Бинарная форма. 4

by individ, Jun 10, 2014, 1:17 PM

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 $x+ty$
Forgot to say. The characters inside the brackets do not depend on the sign of the Pell equation. It depends only before $\pm{1}$
This post has been edited 1 time. Last edited by individ, Jun 10, 2014, 1:23 PM

Comment

0 Comments

Archives
+ September 2019
+ October 2017
+ November 2016
+ March 2016
+ March 2015
Shouts
Submit
  • How did you discover these parametric solutions to diophantine equations?

    by fanzhuyifan, Dec 31, 2016, 9:25 AM

  • Russian? are you sure it ain't greek?

    by Mathisfun04, Dec 27, 2016, 4:03 PM

  • yep i agree

    by Eugenis, Oct 31, 2015, 2:40 AM

  • Best blog ever

    by FlakeLCR, Oct 13, 2015, 8:07 PM

  • too much russian.

    by rileywkong, Aug 21, 2015, 6:10 PM

  • Decided the equation.

    by individ, Aug 20, 2015, 5:05 AM

  • Some insight into how you figured it out?

    by Not_a_Username, Aug 19, 2015, 3:52 PM

  • I figured it out. Decided equation.

    by individ, Aug 19, 2015, 5:01 AM

  • Yes, how do you come up with the formula? :P

    by Not_a_Username, Aug 18, 2015, 10:29 PM

  • I don't understand. There are the equation and there is a formula to it solutions. What is the problem?

    by individ, Aug 13, 2015, 4:22 PM

  • What? Lol you are substituting solutions with literally no motivation

    by Not_a_Username, Aug 13, 2015, 12:59 PM

  • What replacement? Where?

    by individ, Aug 8, 2015, 5:37 AM

  • Darn, what are the motivation for these substitutions???

    by Not_a_Username, Aug 5, 2015, 10:44 AM

  • Are you greek?

    by beanielove2, Dec 24, 2014, 6:31 PM

  • So, a purely mathematical blog?

    by Lionfish, Dec 2, 2014, 1:20 PM

  • To prove that it is necessary to show the method of calculation. I do not want to do yet.

    by individ, Mar 28, 2014, 6:14 AM

  • I can't understand these posts....What language are they written in? I don't recognize it.

    I like your avatar! :P

    by 15cjames, Mar 11, 2014, 1:57 PM

17 shouts
Tags
About Owner
  • Posts: 494
  • Joined: Feb 12, 2014
Blog Stats
  • Blog created: Feb 13, 2014
  • Total entries: 332
  • Total visits: 100492
  • Total comments: 24
Search Blog