### Tuesday, January 27, 2004

## JSH: Mathematical clarity

What I find interesting about this current situation is the willingness of some people on this newsgroup to question rather basic mathematics.

That's easy to show with basic mathematics using the Decker example.

Decker, a professor at Hamilton College, put forward a quadratic, and you can find his original post with the following headers:

Message-ID: <3FF47C4C.6080109@hamilton.edu>

>Date: Thu, 01 Jan 2004 15:00:12 -0500

>From: Rick Decker <rdecker@hamilton.edu>

Newsgroups: sci.math

>Subject: Re: Mathematical consistency, courage

Decker put forward the quadratic

(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)

where his a's are roots of

a^2 - (x - 1)a + 7(x^2 + x).

A simple linear transformation helps balance the factorization, as using

a_2(x) = b_2(x) - 1, and making the substitution gives

(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2).

The point I've made is that if you divide both sides by 7, and wish to try and remain the ring of algebraic integers then necessarily you have

(5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2

which follows rather easily from some basic algebra.

If follows because algebra gives a

That way of multiplying out means that on the left you have 1 and 2 multiplying times each other to get 2 on the right.

It's easy to demonstrate that multiplying out:

25 a_1(x) b_2(x)/7 + 10 a_1(x)/7 + 5 b_2(x) + 2 = 25x^2 + 30x + 2

and you can see the two 2's balanced to each side.

It's basic algebra. Now sure you have some people who let their egos get in the way, but why keep protecting them? What's in it for you?

Now I know, and you know that if you try to divide 7 out any other way then you can't remain in the ring of algebraic integers. Like if you have f_1 and f_2 where f_1 f_2 = 7, with non-unit f_1 and f_2, where

(5a_1(x)/f_1 + f_2)(5b_2(x)/f_2 + 2/f_2) = 25x^2 + 30x + 2

then the algebra tells you that the factors of 2 on the right are f_2 and 2/f_2, and if f_2 is a non-unit factor of 7, then you're not in the ring of algebraic integers.

Now then, you have a choice: ignore basic algebra assuming you can keep this fight up indefinitely for social reasons, or tell the truth.

It seems to me that some of you don't realize that even if you succeed here for a few days, or a few years, or even decades, some day the truth comes out.

It doesn't matter if you're dead. I've made certain that your name will live in infamy if it's known at all: Wiles, Ribet, Granville, or anyone else from this generation.

It doesn't matter, as you're all in it together.

It's kind of funny really. Sure Wiles can appreciate whatever approbation he gets today, but in the future, he's just another loser.

I'm the winner. It's about the

Pump each other up now. Tell yourselves how great you are. Congratulate yourself on your supposed accomplishments while you can.

History will hate you and love me.

I'm the misunderstood and persecuted genius.

You're the assholes.

That's easy to show with basic mathematics using the Decker example.

Decker, a professor at Hamilton College, put forward a quadratic, and you can find his original post with the following headers:

Message-ID: <3FF47C4C.6080109@hamilton.edu>

>Date: Thu, 01 Jan 2004 15:00:12 -0500

>From: Rick Decker <rdecker@hamilton.edu>

Newsgroups: sci.math

>Subject: Re: Mathematical consistency, courage

Decker put forward the quadratic

(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)

where his a's are roots of

a^2 - (x - 1)a + 7(x^2 + x).

A simple linear transformation helps balance the factorization, as using

a_2(x) = b_2(x) - 1, and making the substitution gives

(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2).

The point I've made is that if you divide both sides by 7, and wish to try and remain the ring of algebraic integers then necessarily you have

(5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2

which follows rather easily from some basic algebra.

If follows because algebra gives a

**rigid**way to multiply out.That way of multiplying out means that on the left you have 1 and 2 multiplying times each other to get 2 on the right.

It's easy to demonstrate that multiplying out:

25 a_1(x) b_2(x)/7 + 10 a_1(x)/7 + 5 b_2(x) + 2 = 25x^2 + 30x + 2

and you can see the two 2's balanced to each side.

It's basic algebra. Now sure you have some people who let their egos get in the way, but why keep protecting them? What's in it for you?

Now I know, and you know that if you try to divide 7 out any other way then you can't remain in the ring of algebraic integers. Like if you have f_1 and f_2 where f_1 f_2 = 7, with non-unit f_1 and f_2, where

(5a_1(x)/f_1 + f_2)(5b_2(x)/f_2 + 2/f_2) = 25x^2 + 30x + 2

then the algebra tells you that the factors of 2 on the right are f_2 and 2/f_2, and if f_2 is a non-unit factor of 7, then you're not in the ring of algebraic integers.

