A system of similar nonlinear equations

by individ, Dec 29, 2022, 12:45 PM

There is a whole group of such systems of equations.

$\left\{\!\begin{aligned} & A^2+AB+B^2=C^2 \\ & F^2+FZ+Z^2=C^2 \end{aligned}\right.$
Such a system is solved as standard. First, we write down the parametrization of one equation.

$A=p^2-s^2$
$B=s(s+2p)$
$C=p^2+ps+s^2=x^2+xy+y^2$
$F=x^2-y^2$
$Z=y(y+2x)$
And then we find the parameterization for the necessary parameters.

$p=2q^2+2qt-t^2-3qk+k^2$
$s=-q^2+2qt+2t^2-3tk+k^2$
$y=q^2+qt+t^2-k^2$
$x=q^2+qt+t^2-3(q+t)k+2k^2$

Diophantine equation

system of equations

number theory

The problem of 4 cubes

by individ, Dec 24, 2022, 9:56 AM

$A^3+B^3=C^3+Q^3$
$A=(3k-t)((81k^5-108k^4t+63k^3t^2-27k^2t^3+6kt^4-t^5)x^2+3(9k^3-9k^2t+3kt^2-t^3)xy+(3k-2t)y^2)$
$B=t(3k(27k^4-36k^3t+21k^2t^2-6kt^3+t^4)x^2+t^3xy-(3k-2t)y^2)$
$C=(3k-t)(3k(27k^4-36k^3t+21k^2t^2-6kt^3+t^4)x^2+(3k-t)^3xy+(3k-2t)y^2)$
$Q=t((-162k^5+216k^4t-126k^3t^2+45k^2t^3-9kt^4+t^5)x^2-3(18k^3-18k^2t+6kt^2-t^3)xy-(3k-2t)y^2)$

number theory

Diophantine equation

The problem of 4 cubes

by individ, Dec 22, 2022, 12:01 PM

$A^3+B^3=C^3+Q^3$
$A=k(3(k^3-2t^3)(k^2+kt+t^2)x^2+3(k+t)(k^3-2t^3)xy+(k+t)(k^2-t^2)y^2)$
$B=t(k+t)(3(k^4+k^2t^2+t^4)x^2+3t^3xy+(t^2-k^2)y^2)$
$C=k(k+t)(3(k^4+k^2t^2+t^4)x^2+3k^3xy+(k^2-t^2)y^2)$
$Q=t(-3(2k^3-t^3)(k^2+kt+t^2)x^2-3(k+t)(2k^3-t^3)xy+(k+t)(t^2-k^2)y^2)$

Diophantine equation

number theory

Sequences of cubes

by individ, Dec 21, 2022, 3:24 PM

$(A-B)^3+A^3+(A+B)^3=(C-Q)^3+C^3+(C+Q)^3$
$A=64k^{4}t^{3}x^{2}+2ty^2$
$B=2k(27k^6-18k^{4}t^2+20k^2t^4+8t^6)x^2+(9k^2+2t^2)(3k^2-2t^2)xy+3ky^2$
$C=8tk^2(9k^4-4k^2t^2+4t^4)x^2+8kt(3k^2-2t^2)yx+2ty^2$
$Q=64k^3t^4x^2+(3k^2-2t^2)^2xy+3ky^2$

This post has been edited 1 time. Last edited by individ, Dec 21, 2022, 5:03 PM

number theory

Diophantine equation

Sequences of cubes

by individ, Dec 19, 2022, 8:11 AM

$(A-B)^3+A^3+(A+B)^3=(C-Q)^3+C^3+(C+Q)^3$
There was a question. When there are solutions A−B>0 and C−Q>0

Each attempt at a solution gives a new formula for some reason. It didn't fit.

$A=2(36a^2-4ab+b^2)$
$B=64a^2-ab+3b^2$
$C=2(32a^2+b^2)$
$Q=74a^2-11ab+3b^2$
The following formula gives the necessary solutions.

$A=528a^2-40ab+b^2$
$B=64a^2-25ab+3b^2$
$C=128a^2+b^2$
$Q=764a^2-95ab+3b^2$
The general formula should contain 3 more parameters.

number theory

Diophantine equation

Sequences of cubes

by individ, Dec 16, 2022, 8:24 AM

https://math.stackexchange.com/questions/4599457/sum-of-cubes-of-three-positive-integers-in-arithmetic-progression-in-four-ways/4599816#4599816

$(A-B)^3+A^3+(A+B)^3=(C-Q)^3+C^3+(C+Q)^3$
$A=4(24a^2-8ab+b^2)$
$B=177a^2-59ab+6b^2$
$C=4(33a^2-10ab+b^2)$
$Q=132a^2-49ab+6b^2$

