, 2023

Answer: Yes, there exists.

*Observation 1:* If $p$ is the smallest prime divisor of $n = px$, then $f(n) = n-x = x(p-1)$.

*Observation 2:* Let $p$ be a prime number and $x$ be a positive integer. If $x=1$ or none of the prime divisors of $x$ is less than $p$, then $f(xp) = x(p-1)$.

*Lemma 1:* Let $r$ be a positive integer. Then there exist positive integers $x_1 > x_2 > \dots > x_r$ such that $f(x_1) = f(x_2) = \dots = f(x_r)$

and $x_r$ is prime.

*Proof:* We will prove the claim by induction on $r$. The case $r=1$ is trivial. Assume that the claim is true for $r \ge 1$. Thus, there exist such $x_1, \dots, x_r$. For $i=1, \dots, r$ let $p_i$ be the smallest prime divisor of $x_i$ and $x_i = p_i y_i$. Then by Observation 1 we have
$$y_1(p_1-1) = y_2(p_2-1) = \dots = y_{r-1}(p_{r-1}-1) = p_r-1$$
and $p_1 < p_2 < \dots < p_r$. Let $A$ be the product of all prime numbers less than $p_r$. Therefore, we have $\gcd(A(p_r-1), p_r) = 1$ and thus, by Dirichlet's theorem there exists a prime number $q > p_r$ of the form $A(p_r-1)x + p_r$. Then, $q-1 = A(p_r-1)x + p_r-1 = (p_r-1)(Ax+1)$. Note that $B = Ax+1$ has no prime divisor less than $p_r$. Thus, we get $B = Ax+1$ for each $i=1, \dots, r$ and hence, $Bx_1 > Bx_2 > \dots > Bx_r > q$ satisfy the conditions for $r+1$.

*Lemma 2:* Let $q_1 < q_2 < \dots < q_k$ be prime numbers and $m = \prod_{i=1}^k q_i^{\alpha_i}$. If $f(n) = m$, then $n = 2m$ or $n = m+1$ is a prime number or there exists a $1 \le j \le k-1$ such that $1 + \prod_{i=1}^j q_i^{\alpha_i}$ is a prime number less than $q_{j+1}$ and it is the smallest prime divisor of $n$.

*Proof:* Let $p$ be the smallest prime divisor of $n$ and $n = px$. Then $x(p-1) = \prod_{i=1}^k q_i^{\alpha_i}$. If $x=1$, then $p-1=m$ and $n=p=m+1$ is a prime number. If $p=2$, then $x=m$ and $n=2m$. Now, let $p>2$ and $x>1$. By the observations above we see that $x$ and $p-1$ are relatively prime. Therefore, for every $i=1, \dots, k$ $q_i^{\alpha_i}$ divides exactly one of $x$ and $p-1$. Moreover, if $q_s|p-1$, then $p > q_s$ and none of $q_1, \dots, q_s$ divides $x$. So, for the largest $j$ satisfying $q_j|p-1$ (there exists such $j$ and $j \ne k$)) we obtain $p-1 = \prod_{i=1}^j q_i^{\alpha_i}$ and $q_{j+1} < p$.

By Lemma 1 there exists a positive integer $m = \prod_{i=1}^k q_i^{\alpha_i}$ with $q_1 < q_2 < \dots < q_k$ such that $f(n) = m$ has exactly $r \ge 2023$ solutions. Then by Lemma 2 we see that

$$f(n) = m/q_k^{\alpha_k} = \prod_{i=1}^{k-1} q_i^{\alpha_i}$$
has either *r* or *r* - 1 solutions. Indeed, if *m* + 1 is prime then the number of solutions decreases by 1, otherwise it remains the same. Consequently, we reach a number *m'* where *f*(*n*) = *m'* has exactly 2023 solutions.