Trinary Quadratic Iterator

Finally looked at not setting z=1 to what I now know is a general way to reduce a trinary quadratic equation like:

c₁x² + c₂xy + c₃y² = c₄z² + c₅zx + c₆zy

where the c's are constants, where was able to prove can be generally reduced.

Shows yet another type of general reduced form: u² - Dv² = Fw²

Which has me of course wondering what happens if THAT form is so reduced, which have done in the past to get what I decided to call a binary quadratic Diophantine iterator or BQD Iterator for short. So will use the reduction method on it, copying over the base system.

A = (c₂ - 2c₁)² + 4c₁(c₂ - c₁ - c₃), B = (c₂ - 2c₁)(c₆ - c₅) + 2c₅(c₂ - c₁ - c₃)

and

C = (c₆ - c₅)² - 4c₄(c₂ - c₁ - c₃)

Base result is: A(x+y)² - 2B(x+y)z + Cz² = m²

And some simple algebra gives:

[A(x+y) - Bz]² - Am²   = (B²  - AC)z²

With: u² - Dv² = Fw²

So: c₁ = 1, c₂= 0, c₃ =  -D, c₄ = F, c₅ = 0, c₆ = 0,

x = u, y = v, z = w

So, A = 4 +4(-1+D) = 4D, B = 0, C = -4F(-1+D)

Gives: [4D(u+v)]² - 4Dm²   = (16DF(-1+D))w²

Which is: m²/4- D(u+v)²  = -F(D-1)w²

Where now need m. 4D(u+v)² - 4F(-1 + D)w² = m²

So: m² = 4(Du² + 2Duv +Dv²  + Fw² - DFw²),

and m² = 4(Du² + 2Duv +Dv² + u² - Dv² - DFw²)

Where showing all the detail for once. Helps keep me from making mistakes.

So: m² = 4(Du² + 2Duv + u² - Du² + D²v²), and  m² = 4(u²  + 2Duv + D²v²) = 4(u+Dv)²

Which is: (u+Dv)² - D(u+v)²  = F(-D+1)w²

So the trinary quadratic iterator is just the BQD Iterator with a w² on the end.

Shall call it the TQ Iterator for short.


James Harris