Find all real numbers $x, y, z$ such that

$$
x+y+z=3, \quad x^{2}+y^{2}+z^{2}=3, \quad x^{3}+y^{3}+z^{3}=3
$$

## - Polynomials with integer coefficients -

We now present some specific properties of polynomials with integer coefficients:

* We have already seen that if $P, Q \in \mathbb{Z}[X]$ and $\operatorname{deg} Q \geqslant 1$, we can always perform the Euclidean division of $P$ by $Q$ provided that the leading coefficient of $Q$ is equal to 1.
* An extremely useful property is that if $P \in \mathbb{Z}[X]$, then for all integers $a \neq b, a-b$ divides $P(a)-P(b)$. This is a simple consequence of the fact that $a-b$ divides $a^{n}-b^{n}$ for $n \geqslant 1$.

$\star$ If $P(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0} \in \mathbb{Z}[X]$, and if $p / q$ is a rational root of $P$ in irreducible form, then $p$ divides $a_{0}$ and $q$ divides $a_{n}$. This simple fact allows us to restrict the search for rational roots of a polynomial with integer coefficients.

* A polynomial $P \in \mathbb{Q}[X]$ satisfies $P(k) \in \mathbb{Z}$ for all $k \in \mathbb{Z}$ if and only if there exist integers $a_{0}, a_{1}, \ldots a_{n}$ such that

$$
P(x)=a_{0}\binom{x}{0}+a_{1}\binom{x}{1}+\cdots+a_{n}\binom{x}{n}
$$

where we denote $\binom{x}{k}=\frac{x(x-1) \cdots(x-k+1)}{k!}$ if $k \neq 0$ and $\binom{x}{0}=1$. This property follows from Exercise $10(2)$.

Example 3.42. Let's find all real numbers $x, y, z$ such that

$$
x+y+z=17, \quad x y+y z+x z=94, \quad x y z=168
$$

According to Viète's formulas, $x, y, z$ are roots of $P(x)=x^{3}-17 x^{2}+94 x-168=0$. Let's look for "obvious" roots of $P$. It is natural to first look for integer roots, which must be divisors of 168 (to test if, for example, 2 is a root, we perform the Euclidean division of $P$ by $x-2$ and check if the remainder is zero). We notice that $x=4$ is a root, and without difficulty, we find that $x=6$ and $x=7$ are roots of the polynomial $P(x) /(x-4)$. Thus, the solutions $(x, y, z)$ are the six possible permutations of $(4,6,7)$.
We write Newton's relations:

$$
S_{1}-\sigma_{1}=0, \quad S_{2}-\sigma_{1} S_{1}+2 \sigma_{2}=0, \quad S_{3}-\sigma_{1} S_{2}+\sigma_{2} S_{1}-3 \sigma_{3}=0
$$

Thus, $\sigma_{1}=3, \sigma_{2}=3, \sigma_{3}=1$. From this, we deduce that $x, y, z$ are roots of $t^{3}-3 t^{2}+3 t-1=0$. Now, $t^{3}-3 t^{2}+3 t-1=(t-1)^{3}$. Therefore, $x=y=z=1$.