JSH: Your corrupt discipline

So now some of you know, the mathematical field is corrupt.

I don't know exactly when it happened but at some point over the last hundred years the discipline was corrupted and at this point, spectacular results can be blocked by most mathematicians simply doing
nothing.

And they know that they can just do nothing.

So why am I talking to undergrads?

Because the grads are corrupted as well.

You are on your way to being corrupted as that's part of the training of being a modern mathematician, but at this crucial point, learning that those professors you trust are not to be trusted, and realizing that your efforts at learning what you think are important mathematical theorems are a waste of time, some of you won't be able to do it.

That's the point.

SOME of you will find that knowing you're getting fed b.s. you just can't force yourself to just play along.

JSH: Clock is ticking

This time, unlike before, years will not go by, if I am right about the factoring piece of my research, while you can see that I've shutdown the specious objections that worked on these newsgroups with a result from the complex plane.

But I never got objections from outside the newsgroups.

No math journal has ever put forward those objections.

Top mathematicians don't need that trivial result from the complex plane to know that I am right.

So I am not that hopeful that the truth will win out just yet based on my ability to directly show that some of you were dumb enough to fight the distributive property.

Barry Mazur didn't fight it. Neither did Andrew Granville. And Ralph McKenzie didn't either.

What puts you in the real bind is the question, is the factoring piece correct?

I used the same set of equations with which I get my factoring results to put up the quadratic generator that follows from my reseearch:

a^2 -(1+fx)a + (f^2 x^2 + 2fx) = 0

And notice the quiet, as when I made changes before, posters leapt at the chance to put up "counterexamples" with equations that didn't follow from my research, but were just being made up in what turns out to be a naive and clumsy way.

The mathematics here requires a special form for you to remain in what I call the ring of objects.

And with that form, counterexamples cannot be found, as they would require refutation of, the distributive property.

So the research overturns over a century of mathematical ideas WITHOUT the factoring piece, but you people don't seem to like reason, so you may only respond to force.

The paper by Plotnikov is from 1996. Mathematicians may have deliberately avoided key mathematical results through all that time in some short-sighted need to promote themselves or something as I think about it and wonder exactly what you people have been thinking.

If it does work, then you set up the world for some solution like mine, but why?

If people had known the system could just be broken then protections could have been put in place (hopefully they have been, and I am hoping on that, just in case).

Why set things up as you did? Who has been pulling your strings?

The mystery is still one I'm puzzling over.

SF: I hope I am wrong

At this point as I make progress explaining my non-polynomial factorization research and ending some specious objections by posters by going to the complex plane with a rather basic result, I am concerned that, well, maybe I did solve the factoring problem, but hopefully those who disagree with me, are right.

The equations with T the target composite:

T = (x+y+vz)(vz-x)

where x, y and z are given by

x^2 + xy + k_1 y^2 = k_2 z^2

and

(2(v^2 - k_2)z + vy)^2 = ((1-4k_1)y^2+4T)v^2 + 4k_2(k_1y^2 - T)

where you pick y, k_1 and k_2 to get what I call the surrogate, which is

k_2(k_1y^2 - T)

and by factoring it you can solve v, z and then x, are from research which I easily show overturns over a century of mathematics, as also from those same equations, I can prove that the quadratic

a^2 -(1+fx)a + (f^2 x^2 + 2fx) = 0

for integer x and f, where f is not 1 or -1 and is coprime to x that its roots must have f as a factor, where there's an odd quirk of the ring of algebraic integers where if the roots are non-rational then provably neither of them can have f as a factor IN THAT RING.

(An analogy is with evens where with just evens 2 is not a factor of 6 because 3 is not an even, so my result is that the ring of algebraic integers is incomplete.)

The proper interpretation of the result is that the ring of algebraic integers is flawed in that it is incomplete, showing over a century of mathematics to be false.

Those who just like to play with math programs can have their software find integer solutions and see the result directly.

So, if the factoring piece doesn't work to solve the factoring problem, whew!