Now then, you have a choice: ignore basic algebra assuming you can keep this fight up indefinitely for social reasons, or tell the truth.

It seems to me that some of you don't realize that even if you succeed here for a few days, or a few years, or even decades, some day the truth comes out.

It doesn't matter if you're dead. I've made certain that your name will live in infamy if it's known at all: Wiles, Ribet, Granville, or anyone else from this generation.

It doesn't matter, as you're all in it together.

It's kind of funny really. Sure Wiles can appreciate whatever approbation he gets today, but in the future, he's just another loser.

I'm the winner. It's about the

**legacy**people. I'm destroying yours now and guaranteeing mine.Pump each other up now. Tell yourselves how great you are. Congratulate yourself on your supposed accomplishments while you can.

History will hate you and love me.

I'm the misunderstood and persecuted genius.

You're the assholes.

### Tuesday, January 13, 2004

## JSH: Limited intellects

I've concluded that the problem I'm facing is that the human brain isn't built to handle Mathematics. Human beings have done well with some simple mathematics and basic concepts, but it only goes so far, and what have, is beyond those limits.

For a while I thought that at least my simplest results, like the prime counting function, or that funny little error with algebraic integers might be generally accessible, but after years of arguing I realize that your intellects are too limited to fully grasp my work.

So I think I'll look for a way to further simplify, though I wonder what could be much simpler than what I've shown. Still, no matter how child-like your minds are, no matter how simple your mental processing, since you have language, at some point there's a chance that I'll be able to find something that your minds can handle.

Well, at least, I hope I can.

For a while I thought that at least my simplest results, like the prime counting function, or that funny little error with algebraic integers might be generally accessible, but after years of arguing I realize that your intellects are too limited to fully grasp my work.

So I think I'll look for a way to further simplify, though I wonder what could be much simpler than what I've shown. Still, no matter how child-like your minds are, no matter how simple your mental processing, since you have language, at some point there's a chance that I'll be able to find something that your minds can handle.

Well, at least, I hope I can.

### Wednesday, January 07, 2004

## Algebraic integers result with independent variables

I don't know if you noticed or not, but I've been in an argument that has gone on for quite some time about a problem with the ring of algebraic integers. Luckily for me, I've found the way to end the arguing, by relying on your understanding of

I've been using a modification of an example put forward by Rick Decker, a professor at Hamilton College.

Consider,

7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7(xy - y^2)(5) + 7^2 y^2

so

(5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2)

where the a's are roots of

a^2 - (x - y)a + 7(x^2 + xy).

Notice that x and y are independent variables.

Now letting x=0, gives

a(a + y) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, I let

a_2(x,y) = b_2(x,y) - y, so

(5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2)

Which shows that dividing both sides by 7 must result in

(5a_1(x,y)/7 + y)(5b_2(x,y) + 2y) = 25x^2 + 30xy + 2y^2

as long as x does not equal y, as then some of the terms with y as a factor that do NOT have x as a factor are clipped out.

The point is that y is an independent variable, so it and its coefficients CANNOT vary based on the value of x, so my setting x=0 is NOT a special case!!! It's a way to clear x out of the picture to see what the coefficients are for the terms with y as a factor that do NOT have x as a factor as well.

Now then, on to what happens if x=y. Then you have

a^2 + 7(x^2 + xy),

where I've used x-y=0 to cancel out the middle term but left the variable names as they were.

That case IS a special case as notice that it is indistinguishable from the case if in general

a^2 + 7(x^2 + xy) = 0,

with independent x, and y.

Previous discussions were basically looking over the special case where y=1.

The implications of the result are quite huge, as now you can put in some numbers like y=1, x=2, to dodge the special case at x=1, to get

a^2 - a + 42 = 0, which solves to

a = (1+/-sqrt(-167))/2

and, now you know that

One thing worth noting is that operator ambiguity means that sqrt(-167) represents two numbers, as that operator gives

Therefore, while the sqrt() operator is in place you cannot further resolve the expression to see

Those willing to go looking can try to find their own quadratics like Decker did, and find one with integer roots, for integer x, and then you can see the result directly.

