a' gives

a^11 + 6*a^10 - 181*a^9 - 960*a^8 + 7963*a^7 + 37678*a^6 - 117533*a^5 -

488570*a^4 + 686101*a^3 + 2232396*a^2 - 1726811*a - 2931010 + 987100/a=0

where you can see that if 'a' is a root that has non unit algebraic integer factors of 5, then 987100/a, must STILL have non unit algebraic integer factors of 5, which means that a/5 does not.

For instance, if 'a' has a factors of 5sqrt(5), then because 987100 has a factor of 25 that would leave only sqrt(5), but a full factor of 5 has to still divide across because of 2931010 and all the other terms with a's, so there's be a contradiction.

Now looking at r1, from Fred's post I have

tr = 176866667462880

r1 = (9862453210*a^11 + 54243492655*a^10 - 1762144506621*a^9 -

8336476474707*a^8 + 73634871906622*a^7 + 297005313003728*a^6 -

915848551626666*a^5 - 3051857657480243*a^4 + 3027640866778493*a^3 +

7747460630371967*a^2 + 403832110132962*a - 1198409518887700) / tr

Putting in tr I have

176866667462880 r1 = 9862453210*a^11 + 54243492655*a^10 -

1198409518887700 - 1762144506621*a^9 - 8336476474707*a^8 + 73634871906622*a^7 +

297005313003728*a^6 - 915848551626666*a^5 - 3051857657480243*a^4 +

3027640866778493*a^3 + 7747460630371967*a^2 + 403832110132962*a

and dividing through by 5, as well as collecting a bit gives

35373333492576 r1 = 1972490642*a^11 + 10848698531*a^10 - 239681903777540 -

a/5( 1762144506621*a^8 + 8336476474707*a^7 - 73634871906622*a^6 -

297005313003728*a^5 + 915848551626666*a^4 + 3051857657480243*a^3 -

3027640866778493*a^2 - 7747460630371967*a + 403832110132962)

Let k = 1762144506621*a^8 + 8336476474707*a^7 - 73634871906622*a^6 -

297005313003728*a^5 + 915848551626666*a^4 + 3051857657480243*a^3 -

3027640866778493*a^2 - 7747460630371967*a + 403832110132962

and note that if 'a' has factors of 5, that k does not because

403832110132962

is, of course, NOT divisible by 5.

So I have

35373333492576 r1 = 1972490642*a^11 + 10848698531*a^10 - 239681903777540 - ak/5

Now then, if r1 shares any non unit algebraic integer factors of 5 with 'a', then they can divide across because of

- 239681903777540

which, of course, has a factor of 5.

Then ak/5

**must**also have those factorsBut as I've shown above ak/5 cannot have non unit algebraic integer factors of 5, which forces r1 to be a unit, as I quote from Fred:

Note that r1, r2, r3 are all algebraic integers, as they are roots of the (irreducible) monic polynomial x^6 - 1488*x^4 + 4176*x^2 - 5.

What makes that fascinating is that if you invert the polynomial using y=1/x, you get

5x^5 + 4176 x^3 - 1488 y^2 + 1

which has now been proven to have a unit root, which is, NOT an algebraic integer, but its inverse IS an algebraic integer.

That settles the argument I've had for several years now.

I think the math is nifty.

Here's the background if you're wondering.

I demonstrated recently a key result using the polynomial-like expression

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

with v=-1 + mf^{2j}, where m=1, j=1, with the factorization

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

that ONLY TWO of the a's could have a factor of f.

To make that simple enough, I showed f=0, which I'll give here quickly as it is

3 x^2 y^2 z^2 + x^3 y^3 = (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy)

and it's easy enough to see that ONLY two of the a's are 0.

Then I went to f=sqrt(2), which gives the polynomial

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

where it's easy to demonstrate that ONLY TWO of the a's have a factor of sqrt(2), but then I went to f=sqrt(5), and ran into a problem.