But if it does then reasonable people may conclude that you people betrayed the world to protect yourselves from both results, trying to protect your careers.

So, your careers are probably the first thing that would be taken away.

There's a lot riding on this research now. I proved a key piece which posters have been claiming is false actually depends on the distributive property, thus proving they have been arguing against the distributive property.

The proof is easy.

In the complex plane, given

7C(x) = (A(x) + 7)(B(x) + 1)

true for all x, where A(0) = B(0) = 0

let

C(x) = (A'(x) + 1)(B'(x) + 1)

where A'(0) = B'(0) = 0

and making that substitution, gives

7(A'(x) + 1)(B'(x) + 1) = (A(x) + 7)(B(x) + 1)

and by the distributive property

A(x) = 7A'(x) and B'(x) = B(x).

That result valid over the complex plane allows me to cement the case for the full argument proving I have been right with my research where you can directly SEE the result with integer roots of

a^2 -(1+fx)a + (f^2 x^2 + 2fx) = 0

and if this research, which is so huge that it overturns over a century of mathematics does not solve the factoring problem as well, then we can all breathe a sigh of relief.

But if it does, you people are betraying a lot, and I think you're naively doing it to protect your careers in a short-sighted way.

If you're wrong, you're not the only ones who lose a lot, as you're costing people who don't know a lot about mathematics or who don't know anything about it, who just trust their society to protect them.

Too bad they're not like us, eh? We know not to trust anyone.

Not in this world, as when you trust, someone is going to make you pay.

And this time, it'll be the mathematical community betraying the entire world.

Control and Usenet

One of the basic lies about Usenet is that you can't control other posters, when anyone who looks at newsgroups like sci.crypt or especially sci.math for a while will see control of posters demonstrated, primarily by use of insults.

The most effective tool on Usenet for controlling other posters is insults.

The second most effective tool is volume, as in repeated replying in a person's threads to control the flow of conversation, and possibly drown out their message.

Using those tools a group of four to five posters can completely dominate a newsgroup, controlling who posts, as in, if they don't like a particular person, they can pick them out, and drive them from the newsgroup—if they're like most people.

I am a special case as, for one thing, I study the behavior, and for another, I say it's a free speech area, and I don't like being controlled by insults.

But they keep trying as the behavior is effective in other ways, as you can create a certain kind of atmosphere by doing it, which I don't exactly work to stop, as if you read my postings you can see me working to create a certain atmosphere as well.

The harsh negativity is for a reason on my side, whereas the other side is doing it for their own reasons, and so you have these two sides battling it out on these newsgroups, where my point is that the truth is being drowned out, while the point of those arguing with me is that I am not worth listening to—but somehow worth group efforts at censoring, though that's a hidden point which you have to figure out.

The mathematical ideas I have rely on mostly simple algebra, yet arguments over them have raged for years.

ALL of the recent arguments are around tools developed by using identitites like

x + y + vz = x + y + vz

which I used to get my latest surrogate factoring equations, and the

x^2 + y^2 + vz^2 = x^2 + y^2 + vz^2

which I used to get MY non-polynomial factorization equations.

I call identities like those tautological spaces, and you can generate equations of what I consider only moderate complexity with minimal effort using that technology, BUT mathematicians hate my research, and are fighting to, well, to ignore it!!!

BUT my research is all about factoring. Yup. The analysis tools I use tend to give factorizations as the way I use tautological spaces generates factorizations, all over the place.

So I have these tools I say are highly effective, but they are new analysis tools.

Mathematicians work to ignore them, while posters on these newsgroups work to poison the discussion, and I push the negativies of allowing people to censor as they do—and successfully most of the time I might add—as well as block new research tools.

The battles now are playing out because my research is such a kick in the groin for modern number theorists in all their cherished areas.

They're fighting back, and what you see on the newsgroups is just one aspect of a long on-going war.

I call them, the math wars.

Factoring problem, solution and proof

This post outlines a solution to the factoring problem with working equations and a proof that it is a solution.

