Peter wrote: Find all functions $f: \mathbb{N}\to \mathbb{N}$ such that for all $n\in \mathbb{N}$: \[f(f(f(n)))+f(f(n))+f(n)=3n.\] We have that $f(f(f(1)))+f((1))+f(1)=3$,but every positive integer number is greater or equal to $1$,therefore $f(1)=1$. Let's suppose that we have proved that $f(k)=k$ for all $n\geq{k}$, and let's prove that $f(n+1)=n+1$. If $f(a)=f(b)$ then $3a=f(f(f(a)))+f(f(a))+f(a)=f(f(f(b)))+f(f(b))+f(b)=3b$ so $a=b$. We also have that $f(f(f(n+1)))+f(f(n+1))+f(n+1)=3(n+1)$,but if $n\geq{t}=f(n+1)$ then $f(n+1)=t=f(t)$ $\Rightarrow$ $n+1=t$,contradiction. Therefore, $f(n+1)\geq(n+1)$. Same way if $n\geq{t}=f(f(n+1))$ $\Rightarrow$ $f(f(n+1))=t=f(t)$ $\Rightarrow$ $f(n+1)=f(t)$ $\Rightarrow$ $n+1=t$ contradiction. And again $f(f(n+1))\geq{n+1}$. In the similar way we can prove that $f(f(f (n+1)))\geq{n+1}$. So we get that \[f(n+1)\geq{n+1}\] \[f(f(n+1))\geq{n+1}\] \[f(f(f(n+1)))\geq{n+1}\] and consequently $3(n+1)=f(f(f(n+1)))+f(f(n+1))+f(n+1)\geq3(n+1)$ $\Rightarrow$ $f(n+1)=n+1$. $\Rightarrow$ \[f(n)=n\] for all $n\in{N}$.