This post has been edited 1 time. Last edited by individ, Dec 19, 2022, 8:12 AM

number theory

Diophantine equation

The problem of 4 cubes

by individ, Dec 14, 2022, 12:13 PM

$A^3+B^3=C^3+Q^3$
$A=1144a^2-2024ab-11176ac+900b^2+9896bc+27300c^2$
$B=-559a^2+989ab+5461ac-435b^2-4826bc-13335c^2$
$C=273a^2-547ab-2731ac+269b^2+2726bc+6825c^2$
$Q=1092a^2-1928ab-10664ac+856b^2+9424bc+26040c^2$

number theory

Diophantine equation

The problem of 4 cubes

by individ, Dec 10, 2022, 9:15 AM

https://math.stackexchange.com/questions/4591666/solving-cubic-systems-of-diophantine-equations

$A^3+B^3=C^3+Q^3$
You can write such a parametrization.

$A=28a^2+12ab-68ac+2b^2-16bc+42c^2$
$B=21a^2+ab-43ac-b^2+bc+21c^2$
$C=42a^2+16ab-100ac+2b^2-20bc+60c^2$
$Q=-35a^2-15ab+85ac-b^2+17bc-51c^2$

This post has been edited 1 time. Last edited by individ, Dec 10, 2022, 9:48 AM

number theory

Diophantine equation

Prove that a+b+c+d is a composite number

by individ, Oct 30, 2022, 8:33 AM

[url]http://math.hashcode.ru/questions/249507/теория-чисел-доказать-что-a-b-c-d-составное-число?страница=1&focusedAnswerId=249550#249550[/url]

$X^2+aXY+Y^2=Z^2+aZV+V^2$
Решений можно записать два. Одно когда взаимно простые....

$X=(a+1)k^2+2tk-t^2-(a+2)qk+q^2$
$Y=(a+1)t^2+2tk-k^2-(a+2)qt+q^2$
$Z=k^2+akt+t^2-q^2$
$V=k^2+akt+t^2+(a+1)q^2-(a+2)(t+k)q$
Сумма их имеет вид.

$X+Y+Z+V=(a+2)(t+k-q)^2$
И когда нет...

$X=k^2+akt+t^2-q^2$
$Y=q(aq-at-2k)$
$Z=t(aq-at-2k)$
$V=t^2-k^2-q^2+aqk$
$X+Y+Z+V=(a-2)(q+t)(q+k-t)$
Ну и дальше всё совсем просто... взаимно простые когда скобки некоторые равны 1... ну или -1

number theory

Diophantine equation

The Pell equation

by individ, Oct 11, 2022, 12:06 PM

https://mathoverflow.net/questions/432091/how-to-describe-all-integer-solutions-to-x2y2-3z21/432221#432221

So the problem boils down to finding solutions to the general Pell equation. Then the problem is just in writing a convenient formula for describing solutions.
In addition to a convenient formula, there is another problem - through which Pell equations these solutions are described. Oddly enough, the usual Pell equation appears everywhere. Of this kind. $X^2-qY^2=1$

And there are two possible options for describing solutions. The first option is to use the standard Pell equation. And use such formulas, for example.

$ax^2-by^2+cx-dy+q=0$
https://artofproblemsolving.com/community/c3046h1049910

$aX^2+bXY+cY^2=f$
https://artofproblemsolving.com/community/c3046h1048219

I don't particularly like these formulas. In order to use them, it is necessary to solve the Pell equation. Therefore, it was a question of using solutions of the general and standard Pell equations. That's strange. that it is always necessary to use the standard Pell equation to find solutions.

Since the square shape can always be reduced to the general Pell equation.

$qX^2-nY^2=f$
Using solutions of the standard Pell equation. $a^2-nqb^2=1$

And using solutions of the general Pell equation. $qc^2-nd^2=f$

You can write a formula for finding the next solution using the previous ones.

$X=ac+nbd$
$Y=ad+qbc$
It was interesting, and if you don't do transformations, write an explicit formula. Will the same pattern persist in this case?

$aX^2+bXY+cY^2=f$
You still need a general equation. $ak^2+bkt+ct^2=f$

And the standard equation. $ap^2+bps+cs^2=1$

Solutions can be written like this.

$X=-(ka+bt)p^2-2tcps+cks^2$
$Y=atp^2-2akps-(bk+ct)s^2$
To obtain all solutions sequentially. They are usually used in the formula of the first solutions and going consistently to the big ones. Although formally, any solutions can be used in these formulas. For any solutions, the formula works.

The interesting thing is that no matter how the form of the general Pell equation changes. The form of solutions to this equation will be given by the standard Pell equation.

number theory

Diophantine equation

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

17 shouts

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