Given a composite T to be factored, a factorization algorithm follows from the system of equations:

T = (x-n+vz)(vz-x)

when

k_2 z^2 + nx - x^2 = T

and

z = nv/(v^2 - k_2)

where v does not equal sqrt(k_2) to prevent division by zero.

>From the second equation in the system you have

x^2 - nx + T - k_2 z^2 = 0

and solving for x gives

x = (n +/- sqrt(n^2 - 4(T - k_2 z^2)))/2

and now you pick some nonzero integers k_2 and z, for instance

k_2 = z = 1

and then factor

4(T - k_2 z^2)

to get integer n, where with g_1 and g_2 its integer factors, that is
triival as

n = g_1 + g_2

where

g_1 g_2 = 4(T - k_2 z^2).

Given integer n, you have integer x or 2x is an integer, and you next solve for v, using the third equation in the system, which gives

v = (n +/- sqrt(n^2 + 4z^2 k_2))/2z

which may be non-rational for a given g_1 and g_2, but provably will be rational for some g_1 and g_2.

Given v, you now have the factorization as

f_1 = x-n+vz

and

f_2 = vz-x.

The proof that the system must factor is trivial. Note that for any integer factors f_1 and f_2, given any integer value for z, there must exist rationals x, v and n that will satisfy those equations for f_1 and f_2.

But, any integer value of n, must be a solution of the square root in

x = (n +/- sqrt(n^2 - 4(T - k_2 z^2)))/2

or x could not be an integer.

Therefore, given that T is composite, and an integer n and x must exist, some value of g_1 and g_2 must give that n, and that x, which will give integer v, so a factoring solution must be determinable by this method.

Note that

f_1 + f_2 = 2vz + n

and

f_1 - f_2 = 2x - n

so again, given just z, there must exist rationals x, v and n for f_1 and f_2, completing the proof.

For example, if you pick z=1, with T = 15, and f_1 = 3, and f_2 = 5, you have

3 = x - n + v and 5 = v - x

so

8 = 2v + n and 5 = 2x - n

and rational solutions can readily be found. Notice there are an infinity of them, but your choice of factors g_1 and g_2 from

g_1 g_2 = 4(T - k_2 z^2).

narrows the set down to a finite one.

The proof is done, but the social consequences haven't started yet. I do wonder why reason failed here.

Notice that it has already been some time since I started explaining what I'd do, and began the implementation, but so far you still have no movement from the mathematical community.

The evidence clearly shows that they are going to push this to the limit, and wait until society comes in and forces them to face the truth.

I doubt it'll take long. My decision is, again, for the future. Discoverers who stop to let the truth lose for some small group of people are putting the future in jeopardy for the comfort of a few, against the welfare of all those yet to be born.

The good of the many, outweighs the needs of the few.

I do hope there are many more to be born, but some of you may disagree.

However, it was my decision.

Good night, and good luck.

JSH: My patience has ended

Ok, next post is a post of the solution to the factoring problem with a proof that it is a solution, and I remind tha you are NOT to touch your stocks or savings as it will only make it worse for you.

I made my decision by considering the reality that major discoveries have a value to FUTURE civilizations and that if the discoverers before me had ever decided to allow one group to preserve its social needs rather than force the issue, then we would not have our science and technology of today.

Still I worked diligently at finding reasonableness within the mathematical community.

It has been over three years since I had my first major results, and even publication of a paper could not budge the bulk of the mathematical community, which then also sat back, when that paper was censored by social pressure, and even sat back when the math journal folded.

It is my conclusion that this mathematical community will not accept the truth, but instead will try to compromise our future by preserving its current social order, and the rest of humanity be damned.

It is my responsibility as a discoverer to make the decision that I feel is in the best interests of the entire human race, and my decision is for truth, and further scientific progress, versus the great likelihood of stagnation from flawed mathematical ideas and flawed behaviors that are spreading through the scientific world.

The problems I'm facing are not mine alone as Dr. Halton Arp has problems with the astronomical community, and I feel certain that the spread of disease through the mathematical world is deep and pervasive requiring that I use the best, most powerful tool left to me.