Of course the problem with the ring of algebraic integers is the implication from previous interpretations (here I rely on what I've heard primarily from sci.math posters like Arturo Magidin, "Nora Baron", and Dik winter) is that

Want more advanced polynomial factorization?

Then check out my blog archives:

http://mathforprofit.blogspot.com/2003_10_01_mathforprofit_archive.html

**independent**variables. Note that you have x and y below where x is INDEPENDENT of y, and vice versa. My hope is that such a simple concept that should be familiar to all, will end the arguing.I've been using a modification of an example put forward by Rick Decker, a professor at Hamilton College.

Consider,

7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7(xy - y^2)(5) + 7^2 y^2

so

(5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2)

where the a's are roots of

a^2 - (x - y)a + 7(x^2 + xy).

Notice that x and y are independent variables.

Now letting x=0, gives

a(a + y) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, I let

a_2(x,y) = b_2(x,y) - y, so

(5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2)

Which shows that dividing both sides by 7 must result in

(5a_1(x,y)/7 + y)(5b_2(x,y) + 2y) = 25x^2 + 30xy + 2y^2

as long as x does not equal y, as then some of the terms with y as a factor that do NOT have x as a factor are clipped out.

The point is that y is an independent variable, so it and its coefficients CANNOT vary based on the value of x, so my setting x=0 is NOT a special case!!! It's a way to clear x out of the picture to see what the coefficients are for the terms with y as a factor that do NOT have x as a factor as well.

Now then, on to what happens if x=y. Then you have

a^2 + 7(x^2 + xy),

where I've used x-y=0 to cancel out the middle term but left the variable names as they were.

That case IS a special case as notice that it is indistinguishable from the case if in general

a^2 + 7(x^2 + xy) = 0,

with independent x, and y.

Previous discussions were basically looking over the special case where y=1.

The implications of the result are quite huge, as now you can put in some numbers like y=1, x=2, to dodge the special case at x=1, to get

a^2 - a + 42 = 0, which solves to

a = (1+/-sqrt(-167))/2

and, now you know that

**one**of the solutions is coprime to 7.One thing worth noting is that operator ambiguity means that sqrt(-167) represents two numbers, as that operator gives

**two**solutions.Therefore, while the sqrt() operator is in place you cannot further resolve the expression to see

**which**of the solutions is coprime to 7.Those willing to go looking can try to find their own quadratics like Decker did, and find one with integer roots, for integer x, and then you can see the result directly.

Of course the problem with the ring of algebraic integers is the implication from previous interpretations (here I rely on what I've heard primarily from sci.math posters like Arturo Magidin, "Nora Baron", and Dik winter) is that

**both**roots have non-unit factors in common with 7 in the ring of algebraic integers.Want more advanced polynomial factorization?

Then check out my blog archives:

http://mathforprofit.blogspot.com/2003_10_01_mathforprofit_archive.html

### Monday, January 05, 2004

## JSH: Maybe I'm wrong

The more I think about it the more Decker's example bugs me, despite my thinking that I had a handle on it that actually suited my purposes!

Maybe I'm just wrong here.

I think I'll think about it some more and see if I can find an answer other than just being wrong.

But being wrong wouldn't be so bad, after all, if I have it wrong then the sooner I find that out, the better I think.

That doesn't change the fact that there are quite a few annoying posters here, but they can keep posting all they want, and I doubt I'll stop posting, even if I'm wrong with this argument here.

I find myself at peace with posting. I like doing it, so there's no reason to stop just for being wrong. Then again, I haven't decided I am just yet.

It'll take some more thinking.

Maybe I'm just wrong here.

I think I'll think about it some more and see if I can find an answer other than just being wrong.

But being wrong wouldn't be so bad, after all, if I have it wrong then the sooner I find that out, the better I think.

That doesn't change the fact that there are quite a few annoying posters here, but they can keep posting all they want, and I doubt I'll stop posting, even if I'm wrong with this argument here.

I find myself at peace with posting. I like doing it, so there's no reason to stop just for being wrong. Then again, I haven't decided I am just yet.

It'll take some more thinking.

## Decomposition in algebraic integers

Thanks to a post by Rick Decker, a professor at Hamilton College, I can talk about the problem with conventional thinking on algebraic integers using someone else's example.

In his post Decker claimed to mirror my argument using a quadratic instead of a cubic, where he has

(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)

where his a's are roots of

a^2 - (x - 1)a + 7(x^2 + x).

