Romanian Mathematical Olympiad

For any $k \in \mathbb{N}^*$ we denote $P_k(A) = \{a^k \mid a \in A\}$. The condition $CP(k)$ is then equivalent with
$$x - y \in P_k(A), \quad \text{for any } x, y \in P_k(A),$$
meaning that $P_k(A)$ is a subgroup of the additive group $(A, +)$.

a) For $A = \mathbb{Z}_4$, we have that $P_{2k}(A) = \{\hat{0}, \hat{1}\}$, respectively $P_{2k+1}(A) = \{\hat{0}, \hat{1}, \hat{3}\}$, for any $k \in \mathbb{N}^*$. These are not subgroups of the group $(\mathbb{Z}_4, +)$. Hence, the ring $(\mathbb{Z}_4, +, \cdot)$ does not have the property $CP(k)$ for any $k \in \mathbb{N}$, with $k \ge 2$.

b) For $n \in \mathbb{N}$, $n \ge 3$, we consider the ring $(\mathbb{Z}_n, +, \cdot)$. Because $\hat{1} \in P_k(\mathbb{Z}_n)$ for any $k \in \mathbb{N}$, $k \ge 2$, and $(\mathbb{Z}_n, +)$ is cyclic, generated by $\hat{1}$, it follows that $CP(k) \iff P_k(\mathbb{Z}_n) = \mathbb{Z}_n$. Equivalently, $CP(k) \iff$ the function $p_k : \mathbb{Z}_n \to \mathbb{Z}_n$, defined by $p_k(x) = x^k$ for any $x \in \mathbb{Z}_n$, is bijective.
Then we can rewrite $M(n) = \{m \in \mathbb{N}^* \mid p_m$ is bijective\}.
Because for even $k$ we have that $p_k(\hat{1}) = p_k(-\hat{1})$, and $\hat{1} \ne -\hat{1}$, it follows that any $m \in M(n)$ is odd, so that $M(n) \subseteq 2 \cdot \mathbb{N} + 1$.
Because $p_1 = \text{id}_{\mathbb{Z}_n}$ is bijective, we have that $1 \in M(n)$.
Let $m_1, m_2 \in M(n)$ be arbitrary. Since the functions $p_{m_1}$ and $p_{m_2}$ are bijective, the function $p_{m_1 m_2} = p_{m_1} \circ p_{m_2}$ is also bijective, as the composition of two bijective functions. We deduce that $m_1 \cdot m_2 \in M(n)$.
It follows that $M(n)$ is a submonoid of the monoid $(\mathbb{N}^*, \cdot)$.