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)
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.