Reducing binary quadratic Diophantines

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)

and

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

Which gives:

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

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 + 2s² = 166

which is:

(-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.


James Harris