It is easy to chek, that if $f(n)\not =k^{2}\ \forall k\in N$ and if $m\not =k^{2}$ then exist unique n, suth that $f(n)=m$, $n=m-[\sqrt{m-1}]$. Therefore $f(n)-n=[\sqrt{f(n)-1}]=[\sqrt{f(n)]}$, and equation equavalent to $k^{2}<f(f(n)))=m<(k+1)^{2}$. It give $f(n)=m-k$, t $m\not =k^{2}+k$. All solutions are $n=g(g(m)), m=k^{2}+1,...,k^{2}+k-1,k^{2}+k+1,....k^{2}+2k, \ g(m)=m-[\sqrt m ].$