So after three years plus of considering all the issues, and debating my responsibilities as a discoverer, as well as all the moral issues I can think of, and every possible avenue to resolution, I have determined that solving the factoring problem is the only means left to me to bring about a resolution and the triumph of truth.

The world around us is about to change, to be changed, by my decision.

I hope that when they have all the facts, most people will understand, and agree with me that the future of our world is too important to be held in check so that some small society of mathematicians can preserve their social order and hold on to flawed mathematical ideas.

I vote for the future. My decision is for the future.

I wonder that one man had to make such a decision, but here we are, and I am making it.

God help us all.

Latest surrogate factoring, complete

I just hate this, so I guess I still hedged a bit, but now, finally, here's the full thing. I feel forced into giving this theory.

I did keep working on surrogate factoring after the earlier failures on this newsgroup, but just didn't post about it here, while I put some things on my blog. It turns out that I found a simple approach where hey, I can see reason to doubt, but I'll put it up anyway.

I have concluded that I cannot justify withholding this research, as it relies on advanced mathematical techniques which should be of crucial value in the sciences, while economic reasons may seem important today, but in the big scheme of things, scientific advancement is most important.

With T the target integer to be factored I found the amazingly simple system:

T = (x-n+vz)(vz-x)

when

k_2 z^2 + nx - x^2 = T

and

z = nv/(v^2 - k_2)

where v does not equal sqrt(k_2) to prevent a problem with division by zero.

Of course the problem then is, how do you pick all those variables so that you can non-trivially factor T?

It's easy, as from the second equation in the system you have

x^2 - nx + T - k_2 z^2 = 0

so you can just solve for x:

x = (n +/- sqrt(n^2 - 4(T - k_2 z^2)))/2

and substitute out k_2 using the last equation, and then solve to get

x = (n +/- sqrt((n-2vz)^2 + 4(T - 2v^2 z^2)))/2

and you just pick some nonzero integer v and z, like v = z = 1, plug them in and factor the surrogate to get n, which gives you rational x, and now from the first equation in the system, you have

f_1 = x-n+vz

and

f_2 = vz-x

and that's it. Of course if T is prime then you'll just get T back, but otherwise, I don't see why it shouldn't work, and notice the algorithm that follows is lightweight and easily implemented.

That it's this easy is not actually a surprise as an Anatoly Plotnikov wrote a peer reviewed and published paper back in the late 1990's showing that P=NP, which was mostly ignored by the mathematical community.

I don't know why.

Latest surrogate factoring theory, algorithm

I did keep working on surrogate factoring after the earlier failures on this newsgroup, but just didn't post about it here, while I put some things on my blog. It turns out that I found a simple approach where hey, I can see reason to doubt, but I'll put it up anyway.

I have concluded that I cannot justify withholding this research, as it relies on advanced mathematical techniques which should be of crucial value in the sciences, while economic reasons may seem important today, but in the big scheme of things, scientific advancement is most important.

With T the target integer to be factored I found the amazingly simple system:

T = (x-n+vz)(vz-x)

when

k_2 z^2 + nx - x^2 = T

and

z = nv/(v^2 - k_2)

where v does not equal sqrt(k_2) to prevent a problem with division by zero.

Of course the problem then is, how do you pick all those variables so that you can non-trivially factor T?

It's easy, as from the second equation in the system you have

x^2 - nx + T - k_2 z^2 = 0

so you can just solve for x:

x = (n +/- sqrt(n^2 - 4(T - k_2 z^2)))/2

and you just pick some nonzero integer k_2 and z, like k_2 = z = 1, plug them in and factor the surrogate to get n, which gives you rational x, and now from the first equation in the system, you have

f_1 = x-n+vz

and

f_2 = vz-x

and that's it. Of course if T is prime then you'll just get T back, but otherwise, I don't see why it shouldn't work, and notice the algorithm that follows is lightweight and easily implemented.

That it's this easy is not actually a surprise as an Anatoly Plotnikov wrote a peer reviewed and published paper back in the late 1990's showing that P=NP, which was mostly ignored by the mathematical community.

