System of equations; 3 equations - 2-nd degree

[ins-]

Solve the system:

$y^2+z^2-2 a y z = 1 - a^2$

$z^2+x^2-2 b z x = 1 - b^2$

$x^2+y^2-2 c x y = 1 - c^2$

where $a$, $b$, $c$ are real parameters.

[pogulyat_vyshel]

В общем случае можно так подобрать числа $A,B,C$, что замена координат $x\mapsto Ax,\quad y\mapsto By,\quad z\mapsto Cz$ приведет систему уравнений к виду, когда в правой части стоят $\pm 1$. Левая часть при этом не изменится. Приравнивая соответствующие левые части мы получим два уравнения конусов с общей вершиной в нуле и третье уравнение поверхности второго порядка. Конусы пересекаются по нескольким прямым. Долго, нудно.

[ins-]

I received this problem from a Vietnamese guy. The idea was to show the application of hyperbolic trigonometric functions in algebra. I think the problem can be solved also using pure trigonometry. Posted it here to see different ideas and approaches as well as to see if there is a pure algebraic solution.

[nnosipov]

ins- в сообщении #1477662 писал(а):

if there is a pure algebraic solution

Sorry, but for this system, we have no a "good" (elementary or pure algebraic) answer. In general, each of unknowns $x$, $y$, $z$ is a root of a quartic univariable equation. Thus, we need to use the Ferrari method (for instance) or more complicated methods.

[ins-]

I observed something interesting about this system. If any of the variables or parameters is > 1, all the remaining are greater than 1. If it is < 1 all the remaining are < 1.

[ins-]

https://www.wolframalpha.com/input/?i=E ... 2C+z%7D%5D according to the software $x$ can be found after solving an 8-th degree equation, that can be transformed to a 4-th degree one. For this equation is possible to avoid Ferrari's formula (there is no a 3-rd root in the result) but requires lots of calculations and without software it is possible to be not easy for doing. I suspect there'd be something easier.

[Rak so dna]

ins- в сообщении #1477723 писал(а):

I suspect there'd be something easier.

You're right. The solution is elementary (the answer can be expressed in terms of square roots)

[ins-]

In the Modenov's book I found the following system:
$x^2+yz=a$,
$y^2+zx=b$,
$z^2+xy=c$
I'm posting it in this problem topic because of the following - in the book is given a "nice" answer. While if I put it directly in the software some terrible expressions appear. The case with the system I posted at the beginning is possible to be the same. (I don't know how to solve both systems, but it doesn't mean they are not solvable by a person with more experience.)

[nnosipov]

ins- в сообщении #1477723 писал(а):

For this equation is possible to avoid Ferrari's formula (there is no a 3-rd root in the result)

It happens that you are lucky Indeed, the Galois group for that equation is $\mathbb{Z}^2 \times \mathbb{Z}^2$. In more elementary terms: the Cardano's resolvent has a rational root.

ins- в сообщении #1477740 писал(а):

While if I put it directly in the software some terrible expressions appear.

In this case, we have $S_4$ as the Galois group for the corresponding equation of 4th degree. So, we need some tricks if we believe that there exists a "good" answer.

[ins-]

It looks like the answer or the statement in the Modenov's book for the second problem is wrong. I put some of the answers in the equation and they are not identities.

[nnosipov]

ins- в сообщении #1477850 писал(а):

I put some of the answers in the equation and they are not identities.

Can you write the answer here? What is the mentioned Modenov's book?

[ins-]

The statement:
Solve the system -
$x^2+yz=a$
$y^2+zx=b$
$z^2+xy=c$
One of the answers is:
If $a^3+b^3+c^3 - 3abc \ne 0$ then there are two solutions: $\frac{a^2-bc}{\sqrt{a^3+b^3+c^3 - 3abc}}$, $\frac{b^2-ca}{\sqrt{a^3+b^3+c^3 - 3abc}}$, $\frac{c^2-ab}{\sqrt{a^3+b^3+c^3 - 3abc}}$, where the radical can have any of the two possible signs, but the same for the expressions for $x$, $y$, $z$.

If the signs are "-" instead of "+" in all the system's equations the answer they mention is correct.

This makes me to think systems like these requires an extra attention and more time to be solved.

[nnosipov]

ins- в сообщении #1477854 писал(а):

If the signs are "-" instead of "+" in all the system's equations the answer they mention is correct.

Oh, yes! Как у нас говорят, это две большие разницы. Obviously, we have an errata.

In any case, I recommend to use the Groebner bases technique. (If we want to establish the correct result.)

-- Пт авг 07, 2020 21:56:06 --

I can propose for further experiences the following "more simple" system:

$$x^2-y=a, \quad y^2-z=a, \quad z^2-x=a.$$

For $a=2$ we have a nice "trigonometric" solution. What about other values of $a$? I don't know.

[Rak so dna]

nnosipov в сообщении #1477857 писал(а):

What about other values of $a$?

Пусть $f(x)=x^2-a$ тогда имеем уравнение $f(f(f(x)))=x$
$f(f(f(x)))-x = (x^2-x-a)$$ \left(x^3-(\sqrt{a-\frac{7}{4}}-\frac{1}{2})x^2-(a+\sqrt{a-\frac{7}{4}}+\frac{1}{2})x+a\sqrt{a-\frac{7}{4}}+\frac{a}{2}-1\right)$$ \left(x^3+(\sqrt{a-\frac{7}{4}}+\frac{1}{2})x^2-(a-\sqrt{a-\frac{7}{4}}+\frac{1}{2})x-a\sqrt{a-\frac{7}{4}}+\frac{a}{2}-1\right)$

[nnosipov]

А что, $7/4$ --- это забавно. Пусть теперь $a=7/4+t^2$, где $t>0$ --- рациональное число. Найти все $t$, при которых оба кубических уравнения имеют "красивые" тригонометрические корни. (Надо посчитать их дискриминанты и потребовать, чтобы они оба были точными квадратами.)

Upd. О, да таких $t$ дофига очень много.
Upd-2. Да они (вот ведь бывает!) все такие. Так что красиво будет не только при $a=2$ (хорошо известный случай), но и, например, при $a=4$.