MOPSS, 31st January, 2026

Let \(n \geq 3\) be an integer. Note that the integers \[ p_1p_2 \ldots p_{n - 1} - 1, 2p_1p_2 \ldots p_{n - 1} - 1, 3p_1p_2 \ldots p_{n - 1} - 1, \ldots, p_n p_1p_2 \ldots p_{n - 1} - 1 \] are pairwise relatively prime. Indeed, if a prime \(p\) divides two of them, say \(ap_1p_2 \ldots p_{n - 1} - 1\) and \(bp_1p_2 \ldots p_{n - 1} - 1\) with \(1 \leq a < b \leq p_n\), then \(p\) divides their difference \((b - a)p_1p_2 \ldots p_{n - 1}\). Since \(p\) cannot be any of the primes \(p_1, p_2, \ldots, p_{n - 1}\), it follows that \(p\) divides \(b - a\). Since \(1 \leq b - a < p_n\), we conclude that \(p < p_n\), that is, \(p\) is one of the primes \(p_1, p_2, \ldots, p_{n - 1}\). However, this is impossible since none of these primes divides \(ap_1p_2 \ldots p_{n - 1} - 1\). Thus, the numbers \[ p_1p_2 \ldots p_{n - 1} - 1, 2p_1p_2 \ldots p_{n - 1} - 1, 3p_1p_2 \ldots p_{n - 1} - 1, \ldots, p_n p_1p_2 \ldots p_{n - 1} - 1 \] are pairwise relatively prime. Since \(n \geq 3\), we obtain \[ p_1p_2 \ldots p_{n - 1} - 1 \geq 2 p_{n - 1} - 1 > p_{n - 1} . \] This implies that the numbers

\(\begin{align*} & p_1, p_2, \ldots, p_{n - 1}, \\ & p_1p_2 \ldots p_{n - 1} - 1, 2p_1p_2 \ldots p_{n - 1} - 1, \\ & 3p_1p_2 \ldots p_{n - 1} - 1, \ldots, p_n p_1p_2 \ldots p_{n - 1} - 1 , \end{align*} \\)

are all distinct and greater than \(1\). Moreover, these \(p_n + n - 1\) numbers are pairwise relatively prime. Therefore, each of them has at least one prime divisor, which does not divide any of the other numbers. Since all these numbers are less than or equal to \(p_1p_2 \ldots p_n\), we conclude that there are at least \(p_n + n - 1\) distinct prime numbers that do not exceed \(p_1p_2 \ldots p_n\).

Note that \(f(n) \leq f(n + 1)\) holds for any \(n \geq 2027\), which shows that \(f\vert_{\{2027, 2028, \ldots\}}\) is a non-decreasing function. For any \(m \geq 2027\), note that \(f(m)\) is a positive integer. Observe that for any \(m \geq 2027\), if \[ f(m), f(m + 1), f(m + 2), \dots, f(m + f(m)) \] are all equal, then \(f(m + f(m) + 1)\) is greater than or equal to \(f(m) + 1\). This shows that \[ f(m) \geq 1, \quad f(m) < f(m + f(m) + 1) \] hold for any \(m \geq 2027\). This shows that \(f\) takes arbitrarily large values.

Claim. Let \(r \geq 2027\) be an integer such that \[ f(r) > \max \{f(1), f(2), \ldots, f(r - 1)\} . \] Then \(f\) takes equal values at the integers \[ r, r + 1, r + 2, \dots, r + f(r) \] and \(f(r + f(r) + 1) = f(r) + 1\).

Proof of the Claim.
The maximum number of times that a number appears in the list \(f(1), f(2), \dots, f(r - 1)\) is equal to \(f(r)\). Using the inequality \(f(r) > \max \{f(1), f(2), \ldots, f(r - 1)\}\), we see that any such number is less than \(f(r)\). Therefore, the numbers \(f(r), f(r + 1), \dots, f(r + f(r))\) are equal. It follows that \(f(r + f(r) + 1) = f(r) + 1\). This completes the proof of the Claim.

Since \(f\) takes arbitrarily large values, there exists an integer \(r \geq 2027\) such that \(f(r) > \max \{f(1), f(2), \ldots, f(r - 1)\}\). Using the Claim and applying induction, it follows that the images of \(r, r + 1, r + 2, \dots\) under \(f\) are equal to

\(\begin{align*} & t, t, \dots, t, & \quad (t \text{ occurs } t + 1 \text{ times}), \\ & t + 1, t + 1, \dots, t + 1, & \quad (t + 1 \text{ occurs } t + 2 \text{ times}), \\ & t + 2, t + 2, \dots, t + 2, & \quad (t + 2 \text{ occurs } t + 3 \text{ times}), \\ & \vdots \end{align*} \\)

respectively, where \(t = f(r)\). This implies that \(f(n) = f(n + f(n))\) for infinitely many \(n \in \mathbb{N}\), as desired.