Квадрат с кубиком.

by individ, Apr 29, 2014, 3:34 PM

In the equation: $X^2+Y^2=qZ^3$

If the ratio is such that the root of an integer: $c=\sqrt{q-1}$

Then the solution is:

$X=-2(c+1)p^6+4(2c(q-2)-3q)p^5s+2(c(5q^2-2q-8)-q^2-22q+8)p^4s^2+$

$+8q(5q^2-14q+4)p^3s^3+2(c(5q^2-2q-8)+q^2+22q-8)q(q-2)p^2s^4-$

$-4(2c(q-2)+3q)q^2(q-2)^2ps^5-2(c-1)q^3(q-2)^3s^6$

...........................................................................................................................................................

$Y=2(c-1)p^6+4(2qc+q-4)p^5s+2(c(-5q^2+18q-8)+15q^2-22q-8)p^4s^2+$

$+8q(q^2+6q-12)p^3s^3-2(c(5q^2-18q+8)+15q^2-22q-8)q(q-2)p^2s^4-$

$-4(2qc-q+4)q^2(q-2)^2ps^5+2(c+1)q^3(q-2)^3s^6$

............................................................................................................................................................

$Z=2p^4+8p^3s+4(q^2-2q+4)p^2s^2-8q(q-2)ps^3+2q^2(q-2)^2s^4$

And more.

$X=-2(c-1)p^6-4(2c(q-2)+3q)p^5s+2(c(5q^2-2q-8)+q^2+22q-8)p^4s^2+$

$+8q(5q^2-14q+4)p^3s^3+2(c(5q^2-2q-8)-q^2-22q+8)q(q-2)p^2s^4+$

$+4(2c(q-2)-3q)q^2(q-2)^2ps^5-2(c+1)q^3(q-2)^3s^6$

.............................................................................................................................................................

$Y=2(c+1)p^6-4(2qc-q+4)p^5s+2(c(-5q^2+18q-8)-15q^2+22q+8)p^4s^2+$

$+8q(q^2+6q-12)p^3s^3+2(c(-5q^2+18q-8)+15q^2-22q-8)q(q-2)p^2s^4+$

$+4(2qc+q-4)q^2(q-2)^2ps^5+2(c-1)q^3(q-2)^3s^6$

.............................................................................................................................................................

$Z=2p^4-8p^3s+4(q^2-2q+4)p^2s^2+8q(q-2)ps^3+2q^2(q-2)^2s^4$
number theory
Diophantine equation

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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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