Problem 1830

(Base on the solution of Hadi Alaithan, IMO 2023's team member)
Note that for $x, y$ are coprime positive integers, we have
$$x - y \mid x^{f(x)} - y^{f(y)} \Leftrightarrow x - y \mid x^{|f(x) - f(y)|} - 1.$$
Now we state the following lemma:
Lemma (Zsigmondy's Theorem): for $a > b \ge 1$ be coprime integers and $n \ge 2$, there exists a prime divisor of $a^n - b^n$ that does not divide $a^k - b^k$ for all $1 \le k < n$, except when $n = 2$ and $a + b$ is a power of $2$ or $(a, b, n) = (2, 1, 6)$.

Back to the original problem, for any coprime integers $x$ and $y$, we show that $|f(x) - f(y)| \le 1$. Suppose that the difference is larger than $1$, then there exists a prime $p$ divides $f(x) - f(y)$. Because $x$ and $y$ are relatively prime, one of these numbers is not a multiple of $p$. Assume that $\gcd(x, p) = 1$. By choosing a suitable number $z$, we will show that this is a contradiction.

Denote $k = v_p(f(x) - f(y))$. By the lemma, for some $N_1 > k$, there exists a prime $q_1$ such that
$$\operatorname{ord}_{q_1}(x) = p^{N_1}, \quad q_1 > x y.$$
Similarly, for some $N_2 > N_1 > k$ such that there exists a prime $q_2 > q_1$ and
$$\operatorname{ord}_{q_2}(y) = p^{N_2}$$
It is obvious that $q_1, q_2, x$ and $y$ are pairwise relatively prime. By Chinese remainder theorem, there exists a positive integer $z$ such that
$$\begin{cases} z \equiv 1 \pmod{x}, & z \equiv x \pmod{q_1}, \\ z \equiv 1 \pmod{y}, & z \equiv y \pmod{q_2}. \end{cases}$$
These conditions give
$$\begin{cases} q_1 \mid (z-x) \mid x^{|f(z) - f(x)|} - 1 \\ q_2 \mid (z-y) \mid y^{|f(z) - f(y)|} - 1. \end{cases}$$
Hence $p^{N_1} \mid f(x) - f(z)$ and $p^{N_2} \mid f(x) - f(y)$. These give $p^{N_1} \mid f(x) - f(y)$. This is a contradiction.

Now, we show that $f(x) = f(y)$ for all coprime $x, y > 1$. Choose $z > 2x, 2y$ such that $\gcd(z, x y) = 1$. Then we have
$$z - x > x - 1, \quad z - y > y - 1.$$
Then both $f(z) - f(x) = 0$ and $f(z) - f(y) = 0$ due to the divisibility. We obtain that $f(x) = f(y)$ if $x$ and $y$ are relatively prime.

In conclusion, $f$ is constant on $\{2, 3, \dots\}$ and $f(1) \in \mathbb{N}$.