https://aops.com/community/p11412122 liekkas wrote: It's easy to derive that $f$ is injective. Induct to prove $f(n)=n$. Suppose we have prove it for $1,2,c\dots,n-1$, consider $n$ (the case $n=1$ also satisfies) By the injectivity, $f(n) \ge n$. But $P(1,n)$ implies $1+f(n) \mid 2n+2$ thus $f(n)=n$ or $f(n)=2n+1$. If $f(n)=2n+1$, by $P(1,2n+1)$ and $P(2,n)$ we get $1+f(2n+1) \mid 4n+4$, $2+f(2n+1) \mid 2n+4$. With $f(2n+1) \ge n$, we get $f(2n+1)=n$. But then $P(3,n)$ fails. Thus $f(n)=n$.