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Monday, February 05, 2018

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 - Am  = (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 - 4Dm  = (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)- 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

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