Decker wanted to cast doubt on my reliance on using the constant terms to see how 7 divides both sides, by using an example where you can check at x=1, to see what the factors are, as then you have both a_1(1) and a_2(1) equal to sqrt(14).

However, Decker may have naively thought he was refuting my argument, when he didn't follow through to the logical conclusion from his own analysis, which supports it.

You see, the conventional thinking is that you can divide 7 from both sides and still be in the ring of algebraic integers, because algebraic integer are infinitely decomposable, so from a common sense perspective you might think that you can always find two algebraic integer factors of 7 to divide the two algebraic integer factors (5a_1(x) + 7) and (5a_2(x) + 7) on the left hand side.

So assume there exists algebraic integer functions w_1(x) and w_2(x), such that

w_1(x) w_2(x) = 7, and

(5a_1(x)+ 7)/w_1(x) (5a_2(x) + 7) /w_2(x) = 25x^2 + 30 x + 2.

The assumption is that you're still in the ring of algebraic integers with an algebraic integer x, so consider algebraic integer functions f_1(x), and f_2(x), such that

f_1(x) f_2(x) = 25x^2 + 30x + 2,

and

f_1(x) = (5a_1(x) + 7)/w_1(x)

and

f_2(x) = (5a_2(x) + 7)/w_2(x).

Checking at x=0, gives that f_1(0) = 1, and f_2(0) = 2.

Now then, replacing them with

f_1(0) = g_1(x) + 1, and f_2(0) = g_2(x) + 2, I have

(g_1(x) + 1)(g_2(x) + 2) = 25x^2 + 30x + 2

and letting

g_1(x) = 5 b_1(x), and g_2(x) = 5b_2(x), I have

(5 b_1(x) + 1)(5 b_2(x) + 2) = 25x^2 + 30x + 2.

Pushed to reply further on his original post by me, Decker actually went about calculating b_1(x), and his result is

2b_1(x)^2 - x b_1(x) + x^2 + x = 0.

See:

Message-ID: <3FF8C21F.4090409@hamilton.edu>

>Date: Sun, 04 Jan 2004 20:47:11 -0500

>From: Rick Decker

Newsgroups: sci.math

>Subject: Re: JSH: Rick Decker's example

But

2b_1(x)^2 - x b_1(x) + x^2 + x = 0,

is a non-monic, and not generally reducible over Q, with an integer x, proving that b_1(x) is not in general an algebraic integer.

Now then, what conventional wisdom has clearly fallen?

Well consider that the factorization I have looks something like

AB = 7C

where A, B, C, and of course, 7 are algebraic integers.

But divide both sides by 7, and because of that non-monic, you must have cases where

DE = C, where D and E cannot be algebraic integers!!!

Dividing 7 from both sides results in factors that are NOT in general algebraic integers as shown by Decker's own result, proving that there does NOT always exist a decomposition of 7 in the ring of algebraic integers that will do the job, since the quadratic he found isn't always reducible over Q, for an algebraic integer x.

It is my hope that my use of Decker's own example, and his own analysis to get that non-monic quadratic might in some small way break through the logjam created by various posters who never back down no matter how often their positions are proven wrong.

Want more? See my blog archives:

http://mathforprofit.blogspot.com/2003_10_01_mathforprofit_archive.html

If you wish to interpret Decker's result some different way, then you are free to try, but the gist of it is that you're pushed out of the ring of algebraic integers by dividing both sides by 7, defying conventional thinking that algebraic integer factors of 7 would always exist that can be divided from both of the algebraic integer factors on the left hand side.

[A reply to David C. Ullrich.]

You stupid shit head!!! What the fuck is wrong with you Ullrich?

No matter how many fucking times I tell you to fuck off, you keep replying to me!!!

What the fuck is your problem you shithead?

You Ullrich are a stupid piece of dumb shit who refuses to get the message when someone does NOT want to talk to you, you stupid fucking shitty asshole.

You are an ASSHOLE Ullrich!!! Now why don't you take your dumb ass stupid self somewhere to GET A FUCKING CLUE and QUIT FUCKING REPLYING TO ME AS IF I EVER WANT TO TALK TO YOU!!!!!!!!!!!!!

FUCK OFF!!!!

Can't you get it through your stupid head?

FUCK OFF!!!!!!!!!!!!!!!!!

In his post Decker claimed to mirror my argument using a quadratic instead of a cubic, where he has

(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)

where his a's are roots of

a^2 - (x - 1)a + 7(x^2 + x).

