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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
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