## Thursday, October 30, 2008

### Easy proof of problem with ring of algebraic integers

Some simple algebra with a basic polynomial reveals a serious problem with the naive use of the ring of algebraic integers, bringing into question over a hundred years of algebraic number theory.

In an integral domain, consider a simple polynomial

P(x) = 175x2 - 15x + 2.

Multiply it times 7, to get

7*P(x) = 1225x2 - 105x + 14. Cleverly re-group terms:

1225x2 - 105x + 14 = (49x2 - 14x)52 + (7x-1)(7)(5) + 72

and now factor into non-polynomials:

7*(175x2 - 15x + 2) = (5a1(x) + 7)(5a2(x)+ 7)

where you'll note that the a's are functions of x that are roots of

a2 - (7x-1)a + (49x2 - 14x) = 0.

But now consider our polynomial again with a factorization before any
multiplying by 7:

175x2 - 15x + 2 = (5b1(x) + 2)(5b2(x)+ 1)

Now multiply by 7, to get

7*(175x2 - 15x + 2) = (5b1(x) + 2)(5(7*b2(x))+ 7)

and use the substitutions b1(x) = c1(x) + 1, and 7*b2(x) = c2(x), and you have

7*(175x2 - 15x + 2) = (5c1(x) + 7)(5c2(x)+ 7)

and of course if c1(x) = a1(x) and c2(x) = a2(x), I have my original factorization, but in so doing I'm PICKING that 7 multiplies times just one of the factors of 175x2 - 15x + 2, but what if I picked wrong?

For instance, consider again

175x2 - 15x + 2 = (5b1(x) + 2)(5b2(x)+ 1)

and again multiply times 7, but split it up so that each factor is multiplied times sqrt(7):

7*(175x2 - 15x + 2) = (5*sqrt(7)*b1(x) + 2*sqrt(7))(5*sqrt(7)b2(x)+ sqrt(7))

but there's an immediate problem!

If you let x=0, then you have the factorization:

7*(2) = (5*sqrt(7)*b1(0) + 2*sqrt(7))(5*sqrt(7)b2(0)+ sqrt(7))

which contradicts at x= 0 with

7*(175x2 - 15x + 2) = (5a1(x) + 7)(5a2(x)+ 7)

where the a's are functions of x that are roots of

a2 - (7x-1)a + (49x2 - 14x) = 0

unless b_2(0) divides off sqrt(7), or b1(0) divides off sqrt(7), as

7*(2) = (5a1(0) + 7)(5a2(0)+ 7) = (0 + 7)(-5+ 7)

because then the a's are roots of

a2 + a = 0.

Therefore, there is no other way to multiply

175x2 - 15x + 2 = (5b1(x) + 2)(5b2(x)+ 1)

by 7, and get the factorization

7*(175x2 - 15x + 2) = (5a1(x) + 7)(5a2(x)+ 7)

where the a's are functions of x that are roots of

a2 - (7x-1)a + (49x2 - 14x) = 0

as ANY other way other than multiplying (5b2(x)+ 1) by 7, will contradict with

7*(175x2 - 15x + 2) = (5a1(x) + 7)(5a2(x)+ 7)

at x=0, as demonstrated above with sqrt(7).

Therefore, one of the a's must have 7 as a factor for all x, but it is trivial to show that NEITHER of them can have 7 as a factor for any integer x, for which the a's are not rational, in the ring of algebraic integers, so there is proven a problem with that ring.

James Harris

## Tuesday, October 07, 2008

### Diophantine supermap for binary quadratic equations

IN a previous post I noted the discovery of an infinite series. I decided I should name it and came up with the name: Diophantine supermap for binary quadratic equations

The series starts with

1. x2 + Dy2 = F

2. (x-Dy)2 + D(x+y)2 = F(D+1)

3. ((1-D)x-2Dy)2 + D(2x + (1-D)y)2 = F(D+1)2

4. ((1-3D)x + (D2 - 3D)y)2 + D((3-D)x + (1-3D)y)2 = F(D+1)3

5. ((D2 - 6D + 1)x + (4D2 - 4D)y)2 + D((4-4D)x + (D2 - 6D + 1)y)2 = F(D+1)4

and that goes out to infinity. To get successive terms in the series you use the algebraic result that given:

u2 + Dv2 = C

it must be true that

(u-Dv)2 + D(u+v)2 = C(D+1).

And where whenever the exponent of (D+1) is even, you can have a case where you just have a multiple of x and y, so you can solve for D, which defines possible values for F in terms of x or y.

For instance with

((1-D)x-2Dy)2 + D(2x + (1-D)y)2 = F(D+1)2

I have six possibilities:

(1-D)x-2Dy = +/-(D+1)x or +/-(D+1)y

and

2x + (1-D)y = +/-(D+1)y or +/-(D+1)x

but I will only work through four.

Considering the first gives:

(1-D)x-2Dy = (D+1)x and 2x + (1-D)y = (D+1)y, so

x = -y, from the first and x = Dy, so D=-1. Which gives F=0.

The second gives

(1-D)x-2Dy = -(D+1)x, so x = y,

and 2x + (1-D)y = -(D+1)y, so x = -y, so x=y=0.