Decker wanted to cast doubt on my reliance on using the constant terms to see how 7 divides both sides, by using an example where you can check at x=1, to see what the factors are, as then you have both a_1(1) and a_2(1) equal to sqrt(14).

However, Decker may have naively thought he was refuting my argument, when he didn't follow through to the logical conclusion from his own analysis, which supports it.

You see, the conventional thinking is that you can divide 7 from both sides and still be in the ring of algebraic integers, because algebraic integer are infinitely decomposable, so from a common sense perspective you might think that you can always find two algebraic integer factors of 7 to divide the two algebraic integer factors (5a_1(x) + 7) and (5a_2(x) + 7) on the left hand side.

So assume there exists algebraic integer functions w_1(x) and w_2(x), such that

w_1(x) w_2(x) = 7, and

(5a_1(x)+ 7)/w_1(x) (5a_2(x) + 7) /w_2(x) = 25x^2 + 30 x + 2.

The assumption is that you're still in the ring of algebraic integers with an algebraic integer x, so consider algebraic integer functions f_1(x), and f_2(x), such that

f_1(x) f_2(x) = 25x^2 + 30x + 2,

and

f_1(x) = (5a_1(x) + 7)/w_1(x)

and

f_2(x) = (5a_2(x) + 7)/w_2(x).

Checking at x=0, gives that f_1(0) = 1, and f_2(0) = 2.

Now then, replacing them with

f_1(0) = g_1(x) + 1, and f_2(0) = g_2(x) + 2, I have

(g_1(x) + 1)(g_2(x) + 2) = 25x^2 + 30x + 2

and letting

g_1(x) = 5 b_1(x), and g_2(x) = 5b_2(x), I have

(5 b_1(x) + 1)(5 b_2(x) + 2) = 25x^2 + 30x + 2.

Pushed to reply further on his original post by me, Decker actually went about calculating b_1(x), and his result is

2b_1(x)^2 - x b_1(x) + x^2 + x = 0.

See:

Message-ID: <3FF8C21F.4090409@hamilton.edu>

>Date: Sun, 04 Jan 2004 20:47:11 -0500

>From: Rick Decker

Newsgroups: sci.math

>Subject: Re: JSH: Rick Decker's example

But

2b_1(x)^2 - x b_1(x) + x^2 + x = 0,

is a non-monic, and not generally reducible over Q, with an integer x, proving that b_1(x) is not in general an algebraic integer.

Now then, what conventional wisdom has clearly fallen?

Well consider that the factorization I have looks something like

AB = 7C

where A, B, C, and of course, 7 are algebraic integers.

But divide both sides by 7, and because of that non-monic, you must have cases where

DE = C, where D and E cannot be algebraic integers!!!

Dividing 7 from both sides results in factors that are NOT in general algebraic integers as shown by Decker's own result, proving that there does NOT always exist a decomposition of 7 in the ring of algebraic integers that will do the job, since the quadratic he found isn't always reducible over Q, for an algebraic integer x.

It is my hope that my use of Decker's own example, and his own analysis to get that non-monic quadratic might in some small way break through the logjam created by various posters who never back down no matter how often their positions are proven wrong.

Want more? See my blog archives:

http://mathforprofit.blogspot.com/2003_10_01_mathforprofit_archive.html

If you wish to interpret Decker's result some different way, then you are free to try, but the gist of it is that you're pushed out of the ring of algebraic integers by dividing both sides by 7, defying conventional thinking that algebraic integer factors of 7 would always exist that can be divided from both of the algebraic integer factors on the left hand side.

[A reply to David C. Ullrich.]

You stupid shit head!!! What the fuck is wrong with you Ullrich?

No matter how many fucking times I tell you to fuck off, you keep replying to me!!!

What the fuck is your problem you shithead?

You Ullrich are a stupid piece of dumb shit who refuses to get the message when someone does NOT want to talk to you, you stupid fucking shitty asshole.

You are an ASSHOLE Ullrich!!! Now why don't you take your dumb ass stupid self somewhere to GET A FUCKING CLUE and QUIT FUCKING REPLYING TO ME AS IF I EVER WANT TO TALK TO YOU!!!!!!!!!!!!!

FUCK OFF!!!!

Can't you get it through your stupid head?

FUCK OFF!!!!!!!!!!!!!!!!!

