The set $\mathbb{Z}$ is clearly beautiful and strong. We will prove that it is the only one. To do this, consider a set $S$ that is beautiful and strong: we will actually prove, by strong induction on $n$, that the integers $n$ and $-n$ necessarily belong to $S$.
First, since $S$ is beautiful, it contains the integers $2^{1}-2^{1}=0, 2^{2}-2^{1}=2$, and $2^{1}-2^{2}=-2$. It also contains the integers 1 and -1, which are roots of the polynomials $2-2X$ and $2+2X$, respectively.
We now consider an integer $n \geqslant 3$ such that $-n-1, \ldots, n-1$ all belong to $S$. Let $\alpha$ be the 2-adic valuation of $n$, and $m$ be the odd integer such that $n=2^{\alpha} m$. By noting $\varphi(m)$ as the Euler's totient function of $m$, we observe that the integer $k=2^{\alpha+\varphi(m)+1}-2^{\alpha+1}$, which clearly belongs to $S$, is also a multiple of $2^{\alpha}$ and $m$, and thus of $n$.
Let $\overline{a_{\ell} a_{\ell-1} \ldots a_{0}}$ be the base-$n$ representation of $k / n$. All the integers $\pm a_{0}, \ldots, \pm a_{\ell}$ are between $1-n$ and $n-1$, and thus belong to $S$. By construction, $n$ is an integer root of the polynomial $P(X)=k-\sum_{i=0}^{\ell} a_{i} X^{i+1}$, whose coefficients are all in $S$, so $n$ is in $S$ as well. Similarly, $-n$ is an integer root of the polynomial $Q(X)=k-\sum_{i=0}^{\ell} a_{i}(-X)^{i+1}$, so $-n \in S$, which concludes the induction and the proof.

Remark: Given the statement, one could suspect that using polynomials of degree $d \geqslant 2$ might be useful, otherwise the creators of the statement would have directly chosen to define strong sets as sets stable under division. The remark below is mainly intended to illustrate the fact that if one seeks to drastically simplify a hypothesis in a mathematical statement, here with the goal of showing that $\mathbb{Z}$ would be the only beautiful set stable under division, it is very important to look for simple constructions that could invalidate this simplification.
We will in fact prove that there exists a set $S$ that is beautiful and stable under division, but different from $\mathbb{Z}$: this demonstrates that using polynomials of degree $d \geqslant 2$ was actually necessary to solve this problem. To construct this set, we need Zsygmondy's theorem and the known property that there exist Fermat numbers (i.e., integers of the form $2^{p}-1$) that are not prime, even if $p$ is large enough (below, we will need the inequality $p \geqslant 7$). For example, when $p=11$, $2^{11}-1=23 \times 89$. Indeed, write $2^{p}-1$ as the product $2^{p}-1=q \times r \times m$, where $q$ and $r$ are two primes, and $m$ is any integer. We then consider the set $S$ consisting of 0 and integers of the form

$$\pm 2^{a} \prod_{k \geqslant 2}\left(2^{k}-1\right)^{\alpha_{k}}$$

where the $\alpha_{k}$ are integers, only finitely many of which are non-zero. The set $S$ is clearly beautiful and stable under division.
Suppose now that $q \in S$: we then write $q$ as

$$q=\prod_{k \geqslant 2}\left(2^{k}-1\right)^{\alpha_{k}}$$

and let $\ell$ be the maximal index such that $\alpha_{\ell} \neq 0$. Since the order of 2 modulo $q$ and $r$ divides $p$, it is equal to $p$, so $\ell \geqslant p \geqslant 7$.

Zsygmondy's theorem then indicates that there exists a prime $s$ that divides $2^{\ell}-1$ and no integer $2^{k}-1$ for $1 \leqslant k \leqslant \ell-1$. We then note that

$$v_{s}(q)=\sum_{k \geqslant 2} \alpha_{k} v_{s}\left(2^{k}-1\right)=\alpha_{\ell} v_{s}\left(2^{\ell}-1\right) \neq 0 .$$

We deduce that $s=q$, so $\ell=p$, and that $v_{q}\left(2^{\ell}-1\right)=1$. This means in particular that $r \neq q$. But then, even for $\ell=p$, and instead of choosing $s=q$, we could have satisfied Zsygmondy's theorem by choosing $s=r$, leading to a contradiction. We therefore conclude that $q \notin S$, and thus that $S \neq \mathbb{Z}$.

Comment from the graders: This number theory problem was quite difficult as it required a trick: the simplest solution involved the base-$b$ decomposition of an integer. Only five students thought of this and they all received the maximum score. The grading scale valued this trick, making it impossible to score more than 3 if this base decomposition was not mentioned.
Almost all students realized that the only beautiful and strong set would be $\mathbb{Z}$, without necessarily knowing how to prove it.
Many students tried to show that every integer (sometimes restricted to odd or prime integers) had a multiple in $S$, which was essential for the subsequent steps. Many also showed that $S$ was symmetric with respect to 0 (i.e., if $n \in S$, then $-n \in S$), which allows avoiding dealing with negative integers in the subsequent steps.
Let's finally address some common errors:
$\triangleright$ Managing negative integers sometimes led to a significant error, especially when students attempted to prove by induction that $\llbracket 1, n \rrbracket \subseteq S$ for all $n \geqslant 1$. Some students introduced an integer $k \geqslant 1$ such that $n k \in S$, and they claimed that $n-k$ belonged to $S$ by induction hypothesis as an integer strictly less than $n$. However, these students did not rule out the case where $k \geqslant 2 n$, so it was possible to have $n-k \leqslant -n$, thus undermining their entire reasoning.
$\triangleright$ Some students noted that it sufficed for all primes to be in $S$ to conclude. This is true, but no known solution immediately proves this result. In this case, several students thought they had solved the problem by showing this result. They based their reasoning on the belief (erroneous, as indicated in the remark above) that if $n$ and $m$ are two distinct odd natural numbers, then the order of 2 modulo $n$ is different from the order of 2 modulo $m$. This error is equivalent to the misconception of Zsigmondy's theorem, which states that for $n \notin\{1,6\}$, $2^{n}-1$ has exactly one primitive prime divisor.
$\triangleright$ Finally, vague arguments like "we construct more and more integers by taking the integer roots of polynomials with coefficients in an increasingly large set" can help form an idea, but obviously do not earn any points.
It is worth noting that in a problem like this, explicitly verifying that small values of $n$ (e.g., $n=0, n=1$, and $n=2$) are in $S$ always earns a point, yet some students did not have this reflex.
Finally, many students proved that if $a b \in S$ and $a \in S$ is non-zero, then $b \in S$. This result, while correct, is misleading because it invites the use of only the degree 1 polynomials in the statement, which are clearly insufficient to conclude, as indicated in the remark above.
! if the equality

$$f(f(x+y)+y)=f(f(x)+y)$$

is satisfied for all integers $x$ and $y$. An integer $v$ is said to be $f$-rare if the set of integers $x$ such that $f(x)=v$ is a finite and non-empty set.
a) Prove that there exists a Russian function $f$ for which there exists an $f$-rare integer.
b) Prove that for any Russian function $f$, there is at most one $f$-rare integer.