Romanian Mathematical Olympiad - Final Round

a.
For any polynomial $P \in \mathbb{K}[X]$ we shall denote by $Z_P$ the set of roots of $P$ in the field $\mathbb{K}$. Because $|\mathbb{K}| = p^n$, the characteristic of the field $\mathbb{K}$ is $\text{char}(\mathbb{K}) = p$. Then $\hat{m} = \hat{0}$ for any multiple $m$ of $p$. In particular, since $k^p \equiv k \pmod p$, for any $k \in \{0, 1, \dots, p-1\}$ we have
$$f_1(\hat{k}) = (\hat{k})^p - \hat{k} = \widehat{k^p} - \hat{k} = \widehat{k^p - k} = \hat{0},$$
Hence, $\{\hat{k} \mid k = \overline{0, p-1}\} \subseteq Z_{f_1}$. Also, because $\mathbb{K}$ is a field, $|Z_{f_1}| \le \text{grad}(f_1) = p$. It follows that $Z_{f_1} = \{\hat{k} \mid k = \overline{0, p-1}\}$.

b.
Because $p|C_p^k$, for any $k = \overline{1, p-1}$, the identity $(a+b)^p = a^p + b^p$ holds for any $a, b \in \mathbb{K}$, and inductively we have $(a+b)^{p^k} = a^{p^k} + b^{p^k}$, for any $a, b \in \mathbb{K}$ and any $k \in \mathbb{N}$. Then for any $m \in \mathbb{N}^*$ we have:
$$\begin{align*} f_m(f_1(X)) &= \sum_{k=0}^{m} (-1)^{m-k} \widehat{C_m^k} (X^p - X)^{p^k} = \sum_{k=0}^{m} (-1)^{m-k} \widehat{C_m^k} (X^{p^{k+1}} - X^{p^k}) = \\ &= \sum_{k=0}^{m+1} (-1)^{m+1-k} (\widehat{C_m^k} + \widehat{C_m^{k-1}}) X^{p^k} = \sum_{k=0}^{m+1} (-1)^{m+1-k} \widehat{C_{m+1}^k} X^{p^k} = f_{m+1}(X). \end{align*}$$

We shall prove by induction with respect to $m \in \mathbb{N}^*$ that $Z_{f_m} = \{\hat{k} \mid k = \overline{0, p-1}\}$ for any $m \in \mathbb{N}^*$, which will solve the problem. For $m = 1$ we have shown this in part a.

Assume now that the property holds for some arbitrary $m \in \mathbb{N}^*$. We prove now that it will hold also for $m + 1$:
For any $k \in \{0, 1, \dots, p-1\}$ we have
$$f_{m+1}(\hat{k}) = f_m(f_1(\hat{k})) = f_m(\hat{0}) = \hat{0},$$
so that $\{\hat{k} \mid k = \overline{0, p-1}\} \subseteq Z_{f_{m+1}}$.
Let $\alpha \in Z_{f_{m+1}}$ be arbitrary. Then $f_m(f_1(\alpha)) = f_{m+1}(\alpha) = \hat{0}$, so that $f_1(\alpha) \in Z_{f_m}$. Hence, there is a $k \in \{0, 1, \dots, p-1\}$ such that $f_1(\alpha) = \hat{k}$. We obtain
$$\alpha^p = \alpha + \hat{k}, \quad \alpha^{p^2} = (\alpha + \hat{k})^p = \alpha^p + \hat{k}^p = (\alpha + \hat{k}) + \hat{k} = \alpha + 2 \cdot \hat{k},$$
and, inductively, if $\alpha^{p^m} = \alpha + m \cdot \hat{k}$, then $\alpha^{p^{m+1}} = (\alpha + m \cdot \hat{k})^p = \alpha + (m+1) \cdot \hat{k}$.
In the multiplicative group $(\mathbb{K}^*, \cdot)$ we have $x^{p^n-1} = 1$, for any $x \in \mathbb{K}^*$, so that $x^{p^n} = x$ holds for any element $x \in \mathbb{K}$. Then $\alpha = \alpha^{p^n} = \alpha + n \cdot \hat{k}$, and $n \cdot \hat{k} = \hat{0}$.
Since $n$ is not divisible by the characteristic $p$, it follows that $\hat{k} = \hat{0}$. But then $f_1(\alpha) = \hat{0}$ and $\alpha \in Z_{f_1} = \{\hat{k} \mid k = \overline{0, p-1}\}$. Thus we obtained the reverse inclusion $Z_{f_{m+1}} \subseteq \{\hat{k} \mid k = \overline{0, p-1}\}$, so that the equality $Z_{f_{m+1}} = \{\hat{k} \mid k = \overline{0, p-1}\}$ holds. The claim holds then for any $m \in \mathbb{N}^*$.