Sunday, November 14, 2004


JSH: Contradiction shown

Now it turns out that having thought it all through I finally realized that I could show a direct contradiction using the objections raised against my work.

Notice here I'll actually use constants where before I talked about variables being held constant.

I start with

P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078

which can be factored into non-polynomial factors by using

P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3

which looks complicated but if you multiply it all out and simplify, you get the first, and I just have an analysis method for breaking up a polynomial in a certain way to factor it—into non-polynomial factors—and here the factorization is

P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7)

where the a's are roots of

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

And it's easy to get that cubic by solving for one of the a's using

(5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3.

Now I've pointed out that setting x=0, will show that two of the a's equal 0 at that point, while one equals 3, which others have argued is a "special case".

The cubic with x=0 is

a^3 -3a^2 = 0

so two of the a's equal 0, while one equals 3.

Usually I arbitrarily select a_1, and a_2 to equal 0, at x=0, which is what I'll do now. Then, with

P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7)

at x=0, you have two of the a's equal 0, while one equals 3, to give

P(0) = 7(7)(22) = 1078

which fits with

P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078.

Then for the first two factors (5a_1(x) + 7) and (5a_2(x) + 7) the 7's visible in the expressions are factors of the constant term of P(x).

But P(x) is a multiple of 49 as each coefficient has 49 as a factor, so I divide out the 49 to get

P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22.

Now assume each of the a's has some non-unit factor in common with 7, for a given x, which in fact they do in the ring of algebraic integers whenever all the a's are irrational, which is a point that has been brought up repeatedly by people arguing with me. For a while I resisted that fact, but now as I've said before I concede that they are in fact correct—each of the a's, in the ring of algebraic integers does in fact share a non-unit factor with 7 when all of the a's are irrational.

So let w_1(x) w_2(x) w_3(x) = 49, where the w's are those factors, so

a_1(x) = w_1(x) b_1(x),

a_2(x) = w_2(x) b_2(x),


a_3(x) = w_3(x) b_3(x)

and where

w_1(x) v_1(x) = 7, w_2(x) v_2(x) = 7, and w_3(x) v_3(x) = 7,

and divide through by 49 to get

P(x)/49 =

(5 b_1(x) + v_1(x))(5 b_2(x) + v_2(x))(5 b_3(x) + v_3(x))

and if you allow that the factors are each factors of the constant term as before, then you have

v_1(0) v_2(0)(15 + v_3(0)) = 22

at x=0, and when x does not equal 0, you have

v_1(x) v_2(x)(5u_3(x) + v_3(x)) = 22

where I introduce u_3(x) to handle any further weirdness with how a_3(x) behaves, where u_3(0) = 3, to agree with previous results, and remember

P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22

and the constant term doesn't change, as, well, it's constant, as it's 22.

So, notice, I now have that v_1(x) and v_2(x) are factors of 22.

Provably, they cannot be units in the ring of algebraic integers when a_1(x), a_2(x), and a_3(x) are all irrational as that's what people argued with me about for so long.

The full result was just so weird that it escaped even me for a while, as following their arguments to their logical conclusion 22 and 7 must share non-unit factors in the ring of algebraic integers.

But you can also appear to prove that 22 and 7 are coprime in the ring of algebraic integers using various accepted definitions of coprimeness, like you can find algebraic integers x, and y such that

22x + 7y = 1

and claim to have proven that they are coprime.

So, in the ring of algebraic integers, using what's commonly accepted you can prove that 22 and 7 are both coprime and that they share non-unit algebraic integer factors.

That's the full result.

Now, of course, posters arguing with me on Usenet wouldn't follow through to the full result, probably because they weren't smart enough to see it, as make no mistake, that's the kind of result that's quite big.

That's a career maker, or could have been one. For me, it's just one of my results. Nice, sure, but still just one, with techniques I introduced in a couple of lines in another paper.

The ring of algebraic integers is quirky. That arbitrary selection of roots of monic polynomials with integer coefficients allows you to do all kinds of wacky things, including believing that you're doing all kinds of great math, when you have nothing at all.

Make no mistake, the big names in math today have just about zero reason to accept these results. Would you if you were them?

But you're not them.

You are students. You have the future to make your names, with real math that is actually correct.

I'm offering you your future. Follow the math, and then if you have what it takes, then accept what is true without holding your breath waiting for old men who are trying to stave off ruin to do the right thing.

They are cowards. They will admit the truth when they no longer believe they can get away with lying.

Come on, do you really think results like these could have been missed by the top "mathematicians" of the world?

Well, maybe, maybe they don't know. But I think they do. After all, I've contacted people all over the math world, as I've been quite bold in pushing forward this issue, in warning about the error.

I think they do know and they are waiting to see what you do.

I think they are waiting to see if you will let them get away with the lies.

Maybe you will—or at least, maybe you'll try. LOL. Some of you probably will try because you're young, and foolish, and so eager to please them.

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