https://onemperu.files.wordpress.com/2021/05/selectivos-imo.pdf

Hopefully, this works Let $P(a,b)$ denote the given assertion. $P(1,b)$ gives that $f(1)^{\text{something}} \le 1$ and so $f(1) = 1$ $P(a,1)$ gives that $f(a) \le a$. So in particular, $f(2) \le 2$ So now, consider the following two cases Case 1: $f(2) = 1$ $P(a,2)$ gives that $a \ge f(a)^{2f(4)} \ge f(a)^2$. Putting $a=3$, we get $3 \ge f(3)^2$ and so $f(3) = 1$. Now, induct, suppose $f(x) = 1$ for all $1 \le x \le n$ and we need to prove $f(n+1) = 1$ $P(n+1,n)$ gives $n+1 \ge f(n+1)^{nf(n)^2} \ge f(n+1)^n$. If $f(n+1) > 1$, then we get $n+1 \ge f(n+1)^n \ge 2^n$, which is impossible for $n \ge 2$. So, by induction, we get that $f(x) = 1$ for all $x \in \mathbb{N}$ Case 2: $f(2) = 2$ $P(2,b)$ gives that $2^{f(b)^3} \ge 2^{bf(b^2)}$ which gives $f(b)^3 \ge bf(b^2) \implies f(b^2) \le \frac{f(b)^3}{b}$ Suppose $f(k) \le k-1$ for some $k$. Then, we have that $f(k^2) \le \frac{(k-1)^3}{k}$ and $f(k^4) \le \frac{(k-1)^9}{k^5}$ and similarly, we can get that $f(k^{2^n}) \le \frac{(k-1)^{3^n}}{k^{3^n-2^n}}$ Now, I claim that $\frac{(k-1)^{3^n}}{k^{3^n-2^n}} < 1$ for sufficiently large $n$, which will mean that $f(k^{2^n}) < 1$, which is impossible. To do this, it suffices to prove that $\left(\frac{b}{b-1}\right)^{3^n} > b^{2^n}$ for sufficiently large $n$. Suppose $p$ is the smallest number such that $\left(\frac{b}{b-1}\right)^p > b$, which obviously exists. Now, pick a large $n$ such that $\left(\frac{3}{2}\right)^n > p$, such an $n$ obviously exists. Then, its easy to see that this $n$ works for the claim. So this means that we have proved that it is impossible for $f(x) \le x-1$ and so $f(x) \ge x$ but since we had $f(x) \le x$, this means that $f(x) = x$ for all $x$ So, the only possible functions are $f(x) = 1$ and $f(x) = x$ for all $x \in \mathbb{N}$