INMO 2026 Questions, Solutions, Discussions

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.