Given an equation of the form

c

_{1}x

^{2}+ c

_{2}xy + c

_{3}y

^{2}= c

_{4}+ c

_{5}x + c

_{6}y

I've proven that you can reduce to

(A(x+y) - B)

^{2}- As

^{2}= B

^{2}- AC

which is itself a binary quadratic with (x+y) and s unknowns, where

A = (c

_{2}- 2c

_{1})

^{2}+ 4c

_{1}(c

_{2}- c

_{1}- c

_{3})

B = (c

_{2}- 2c

_{1})(c

_{6}- c

_{5}) + 2c

_{5}(c

_{2}- c

_{1}- c

_{3})

and

C = (c

_{6}- c

_{5})

^{2}- 4c

_{4}(c

_{2}- c

_{1}- c

_{3})

when neither A nor B equals 0.

As a first example let c

_{1}= 1, c

_{2}= 1, c

_{3}= 1, c

_{4}= 1, c

_{5}= 1, c

_{6}= 1, so:

x

^{2}+ xy + y

^{2}= 1 + x + y

Which gives:

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

So the equation reduces to:

(-3(x+y) + 2)

^{2}+ 3s

^{2}= 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 c

_{1}= 1, c

_{2}= 2, c

_{3}= 3, c

_{4}= 4, c

_{5}= 5, c

_{6}= 6, so:

x

^{2}+ 2xy + 3y

^{2}= 4 + 5x + 6y

Which gives:

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

And substituting and dividing 4 from both sides the equation reduces to:

(-4(x+y) + 10)

^{2}+ 2s

^{2}= 166

which is:

(-4(x+y) + 10)

^{2}= 166 - 2s

^{2}= 2(83 - s

^{2})

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 = s

^{2}

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:

c

_{2}- 2c

_{1}= c

_{2}- c

_{1}- c

_{3}= c

_{6}- c

_{5}= 0

However that special case is easily handled, for example:

x

^{2}+ 2xy + y

^{2}= c

_{4}+ x + y

is: (x+y)

^{2}- (x+y) - c

_{4}= 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.

James Harris