jstevh

Now concentrating on the expression

(v^3+1)x^3 - 3vxy^2 + y^3

let v= -1 + mf^2.

If you have to have a ring, assume it's algebraic integers, m is a
nonzero integer, and f is a nonzero algebraic integer coprime to 3 and
m.

Now then substituting for v gives

((-1 + mf^2)^3 + 1)x^3 - 3(-1 + mf^2)xy^2 + y^3

which is

(m^3 f^6 - 3m^2 f^4 + m f^2 ) x^3 - 3(-1 + mf^2 )xy^2 + y^3

and now let y = uf, where u is an algebraic integer coprime to f, and
using that substitution gives

(m^3 f^6 - 3m^2 f^4 + m f^2 ) x^3 -

3(-1 + mf^2)x u^2 f^2 + (uf)^3

which is

f^2 ((m^3 f^4 - 3m^2 f^2 + m) x^3 -

3(-1 + mf^2 )x u^2 + u^3 f).

Now let

P(m) = f^2((m^3 f^4 - 3m^2 f^2 + m) x^3 -

3(-1+mf^2 )x u^2 + u^3 f)

and consider that the constant term is

P(0) = f^2(3x u^2 + u^3 f).

Then

P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + m) x^3 -

3(-1+mf^2 )x u^2 + u^3 f.


Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization

P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)

where w_1 w_2 w_3 = f, and

b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + m),

and at m=0

P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),

so two of the b's must equal 0, which means

P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)

which is

P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf)

proving that w_1 w_2 must equal 1, as f is coprime to 3 from before,
which leaves b_3 = 3.

Here notice that it's clear that two of the b's must go to 0 from the
"u" symbols, as that's the only way to get that factor of u^2 in
P(0)/f^2.

Given that w_1 and w_2 are coprime to f, then the factorization is

P(m)/f^2 = (b_1 x + u)(b_2 x + u)(b_3 x + uf).

Now looking at the factorization

P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)

it's clear that a_1 = f b_1, and a_2 = f b_2, which proves that a_3 is
coprime to f, since

a_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + m),

so

b_1 b_2 a_3 = (m^3 f^4 - 3m^2 f^2 + m),

which is coprime to f as m is coprime to f, notice that a_3 = b_3.


James Harris

virgil

Thank Heavens he's demonstrating again.

Since James hadn't demonstrated for a couple of months, I was afraid
he was stagnant.

w. dale hall

Still looking for that promised follow-on to this from his preceding
article:


Note the cavalier "if you have to have a ring", nearly rubbing it in
by following with "you babies", as though *real* men don't need no
stinking rings. REAL MEN do algebra with objects devoid of ANY connection to mathematics.


I keep seeing these expressions, without any reference to *what*
is an integer, *what* is an indeterminate, or *where* all these
manipulations take place. I know he said "algebraic integers", but
some of these things *surely* must be variables.


So x is a number, m is a variable. However, he says above m is
a nonzero integer, and f is coprime to m. So m can't be a variable.


So, b_3 = 3? Are we talking about the polynomial


Aren't the b's coefficients of the x's? Is this as dumb as it looks?

I mean, does 3 divide 65? I didn't think so.

JSH likes to argue from "it's clear that" to "you lying incompetent
bastards", I guess. I suppose that's the only way a person of his
monumental absence of honor could produce the results he, er, has.

Oh, no, wait, he HASN'T produced any results, has he?


Still not seeing that polynomial you promised:


Were you still on that thing about two of the coefficients being coprime
to 5?

Didn't I show you those coefficients, as well as the common factor each
one has with 5?

C'mon, stop being such a girly-man.

Show some gumption. Say what it is you claim with respect to the
*actual* coefficients of that polynomial, without hiding it behind
your mama's skirts, or behind the fig leaf of m and u and v and f and
all this other distraction.

All you're likely to prove is that you're full of hot air.

Dale.

jstevh

I took that out to *lessen* confusion. Looking at this poster's
reply, it's clear he put it back in to add it.