I don't know why.

JSH: Final solution

Unfortunately, I think I've run out of options. My mistakes with working through the Decker example have taken away the reasoning option, as I now think I have some understanding of why the latest counterexample isn't quite a counterexample, but how do I come back from that?

The errors have given me no choice but to use a final solution which does not require agreement.

That solution is to release fully the information on solving the factoring problem, and post the algorithm that leads to ways to crack RSA encryption, thus ending the Internet as it currently operates, and impacting nations and businesses around the world.

I, at this point, see no other choice, as mathematicians rather than trying to work through the issues to get a complete answer, are simply ignoring or fighting me, and I keep making mistakes that take away the reasoning option.

The impact will be almost immediate, soon enough economies will be changed, and even nations will shift, as I make a decision for the future.

Of course that sounds nutty now, with me just having admitted a major failure, but in just a few days, you will wish you could go back in time to now, to give me another option.

That's it. I'm off now to implement the final solution.

God help us all.

JSH: Generalized Decker example, examples

Decker's example can be generalized to help in finding rational solutions, where you will find that the mathematics follows the distributive property:

f Q(x) = f((x^2 + x)(5^2) + (-1 + x)(5) + f) = f(25 x^2 + 30 x + 2)

and

f Q(x) = (5a_1(x) + f)(5a_2(x) + f)

where the a's are defined by

a^2 - (x - 1)a + f(x^2 + x) = 0

and you can let f be a rational number. Now there will be posters who will loudly declare coprimeness means nothing with rationals, but consider

f(x^2 + 3x + 2) = (fx + f)(x + 2)

and solutions with rationals, as guess what? You can STILL see that one factor is multiplied by f, even with rationals. If you don't think so, play with that example with some rational f's and rational x and see if the factor multiplied by f doesn't betray that it was.

Now then, if I am wrong, some rational f can be found with a rational x that shows it.

Like let f=32 with the generalized Decker example, and find some rational solutions and see if that f gets split up. Or let f=1024 or anything you want!!!

You see, no counterexample exists, as the distributive property is right!!!

So posters here at best can loudly proclaim that Galois Theory doesn't work with rationals but only with non-rationals, which is the dodge because it actually doesn't work, but you can't see that with non-rationals.

The proof I've given relies on the distributive property.

Even lower rung mathematicians cannot be incapable of quickly seeing it MUST be correct, but clearly as this impasse continues they are running.

And them running means they are hoping that none of you who are not already established in careers, who are just learning as you're still in school, will stop protecting them by ignoring this result.

They are in the weak positon of needing your protection so that they can teach you wrong mathematical ideas, as if some students start protesting, they will collapse like the cowards they are, running the other way, selling each other out to protect themselves.

First mathematicians on the block will be the ones who are posters on sci.math, and their own will destroy their careers. People like Magidin and Ullrich will be out of their universities so fast your head will spin, as they are tossed to the wolves.

So they sit and wait, checking each day to see if any of you are breaking out of the wall of silent acceptance, or irrational denial in the face of a simple proof that relies on the distributive property at the point of dispute.

That check may be the best evidence against those mathematicians as they leave cyber clues to prove they knew, but were checking to see if they could keep getting away with lying.

JSH: Headlines around the world

Let's just say for the sake of argument that mathematicians just came out and acknowledged that through accidents of history and subtlety of some difficult concepts an erroneous group of techniques became dominant in number theory.

That news would make headlines around the world.

Now though, there is quiet, so let's look at the other way this can work out, as mathematicians can instead try to ignore the result.

Then, as history shows, the result will emerge eventually as there is over 10,000 years of human civilization where these kinds of battles have played out, and the side opposing the truth, has always lost.

If the story emerges within a couple of years then clearly mathematicians will have been gambling on it not coming out within their careers, and will have lost that gamble.

Why is it a gamble?

Well these things have happened before. In the physics field recently there was the emergence of quantum mechanic and relativity, where adjustments had to be made with dramatic consequences for the entire world.

