Here's my solution.

Attachments:
JBMO_2013.docx (13kb)

P. S. In first system you express y, then x and at the end you get x=y=z.

It is not $$x=y=z$$, for example $$x=1,y=1,z=-1$$Sonnhard.

Excuse me, at last system is: x=-y or x=y and x=-z or x=z. So only answers are (x, y, z)=(1, 1, -1) and every permutation of that answer.

My apologise to Sonnhard.

And (x, y, z)=(-1, -1, -1).

Can you explain it?

Solutions in complex numbers?

There writes just real numbers.

The equations rearrange to \[ x^3yz=z^2-2y^2, \ xy^3z=x^2-2z^2, \ xyz^3=y^2-2x^2. \]Summing gives $xyz(x^2+y^2+z^2)=-(x^2+y^2+z^2)$, so $(xyz+1)(x^2+y^2+z^2)=0$. Hence either $x=y=z=0$ or $xyz=-1$. If $xyz=-1$, then $z=-\frac{1}{xy}$. So the first two equations become \[ -x^2 = \frac{1}{x^2y^2} - 2y^2, \ -y^2 = x^2-\frac{2}{x^2y^2}. \]Multiplying through by $x^2y^2$ gives \[ -x^4y^2=1-2x^2y^4, \ -x^2y^4=x^4y^2-2 \implies x^4y^2=1, \ x^2y^4=1. \]Hence $|x^2y|=1$ and $|xy^2|=1$, so dividing gives $|x/y|=1$, i.e. $|x|=|y|$. Therefore, $|x|=|y|=|z|=1$. The solutions are $(x,y,z)=(0,0,0),(-1,-1,-1),(-1,1,1),(1,-1,1),(1,1,-1)$.

@above, $(0,0,0)$ is not allowed, so I think they are $(x,y,z)=(-1,-1,-1),(-1,1,1),(1,-1,1),(1,1,-1)$ which follows from the fact that \[xyz=-1\]and \[-x^2=z^2-2y^2\]\[-y^2=x^2-2z^2\]\[-z^2=y^2-2x^2\]adding the first two gives $y^2=z^2$ and doing the same thing again gives $x^2=y^2=z^2$ and $xyz=-1$ so that means they are either $3$ or $1$ of them is $-1$.

tenplusten wrote: $\boxed{A1}$ Find all ordered triplets of $(x,y,z)$ real numbers that satisfy the following system of equation $x^3=\frac{z}{y}-\frac {2y}{z}$ $y^3=\frac{x}{z}-\frac{2z}{x}$ $z^3=\frac{y}{x}-\frac{2x}{y}$ $x^3=\frac{z}{y}-\frac {2y}{z}$, we multiply this by $yz$ we get $x^3yz=z^2-2y^2$, similarly with others we get $xy^3z=x^2-2z^2$ $xyz^3=y^2-2x^2$ Now summing them we get $xyz(x^2+y^2+z^2)=-(x^2+y^2-z^2)$ $xyz=-1$ which means that $x^3yz=-x^2$ $x^3yz=z^2-2y^2$ $-x^2=z^2-2y^2$ or $x^2=2y^2-z^2$...(1) similarly we have $y^2=x^2-2z^2$ Putting ...(1) here we get that $y^2=z^2$ similarly we get that $x^2=y^2=z^2$ and also we have $xyz=-1$. so the set of solutions are $(x,y,z)=(0,0,0),(-1,-1,-1),(-1,1,1),(1,-1,1),(1,1,-1)$

The idea is to rewrite $$x^3 = \frac zy - \frac {2y}z \iff x^3yz = z^2-2y^2.$$Summing cyclically, $$xyz(x^2+y^2+z^2) = -(x^2+y^2+z^2) \implies xyz=-1.$$Substituting this back in, each of our cyclic equations becomes $$-x^2 = z^2 - 2y^2 \implies x^2+z^2=2y^2.$$Combining this with cyclic permutations gives us $x^2=y^2=z^2$, which means that the answers are $\boxed{(-1, -1, -1)}$ and $\boxed{(-1, 1, 1)}$, including cyclic permutations.

We will start off by dividing the first equation by the second equation for this problem and we get that $x^3(\frac xz -\frac {2z}{x} = y^3(\frac zy - \frac{2y}{z}$ and after some rearrangment we get that $x^4+2y^4=2x^2z^2+y^2z^2$ if we now add the cyclic variants this tells us that $x^2=y^2=z^2$ and this gives us that the only solutions are $\boxed{(-1, -1, -1)}$ and $\boxed{(-1, 1, 1)}$ and this includes the cyclic permutations of these solutions.

