Sunday, November 20, 2016

Infinite Diophantine quadratic progression

One result I stared at for quite a bit is the start of an infinite progression using what I now call the BQD Iterator:

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.

Here is my reference post discussing it, wow back in 2008. So yeah back then hadn't named the BQD Iterator. Just kind of informally talked it for a long time. Naming it worked better, and think I came up with a cool one.

It's interesting looking through the post seeing my efforts to try and use that series to solve for x and y. Through the years have had mixed feelings about such efforts.

James Harris
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