There's a problem with the ring of algebraic integers, which this
argument shows.

No histrionics are necessary.

It's algebra, they're all variables, as that's what "x" or "y" or any
letter in an algebraic expression is.

Notice the attack on the basis of algebra itself!!!

After all, algebra is about symbols in place of numbers, and those
symbols are all called variables.

It's algebra; they're all variables.


It is shown that at m=0, b_3 = 3.


Now readers can see why the poster introduced 65x^3 - 12x + 1, though
it wasn't in the post to which he replied.


Apparently, the idea is to keep throwing up smoke to interfere with
readers like yourself being able to follow.

And readers can see the point of this person's post: telling them what
to think.

Here's some person on Usenet trying to control other readers on the newsgroup.


I took that out to *lessen* confusion. Looking at this poster's
reply, it's clear he put it back in to add it.


Histrionics, personal attacks, and emotional appeals from a poster
trying to tell you what to think, should worry the reader.

If I were wrong, with such an approach laid out as it was then instead
of a personal attack, posters could actually find an error.

Ultimately, it's the reader's choice. You can be controlled by
posters, or you can follow the math.


James Harris

the last danish pastry

that you have no conception as to what it is.

w. dale hall

Yes, I suppose it's *confusing* to see actual numbers. Your argument is
a huge pile of, er, error and confusion itself, and it does wonders to
see what it *means* when actual numbers are in place of the symbols you
strew about.

I find it somewhat surprising that you're willing to respond to this (my
preceding) posting, which was evidently *not* about the details of this
exceedingly flawed process you insist on parading about, but rather on
the absence of any clear association to the polynomial you *said* you
were talking about.

When I have pointed out (on too many occasions to mention here, but
I did post a count, and it was embarassingly large) that your method
produces an incorrect conclusion, you have been unwavering in your
willingness to ignore the evidence. You have claimed obliquely to
Nora Baron that I haven't shown anything that contradicts your case,
but have simply ignored the direct, ******** calculations that even
an elementary algebra student could follow. Those calculations show
that your claim regarding coefficients of the relevant factorization
of the polynomial

65x^3 - 12x + 1

is mistaken. I didn't need any extra variables. I didn't need to
claim to minimize confusion. I just went there and found what you
said wasn't so, and did it in a way that ANYONE can verify.

You have chosen to ignore a simple and direct argument, requiring
nothing more than algebraic manipulation. I don't need to assume
anything about factorizations, coefficients, divisibility, or
common factors or their absence. Yet you ignore that, in preference
for the current thread, in which extra variables exist in multitudes,
for (what you say) is "to *lessen* confusion".

I say you're a coward, and hide behind these since it's easy for you
to maintain the pretense that you're doing *something* right.

It's a charade, and you're the charlatan.

BTW, It's plain rude to say you're talking about a particular problem,
and then proceed to ignore said problem. I doubt that anyone could tell
what claims you're making now about the factorization of the polynomial
you *SAID* you were talking about. I know I couldn't.

I, for one, refuse to follow an argument once it is SHOWN to produce an
incorrect conclusion. It's up to you to clean up your own mess, not up
to anyone else to wade through the raw sewage and point out the floating turds.


No, you have yet to point out an ********** problem with that ring. Up
until recently, you have denied it's a ring, after all. Perhaps the
problem lies entirely between your ears.

Oh, so the "if you have to have a ring" means something *other* than
what I pointed out?

Can you show me an argument involving polynomials, that DOES NOT require
the existence of a ring? Just one?

I didn't think so.

SO, what else could your taunting remark mean?

... stuff deleted ...

A variable isn't an integer.

Notice the attack is on the audacity of someone asking what JSH means!!!


Yes, and freedom's just another word for nothing left to lose. You think
that everything is all about your little world and your little ideas.

Variables are different from numbers, which are different from parameters.


This is where you are absolutely clueless.

Oh, I see. Your preceding post, titled

"Advanced polynomial factorization, demonstration"

and this thread, titled

"Algebraic integer factorization, demonstration"

were *not* talking about the cubic I mentioned?

