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

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

Distinction between polynomial and polynomial function Here, we explain why it is necessary to make this distinction, starting by defining a polynomial in another way. Here, $\mathbb{K}=\mathbb{Q}, \mathbb{R}, \mathbb{C}$ or $\mathbb{Z} / p \mathbb{Z}$ equipped with the usual addition and multiplication laws.

Definition. A polynomial with coefficients in $\mathbb{K}$ is an infinite sequence of elements of $\mathbb{K}$ that is zero from a certain rank.

Example. For example, $(0,1,2,3,0,0, \ldots, 0, \ldots)$ is a polynomial, as is $(0,0, \ldots, 0, \ldots)$. However, $(1,1, \ldots, 1, \ldots)$ is not one.

Definition. Let $P=\left(u_{n}\right)_{n}$ and $Q=\left(v_{n}\right)_{n}$ be two polynomials. We define the polynomial $P+Q$ by the sequence $w_{n}=u_{n}+v_{n}$ (which is indeed zero from a certain rank) and the polynomial $P \times Q$ by the sequence $\left(z_{n}\right)$, where $z_{n}=\sum_{i+j=n} u_{n} v_{n}$ (verify that $\left(z_{n}\right)$ is zero from a certain rank). We identify the elements of $\mathbb{K}$ with the constant polynomials via the application that associates to an element $\lambda \in \mathbb{K}$ the polynomial $(\lambda, 0,0, \ldots, 0, \ldots)$. Note that this is consistent with the intuitive notion of multiplying a polynomial by an element of $\mathbb{K}$: if $\left(u_{n}\right)$ is a polynomial and $\lambda \in \mathbb{K}$, then the polynomial $\lambda \times\left(u_{n}\right)$ is the polynomial $\left(\lambda u_{n}\right)$.

We now introduce the indeterminate $X$.

Definition. Let us denote by $X$ the polynomial $(0,1,0,0, \ldots)$.

Proposition. Every polynomial $P$ can be expressed in the form $P=\sum_{i=0}^{n} a_{n} X^{n}$. We denote $P$ or $P(X)$ indifferently to remind that we denote $X$ the indeterminate (we could very well denote it $Y!$ ).

Proof. If $P=\left(a_{0}, a_{1}, a_{2}, \ldots\right)$, let $N$ be an integer such that $i \geq N$ implies $a_{i}=0$. Then $P(X)=a_{0}+a_{1} X+\cdots+a_{N} x^{N}$. This is an immediate consequence of the definition of $X$ and the multiplication between polynomials.

Here is now the link between polynomial and associated polynomial function. Recall that, for now, a polynomial is just a sequence of numbers that is zero from a certain rank and is not seen as an application.

Proposition. Let $P(X)=a_{0}+a_{1} X+\cdots+a_{n} X^{n} \in \mathbb{K}[X]$ be a polynomial. We denote by $\widetilde{P}$ the application defined by $\widetilde{P}(x)=a_{0}+a_{1} x+\cdots+a_{n} x^{n}$ for $x \in \mathbb{K}$, which we call the polynomial function associated with $P$. The application $P \mapsto \widetilde{P}$ is injective if $\mathbb{K}$ is infinite. If $\mathbb{K}$ is finite, this application is not necessarily injective.

Proof. First, consider the case where $\mathbb{K}$ is infinite. Let $P, Q \in \mathbb{K}[X]$ such that $\widetilde{P}=\widetilde{Q}$. Write $P(X)=\sum_{i} a_{i} X^{i}$ and $Q(X)=\sum_{i} b_{i} X^{i}$. Then the polynomial $P(x)-Q(x)$, in the sense of the previous sections, has an infinite number of roots, so it is zero. Therefore, $a_{i}=b_{i}$ for all $i$.

However, in the case where $\mathbb{K}$ is finite, the previous reasoning does not apply. Moreover, let us exhibit a counterexample. Consider $\mathbb{K}=\mathbb{Z} / p \mathbb{Z}$ and $P(X)=X^{p}-X$. According to Fermat's little theorem, for all $x \in \mathbb{K}$, we have $P(x)=0$. Thus, $P$ is not the zero polynomial, but the two associated polynomial functions are the same.

In other words, when $\mathbb{K}$ is infinite (for example, $\mathbb{K}=\mathbb{Q}, \mathbb{R}, \mathbb{C}$, which explains why we did not lose generality in the first sections), we can talk without distinction about a polynomial or the associated polynomial function. However, in other cases, one must be very careful!

Elements of the answer to the exercises
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, $x=y=z=1$.

## 2 Thursday afternoon, February 21: Guillaume Conchon-Kerjan

## Exercises

Warning: The present sheet has a purely mathematical interest and is not intended to provide dietary advice.

## Aperitif