An easy system with radicals

[ins-]

Solve the system:

$a \sqrt{x+z-y} . \sqrt{x+y-z} = x \sqrt{yz}$

$b \sqrt{x+y-z} . \sqrt{y+z-x} = y \sqrt{xz}$

$c \sqrt{y+z-x} . \sqrt{x+z-y} = z \sqrt{xy}$

[wrest]

Все буквы неотрицательные и равны.

[ins-]

What about: $(0, t, t)$ ?

[Dendr]

What are $(a, b, c)$? Are they parameters?

Presumably, we can define $z=z(a,x,y)$ from the first equation, then $y=y(a,b,x)$ from the second one and eventually get one equation for x, which is just an implicit function $x=x(a,b,c)$. Still, even a domain here is twisted drastically, so it's very hard to find a common solution.

Although, in a special case $b=c$ there's a nontrivial solution $y=z=a, x=2a-\frac{a^3}{b^2}$

For example, if $a=4, b=c=8: x=7, y=z=4$.

[ins-]

$a$, $b$, $c$ are parameters. Is that system harder than I expected?
(Saw it in an old source and had an idea for solution. Then decided to share the problem with more people.)

[StaticZero]

ins- в сообщении #1475471 писал(а):

Is that system harder than I expected?

Seems so.

As you see, it is not homogeneous, and I expect it not to be solvable with any "clever" coodinate transformations as many similar problems are.

[ins-]

I thought about the substitutions: $u=x+y-z, v=x-y+z, z=-x+y+z$. Then to square and divide the system's equations.

[StaticZero]

The problem suggests to write

$$ \begin{cases} \sqrt{x^2 - \xi^2} &= a x \sqrt{yz} \\ \sqrt{y^2 - \eta^2} &= b y \sqrt{zx} \\ \sqrt{z^2 - \zeta^2} &= c z \sqrt{xy} \end{cases} $$

where $\xi + \eta + \zeta = 0$ (infer them by yourself ) (here I declare $a, b, c$ to be movable back and forth through the $=$ sign, so don't bother to write $a^{-1}$).

I surrendered going this way.

[Rak so dna]

ins- в сообщении #1475276 писал(а):

Solve the system:

$a \sqrt{x+z-y} . \sqrt{x+y-z} = x \sqrt{yz}$

$b \sqrt{x+y-z} . \sqrt{y+z-x} = y \sqrt{xz}$

$c \sqrt{y+z-x} . \sqrt{x+z-y} = z \sqrt{xy}$

ins- в сообщении #1475471 писал(а):

Is that system harder than I expected?

No i consider it a normal school problem
$(t, t, 0), (t, 0, t), (0, t, t), (\frac{a(b^2+c^2-a^2)}{bc}, \frac{b(c^2+a^2-b^2)}{ca}, \frac{c(a^2+b^2-c^2)}{ab})$

[novichok2018]

Должна однородность сработать.

[ins-]

If I'm not wrong, I ovserved something interesting about this problem. Wolfram Alpha online vesrion cannot solve it.

Rak so dna did you use something similar to my suggestion to solve the system?

[nnosipov]

ins- в сообщении #1475523 писал(а):

Wolfram Alpha online vesrion cannot solve it.

I used the "Groebner" module from Maple. No problems, it's easy. Remark: previously, I squared both sides of the equations.

[ins-]

nnosipov в сообщении #1475537 писал(а):

ins- в сообщении #1475523 писал(а):

Wolfram Alpha online vesrion cannot solve it.

I used the "Groebner" module from Maple. No problems, it's easy. Remark: previously, I squared both sides of the equations.

What is "Groebner" module? Is it named after Sonhard Groebner?

[Rak so dna]

ins- в сообщении #1475523 писал(а):

Rak so dna did you use something similar to my suggestion to solve the system?

square and divide
$\begin{cases} \frac{a^2}{b^2}\frac{x+z-y}{y+z-x}=\frac{x}{y}\\ \frac{c^2}{b^2}\frac{x+z-y}{x+y-z}=\frac{z}{y} \end{cases}$
or
$\begin{cases} \frac{a^2}{b^2}(\frac{x}{y}+\frac{z}{y}-1)=\frac{x}{y}(1+\frac{z}{y}-\frac{x}{y})\\ \frac{c^2}{b^2}(\frac{x}{y}+\frac{z}{y}-1)=\frac{z}{y}(\frac{x}{y}+1-\frac{z}{y}) \end{cases}$
subtract
$(\frac{x}{y}+\frac{z}{y}-1)(\frac{a^2}{b^2}-\frac{c^2}{b^2}+\frac{x}{y}-\frac{z}{y})=0$
from
$\begin{cases} \frac{x}{y}=\frac{a^2(b^2+c^2-a^2)}{b^2(a^2-b^2+c^2)}\\ \frac{z}{y}=\frac{c^2(a^2+b^2-c^2)}{b^2(a^2-b^2+c^2)} \end{cases}$

[nnosipov]

ins- в сообщении #1475542 писал(а):

What is "Groebner" module? Is it named after Sonhard Groebner?

No, see https://en.wikipedia.org/wiki/Groebner_basis