first prove that $ f(3^{n}) = 2\times 3^{n}$ for $ n\geq 0$... then we have: $ f(f(3^{n})) = f(2\times 3^{n}) = 3^{n+1}$ thus: $ \boxed{f(3^{n}) = 2\times 3^{n},f(2\times 3^{n}) = 3^{n+1}}$ now note that there are $ 3^{n}-1$ numbers $ k$ such that $ 3^{n}< k < 2\times 3^{n}$. and also there are $ 3^{n}-1$ numbers $ t$ such that $ f(3^{n}) = 2\times 3^{n}< t < 3^{n+1}= f(2\times 3^{n})$. so for $ k$ such that $ 3^{n}< k < 2\times 3^{k}$ and $ k = 3^{n}+m$ we have: $ f(k) = 2\times 3^{n}+m$ and also if $ 2\times 3^{n}< k < 3^{n+1}$ and $ k = 2\times 3^{n}+m$ then: $ f(k) = 3(3^{n}+m)$ so we have: $ \boxed{f(3^{n}) = 2\times 3^{n},f(2\times 3^{n}) = 3^{n+1}}$ $ \boxed{0\leq m\leq 3^{n},k = 3^{n}+m: f(k) = 2\times 3^{n}+m}$ $ \boxed{0\leq m\leq 3^{n},k = 2\times 3^{n}+m: f(k) = 3(3^{n}+m)}$

BaBaK Ghalebi wrote: first prove that $ f(3^{n}) = 2\times 3^{n}$ for $ n\geq 0$... Is it because I am sleeping, or is this not so obvious?

I don't think I'd call it obvious... the fastest proof I have is this. First, since $ f$ is strictly increasing, we have $ f(1)\leq 3$. If $ f(1) = 3$, then $ f(f(1)) = f(3) > 3$, which is bad. But if $ f(1) = 1$ then $ f(f(1)) = 1$, which is bad. This leaves only $ f(1) = 2$. By taking $ f$ of both sides of the original equation, we get that $ f(3n) = f(f(f(n))) = 3f(n)$. We conclude that $ f(3^{k}) = 3^{k}f(1) = 2\cdot 3^{k}$ for all nonnegative $ k$.

or you can just prove it usung induction on $ n$... first note that $ f(1)\not =1$ because otherwise we would have: $ f(f(1))=f(1)$ and on the other hand we have $ f(f(1))=3$ so we have $ 3=1$ contradiction... so we have: $ f(1)>1$ (*) now $ f$ is strictly increasing so for $ x>y$ we have $ f(x)>f(y)$ so from (*) we get that $ f(f(1))>f(1)$ so $ f(1)<3$... thus we have: $ 1<f(1)<3$ so we must have $ f(1)=2$ as MellowMelon said... so for $ n=0$ we have $ f(3^{0})=2\times 3^{0}$ as wanted... now assume that the statement is true for $ n$ i.e. $ f(3^{n})=2\times 3^{n}$... now we want to prove that $ f(3^{n+1})=2\times 3^{n+1}$... note that we have: $ f(3^{n})=2\times 3^{n}\Rightarrow f(f(3^{n}))=f(2\times 3^{n})\Rightarrow f(2\times 3^{n})=3\times 3^{n}=3^{n+1}$ $ \Rightarrow f(3^{n+1})=f(f(2\times 3^{n}))=3\times 2\times 3^{n}=2\times 3^{n+1}$ as wanted... so by induction our claim is proved...

Peter wrote: Find all strictly increasing functions $ f: \mathbb{N}\to \mathbb{N}$ such that \[ f(f(n)) = 3n. \] and more Find all strictly increasing functions $ f: \mathbb{N}\to \mathbb{N}$ such that $ f(f(n)) = pn.$ wher $ p$ is a prime