William Hughes <wpihughes@hotmail.com> wrote:

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

This is different. His claim seems to be that *if* two of the terms in
the product *are* divisible by 7 for a *specific* value of x, then they
are divisible by 7 for *all* values of x. In your example neither is ever
divisible by 3.

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

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

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

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

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

William Hughes <wpihughes@hotmail.com> wrote:

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

This is different. His claim seems to be that *if* two of the terms in
the product *are* divisible by 7 for a *specific* value of x, then they
are divisible by 7 for *all* values of x. In your example neither is ever
divisible by 3.

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)

(I will call the above equation *)

: 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)

Okay. I'm going to assume that your algebraic manipulation is accurate.

: 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.

Please forgive me, could you elaborate on what you mean by "divides off"?

It's true that 49 divides the right-hand side of * so it divides the
left-hand side of *. Am I correct in assuming that's what you mean? Or
perhaps you are investigating _the way_ it divides the left-hand side?

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

Okay. So far so good. Where do we go from here? Are you claiming that

since, when x=0, * becomes (7)(7)(22)=49(22), that the 7s must distribute
as such on the left-hand side for all x? In other words, are you
claiming that (5 a_1(x) + 7) and (5 a_2(x) + 7) are divisible by 7 for
all x? If so, could you please explain why the fact that it's true for
x=0 implies it's true for other x also?

Thank you,
Justin

"Me" <no@spam.com> wrote in message
news:<brk4rd$c20$1@grapevine.wam.umd.edu>...
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)

(I will call the above equation *)

: 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)

Okay. I'm going to assume that your algebraic manipulation is accurate.

: 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.

Please forgive me, could you elaborate on what you mean by "divides
off"?

To divide off 49 with

(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)

you divide both sides by 49.


It's true that 49 divides the right-hand side of * so it divides the
left-hand side of *. Am I correct in assuming that's what you mean? Or
perhaps you are investigating _the way_ it divides the left-hand side?

To divide off 49 with

(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)

you divide both sides by 49.


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

Okay. So far so good. Where do we go from here? Are you claiming that

My original post was quit specific. Now that I've explained "divides
off" possibly you simply need to reread it.

If you still do not understand, I recommend you turn to something
else.

since, when x=0, * becomes (7)(7)(22)=49(22), that the 7s must
distribute
as such on the left-hand side for all x? In other words, are you
claiming that (5 a_1(x) + 7) and (5 a_2(x) + 7) are divisible by 7 for
all x? If so, could you please explain why the fact that it's true for
x=0 implies it's true for other x also?

Thank you,
Justin

My original post was quite specific. Now that I've explained "divides
off" you should return to that post. If you're still confused,
possibly you should move on to another subject.


James Harris

Speak your minds.

James Harris

Please forgive me, could you elaborate on what you mean by "divides off"?

To divide off 49 with

(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)

you divide both sides by 49.


It's true that 49 divides the right-hand side of * so it divides the
left-hand side of *. Am I correct in assuming that's what you mean? Or
perhaps you are investigating _the way_ it divides the left-hand side?

To divide off 49 with

(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)

you divide both sides by 49.

On Mon, 15 Dec 2003 15:57:12 +0000 (UTC), magidin@math.berkeley.edu
(Arturo Magidin) wrote:

[what he wrote snipped]

You know, I thought it was a big mistake for you to agree to
stop replying to James's posts. Never mind why.

You probably haven't noticed that he's replying to _your_
posts,

calling you a liar and including the usual blather
about how you're still including your email address in your
sig. Seems to me he misses you. (Or he's trying to bait
you into replying so he can score points by whining about
how you don't keep your promises.