As much as I talk about other key discoveries, I have talked less about my results about binary quadratic Diophantine equations, though a key one allows a one-step way to generally reduce binary quadratic Diophantine equations.
Given an equation of the form
c1x2 + c2xy + c3y2 = c4 + c5x + c6y
I've proven that you can reduce to
(A(x+y) - B)2 - As2 = B2 - AC
which is itself a binary quadratic with (x+y) and s unknowns, where
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)
when neither A nor B equals 0.
As a first example let c1 = 1, c2 = 1, c3 = 1, c4 = 1, c5 = 1, c6 = 1, so:
x2 + xy + y2 = 1 + x + y
A = -3, B = -2, and C = 4
So the equation reduces to:
(-3(x+y) + 2)2 + 3s2 = 16
which works with s=0. So -3(x+y) + 2 = 4 or -4 should work, and the latter gives x+y=2, and you can substitute out x or y in the original equation to solve that way, where x=y=1 is a solution.
For a second example let c1 = 1, c2 = 2, c3 = 3, c4 = 4, c5 = 5, c6 = 6, so:
x2 + 2xy + 3y2 = 4 + 5x + 6y
A = -8, B = -20, and C = 33
And substituting and dividing 4 from both sides the equation reduces to:
(-4(x+y) + 10)2 + 2s2 = 166
(-4(x+y) + 10)2 = 166 - 2s2 = 2(83 - s2)
Running through possible odd s's I notice that s=9 works to give -4(x+y) + 10 = 2 or -2, so x+y = 2, or x+y = 3. And x = 4, y = -2, or x = 5, y = -2 work.
The reducing form involves a completion of the square, as notice you can just multiply it out and also use the form:
A(x+y)2 - 2B(x+y) + C = s2
Note that form is necessary if A or B equals 0, because then you can't complete the square to get the other form.
This approach will work for any binary quadratic Diophantine equation except for if:
c2 - 2c1 = c2 - c1 - c3 = c6 - c5 = 0
However that special case is easily handled, for example:
x2 + 2xy + y2 = c4 + x + y
is: (x+y)2 - (x+y) - c4 = 0
And that form is equivalent to a unary case anyway, which is probably why my equations won't handle it. I worked out exactly why years ago, but don't remember enough to say that for certain.
I look at those equation every once in a while, just kind of wondering.
My method is to my knowledge the only known that turns reducing any binary quadratic Diophantine, except the special case as noted, into a one-step process.