### Saturday, November 01, 2008

## JSH: So yes, it is simple, what now?

I think I've finally stumbled across an irrefutable way to demonstrate that the ring of algebraic integers is severely flawed. You can try to be in that ring for it.

Given

175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)

Step through getting the factorization:

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

where the a's are roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0.

So that you know it CAN be done, algebraically, I'll give you the primary way.

Multiply both sides by 7 in a particular way which is my controversial way:

7*(175x^2 - 15x + 2) = (5b_1(x) + 2)(5*7*b_2(x)+ 7)

and now use the substitutions b_1(x) = a_1(x) + 1, and 7*b_2(x) = a_2(x), to get

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

and you're done. Note that b_1(x) and b_2(x) are given by this method. That is also a requirement for any answers from the mathematical community: give b_1(x) and b_2(x).

Those of you who try to handle the example will find you can't get b_1(x) and b_2(x) in general that are algebraic integers such that

175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)

AND give b_1(x) and b_2(x), as the algebra cannot let you, as if it could, then you could just step through the route I take, and you'd have

7*b_2(x) = a_2(x)

and a contradiction with the ring of algebraic integers.

The rational explanation is that the definition of algebraic integers as roots of monic polynomials with integer coefficients chops off numbers that NEED to be in the ring to prevent apparent contradiction.

The result above is so devastating that you can use it to prove that 7 is a factor of

a^2 - (7x-1)a + (49x^2 - 14x) = 0.

algebraically AND prove that 7 cannot be a factor of either root if you are irreducible over Q IN THE RING OF ALGEBRAIC INTEGERS.

So you can do the mathematically verboten thing of proving two contradictory things, and I showed that several years ago in a paper published in the mathematical journal SWJPAM.

Now I DID pass the paper by Barry Mazur first. He never claimed an error with it, but just asked some general questions. I DID explain non-polynomial factorization in person to Ralph McKenzie at my alma mater Vanderbilt University, having traveled there to be able to speak to him in-person.

So the paper was fully vetted before it went to SWJPAM. And they had it for nine months, and I was in continual contact with two of the editors up until it was published, so it was not mistakenly published.

For years now posters have routinely lied, misrepresented or made up details.

The reason is that the result is so crushing for people established in the field, but less so for undergrads, so what they are doing I think, is trying to pull you into their nightmare: get you to invest years or even decades of your life with the flawed mathematical ideas and be as dedicated as them in preserving them.

They're fighting for their sense of self and dignity. They're fighting to keep believing they're brilliant mathematicians.

And they're trying to sacrifice you to do it.

Don't just believe me. Do the math. Try the demonstration I've mentioned. Read around. Do the research and face the truth.

You are some of the luckiest undergrads ever as if you choose the right path then for the first time in a over a hundred years, math students turned from the dark side with these number theory results.

Life is hard. Mathematics is often harder. Only the best will do. Only the truly best will do the right thing.

Follow the math.

Given

175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)

Step through getting the factorization:

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

where the a's are roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0.

So that you know it CAN be done, algebraically, I'll give you the primary way.

Multiply both sides by 7 in a particular way which is my controversial way:

7*(175x^2 - 15x + 2) = (5b_1(x) + 2)(5*7*b_2(x)+ 7)

and now use the substitutions b_1(x) = a_1(x) + 1, and 7*b_2(x) = a_2(x), to get

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

and you're done. Note that b_1(x) and b_2(x) are given by this method. That is also a requirement for any answers from the mathematical community: give b_1(x) and b_2(x).

Those of you who try to handle the example will find you can't get b_1(x) and b_2(x) in general that are algebraic integers such that

175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)

AND give b_1(x) and b_2(x), as the algebra cannot let you, as if it could, then you could just step through the route I take, and you'd have

7*b_2(x) = a_2(x)

and a contradiction with the ring of algebraic integers.

The rational explanation is that the definition of algebraic integers as roots of monic polynomials with integer coefficients chops off numbers that NEED to be in the ring to prevent apparent contradiction.

The result above is so devastating that you can use it to prove that 7 is a factor of

a^2 - (7x-1)a + (49x^2 - 14x) = 0.

algebraically AND prove that 7 cannot be a factor of either root if you are irreducible over Q IN THE RING OF ALGEBRAIC INTEGERS.

So you can do the mathematically verboten thing of proving two contradictory things, and I showed that several years ago in a paper published in the mathematical journal SWJPAM.

Now I DID pass the paper by Barry Mazur first. He never claimed an error with it, but just asked some general questions. I DID explain non-polynomial factorization in person to Ralph McKenzie at my alma mater Vanderbilt University, having traveled there to be able to speak to him in-person.

So the paper was fully vetted before it went to SWJPAM. And they had it for nine months, and I was in continual contact with two of the editors up until it was published, so it was not mistakenly published.

For years now posters have routinely lied, misrepresented or made up details.

The reason is that the result is so crushing for people established in the field, but less so for undergrads, so what they are doing I think, is trying to pull you into their nightmare: get you to invest years or even decades of your life with the flawed mathematical ideas and be as dedicated as them in preserving them.

They're fighting for their sense of self and dignity. They're fighting to keep believing they're brilliant mathematicians.

And they're trying to sacrifice you to do it.

Don't just believe me. Do the math. Try the demonstration I've mentioned. Read around. Do the research and face the truth.

You are some of the luckiest undergrads ever as if you choose the right path then for the first time in a over a hundred years, math students turned from the dark side with these number theory results.

Life is hard. Mathematics is often harder. Only the best will do. Only the truly best will do the right thing.

Follow the math.