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I am trying to find out how to solve $ax^2 - by^2 + cx - dy + e = 0$ to get integer solutions, failing this the rational solutions.

Thanks!

gautam
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  • When write the equation, the solutions will be determined by the solutions of certain equations Pell. – individ Jun 10 '14 at 05:25

1 Answers1

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Solutions of the equation:

$ax^2-by^2+cx-dy+q=0$

you can record if the root of the whole: $k=\sqrt{(c-d)^2-4q(a-b)}$

Then using the solutions of the equation Pell: $p^2-abs^2=\pm1$

Then the formula of the solution, you can write:

$x=\frac{\pm1}{2(a-b)}(((d-c)\pm{k})p^2+2(bk\mp(bc-ad))ps+b(a(d+c)-2bc\pm{ak})s^2)$

$y=\frac{\pm1}{2(a-b)}(((d-c)\pm{k})p^2+2(ak\mp(bc-ad))ps-a(b(d+c)-2ad\mp{bk})s^2)$

If the root is a need to find out if this is equivalent to the quadratic form in which the root of the whole. This is usually accomplished this replacement: $x$ in such number $x+ty$

individ
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  • Forgot to say. The characters inside the brackets do not depend on the sign of the Pell equation. It depends only before $\pm{1}$ – individ Jun 10 '14 at 13:21
  • I think it's harder than this. Can you tell me, for example, all the integer solutions to $x^2-37y^2=3$? – Jeroen Jan 05 '16 at 19:17
  • @Jeroen it is necessary to investigate this equation using the General formula. http://www.artofproblemsolving.com/community/c3046h1048219 So the root was rational need to make the change. $x=x+6y$ Obtain the equation. $x^2+12xy-y^2=3$ Substitute in the formula and my solution is ready. – individ Jan 06 '16 at 04:48
  • Why don't you answer the question rather than just theorising about how to answer it -- let me know the first few solutions. My point is I suspect that you are missing at least one of the subtleties involved in solving these equations: there is no simple formula for a generalised Pell equation because there are subtle arithmetic issues. – Jeroen Jan 06 '16 at 07:25
  • @Jeroen I don't understand - the formula brought. Substitute and get the answer. You want me brought even arithmetic? I do laziness. Maybe I will if the mood is good. In this issue, the wording and formulation of the problem was different. I responded - wrote the General formula. The rest I'm not interested. – individ Jan 06 '16 at 11:44
  • Your formula is wrong. How can we see this? We can see it if you try and use the formula to solve my problem and then you will see that it is wrong. Solving Pell's equation in integers is not enough to solve the generalised Pell equation $x^2-Dy^2=C$ in integers. The problem is that $C$ may factor into non-principal ideals in $Z[\sqrt{D}]$ and detecting this is a subtle problem in arithmetic. There is no "formula" of the kind you have written, for integer solutions to a generalised Pell equation. If the Hilbert class field of $Q(\sqrt{D})$ is not abelian over $Q$ then life is complicated. – Jeroen Jan 06 '16 at 13:30
  • @Jeroen The formula is correct. If you don't believe use of mathematical packages and check. I don't understand what You do not like? To solve the equation $x^2-Dy^2=C$ the formula reduces to a Pell equation $x^2-Dy^2=\pm1$ exactly what You're talking about. So I do not understand what is not to like? – individ Jan 06 '16 at 13:43
  • no. To solve $x^2-Dy^2=C$ in integers is harder than solving $x^2-Dy^2=1$. Here is the proof: go ahead and tell me the first few solutions to $x^2-37y^2=3$ in positive integers. I will tell you that the first few solutions to $x^2-37y^2=1$ in positive integers are $(x,y)=(73,12),(10657,1752)...$. Now it's your move. You can claim all you like that your formula is correct, but if you do not solve my problem then who will believe you? I do not, and the reason I do not is that I know what I am talking about and I suspect that you do not. – Jeroen Jan 06 '16 at 14:23
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    Perhaps the problem is that the OP is asking for integer solutions and your formula is producing rational ones? – Jeroen Jan 06 '16 at 19:01