I sure though they were. As you said, you left the mention of that cubic
out, "to *lessen* confusion."


I'm sorry if I misinterpreted. I haven't seen a single word to the
effect that something is amiss with the ring of algebraic integers,
save your bald assertions. However, it's ME who wishes to tell others
what and how to think?

Wasn't that posed flatly as a question? A rhetorical question, to boot?

Are you *sure* English is your native language?


Oh, I get it. YOU can threaten:

1. legal action (James S. Harris, sci.math 2003-07-31 05:31:36 PST ):

My warning to universities and mathematicians teaching
at them remains.

Some of you may face civil liabilities, maybe worse--
angry parents--when the full story comes out.

School is about to start, and if universities teach the
flawed mathematics they may also face federal authorities
for deliberate fraud.

2. murder (James S. Harris, sci.math 2003-04-19 16:52:09 PST ):

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.

3. exposure as frauds
(James S. Harris, sci.math 2003-04-23 13:13:17 PST):

It seems to me that your statements here can be deliberately
taken as an attempt at fraud in further attempts to prevent
the world from recognizing that I have a short proof of Fermat's
Last Theorem.

It also seems to me that you must be aware that proper
recognition has monetary value for me, so malice aforethought
can be assumed.

I am also making it clear to you that I deem your statement
and the defense of them by at least one other mathematican to
be evidence of a severe problem within the mathematical
community, which at least in the United States, I feel should
be a matter that is considered by Congress in order to determine
proper distribution for funding of mathematical research in the
United States.

but if I raise the rhetorical (but factual) issue of whether you've
*actually proven anything*, it's I who is trying to control other
readers on the newsgroup?

Does your dictionary contain a mirror instead of a verbal definition,
next to the word "hyprocrite"?

Doesn't there seem to be a bit of an imbalance between the tone of your
excerpts above, and what I wrote that you're whining about? Does that
imbalance point to *me* as being the one whos over the top?
Hardly.

I see. Here's what I imagine you're saying

THIS IS CONFUSING:

Let

65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)

In response to the claim that any of the ai's could
be coprime to 5, I gave these formulas:

q(a) = 8 a^2 + 76 a - 185
r(a) = 8 a^2 + 4 a - 45
s(a) = 4 a^2 + 37 a - 104

and showed that whenever ai is one of the
ai's in the first factorization, then:

Letting qi = q(ai), ri = r(ai), si = s(ai),

ai = qi ri
and 5 = ri si.

This exhibits the common factor of ri that divides
*both* ai and 5. It's what you "proved" could not
exist.

I ask you what about those factors, and how does this
thread's discussion relate to that particular polynomial?

whereas

THIS IS NOT CONFUSING:

Now then substituting for v gives

((-1 + mf2)3 + 1)x3 - 3(-1 + mf2)xy2 + y3

which is

(m3 f6 - 3m2 f4 + m f2 ) x3 - 3(-1 + mf2 )xy2 + y3

and now let y = uf, where u is an algebraic integer
coprime to f, and using that substitution gives

(m^3 f^6 - 3m^2 f^4 + m f^2 ) x^3 -

3(-1 + mf^2)x u^2 f^2 + (uf)^3

which is

f^2 ((m^3 f^4 - 3m^2 f^2 + m) x^3 -

3(-1 + mf^2 )x u^2 + u^3 f).

... and so forth.

Now, tell me. Given the fact that I have given ********** formulas,
with no clever dancing around the point, like this:

"where u is an algebraic integer coprime to f,
and using that substitution ..."

how is it that you claim *I* am making things confusing? My work can
be verified directly. Yours relies on the choice of an algebraic integer
coprime to f, making some substitution, and calling out to the US Army
when no one accepts your bogus argument as holding any water.

I have a direct calculation that shows you're full of shit, and you're
dancing around in your leprechaun hat, saying "it's full of magical surprises!"