The system is also quite readily be solved in the complex numbers as well: Here, two $\pm$ signs mean both signs are the same in any solution, while a $\pm$ and $\mp$ used together means the two signs are opposite in any solution. Real: $(-1,-1,-1)$ and permutations of $(1,1,-1)$ Complex: $\left(e^{\pm\pi i/3},e^{\pm\pi i/3},e^{\pm\pi i/3}\right)$ Cyclic permutations of $\left(e^{\pm\pi i/3},e^{\mp 2\pi i/3},e^{\mp 2\pi i/3}\right)$ $\left(\sqrt[3]{\frac{1\pm 3i\sqrt{3}}{2}},e^{\pm 2\pi i/3}\sqrt[3]{\frac{1\pm 3i\sqrt{3}}{2}},e^{\mp 2\pi i/3}\sqrt[3]{\frac{1\pm 3i\sqrt{3}}{2}}\right)$ $\left(\sqrt[3]{\frac{1\pm 3i\sqrt{3}}{2}},-e^{\pm 2\pi i/3}\sqrt[3]{\frac{1\pm 3i\sqrt{3}}{2}},-e^{\mp 2\pi i/3}\sqrt[3]{\frac{1\pm 3i\sqrt{3}}{2}}\right)$ $\left(-\sqrt[3]{\frac{1\pm 3i\sqrt{3}}{2}},e^{\pm 2\pi i/3}\sqrt[3]{\frac{1\pm 3i\sqrt{3}}{2}},-e^{\mp 2\pi i/3}\sqrt[3]{\frac{1\pm 3i\sqrt{3}}{2}}\right)$ $\left(-\sqrt[3]{\frac{1\pm 3i\sqrt{3}}{2}},-e^{\pm 2\pi i/3}\sqrt[3]{\frac{1\pm 3i\sqrt{3}}{2}},e^{\mp 2\pi i/3}\sqrt[3]{\frac{1\pm 3i\sqrt{3}}{2}}\right)$

This is my sol

Attachments:

Let's clear the denominators first, to get the equivalent system \begin{align*} x^3yz&=z^2-2y^2 \\ xy^3z&=x^2-2z^2 \\ xyz^3&=y^2-2x^2. \end{align*}Summing them, we get \[xyz(x^2+y^2+z^2)=-(x^2+y^2+z^2)\]so we can factor as \[(xyz+1)(x^2+y^2+z^2)=0.\]Note that since $x$, $y$, $z$ are each nonzero, $x^2+y^2+z^2>0$, therefore $xyz=-1$. So our system of equations is now \begin{align*}y^2+z^2&=2x^2 \\ z^2+x^2&=2y^2 \\ x^2+y^2&=2z^2\end{align*}so since $y^2+z^2=2x^2$ and $z^2+x^2=2y^2$, $y^2-x^2=2x^2-2y^2$, so $x=y$ or $x=-y$. Similarly, $x=z$ or $x=-z$. Now, you can substitute these equalities back in to obtain $(-1, -1, -1)$, and the permutations of $(1, 1, -1)$ as the only solutions. $\blacksquare$

Clearing the denominators in the equations gives us \[x^3yz+2y^2-z^2=0\] and the same applies for cyclic variations. Adding them up, we get \[(x^2+y^2+z^2)xyz+(x^2+y^2+z^2)=0\] implying $xyz=-1$. Then, we see that $x^2=y^2=z^2$, so \[(x,y,z)=\boxed{(-1,-1,-1), (1,1,-1) \text{ and permutations}}\]are the only solutions.

The solutions are $(x,y,z)=(-1,-1,-1)$ and $(1,1,-1)$ along with their permutations. Clearly none of them can be $0$. Firstly we get that $x^3 = \dfrac{z}{y} - \dfrac{2y}{z} = \dfrac{z^2-2y^2}{yz}\implies x^2 = \dfrac{z^2 - 2y^2}{xyz}$. Now adding all the cyclic iterations, we get that $xyz = -1$. Thus we now have $x^2 = 2y^2 - z^2 \implies x^2 + z^2 = 2y^2$. Cyclically we also get $y^2 + x^2 = 2z^2$. Subtracting this from the first one, we get $y = \pm z$. In this way, we get that all their absolute values are equal. Taking modulus on both sides of $xyz = -1$, we get that $x,y,z \in \left\{+1,-1\right\}$. Now some case checking gives that $(-1,-1,-1)$ and $(1,1,-1)$ along with their permutations work.

The only solutions are $(x, y, z) = (1, 1, -1)$ and permutations, as well as $(-1, -1, -1)$, which can be checked to work. Multiplying the first equation and rearranging yields $x^3yz + 2y^2 - z^2 = 0$. Similarly $xy^3z + 2z^2 - x^2 = 0$ and $xyz^3 + 2x^2 - y^2 = 0$, so adding these up gives $$x^3yz + xy^3z + xyz^3 + x^2 + y^2 + z^2 = 0 \implies (x^2 + y^2 + z^2)(1 + xyz) = 0,$$so $xyz = -1$ (as other $x = y = z = 0$ and we divide by $0$). Substituting $yz = -\tfrac{1}{x}$ into two of our earlier equations gives $-x^2 + 2y^2 - z^2 = 0$ and $-y^2 + 2z^2 - x^2 = 0$, so $3y^2 = 3z^2$ and $|y| = |z|$. Similarly $|z| = |x|$, so from $xyz = -1$ we get $|x| = |y| = |z| = 1$ and we can therefore extract $(1, 1, -1)$ and permutations and $(-1, -,1 -1)$.

The first equation yields $\newline$ \[x^3yz + 2y^2 - z^2 = 0\]And cyclically summing for the other equations gives \[(xyz + 1)(x^2 + y^2 + z^2) = 0\]which implies that $xyz = -1$. $\newline$ Then by substitution($x = -\frac{1}{yz})$ we get \[-x^2 + 2y^2 - z^2 = 0 \implies x^2 + z^2 = 2y^2\]and cyclic varations. $\newline$ This implies that $x^2 = y^2 = z^2$ and combined with the fact that $xyz = -1$, we get $(x, y, z) = (-1, -1, -1)$, and permutations of $(-1, 1, 1)$.