### Thursday, January 01, 2004

## JSH: Understanding the math

My research can be difficult to understand, so I thought I'd try out yet another way of explaining it. Some of you may have figured out that I test out explanations on Usenet for use elsewhere, to refine my own understanding, or just in case someone out there might finally get it.

Now then, again here's my discovery:

(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)

where b_3(x) = a_3(x) - 3 and the a's are roots of

a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)

and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.

In that form it's hard to understand what follows next unless you pay attention to what you have, specifically that cubic defining the a's.

I can get it because of the symmetry of

(5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)

where I've gone ahead and substituted a_3(x) back in to replace b_3(x), and it's important that you focus on that symmetry.

It's that symmetry which allows the cubic

a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)

to define ALL the a's, but something happens when I divide by 49.

Then the symmetry is broken. Without that symmetry it's impossible to find a SINGLE cubic to handle what results when you divide both sides by 49.

That's important because it's why the functions are NOT algebraic integer functions!!!

Now then, I'll recap. Symmetry allows the a's to be defined by a cubic, which shows them to be algebraic integer functions, but dividing by 49 *breaks* that symmetry, taking away the ability to find some cubic to define the results, which proves that the resulting functions are not algebraic integer functions.

(5 b_1(x) + 1)(5 b_2(x) + 1)(5 b_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22

where the b's are roots of

b^3 + ? b^2 + ? b - (2401 x^3 - 147 x^2 + 3x)

and when x=0, b_1(0) = b_2(0) = b_3(0) = 0.

My point is that the second and third coefficients are impossible to define in general.

You may find them for some particular x, but in general, they are forever hidden from you.

Notice that doing that substitution with a_3(x) for b_3(x) gives me

(5 b_1(x) + 1)(5 b_2(x) + 1)(5 a_3(x) + 7) = 300125 x^3 - 18375 x^2 - 360 x + 22

but you have broken symmetry since the other constant terms are 1 and 1, so you're still stuck.

Now by emphasizing what happens

Courage in mathematics might sound new to you as an idea, but those who came before you had courage. That's how they learned of sqrt(2), and sqrt(-1), and much other mathematics that some refused, fought, and bitterly attacked as they lacked that courage.

Of course, the mathematics should win, assuming that humanity survives long enough, but regardless each of you now faces an individual test of your own courage. It's not yet in the history books this time, where you can read and just imagine that you'd never have fought over sqrt(2) or refused to accept sqrt(-1) as an "imaginary" number.

Here it's the present, and your test of courage is now.

Oh, and be sure to check my blog archives:

http://mathforprofit.blogspot.com/2003_10_01_mathforprofit_archive.html

And hey, if you're moping and miserable because mathematics tests you, then maybe, if you think you're a mathematician, you might want to try a different field.

[A reply to David C. Ullrich.]

Well sure, delete out the argument which settles the issue, and then come back and claim that it's not settled.

Now math history is full of people like David Ullrich, a guy with a title, fighting against some new idea. In the past it was sqrt(-1), as mathematicians fought against an idea they thought of as silly.

The hallmark of such people is a refusal to follow mathematical logic. You give a detailed mathematical position, and they just delete it all out, and claim it's not what it is.

Yet the facts speak for themselves:

My work depends on advanced mathematical tools for factoring polynomials, where I use one polynomial to factor another polynomial into non-polynomial factors.

I'd be curious if anyone can find anything like it in the recorded history of mathematics.

Along with that work, I have my prime number research, where I have a partial difference equation used to count prime numbers—a first in math history.

People like David Ullrich are the anti-knowledge crowd. Those people who know just enough to feel comfortable in a complex world. The kind of people who say, "no more knowledge".

Ullrich probably feels very comfortable with whatever mathematical knowledge he has. He can pay his bills using the money given him by taxpayers, as he's a professor at Oklahoma State University. He probably has lots of feedback from various people in society to make him feel good about himself.

What I represent is the unknowable future. The power that comes no matter how comfortable people are, no matter how satisfied they are with themselves or their positions—the power of change.

In that sense I'm a force of Nature, a force of the Universe, a living emodiment of change itself, so I understand the fear I can induce in people like David Ullrich.

But the spirit of mathematics is ultimately about what's not known, what's beyond what's known today, as that defines the work of the future.

I stand here part of a great tradition of discoverers, people who got past the Ullrich's to push forward knowledge so that today you have sqrt(2), i, e, e^x, planes, trains, automobiles, spacecraft, and computers.

