Diophantine equation - Special form (quadratic)

Question

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)

Answer 1

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

So, what is so wrong about this?

Here is a similar one from yesterday or the day before. Integer solutions of a cubic equation What is this obsession with closed form solutions? If you know that you have a finite number of solutions, and finding all of the solutions comes down to factoring a single number, you call it a good day.

Answer 2

Solutions for $2^{nd}$ degree "variables" may be found using the quadratic equation or WolframAlpha.

Given $$A^2x^2 - C^2y^2 + Dx - Ey + F = 0$$ we find

$$x = \pm\frac{-\sqrt{4 A^2 (y (C^2 y + E) - F) + D^2} + D}{2 A^2} \text{ for } A\ne 0$$

$$x =\frac{y (C^2 y + E) - F}{D} \text{ for } A = 0 \text{ and } D\ne 0$$ Given $$324x^2 - 9y^2 + 101x - 13y + F =0 $$ we find $$x = \frac{\pm\sqrt{-1296 F + 11664 y^2 + 16848 y + 10201} - 101}{648}$$ and this means there my be complex solutions for some combinations of $F,y$ but there are real solutions for all values of $F$ is y is large enough.

In the case of $F=50$, $$11664 y^2 + 16848 y + 10201 - 1296×50 = 0$$

$$y = \frac{\pm\sqrt{60683} - 78}{108}$$ If we substitute the positive $"y"$ into the solution for $x$ above, the radical becomes zero and

$$x=\frac{-101}{648}$$ Any larger values of $y$ will make the radical increasingly greater than zero, eventually making $x>0$.