Saudi Arabian Mathematical Competitions

Firstly, we prove that $f$ is injective. Let two natural numbers $a, b$ such that $a+1 \leq b$ and $f(a)=f(b)$. By the given condition, we have
$$f(a)[f(f(a))]^{2} \leq a^{3}+a^{2}<a^{3}+2a^{2}+a=(a+1)^{3}-(a+1)^{2}.$$
Since $a, b \in \mathbb{Z}^{+}$ and $a+1 \leq b$, we have
$$(a+1)^{3}-(a+1)^{2} \leq b^{3}-b^{2} \leq f(b)[f(f(b))]^{2}.$$
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
$$0 \leq f(1)[f(f(1))]^{2} \leq 2.$$
Since $f: \mathbb{Z}^{+} \longrightarrow \mathbb{Z}^{+}$, we have two cases
- $f(1)=1$.
- $f(1)=2$ and $f(2)=1$.

We shall prove that if $\{f(1), \ldots, f(k)\}=\{1, \ldots, k\}$ 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 $f(k+1) \geq k+1$. If $f(k+1)=k+1$, we have
$$\{f(1), \ldots, f(k+1)\}=\{1, \ldots, k+1\}.$$
Otherwise, if $f(k+1) \geq k+2$, we have $[f(f(k+1))]^{2} \geq (k+1)^{2}$. In given condition, let $n=k+1$, we have
$$(k+1)^{3}+(k+1)^{2} \geq f(k+1)[f(f(k+1))]^{2} \geq (k+2)(k+1)^{2}.$$
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 $\{f(1), \ldots, f(k+2)\}=\{1, \ldots, k+2\}$.

Now we can check both cases satisfy the conditions
$$\begin{aligned} & (k+2)^{3}-(k+2)^{2} \leq f(k+2) f(f(k+2))^{2} \leq (k+2)^{3}+(k+2)^{2} \\ & \Leftrightarrow (k+2)^{2}(k+1) \leq (k+1)(k+2)^{2} \leq (k+2)^{2}(k+3). \end{aligned}$$
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 $f(a)=a$.