Find all $(x, y, z)$ such that:

$$
\left\{\begin{array}{c}
x+y+z=\frac{1}{x}+\frac{1}{y}+{ }^{\prime} \frac{1}{z} \\
x^{2}+y^{2}+z^{2}=\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}
\end{array}\right.
$$

Lemma 2. Let $P$ be a polynomial with integer coefficients, $a, b$ two distinct integers, then $a-b \mid P(a)-P(b)$

Proof. Let's prove the lemma on each polynomial $X^{n}$. Let $a, b$ be distinct integers. For $n=0, a-b \mid 1-1=0$. For $n=1, a-b \mid a-b$. Let $n \geqslant 2$, then $a^{n}-b^{n}=(a-b)\left(a^{n-1}+a^{n-2} b+\ldots+a b^{n-2}+b^{n-1}\right)$. Thus, $a-b \mid a^{n}-b^{n}$. Since a polynomial is a linear combination of monomials, the property holds for any polynomial.
Let $\sigma_{1}=x+y+z, \sigma_{2}=x y+y z+z x, \sigma_{3}=x y z$. By multiplying the first by $\sigma_{3}$, we get $\sigma_{1} \sigma_{3}=\sigma_{2}$. By multiplying the second by $\sigma_{3}^{2}$,

$\sigma_{3}^{2}\left(\sigma_{1}^{2}-2 \sigma_{2}\right)=\sigma_{2}^{2}-2 \sigma_{1} \sigma_{3} \Longleftrightarrow \sigma_{3}^{2}\left(\sigma_{1}^{2}-2 \sigma_{1} \sigma_{3}\right)=\sigma_{1}^{2} \sigma_{3}^{2}-2 \sigma_{1} \sigma_{3} \Longleftrightarrow \sigma_{3}^{2}=1 \Longleftrightarrow \sigma_{3}=1$ or $\sigma_{3}=-1$

Assume that $\sigma_{3}=1, \sigma_{2}=\sigma_{1}=\lambda$. Then $x, y, z$ are roots of $X^{3}-\lambda X^{2}+\lambda X-1=$ $(X-1)\left(X^{2}-(\lambda+1) X+1\right)$

By permutation, $x=1, z=\frac{1}{y}$, according to Viète's formula. It is easy to verify that this is a solution. If $\sigma_{3}=-1$, the polynomial is $X^{3}+\lambda X^{2}+\lambda X+1$ and we find $x=-1, y=\frac{1}{z}$. Thus, by permutation, $(x, y, z)=\left(1, t, \frac{1}{t}\right)$ or $\left(-1, t, \frac{1}{t}\right)$ for $t \in \mathbb{R}^{*}$

## Second Solution

We resume from the beginning up to $\sigma_{3}^{2}=1 \Longleftrightarrow \sigma_{3}=\frac{1}{\sigma_{3}}$.

$$
\begin{aligned}
(x+y+z)^{2}=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^{2} & \Longleftrightarrow x^{2}+y^{2}+z^{2}+2(x y+y z+z x)=\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}+2\left(\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}\right) \\
& \Longleftrightarrow x y+y z+z x=\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}
\end{aligned}
$$

Thus $(X-x)(X-y)(X-z)=\left(X-\frac{1}{x}\right)\left(X-\frac{1}{y}\right)\left(X-\frac{1}{z}\right)$. Therefore, $\{x, y, z\}=\left\{\frac{1}{x}, \frac{1}{y}, \frac{1}{z}\right\}$. There are 3 cases to test: $x=\frac{1}{x}, y=\frac{1}{y}, z=\frac{1}{z} \Longleftrightarrow x, y, z= \pm 1$, which is a solution. If $x=\frac{1}{x}, y=\frac{1}{z}, z=\frac{1}{y^{\prime}}$, we find $\left(1, t, \frac{1}{t}\right), \forall t \in \mathbb{R}^{*}$ which is a solution. Finally, if $x=\frac{1}{y}, y=\frac{1}{z}, z=\frac{1}{x}$, by substituting all the equations, we fall back to case 1.

Thus, by permutation, $(x, y, z) \in\left\{\left(e, t, \frac{1}{t}\right), e \in\{-1,1\}, t \in \mathbb{R}^{*}\right\}$