It's too easy to look at the past as if it were easy. As if new ideas were just accepted because they were right, and could be proven logically, mathematically, or practically, so that you stand by today, forcing yet another discovery to break through.

But that is your loss, as my strength comes from the knowledge I have, the knowledge of those who came before me, the knowledge of those who will come after.

The knowledge that the truth is beautiful.

All of you can resist, fight until you die, as you will die someday. I can't look to any of you for permanence as you're in a sense, dead already. But I can, and you can look at the truth—beautiful, permanent, absolute.

The discoverers are the forces of Nature, born to a grand tradition, tasked with the seeming impossible, fighting throughout time, throughout history, to make history itself.

The discoverers are the living embodiment of the forces of Good against Evil. The people who push the limits of knowledge for the benefit of all, and the love of Truth.

We are the greatest warriors for Good, of all time, through all time, through all worlds and realities where sentient beings live.

We are the discoverers.

[A reply to someone who said that “History is also full of people that proposed ridiculous ideas which now are dead and buried”.]

Most people show worth through effort. For instance, if you value your job, one would assume that you put a lot of effort into it. Similarly, if you value a relationship, you're willing to work to keep it going.

People like Ullrich value trying to attack the new, as evidenced by his

There are a lot of people out there I probably would suppose are nutty, who have ideas that I don't care at all about, and then again there are people who are supposedly very competent, or expert, who have ideas that I don't care about, and you know what?

I don't spend time on them.

Evil is the pursuit of ignorance, challenging new truths to hold on to an old, and comfortable point of view. It's also stupid as change is inevitable, and the energy of the evil person is wasted.

However, in spite of its stupidity evil poses challenges, which the discoverer is ever tasked with fighting through, including handling those who fight for evil in their attempts to maintain their own comfort against knowledge.

That fight is one of the continuing burdens of the Universe's first, greatest, and last fighting force.

[Another reply to David C. Ullrich.]

Fuck off!!! You fucking stupid idiot!!! How many times do I have to tell you that your attention is not wanted David Ullrich?

When will you fucking get the goddamn fucking message to fuck off?

FUCK OFF David Ullrich!!!!!!! Fuck off you stupid fucking idiot!!!!

FUCK OFF!!!!!!!!!!!!

Now then, again here's my discovery:

(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)

where b_3(x) = a_3(x) - 3 and the a's are roots of

a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)

and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.

In that form it's hard to understand what follows next unless you pay attention to what you have, specifically that cubic defining the a's.

I can get it because of the symmetry of

(5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)

where I've gone ahead and substituted a_3(x) back in to replace b_3(x), and it's important that you focus on that symmetry.

It's that symmetry which allows the cubic

a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)

to define ALL the a's, but something happens when I divide by 49.

Then the symmetry is broken. Without that symmetry it's impossible to find a SINGLE cubic to handle what results when you divide both sides by 49.

That's important because it's why the functions are NOT algebraic integer functions!!!

Now then, I'll recap. Symmetry allows the a's to be defined by a cubic, which shows them to be algebraic integer functions, but dividing by 49 *breaks* that symmetry, taking away the ability to find some cubic to define the results, which proves that the resulting functions are not algebraic integer functions.

(5 b_1(x) + 1)(5 b_2(x) + 1)(5 b_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22

where the b's are roots of

b^3 + ? b^2 + ? b - (2401 x^3 - 147 x^2 + 3x)

and when x=0, b_1(0) = b_2(0) = b_3(0) = 0.

My point is that the second and third coefficients are impossible to define in general.

You may find them for some particular x, but in general, they are forever hidden from you.

Notice that doing that substitution with a_3(x) for b_3(x) gives me

(5 b_1(x) + 1)(5 b_2(x) + 1)(5 a_3(x) + 7) = 300125 x^3 - 18375 x^2 - 360 x + 22

but you have broken symmetry since the other constant terms are 1 and 1, so you're still stuck.

Now by emphasizing what happens

**after**49 is divided from both sides I'm trying to get at least some of you to face the mathematical realities here, and I've made other posts pointing it out as well.Courage in mathematics might sound new to you as an idea, but those who came before you had courage. That's how they learned of sqrt(2), and sqrt(-1), and much other mathematics that some refused, fought, and bitterly attacked as they lacked that courage.