The physicists absorbed the impact, fought a bit, yes, but quickly came on the side of truth.

If mathematicians do not, then they will still lose, but they also get a much darker story, and those mathematicians who are in powerful positions when the story comes out, will probably take the worst of it.

Social castigation. Their pictures in the papers. Reporters hounding them with hard questions.

And it won't end during their lifetimes.

These stories are huge on a scale that's hard to comprehend.

They keep coming but people never see them coming. The stories get bigger with time, but while you're living in it, it can seem unimaginable.

Simple self-preservation would make some of you sing like canaries at this point if you had any sense of what is going to come, while otherwise you have to hold on to the belief that you can play the odds, play for time, and hope that the story stays buried long enough for you to have a long career in mathematics, retire, and die with no one ever knowing the truth.

Let the future handle it, you may think.

But you will not get that time. I'll make sure you don't get that time.

You will not get to grow old and die with the world thinking you're something you're not.

The story will come out before then and instead you'll be castigated by world society.

I'll paint you for what you are—a dangerous element in society fighting against the foundations of society and technological progress—and remind that if people like you ever succeed then our Progress as a species, comes to a screeching halt.

What if physicists had tried and succeeded at what some of you may now think you can succeed at? What if they'd shut down Einstein's work, ignored his papers? What if they'd fought quantum mechanics tooth-and-nail?

What if they'd blocked the knowledge?

Well, I wouldn't be typing this up on this computer as computers wouldn't be here. Or maybe we'd have some kind of clunky mechanical computer, but would we have had the transistor, and the technological revolution?

Or might we have physicists fighting to explain odd behavior within the framework of the old knowledge, vigorously attacking "cranks" and "crackpots" who attempted to push through the ideas of quantum mechanics?

You will not succeed in blocking the knowledge. But if you attempt to do so, when you are broken, you lose so much.

So why bother?

Why not just tell the truth now?

Why fight the future?

JSH: Decker example, end of an era

Now I'm going to go over the Decker example and how it refutes the standard teaching of Galois Theory, and it turns out that it shows the theory of ideals to be flawed as well, but it's more convoluted to explain why that's true and I'll leave that explanation as it is somewhat technical to others.

This example because of the use of simple quadratics takes away the objections of posters who have inexplicably refused to accept the distributive property in proofs I've shown with my own more complicated expressions.

And also I'm starting with equations put up by a Rick Decker, so it's not like I picked them personally.

The attempt will be to have everthing below in the ring of algebraic integers, so consider that the ring, but notice at a key point, we will be forced out of that ring.

Consider

7 Q(x) = 7((x^2 + x)(5^2) + (-1 + x)(5) + 7) = 7(25 x^2 + 30 x + 2)

and

7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7)

where the a's are defined by

a^2 - (x - 1)a + 7(x^2 + x) = 0

where Decker chose this example because at x=1, the middle term goes to 0 which he I guess thought was a counterexample to my ideas because at x=0, the a's are 0 and 1, indicating that 7 divides through only one of the expressions, but clearly it does not at x=1, but that special case doesn't change the refutation so I'll leave out the explanation of it

The following disproof focuses directly on Galois Theory while the relation to the theory of ideals is more complex.

The first step in the disproof is to focus on that last quadratic

a^2 - (x - 1)a + 7(x^2 + x) = 0

and write it with the focus on x, which gives a non-monic polynomial that importantly becomes monic if 'a' has 7 as a factor:

7x^2 + (7-a)x + a^2 + a = 0

and using a = 7b, you get

7x^2 + (7 - 7b)x + 49b^2 + 7b = 0

and dividing off 7, gives you

x^2 + (1 - b)x + 7b^2 + b = 0

so for ANY algebraic integer 'a' that has 7 as a factor, you get an algebraic integer x.

Also, notice that 7b^2 in the expression so that if you focus on 'b' you have a non-monic polynomial and as before, 'b' cannot be an algebraic integer if that polynomial has integer coefficients and is irreducible over Q.