No, I have not made any personal attacks. What I *have* attacked is
what you're doing, way it is you're behaving. You're behaving like
a naughty child who wants to get away with having done something that
he shouldn't have done. It's abundantly clear that you *don't* have
anything to say about that *specific* polynomial, since if you did,
I could show that it behaves *precisely* the way that one would surmise
from standard Galois theory and the perspective of the standard world
of algebraic integers.

You wish to have people believe that the algebraic integers are flawed,
and still you wish to have that happen IN THE TOTAL ABSENCE OF SPECIFIC
EXAMPLES OF SHORTCOMINGS!

Show me how wrong I am. Produce a SPECIFIC polynomial, and roots and/or
factors that behave OTHER than how they should. EXHIBIT a flaw in the
ring of algebraic integers. No? Well, you'll just have to put up with
being the butt of all jokes.

If you could read more than just what you hoped to see, you would have
seen that I've asked you to state, in so many words, exactly what it is
that you're claiming about the polynomial that you've excised "to
*lessen* confusion". Don't be afraid of confusing *me* with numbers,
because I know what the roots and relevant factors of the roots of
that polynomial are.

As far as telling the reader what to think, let's let the readers
compare what it is you've said and threatened, to what you're crying
about from my preceding posting.


I'm sorry. I have done that *HOW MANY TIMES*???

Over the past month or so, I believe I have posted on no fewer than 10
occasions, a *direct*, ********** proof that your argument produces an
erroneous conclusion.

You are now acting as though none of those postings have taken place.

Your attitude of evasion will surely catch up with you, if not on
sci.math, or whererever the hell you are throwing cross-postings these
days, then in real life.

A person sows what he reaps.


Show of hands for folks who follow what JSH is terming "the math"
in his articles? Anyone?

To find my argument, find THIS posting of mine:
<3F1C3F01.7010501@farir.com>

Actually, there are many in the June-July 2003 timeframe, with a
similarly-phrased argument.

That article has the full argument I've alluded to, and anyone can
easily tell whether I've got the goods, or not.

Dale.

jstevh

Deleting out mathematics does not make it go away. I've seen this
behavior repeatedly and it's bizarrely childish, as if by making a
post where the information is deleted out...and I'll put key
information back in here as I'm replying...a poster believes they've
refuted a *math* argument!!!

What makes it extraordinary is that apparently some vocal math people
don't believe in mathematics as intrinsically true, but as a body of
social "truth" where a math argument gains validity based on group
acceptance. They clearly are not afraid of showing their belief
system when it comes to posting on the sci.math newsgroup.

That is, math to them is true, if people believe it's true. Here's
some of the deleted information which has taken away room for cogent
objections:

Then

P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 -

3(-1+mf^2 )x u^2 + u^3 f.


Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization

P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)

where w_1 w_2 w_3 = f, and

b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m),

and at m=0

P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),

so two of the b's must equal 0, which means

P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)

which is

P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf)

proving that w_1 w_2 must equal 1, as f is coprime to 3 from before,
which leaves b_3 = 3.

By shifting focus to P(m)/f^2 I shorten the space where posters can
try to confuse. Remember though readers, they are trying to
manipulate you.

My position is that the mathematics speaks for itself, but if it is
incorrect, someone can find a problem with the argument
presented--none of this other nonsense.

Math is simple in that way--if it's wrong, you can find something
wrong with what's presented by showing a break in the logical chain
i.e. show a given statement, which is followed by another statement,
which does NOT follow logically.

So demonstrating an error means doing something like, saying that
while statement A was ok, you follow with statement B, which does not
follow mathematically.


James Harris

virgil

Since you put it in the ****ogy originally, it is your confusion,
not anyone else's.

the last danish pastry

book about math but you can't remember any of it.

will twentyman

Like here? http://groups.google.com/groups?q=g:thl1061915289d&dq=&hl=en&lr=&ie=UTF-8&selm=3f1af19c%241_2%40newsfeed&rnum=52

--
Will Twentyman
email: wtwentyman at copper dot net