Of course, the mathematics should win, assuming that humanity survives long enough, but regardless each of you now faces an individual test of your own courage. It's not yet in the history books this time, where you can read and just imagine that you'd never have fought over sqrt(2) or refused to accept sqrt(-1) as an "imaginary" number.

Here it's the present, and your test of courage is now.

Oh, and be sure to check my blog archives:

http://mathforprofit.blogspot.com/2003_10_01_mathforprofit_archive.html

And hey, if you're moping and miserable because mathematics tests you, then maybe, if you think you're a mathematician, you might want to try a different field.

[A reply to David C. Ullrich.]

Well sure, delete out the argument which settles the issue, and then come back and claim that it's not settled.

Now math history is full of people like David Ullrich, a guy with a title, fighting against some new idea. In the past it was sqrt(-1), as mathematicians fought against an idea they thought of as silly.

The hallmark of such people is a refusal to follow mathematical logic. You give a detailed mathematical position, and they just delete it all out, and claim it's not what it is.

Yet the facts speak for themselves:

My work depends on advanced mathematical tools for factoring polynomials, where I use one polynomial to factor another polynomial into non-polynomial factors.

I'd be curious if anyone can find anything like it in the recorded history of mathematics.

Along with that work, I have my prime number research, where I have a partial difference equation used to count prime numbers—a first in math history.

People like David Ullrich are the anti-knowledge crowd. Those people who know just enough to feel comfortable in a complex world. The kind of people who say, "no more knowledge".

Ullrich probably feels very comfortable with whatever mathematical knowledge he has. He can pay his bills using the money given him by taxpayers, as he's a professor at Oklahoma State University. He probably has lots of feedback from various people in society to make him feel good about himself.

What I represent is the unknowable future. The power that comes no matter how comfortable people are, no matter how satisfied they are with themselves or their positions—the power of change.

In that sense I'm a force of Nature, a force of the Universe, a living emodiment of change itself, so I understand the fear I can induce in people like David Ullrich.

But the spirit of mathematics is ultimately about what's not known, what's beyond what's known today, as that defines the work of the future.

I stand here part of a great tradition of discoverers, people who got past the Ullrich's to push forward knowledge so that today you have sqrt(2), i, e, e^x, planes, trains, automobiles, spacecraft, and computers.

It's too easy to look at the past as if it were easy. As if new ideas were just accepted because they were right, and could be proven logically, mathematically, or practically, so that you stand by today, forcing yet another discovery to break through.

But that is your loss, as my strength comes from the knowledge I have, the knowledge of those who came before me, the knowledge of those who will come after.

The knowledge that the truth is beautiful.

All of you can resist, fight until you die, as you will die someday. I can't look to any of you for permanence as you're in a sense, dead already. But I can, and you can look at the truth—beautiful, permanent, absolute.

The discoverers are the forces of Nature, born to a grand tradition, tasked with the seeming impossible, fighting throughout time, throughout history, to make history itself.

The discoverers are the living embodiment of the forces of Good against Evil. The people who push the limits of knowledge for the benefit of all, and the love of Truth.

We are the greatest warriors for Good, of all time, through all time, through all worlds and realities where sentient beings live.

We are the discoverers.

[A reply to someone who said that “History is also full of people that proposed ridiculous ideas which now are dead and buried”.]

Most people show worth through effort. For instance, if you value your job, one would assume that you put a lot of effort into it. Similarly, if you value a relationship, you're willing to work to keep it going.

People like Ullrich value trying to attack the new, as evidenced by his

**efforts**in that regard.There are a lot of people out there I probably would suppose are nutty, who have ideas that I don't care at all about, and then again there are people who are supposedly very competent, or expert, who have ideas that I don't care about, and you know what?

I don't spend time on them.

Evil is the pursuit of ignorance, challenging new truths to hold on to an old, and comfortable point of view. It's also stupid as change is inevitable, and the energy of the evil person is wasted.

However, in spite of its stupidity evil poses challenges, which the discoverer is ever tasked with fighting through, including handling those who fight for evil in their attempts to maintain their own comfort against knowledge.

That fight is one of the continuing burdens of the Universe's first, greatest, and last fighting force.

[Another reply to David C. Ullrich.]

Fuck off!!! You fucking stupid idiot!!! How many times do I have to tell you that your attention is not wanted David Ullrich?

When will you fucking get the goddamn fucking message to fuck off?

FUCK OFF David Ullrich!!!!!!! Fuck off you stupid fucking idiot!!!!

FUCK OFF!!!!!!!!!!!!