The problem is in demonstrating the factorization, but I stumbled upon a route that worked by accident.

First off the polynomial for f=sqrt(5) is

65 z^6 - 12 x^2 y^2 z^2 + x^3 y^3 = (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy)

but because it's irreducible there's a twist because there have been posters arguing with me for over a YEAR that because it's irreducible, now ALL the a's would have non unit algebraic integer factors of 5. (Don't get confused on the usage here as it's like 3 is a factor of 12, so if I talk about factors of 12, I don't have to mean 12 itself.)

Well I have a VERY simple argument that proves these people wrong, but the arguing continued.

Now one of them had demonstrated a technique to isolate factors of 5, by solving for the a's first, which because b1 b2 b3=1 is easy here as I'd have

u^3 - 12 u^2 + 65 = 0

and the negative of the roots gives the a's, and using the substitution sqrt(5)w, I have

5sqrt(5) w^3 - 12(5) w^2 + 65 = 0, and dividing off 5, I get

sqrt(5) w^3 - 12 w^2 + 13 = 0, so

sqrt(5) w^3 = 12 w^2 - 13

5w^6 = 144 w^4 - 312 w^2 + 169, so

5w^6 - 144 w^4 + 312 w^2 - 169 =0

and the assertion by these people that argue with me is that because that polynomial is non monic primitive and irreducible over Q, none of the I know that the a's from before must all have factors of 5.

Now I erroneously believed that if they were right the factorization

5w^6 - 144 w^4 + 312 w^2 - 169 = (c1 w + d1)…(c6 w + d6)

where the c's and d's are algebraic integers could not be found.

Well two people responded by posting the polynomials defining the c's and d's where they were clearly algebraic integers, and I did a cursory check but haven't seen anything wrong with their factorization.

### Tuesday, April 22, 2003

## Why I can't be wrong, FLT argument key claim

Believe it or not, I have NO interest in continuing with the claim that I have a proof of Fermat's Last Theorem if I do not, and I'm considering what some may try to call irrefutable evidence that I don't (um, even me), but in the meantime, maybe SOMEONE could tell me what's wrong with the simple argument which still has me convinced!!!

And yes I'm desperate. I need to put this behind me if I'm wrong. But I can't let it go if I'm right, as that would really suck.

Ok, I have this expression I can factor as

(a1 z^2 + b1 f^j uy)(a2 z^2 + b2 f^j uy)( a3 z^2 + b3 f^j uy),

where

a1 a2 a3 = (-1+mf^{2j})^3 + 1, and b1 b2 b3 = -2.

But by keeping a lot of variables, deliberately, that same expression is also

f^{6j} z^6 m^3 -3f^{4j} z^6 m^2 + 3(f^{2j} -u^2 f^{3j} y^2 z^2) m + 3 (uf^j)^2 y^2 z^2 - 2 u^3 f^{3j} y^3

and I consider it as a polynomial with respect to m, as m is the only thing I vary in my FLT argument.

(Expression derived in detail at

http://groups.msn.com/AmateurMath/algebraoffactorizations.msnw )

Here's where I need help with the following statement:

Now each factor can be separated into factors of the constant term with respect to m, versus those terms with a factor of m.

If that is true then I can't see how I can be wrong. It's the simple idea that lies at the heart of my claim of proof. I need to see if anyone can refute it.

If it's true then there's nothing wrong with my proof. It is true; therefore, there is nothing wrong with my proof.

At least one of the a's must have a factor of m, from their product, so assuming that a1 does, consider

(a1 z2 + b1 f^j uy)

and since a1 has a factor of m, let's split b1 up into possible factors of m, which I'll call w, and factors of the constant term which I'll call c. Then I have

(a1 z2 + (w+c) f^j uy) = a1 z^2 + w f^j uy + c f^j uy

showing the factor of the constant term times a factor of f, which is f^j, but when f^{2j} is divided off from the original expression it is removed from the constant term, proving that a factor of f^j must be removed here, if a1 has a factor of m, proving that a1 has a factor of f^j if it has a factor of m.

