1

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)

Latin PMB
  • 33
  • I repeat the question. http://math.stackexchange.com/questions/1365516/prove-quadratic-diophantine-has-no-solutions – individ Jul 22 '15 at 09:48
  • 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

2 Answers2

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.

Community
  • 1
Will Jagy
  • 146,052
  • Thanks. I understand this procedure will give you a solution. The reason for the question/requirement is that the coefficients I work with are quite large (especially F - no bounds) and I believe time complexity is almost exp. Thus I wondered whether any analytical methods would be available for an equation of this form. Even methods that don't produce the actual solution - but let you infer some requirements for x,y or that let you determine that the equation has no integer solution. Thanks! – Latin PMB Jul 23 '15 at 05:43
0

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 $x>0$.

poetasis
  • 6,795