# Diophantine equation - Special form (quadratic)

I am dealing with a series of quadratic diophantine equations that all have the same form: $$A^2x^2 - C^2y^2 + Dx - Ey + F = 0$$ ($A,C,D,E >0$ | $A$ and $C$ have a common factor (or $C=1$) | $D,E,F$ coprime)

Do you know if there's an analytical method (ie. one that doesn't require testing all factorizations of a large number) to find a single solution (one is enough) if it exists - or a way to show there are no solutions if that is the case?. Thanks!

Eg. $$324x^2 - 9y^2 + 101x - 13y + F =0$$ ($F= -229$: Sol ($x=1, y=4$))

($F= 50$: No solutions)

• I repeat the question. math.stackexchange.com/questions/1365516/… – individ Jul 22 '15 at 9:48
• Individ, you did not answer that question (provided a magical formula that does not work). This question is also about a method for finding the solution (one) to the quadratic equation. No magic formulas please :) – Latin PMB Jul 22 '15 at 10:54
• If the formula is not good enough for you, that is no reason to open 20 of the same topics. Although still not clear, what is not satisfied. When the squares are then infinitely many solutions may not be. So the formula cannot be. If not - give an example when solutions can be infinite. – individ Jul 22 '15 at 11:00
• Yeah right. Please stay away. – Latin PMB Jul 22 '15 at 11:08
• Why do you need to know whether this has a solution? Apparently this is some sort of programming assignment. What is so bad about completing the squares, factoring a single integer constant that will have many small prime factors? – Will Jagy Jul 22 '15 at 21:55

I get $$\left( 2A^2Cx + 2 A C^2 y + CD + AE \right) \left( 2A^2Cx - 2 A C^2 y + CD - AE \right) = \color{red}{C^2 D^2 - A^2 E^2 - 4 A^2 C^2 F}.$$