Romanian Mathematical Olympiad

We firstly notice that the function is injective. Indeed, if $f(n)=f(m)$, then $3f(f(f(n)))+2f(f(n))+f(n)=3f(f(f(m)))+2f(f(m))+f(m)$, hence $6n=6m$, so $n=m$.

Plugging $n=0$ yields $3f(f(f(0)))+2f(f(0))+f(0)=0$, so $f(0)=0$. We prove, using induction, that $f(n)=n$, for every $n\in \mathbb{N}$.

Suppose $f(0)=0$, $f(1)=1$, ..., $f(n)=n$. From the injectivity, $f(n+1)\ge n+1$, $f(f(n+1))\ge n+1$ and $f(f(f(n+1)))\ge n+1$. On the other hand, $3f(f(f(n+1)))+2f(f(n+1))+f(n+1)=6n+6$, whence $f(n+1)=n+1$.