SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Firstly, we prove that $f$ is injective. Let two natural numbers $a,b$ such that $a+1\le b$ and $f(a)=f(b)$. By the given condition, we have $$f(a)[f(f(a)){]}^2\le a^3+a^2<a^3+2a^2+a=(a+1{)}^3-(a+1{)}^2.$$ Since $a,b\in {\mathbb{Z}}^{+}$ and $a+1\le b$, we have $$(a+1{)}^3-(a+1{)}^2\le b^3-b^2\le f(b)[f(f(b)){]}^2.$$ Thus $f(a)[f(f(a)){]}^2<f(b)[f(f(b)){]}^2$, a contradiction. This implies $f$ is injective.

In given condition, let $n=1$, we have $$0\le f(1)[f(f(1)){]}^2\le 2.$$ Since $f:{\mathbb{Z}}^{+}⟶{\mathbb{Z}}^{+}$, we have two cases - $f(1)=1$. - $f(1)=2$ and $f(2)=1$.

We shall prove that if $\{f(1),\dots ,f(k)\}=\{1,\dots ,k\}$ then either $f(k+1)=k+1$ or $f(k+1)=k+2,f(k+2)=k+2$. Indeed, Since $f$ is injective, we have $f(k+1)\ge k+1$. If $f(k+1)=k+1$, we have $$\{f(1),\dots ,f(k+1)\}=\{1,\dots ,k+1\}.$$ Otherwise, if $f(k+1)\ge k+2$, we have $[f(f(k+1)){]}^2\ge (k+1{)}^2$. In given condition, let $n=k+1$, we have $$(k+1{)}^3+(k+1{)}^2\ge f(k+1)[f(f(k+1)){]}^2\ge (k+2)(k+1{)}^2.$$ Since $(k+1{)}^3+(k+1{)}^2=(k+2)(k+1{)}^2$, we deduce that $f(k+1)=k+2$ and $f(k+2)=k+1$. Thus $\{f(1),\dots ,f(k+2)\}=\{1,\dots ,k+2\}$.

Now we can check both cases satisfy the conditions $$\begin{array}{rl}& (k+2{)}^3-(k+2{)}^2\le f(k+2)f(f(k+2){)}^2\le (k+2{)}^3+(k+2{)}^2\\ & \iff (k+2{)}^2(k+1)\le (k+1)(k+2{)}^2\le (k+2{)}^2(k+3).\end{array}$$ Therefore, the solutions of this problem are the functions have following properties - There are some disjoint pairs $(k,k+1)$ with $f(k)=k+1$ and $f(k+1)=f(k)$. - For other positive integer $a$, then $f(a)=a$.