**c**

_{1}x^{2}+ c_{2}xy + c_{3}y^{2}= c_{4}z^{2}+ c_{5}zx + c_{6}zywhere the c's are constants, and was able to generally reduce. Subtracting an expression like that which I call the conditional through a complex identity gives what I call the conditional residue, and THAT can be analyzed. Is like the math does all the heavy work for you. And of course, the math does not care.

Which is really cool. And my original research was in 2008, but then set z = 1 and focused on the binary quadratic Diophantine case, for YEARS until now. And this time got general reduced form:

u

^{2}- Dv

^{2}= Fw

^{2}

Which looks much like the binary quadratic reduced form.

With x, y and z of course have two degrees of freedom, but here is an example showing a path to solution with reducing constraint! And find it interesting as quadratics seem to be special.

So you can still generally reduce with three variables for the quadratic case AND will show a situation where also can get integer solutions for all three with setting the variable z.

Which can see with an example.

Let c

_{1}= 1, c

_{2}= 2, c

_{3}= 3, c

_{4}= 4, c

_{5}= 5, c

_{6}= 6, so:

**x**

^{2}+ 2xy + 3y^{2}= 4z^{2}+ 5xz + 6yzWhich with my method for reducing gives:

A = -8, B = -20, and C = 33

So:

**[-4(x+y) + 10z]**

^{2}= 166z^{2 }- 2m^{2}And I notice that m = 9z, gives me a useful substitution:

And [-4(x+y) + 10z]

^{2}= 166z

^{2 }- 162z

^{2}= 4z

^{2}

So now have: -4(x+y) + 10z = +/-2z, so:

**x+y = 2z or 3z**

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

Let's trot through a full solution. Let z = 10, and use first, so: x+y = 20, so y = 20 - x

x

^{2}+ 2x(20-x) + 3(20-x)

^{2}= 400 + 50x + 60(20 - x), and multiplying out:

x

^{2}+ 40x -2x

^{2}+ 1200 - 120x + 3x

^{2}= 400 + 50x + 1200 - 60x

Where get easily enough:

x

^{2}- 35x - 200 = 0, which is: (x-40)(x+5) = 0

And one full solution then is: x = -5, y = 25, z = 10

Where just picked some easy.

James Harris

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