A special quadratic form.

by individ, Dec 13, 2015, 10:38 AM

The same formal record. If we look for a parameterization not in 2 and in 3 option the problem can be solved quite simply.

For the equation.

$$aX^2+bY^2+cZ^2=dXY+eXZ+fYZ$$
If you know any solution $(x,y,z)$ of this equation. Then the formula for the solution of the equation can be written immediately.

$$X=(dy+ez-ax)p^2+(fz-2by)ps+bxs^2+cxt^2+(fy-2cz)pt-fxst$$
$$Y=ayp^2+(ez-2ax)ps+(dx+fz-by)s^2+cyt^2-eypt+(ex-2cz)st$$
$$Z=azp^2-dzps+bzs^2+(ex+fy-cz)t^2+(dy-2ax)pt+(dx-2by)st$$
$p,s,t - $ any integers.

It is seen that such formulas can be written infinitely many. If so like to use the well-known decision, it is better to write like this. This will allow to solve the equation.

$$A(x^2+y^2+z^2)=B(xy+xz+yz)$$
It is easy to see that there are solutions for any $A$. Ask yourself the number $c$. And place into factors.

$$c^2+A=ab$$
Then the coefficient is set to $B$ us so.

$$B=a^2+b^2-2(a+b)c+3c^2$$
Finding all the factors of $c,a,b$ you can write a formula for the parameterization of the solution of this equation.

$$x=c(ab-c^2)(k^2+s^2)+((a+b)(a^2+b^2)-(4a^2+5ab+4b^2)c+7(a+b)c^2-$$
$$-5c^3)p^2-(a^2+b^2-2(a+b)c+3c^2)cks+(b(b^2-a^2)-c(a^2+3b^2)+$$
$$+(4a+5b)c^2-5c^3)pk+(a(a^2-b^2)-c(b^2+3a^2)+(5a+4b)c^2-5c^3)ps$$
$$y=(a-c)(ab-c^2)(p^2+s^2)+(b^3-(a+2b)bc+(a+3b)c^2-c^3)k^2-$$
$$-(a-c)(a^2+b^2-2(a+b)c+3c^2)ps+(-2ab^2+c(a+b)^2-2ac^2+c^3)ks+$$
$$+(b(b^2+a^2)-(a^2+4ab+3b^2)c+(2a+5b)c^2-c^3)kp$$
$$z=(b-c)(ab-c^2)(p^2+k^2)+(a^3-(b+2a)ac+(3a+b)c^2-c^3)s^2-$$
$$-(b-c)(a^2+b^2-2(a+b)c+3c^2)pk+(a(a^2+b^2)-(3a^2+4ab+b^2)c+$$
$$+(5a+2b)c^2-c^3)ps+(-2ba^2+c(a+b)^2-2bc^2+c^3)ks$$
The transition to 3 parameters $p,s,k - $ Allows not to use too much, but to write formulas.
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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