The third and fourth also give D=-1.

The next case where D can be set occurs with

((D2 - 6D + 1)x + (4D2 - 4D)y)2 + D((4-4D)x + (D2 - 6D + 1)y)2 = F(D+1)4

which gives again six possibles:

(D2 - 6D + 1)x + (4D2 - 4D)y = +/-(D+1)2x or +/-(D+1)2y

and

(4-4D)x + (D2 - 6D + 1)y = +/-(D+1)2y or +/-(D+1)2x

but I will only show one:

(D2 - 6D + 1)x + (4D2 - 4D)y = (D+1)2x, so 2x = (D - 1)y

and

(4-4D)x + (D2 - 6D + 1)y = (D+1)2y, so (1-D)x = 2Dy,

which solves as D2 + 2D + 1 = 0, so D=-1 again.

D=-1 will always be one of the solutions.

My theory is that other integers will emerge and that you will get all possible integer values for D somewhere in the supermap, and with other values you can get solutions for F, relative to x and y.

I think though mapping the number theoretic structure is best done by computer.

James Harris

## Saturday, October 04, 2008

### Huge number theoretic structure, major find

One of the more remarkable results from my recent research studying 2 variable quadratic Diophantine equations is a relation from my last post which I used to show a general solution for equations of the form x2 + Dy2 = F.

Here I write that result as, given

x2 + Dy2 = F

it is forced that

z2 + D(x+y)2 = F(D+1)

and you can verify trivially that z=x-Dy, by substituting out F, from the first equation into the second:

z2 + D(x+y)2 = (x2 + Dy2)(D+1)

so, expanding and simplifying:

z2 = (D+1)x2 + D2y2 + Dy2 - Dx2 - 2Dxy - Dy2

gives

z2 = x2 - 2Dxy + D2y2 = (x-Dy)2.

And it may seem like a trivial result, but now you have a chain where the first 3 equations are:

x2 + Dy2 = F

(x-Dy)2 + D(x+y)2 = F(D+1)

and

(x-Dy - D(x+y))2 + D(x-Dy + x+y)2 = F(D+1)2

which is

((1-D)x-2Dy)2 + D(2x + (1-D)y)2 = F(D+1)2.

For each following equation you use the general form that with:

u2 + Dv2 = C

the next equation is given by

(u-Dv)2 + D(u+v)2 = C(D+1)

so the next in the series is

(((1-D)x-2Dy)-D(2x + (1-D)y))2 + D(((1-D)x-2Dy)+(2x + (1-D)y))2 = (F(D+1)2)(D+1).

Which simplifies to

((1-3D)x + (D2 - 3D)y)2 + D((3-D)x + (1-3D)y)2 = F(D+1)3.

And that may not seem remarkable unless you realize that for every solution to the first equation, you have a solution to that last with (D+1)x, and (D+1)y, so what if you go backwards and ask, can

(1-D)x-2Dy = +/-(D+1)x or +/-(D+1)y

and

2x + (1-D)y = +/-(D+1)y or +/-(D+1)x?

But if so, then you have two solutions for x in terms of y, so BOTH must give the same answer or there is a contradiction.

For instance if

(1-D)x-2Dy = (D+1)x, then x=-y, is a solution, and then

2x + (1-D)y = (D+1)y, gives that x = Dy, and only D=-1 can work.

But then I have x2 - y2 = 0 = F, so no solutions.

So, it's now possible to see that all integer solutions to equations of the form x2 + Dy2 = F, are part of this immensely huge number theoretic structure, which is infinitely sized, where depending on the value of D, and the value of F, at various points in the chain you will have situations when you have multiples of x and y.

What's remarkable about that super structure is it allows an explanation for all previously noted behavior and indicates that solutions have to do with linear equations and nothing else.

Intriguingly there is other research noting a lack of a need for non-rational numbers with these equations:

Pell's equation without irrational numbers
Authors: N. J. Wildberger
(Submitted on 16 Jun 2008)

Abstract: We solve Pell's equation in a simple way without continued fractions or irrational numbers, and relate the algorithm to the Stern Brocot tree.

Subjects: Number Theory (math.NT)
Cite as: arXiv:0806.2490v1 [math.NT]

So there is an infinitely sized and static number theoretic structure which is just an infinite expansion where the first four equations are:

1. x2 + Dy2 = F

2. (x-Dy)2 + D(x+y)2 = F(D+1)

3. ((1-D)x-2Dy)2 + D(2x + (1-D)y)2 = F(D+1)2

4. ((1-3D)x + (D2 - 3D)y)2 + D((3-D)x + (1-3D)y)2 = F(D+1)3

and that goes out to infinity. To get successive terms in the series you use the algebraic result that given:

u2 + Dv2 = C

it must be true that

(u-Dv)2 + D(u+v)2 = C(D+1).

And where whenever the exponent of (D+1) is even, you can have a case where you just have a multiple of x and y, so you can solve for D, which defines possible values for F in terms of x or y.

So the previously seen behavior with solutions to these equations can be explained as finding points in the expansion where you have a multiple with a given D, which will allow your F.

James Harris