74th Romanian Mathematical Olympiad

For any $k\in {\mathbb{N}}^{\ast }$ we denote $P_k(A)=\{a^k\mid a\in A\}$. The condition $CP(k)$ is then equivalent with $$x-y\in P_k(A),\text{for any }x,y\in P_k(A),$$ 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 $k\in {\mathbb{N}}^{\ast }$. 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 $CP(k)⟺P_k({\mathbb{Z}}_n)={\mathbb{Z}}_n$. Equivalently, $CP(k)⟺$ 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 $M(n)=\{m\in {\mathbb{N}}^{\ast }\mid p_m$ 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 $({\mathbb{N}}^{\ast },\cdot )$.