BUT

a^2 - (x - 1)a + 7(x^2 + x) = 0

means that for ANY algebraic integer x, 'a' is an algebraic integer, but if 'a' does not have 7 as a factor

7x^2 + (7-a)x + a^2 + a = 0

will not allow an algebraic integer x, if 'a' is an integer and the quadratic is irreducible over Q.

So you are forced out of the ring of algebraic integers.

Algebraically you still have

x^2 + (1 - b)x + 7b^2 + b = 0

but the 'b' is outside of the ring of algebraic integers for certain algebraic integer values of x.

To save Galois Theory and the theory of ideals, the attempted defense of those ideas at this point would require that for some non-rational 'a' with partial factors of 7, as it is a root of

a^2 - (x - 1)a + 7(x^2 + x) = 0

with an integer x other than x=1, you would get a polynomial with an integer solution for x from

7x^2 + (7-a)x + a^2 + a = 0

which would remain non-monic because the factors of 7 not in common with 'a' would remain on the leading coefficient.

However, you can simply pick some x, like x=3, and solve for the a's, and then make the substitution in which will give a seemingly non-monic expression of some complexity as it will have square root radicals, but you can use basic algebraic manipulations to remove the square roots, giving you a quartic, where you will find that the leading coefficient just divides off, as one of the roots will be 3, so it will also be reducible over Q.

So Galois Theory as a way to determine where factors go is refuted, and it is proven that the ring of algebraic integers is incomplete, so that some numbers are excluded simply because they are not roots of monic polynomials with integer coefficients, which leads to the conclusion that the theory of ideals is flawed.

This post marks the end of an era in the world of mathematics.

JSH: Simple key

So yeah, you can take the expression

a^3 + 3(-1+xf^2)a^2 - f^2(x^3 f^4 - 3x^2 f^2 + 3x) = 0

and instead of focusing on the a's multiply it out and focus on x like you normally do anyway to get a polynomial P(x).

It will be non-monic where the leading coefficient is f^2.

But, it will also be clear that if 'a' has f as a factor, the f^2 will divide off, giving you a monic polynomial, and thus, an algebraic integer value for x.

So you can map algebraic integer a to algebraic integer x that way.

You can also do that with an a coprime to f.

But you cannot map algebraic integer a to algebraic integer x if a does not have f as a factor nor is it coprime to it.

So it's easy to check me when you have the key.

I like that way of explaining which I just thought of yesterday to handle an objection from a poster.

It's oddly simple: Focus not on the a's but on solving x in terms of the a's, and then it all just falls into place so easily, you wonder how people could argue about it for years.

I am a bit curious to see if anyone will keep arguing at this point, and I'll check tomorrow.

But remember what I said about brilliance is as brilliance does, and also remember that certain people are getting serious gain from the world thinking that mathematical ideas I can show are false are correct, but their gain is at the expense of others, especially their students who are being taught and tested on the bogus stuff.

If that continues with the information readily available, then the natural thing for world society to do when it does figure it out, is to erase the gains and give penalties.

JSH: Brilliance is as brilliance does

I'm going to make a crucial point here, which goes to your innate brilliance and potential as a mathematician.

If I am right, and it turns out there is this remarkable flaw that has remained hidden for over a hundred years through accidents of history and other stuff, like Galois Theory being said to only work with non-rationals, where I've been posting about it, and you have at all come anywhere close to this story, can you be a brilliant mathematician and not get it?

For some of you the most important part of this tale will be the clear evidence that at best you are a middling mathematician with moderate potential.

Any mathematician of any true brilliance would get it with only hints, let alone full proofs.

There is at least one math grad student at a major university who has shown indications that he gets it.

Necessarily, true brilliance is a rare thing. And, also, it's rare that a super test is given to clear out all the chaff, but that test is here now.

Make no mistake, this story will break soon. And when it does, and you sat or didn't quite get it, or worst of all, thought I was wrong, then don't lie to yourself. At that point, do yourself a favor, and just sort of head on to other areas.

It sounds brutal, but the field of mathematics needs the best and brightest, especially now. 