Cubes with squares.

by individ, Oct 20, 2014, 1:31 PM

For the equation: \[X^2+qY^2=Z^3\]

You can write this simple solution:

\[X=(p^2+qs^2)((p^4-q^2s^4)t^3-3(p^2+qs^2)^2kt^2+3(p^4-q^2s^4)tk^2-\]
\[-(p^4-6qp^2s^2+q^2s^4)k^3)\]

\[Y=2ps(p^2+qs^2)((p^2+qs^2)t^3-3(p^2+qs^2)tk^2+2(p^2-qs^2)k^3)\]

\[Z=(p^2+qs^2)((p^2+qs^2)t^2-2(p^2-qs^2)tk+(p^2+qs^2)k^2)\]

$q - $The ratio is given for the problem.

$p,s,t,k - $ integers asked us.
This post has been edited 1 time. Last edited by individ, Oct 20, 2014, 1:35 PM

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  • 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

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