So can I split b1 up that way or not? Aren't a's that have factors of m forced out of the constant term?

It's so simple. That's it. So if you can prove that all wrong then you'll do me a favor.

Then again, it seems to befuddle people and I think all those variables make people dizzy.

Of course, I'm not wrong, which is why things are really frustrating for me now.

And yes I'm desperate. I need to put this behind me if I'm wrong. But I can't let it go if I'm right, as that would really suck.

Ok, I have this expression I can factor as

(a1 z^2 + b1 f^j uy)(a2 z^2 + b2 f^j uy)( a3 z^2 + b3 f^j uy),

where

a1 a2 a3 = (-1+mf^{2j})^3 + 1, and b1 b2 b3 = -2.

But by keeping a lot of variables, deliberately, that same expression is also

f^{6j} z^6 m^3 -3f^{4j} z^6 m^2 + 3(f^{2j} -u^2 f^{3j} y^2 z^2) m + 3 (uf^j)^2 y^2 z^2 - 2 u^3 f^{3j} y^3

and I consider it as a polynomial with respect to m, as m is the only thing I vary in my FLT argument.

(Expression derived in detail at

http://groups.msn.com/AmateurMath/algebraoffactorizations.msnw )

Here's where I need help with the following statement:

Now each factor can be separated into factors of the constant term with respect to m, versus those terms with a factor of m.

If that is true then I can't see how I can be wrong. It's the simple idea that lies at the heart of my claim of proof. I need to see if anyone can refute it.

If it's true then there's nothing wrong with my proof. It is true; therefore, there is nothing wrong with my proof.

At least one of the a's must have a factor of m, from their product, so assuming that a1 does, consider

(a1 z2 + b1 f^j uy)

and since a1 has a factor of m, let's split b1 up into possible factors of m, which I'll call w, and factors of the constant term which I'll call c. Then I have

(a1 z2 + (w+c) f^j uy) = a1 z^2 + w f^j uy + c f^j uy

showing the factor of the constant term times a factor of f, which is f^j, but when f^{2j} is divided off from the original expression it is removed from the constant term, proving that a factor of f^j must be removed here, if a1 has a factor of m, proving that a1 has a factor of f^j if it has a factor of m.

So can I split b1 up that way or not? Aren't a's that have factors of m forced out of the constant term?

It's so simple. That's it. So if you can prove that all wrong then you'll do me a favor.

Then again, it seems to befuddle people and I think all those variables make people dizzy.

Of course, I'm not wrong, which is why things are really frustrating for me now.

### Monday, April 14, 2003

## Mythology and mathematics

One of the great blessings in my life has been the love of story, and in my own searches for truth I've often relied on the great stories of humanity's past--its mythology and religions.

From King Arthur, to Beowulf, to the Crow of Native American myth, and many other sources from African, to Irish, to Japanese, I've wrapped myself in a story grand enough to handle finding great mathematical truths like a short proof of a famous proposition that eluded mathematicians for 360 years or more.

Part of the necessity of an overarching Story has to do with the dryness that has been put in place by generations of mathematicians who have robbed mathematics of the vibrancy and color that is possible in that language we use to help describe our world.

Mathematics is Herself a Power, and in fact She is absolute power.

In following quests to understand Her, the grandness of myth and religion can be a sustaining force, as I've found it to be an extremely powerful sustaining force.

Besides that way I could get a really cool story versus just amateur guy finds some neat math, mathematicians turn out to be bad guys, but he wins in the end.

Now it's a grand battle between powerful forces as Truth itself becomes a weapon in the proof I call the Hammer, and I remind that it gives me absolute power.

And the greatest battles foretold throughout human history take place before a world with an ability to watch in a way that never was possible in the past.

