Mongolian Mathematical Olympiad

Let's see the given condition as a substitution $P(m,n)$. Adding to both sides $k^3$ and taking $f$ we get:

$$\begin{array}{l}f(k^3+f(m^3+f(n)))=f(k^3+f(m{)}^3+n)\\ P(k,m^3+f(n))\Rightarrow f(k^3+f(m^3+f(n)))=f(k{)}^3+m^3+f(n)\\ \Rightarrow f(k^3+f(m{)}^3+n)=f(k{)}^3+m^3+f(n)(\ast ).\end{array}\}$$

Setting in $(\ast )$ $n=l^3$ we get: $$f(k^3+f(m{)}^3+l^3)=f(k{)}^3+m^3+f(l^3),$$ in $(\ast )$ $k=l$, $n=k^3$ we get: $$f(l^3+f(m{)}^3+k^3)=f(l{)}^3+m^3+f(k^3)$$ Therefore, $$f(k{)}^3-f(k^3)=f(l{)}^3-f(l^3)=\text{const}=c,c\in \mathbb{Z}$$ (1)

Setting in $(\ast )$ $k=f(s)$ we get: $$f(f(s{)}^3+f(m{)}^3+n)=f(f(s){)}^3+m^3+f(n)$$ in $(\ast )$ $k=l$, $n=k^3$ we get: $$f(f(m{)}^3+f(s{)}^3+n)=f(f(m){)}^3+s^3+f(n)$$ Therefore, $$f(f(s){)}^3-s^3=f(f(m){)}^3-m^3=\text{const}=t,t\in \mathbb{Z}$$ In other words, $f(f(m){)}^3=m^3+t$ and for $m$ which is sufficiently large there is no perfect cube of the form $m^3+t$, so $t=0$. I.e. $f(f(m){)}^3=m^3$ and more accurately $f(f(m))=m$. $(\star )$

From (2) $f(f(m){)}^3=m^3$ Set in (1) $l=f(m)$: $$f(f(m){)}^3=f(f(m{)}^3)+c$$ If we put $m=1$ then $0<f(f(1{)}^3)=1-c\Rightarrow c<1$. Therefore, $$f(f(m{)}^3)=m^3-c$$ Let's prove that $c=0$. In the case $c<0$: $$f(l{)}^3-c=f(l^3)$$ $P(l,-c)\Rightarrow f(l{)}^3-c=f(l^3+f(-c))$ Therefore, $$f(l^3)=f(l^3+f(-c))$$ On the other hand, supposing that $f(a)=f(b)$: $$\begin{gathered} P(m,a)\Rightarrow f(m^3+f(a))=f(m{)}^3+a \\ P(m,b)\Rightarrow f(m^3+f(b))=f(m{)}^3+b \end{gathered}$$ Therefore, $a=b$ which means $f$ is injective. Moreover, $f(l^3)=f(l^3+f(-c))\Rightarrow f(-c)=0$ leads to contradiction. From this follows $c=0$. Therefore, $$f(1^3)=f(1{)}^3\Rightarrow f(1)(f(1{)}^2-1)=0\Rightarrow f(1)=1$$ And $$P(1,n)\Rightarrow f(1+f(n))=1+n.$$ Let's prove that $f(n)=n$ by induction. Case $f(1)=1$ is trivial. Supposing that $f(n)=n$ is true, we get $f(1+f(n))=f(n+1)=n+1$ and this completes the proof. Obviously, the function $f(n)=n$ satisfies the given condition.