# Quadratic Diophantine equation – I

Quadratic Diophantine equation is an equation of the form *Ax**2 + Bxy + Cy**2 + Dx + Ey + F = 0* where* A, B, C, D, E,* and *F* are integer constants and x and y being integer variables. Study of this equation has always been an interesting area among number theorists. The famous pell equation is a special case of the above with* delta = B**2-4AC > 0* and delta not being a perfect square. Normally, this equation is broken down into five cases for analytical purposes.

1)* A = B = C = 0* (Linear case): Reduces to a linear Diophantine equation of two variables.

2) *A = C = 0* and* B != 0 * (Simple hyperbolic case): Equation reduces to *(Bx + E) (By + D) = DE – BF, *which can be solved by considering the factors of *DE – BF*.

3) *B**2 – 4AC < 0 *(Elliptical case): In this case, values of the x should lie between the roots of the equation *(B**2-4AC)x**2 + 2(BE – 2CD)x + E**2 – 4CF = 0. *Values for x should be selected so that y is an integer.

4) *B**2 – 4AC = 0 *(Parabolic case): Solution procedure is rather complex in this case.

I will describe these cases in detail in the future posts. I had almost completed the above cases at the start of Week 2.

5) *delta = B**2 – 4AC > 0*: This is split into several subcases.* *

**Case delta = B**2 – 4AC > 0:**

Subcase 1:* D = E = 0*:

This is the homogeneous case and again considered under two cases *F = 0* and *F != 0*. If* F = 0*, then *x = 0* and *y = 0* are solutions. More solutions may exist if* B**2 – 4AC* is a perfect square. Otherwise* x = 0* and *y = 0* is the only solution. I implemented this case in the module. If *F != 0* the solution procedure is rather complex and involves continued fractions. I am currently working on this.

- Posted in: GSoC-2013-SymPy
- Tagged: Diophantine Equations, Number Theory, Quadratic Diophantine equations

There’s a typo in your description of case #2. It should be “A = C = 0, B != 0″/

Thanks Keith, I corrected it.