Quadratics easier with more degrees of freedom

One thing that DID surprise me was when found solutions were actually easier when I focused on more variables with a quadratic case:

c₁x² + c₂xy + c₃y² = c₄z² + c₅zx + c₆zy

where the c's are constants. Years ago, actually shied away from and went simpler by setting z=1, to get my method for generally reducing binary quadratic Diophantines. And FINALLY got curious enough to look at the more general case, where handled in two key posts:

Three variable quadratic reduction

and

Trinary Quadratic Iterator

Where yeah my BQD Iterator is my primary workhorse still, where who knows really what the TQ Iterator adds and for the moment am not working to find out.

What is really cool to me though is how you can end up with quadratics easier to solve for integers. Like copying from my post giving the reduction method, let c₁ = 1, c₂ = 1, c₃ = 1, c₄ = 1, c₅ = 1, c₆ = 1, so:

x² + xy + y² = z² + zx + zy

Which gives:

A = -3, B = -2, and C = 4

[-3(x+y) + 2z]²  + 3m²   = 16z²

Where now have two unknowns m and z determining existence of a rational solution. But by inspection has an infinite number of rational solutions with m = 0. Still shouldn't just assume that then x and y always will be integers. Easy enough to check:

-3(x+y) + 2z = +/- 4z, so: -3(x+y) =  2z or -6z

So there will be a set of integer solutions for every nonzero integer z.

Where a trivial set is: x = y = z

But you can also find others.

Here's one a little more complicated. Let c₁ = 1, c₂ = 2, c₃ = 3, c₄ = 4, c₅ = 5, c₆ = 6, so:

x² + 2xy + 3y² = 4z² + 5xz + 6yz

Where similarly can show that you can always find solutions with: x+y = 2z or 3z

Which means can always have integer solutions with an integer z, with one reduced constraint.

James Harris