wpihughes@hotmail.com (William Hughes) wrote in message news:<4d5e4663.0312150510.86bcfd1@posting.google.com>...
jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0312141928.49c9b250@posting.google.com>...
Given, where x is in the ring of algebraic integers, I've shown the
factorization

(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)

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

I'm curious about the mental processes that allow *some* of you to
claim that 49 divides off as a *variable* dependent on x, so I'm
giving another opportunity for you to speak your minds.

To my knowledge, in the history of mathematics, no one has ever
presented such a proposition, so it is a unique one, and I must say
that I'm intrigued.

Speak your minds.


James Harris

Why does this not count as presenting "such a proposition"?

Let g1(x)=4-sqrt(1+3x) and g2(x)=4+sqrt(1+3x).

then
g1(x)*g2(x) = 3(5-x)

I noted there's a *sign* ambiguity in the sqrt() operator, which
sparked a lot of debate.

One thing I found interesting is that posters ignored that if you
divide both sides by 3, with the convention that you're taking the
positive of the sqrt() operator for *integer* results, you have

1 + (1-sqrt(1+3x))/3

as a factor, which is an integer (remember x is an integer and
remember 1+3x is a square).

but neither g1 nor g2 is divisible by 3 for all x.

-"William Hughes"

Prove it.

wpihughes@hotmail.com (William Hughes) wrote in message news:<4d5e4663.0312150510.86bcfd1@posting.google.com>...
jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0312141928.49c9b250@posting.google.com>...
Given, where x is in the ring of algebraic integers, I've shown the
factorization

(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)

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

I'm curious about the mental processes that allow *some* of you to
claim that 49 divides off as a *variable* dependent on x, so I'm
giving another opportunity for you to speak your minds.

To my knowledge, in the history of mathematics, no one has ever
presented such a proposition, so it is a unique one, and I must say
that I'm intrigued.

Speak your minds.


James Harris

Why does this not count as presenting "such a proposition"?

Let g1(x)=4-sqrt(1+3x) and g2(x)=4+sqrt(1+3x).

then
g1(x)*g2(x) = 3(5-x)

I noted there's a *sign* ambiguity in the sqrt() operator, which
sparked a lot of debate.

One thing I found interesting is that posters ignored that if you
divide both sides by 3, with the convention that you're taking the
positive of the sqrt() operator for *integer* results, you have

1 + (1-sqrt(1+3x))/3

as a factor, which is an integer (remember x is an integer and
remember 1+3x is a square).

but neither g1 nor g2 is divisible by 3 for all x.

-"William Hughes"

Prove it. Readers should note that this poster presented a *later*
post claiming a result that covers integer results of the square root
operator but made a rather simple mistake. In my reply to that post I
noted the sign ambiguity in the square root operator.

However sqrt(x), where x is a square can be taken to be the positive
result since you can *give* the result, but the ambiguity remains if
you see sqrt(x) without a given value.

That is, for instance, sqrt(4) is 2 *or* -2, but by convention, it's
*usually* taken as 2, though if you do enough analysis you will run
into situations where you need the negative!!!

James Harris <jstevh@msn.com> wrote:

: Given, where x is in the ring of algebraic integers, I've shown the
: factorization

: (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)
: so when x=0, a_1(0) = a_2(0) = b_3(0) = 0.
: Now you can divide both sides by 49.

: Some posters have claimed that when you divide by 49, what results
: varies depending on what value x has.
: Do you understand?

I don't understand what you mean by the phrase "what results".

Please consider my question in the following step-by-step:

1) You get your factorization. I agree.
2) 49 divides right-hand side. I agree.
3) 49 divides left-hand side. I agree.
4) So when x=0 we get

(7)(7)(22) = 49(22) I agree.

5) But when x!=0 it's not immediately clear what results from this
equation. Simply because 7 divides the first two terms when x=0 does not
imply anything about when x!=0. Are you suggesting that 7 divides the
first two terms when x!=0? If so, could you please explain why?

: I'm curious to know if you will admit understanding now, or if you
: will continue to ask questions, so I ask again:
: Do you understand?

I will certainly admit understanding after you clear up my question.