From the realms of the mind the battles cross into the magic realm of the world where people kill each other and blow themselves and buildings apart, and so many of you run from your True Nature, as you fight desperately to continue to hide from the truth of who you are.

The Story inspires and scares me as I contemplate powers and Power, gods and God, dragons and the Dragon, while looking at the world that the Forces have wrought.

And you know, I think it's a decent tale.

[A reply to someone who said that James doesn't seem to be a quite sensible person.]

Dude, I am SO bored.

I'd think things would be MORE interesting if my FLT proof were wrong!!!

Then at least I wouldn't have to keep bothering with mathematicians.

It seemed like such a good idea, find a short proof of Fermat's Last Theorem, get famous, get cash, hang out on the beach (my latest addition to the plan).

Trouble is, I never figured that mathematicians were both incompetent and stupid.

Oh well, new plan as now I get to use the Hammer with full force, which is probably what was intended.

Now, no feeling sympathy for mathematicians who start marching with signs like "Will work for food" in the future, as I've given them lots of opportunities to play right.

I will not show mercy going forward. I was trained as a soldier in the United States Army after all.

And when the US Army plays a game, we play to win.

Yup, you guessed it. If worse comes to worse, I

And I think I know the CIA and NSA better than any mathematician.

When push comes to shove, they'll throw you out with the garbage.

They'd personally shoot you themselves, if it were necessary.

If I have to sic Army generals on you, I will be really pissed.

From King Arthur, to Beowulf, to the Crow of Native American myth, and many other sources from African, to Irish, to Japanese, I've wrapped myself in a story grand enough to handle finding great mathematical truths like a short proof of a famous proposition that eluded mathematicians for 360 years or more.

Part of the necessity of an overarching Story has to do with the dryness that has been put in place by generations of mathematicians who have robbed mathematics of the vibrancy and color that is possible in that language we use to help describe our world.

Mathematics is Herself a Power, and in fact She is absolute power.

In following quests to understand Her, the grandness of myth and religion can be a sustaining force, as I've found it to be an extremely powerful sustaining force.

Besides that way I could get a really cool story versus just amateur guy finds some neat math, mathematicians turn out to be bad guys, but he wins in the end.

Now it's a grand battle between powerful forces as Truth itself becomes a weapon in the proof I call the Hammer, and I remind that it gives me absolute power.

And the greatest battles foretold throughout human history take place before a world with an ability to watch in a way that never was possible in the past.

From the realms of the mind the battles cross into the magic realm of the world where people kill each other and blow themselves and buildings apart, and so many of you run from your True Nature, as you fight desperately to continue to hide from the truth of who you are.

The Story inspires and scares me as I contemplate powers and Power, gods and God, dragons and the Dragon, while looking at the world that the Forces have wrought.

And you know, I think it's a decent tale.

[A reply to someone who said that James doesn't seem to be a quite sensible person.]

Dude, I am SO bored.

I'd think things would be MORE interesting if my FLT proof were wrong!!!

Then at least I wouldn't have to keep bothering with mathematicians.

It seemed like such a good idea, find a short proof of Fermat's Last Theorem, get famous, get cash, hang out on the beach (my latest addition to the plan).

Trouble is, I never figured that mathematicians were both incompetent and stupid.

Oh well, new plan as now I get to use the Hammer with full force, which is probably what was intended.

Now, no feeling sympathy for mathematicians who start marching with signs like "Will work for food" in the future, as I've given them lots of opportunities to play right.

I will not show mercy going forward. I was trained as a soldier in the United States Army after all.

And when the US Army plays a game, we play to win.

Yup, you guessed it. If worse comes to worse, I

**will**turn to the Army to help me with mathematicians. And then mathematicians don't think the NSA or CIA can save your asses, as generals LIKE me.And I think I know the CIA and NSA better than any mathematician.

When push comes to shove, they'll throw you out with the garbage.

They'd personally shoot you themselves, if it were necessary.

If I have to sic Army generals on you, I will be really pissed.