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It turned out that Draghilev's method is capable of solving Diophantine equations.
Can be viewed here.

Could that be related to one of the seven millennium problems: Birch and Swinnerton-Dyer conjecture? (In Spanish, hope that can add automatic translations). Can someone tell Grigori Perelman?



Best regards.
Alvaro.

Wrote

Could that be related to one of the seven millennium problems: Birch and Swinnerton-Dyer conjecture?

This question is not for me, I just suggested a simple way to solve equations in integers. Such theories are too complex for me to understand.

Wrote

... Such theories are too complex for me to understand.

For me too, that's why I use an outreach video aimed at a general public and in my mother tongue.

Best regards.
Alvaro.

Rolling without slipping. Once again.

There, small pictures are links to full-fledged animations. The equation of the rolling surface is displayed on the graph. And just a few words in Russian.

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Wrote

Originally Posted by: алексей_алексей [/url]It turned out that Draghilev's method is capable of solving Diophantine equations.
[url=https://www.mapleprimes.com/posts/218880-Diophantine-Equations]Can be viewed here.

We have Diophantine in Smath 30405, works superb from MCD style.
Jean

Wrote

Rolling without slipping. Once again.
There, small pictures are links to full-fledged animations. The equation of the rolling surface is displayed on the graph. And just a few words in Russian.
1.

Можешь выложить анимацию или уравнения с начальными условиями сюда?. Качество очень низкое в превью.

Можешь выложить анимацию или уравнения с начальными условиями сюда?. Качество очень низкое в превью.

Там картинки являются ссылками.

Вот тема на киберфоруме, где все эти рисунки присутствуют. Они в конце.

Вот тема на MaplePrimes, где есть основной текст. Он сильно корявенький, но и я не программист.

On a stationary surface x1^4 + x2^4 + x3^4 - 1 = 0, its reduced copy rolls without slipping, the equation of which at the origin is
x1^4 + x2^4 + x3^4 - 0.01 = 0. It rolls by "sideways ". The equation corresponding to the current position is displayed on the graph.

Rolling without slipping

Rolling without slipping. Another example, but with a very simplified and detailed graph.

[RUS]Рассмотрим движение окружности по параболе без проскальзывания. Центр окружности двигается по эквидистантной кривой к параболе. Для того чтобы движение окружности было без проскальзывания, необходимо чтобы скорость вращательного движения была равна скорости центра окружности, и в точке касания к параболе эти скорости должны быть противоположными по направлению. При выполнении этих условий скорость в точке касания равна нулю и мы наблюдаем "зубья". Полученные кривые являются циклоидными.[/RUS]
[ENG]Consider the motion of a circle along a parabola without slip. The center of the circle moves along an equidistant curve to the parabola. In order for the motion of the circle to be slip-free, it is necessary that the velocity of the rotational motion be equal to the velocity of the center of the circle, and at the point of tangency to the parabola these velocities must be opposite in direction. If these conditions are met, the velocity at the point of tangency is zero and we observe "teeth". The obtained curves are cycloidal.[/ENG]
kol1.avi (1,77 MiB) descargado 550 vez(es).
[RUS]В случае самопересечения эквидистантной кривой картина движения усложняется, тем не менее все условия продолжают выполнятся. В данном случае окружность выходит за границы параболы. Если необходимо, чтобы границы соблюдались, то выполняется "перекат" окружности с одной ветки параболы на другую. Перекатом я называю скачкообразное изменение точки касания окружности к параболе. На рисунке ниже показано "перекат" из точки А в точку В.[/RUS]
[ENG]In the case of self-intersection of the equidistant curve the picture of motion becomes more complicated, nevertheless all conditions continue to be fulfilled. In this case, the circle goes beyond the boundaries of the parabola. If it is necessary that the boundaries are observed, the circle is "rolled" from one branch of the parabola to another. I call a roll a jump-like change of the point of tangency of the circle to the parabola. The figure below shows a "roll" from point A to point B.[/ENG]
kol2.avi (1,6 MiB) descargado 539 vez(es).

kol2a.avi (1,16 MiB) descargado 537 vez(es).

Draghilev's method also allows us to construct equidistant surfaces.

https://en.smath.info/forum/resource.ashx?image=1703
https://en.smath.info/forum/resource.ashx?image=1702
https://en.smath.info/forum/resource.ashx?image=1701
https://en.smath.info/forum/resource.ashx?image=1700

Wrote

Draghilev's method also allows us to construct equidistant surfaces.

https://en.smath.info/forum/resource.ashx?image=1703
https://en.smath.info/forum/resource.ashx?image=1702
https://en.smath.info/forum/resource.ashx?image=1701
https://en.smath.info/forum/resource.ashx?image=1700

Не открывается, поправь пожалуйста.

Wrote

Draghilev's method also allows us to construct equidistant surfaces.

A little more on near this subject. Link to one Russian-language forum.

Wrote

Wrote

Draghilev's method also allows us to construct equidistant surfaces.

A little more on near this subject. Link to one Russian-language forum.

[RUS]Поверхность можно получить разными способами. Один из способов получения - решение нелинейных уравнений при помощи метода Драгилева. Полученные решения представляют собой набор замкнутых кривых, составляющие поверхность. При помощи полученных данных можно построить полигональную поверхность.[/RUS]
[ENG]The surface can be obtained in different ways. One way is to solve nonlinear equations using Dragilev's method. The obtained solutions are a set of closed curves that make up the surface. Using the obtained data, a polygonal surface can be constructed.[/ENG]

pov1.mov (3,16 MiB) descargado 403 vez(es).

Find all solutions to a system of equations

z^2 - 0.1*x^4 - 0.05*y^4 +1 = 0
x^3 + y^3 + 0.05*z^3 - 1 = 0
-2*cos(3*x) + 2*cos(3*y) - 2*cos(3*z) + 1 = 0

Wrote

Find all solutions to a system of equations

z^2 - 0.1*x^4 - 0.05*y^4 +1 = 0
x^3 + y^3 + 0.05*z^3 - 1 = 0
-2*cos(3*x) + 2*cos(3*y) - 2*cos(3*z) + 1 = 0

Hi Alexey. Using Draghilev's method for the Solve Block from this post, with this setup, 40 solutions can be found. By varying the parameters you can find more.



DraghilevSolveBlock - alexey example.sm (273,18 KiB) descargado 1733 vez(es).

Best regards.
Alvaro.

Wrote

40 solutions can be found

Hello Alvaro! Yes, thanks for the work done. This was, one might say, a test example for the Method. This system of equations has 116 solutions. The first two equations describe a closed curve; moving along it, we track the change in the sign of the last equation. There are 116 solutions in total.