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(a) Let the range of $f$ be $A$, where $A \subset \mathbb{Z}$. For any $a \in A$, let $a=f(u), u \in \mathbb{R}$, then $P(a)=P(f(u))=f(P(u)) \in A$, so $P(A) \subset A$. On the other hand, for any $a \in A$, let $a=f(u), u \in \mathbb{R}$. Because $\deg P$ is odd, there exists $v \in \mathbb{R}$ such that $u=P(v)$, then $a=f(u)=f(P(v))=P(f(v)) \in P(A)$, so $A \subset P(A)$. Therefore, $P(A)=A \subset \mathbb{Z}$. Because $\deg P \ge 3$, for every sufficient large $M>0$, we have: if $|x| \ge M$, then $|P(x)| > |x| \ge M$. (*) Construct a graph on $A$ such that if $P(a)=b$, then we draw an arrow from $a$ to $b$. Since $P(A)=A$, every vertice has exactly one out-edge and at least one in-edge. Fix any $a \in A$, we choose a sufficient large $M>|a|$. Let $a=P(b)$ where $b \in A$. If $|b| \ge M$, then from the property (*), $|a|=|P(b)|>|b| \ge M$, a contradiction. Hence $|b| <M$. In the same manner, we can trace back from $a$ like $\cdots \mapsto \cdots \mapsto b \mapsto a$. Every element in this chain is an integer bounded by $M$, hence there must be an integer that appears twice, like $\cdots \mapsto x \mapsto \cdots \mapsto x \mapsto \cdots \mapsto b \mapsto a$, so there is a cycle. But every vertice has exactly one successor, hence $a$ must lies in the cycle. Therefore, any elements in $A$ lies in a (finite) cycle. If $|u|>M$, from the property (*) we have $|P(u)|>|u|>M$, thus $|P(P(u))|>|P(u)|>M$ and so on. It shows that $u$ is never contained in a cycle, hence $u \notin A$. Therefore, any element in $A$ is bounded by $M$, so $A$ is finite. (b) Pick an odd integer $N \ge n$, consider $P(x)=c(x-1)(x-2)\cdots(x-N)+x$, where $c$ is a positive constant to be chosen later. Note that $P'(x)=c \cdot \left( \text{ a polynomial of degree } N-1 \text{ with positive leading coefficient } \right) +1$. Because a polynomial of even degree with positive leading coefficient has global minimum, if we choose $c$ sufficient small, then $P'(x)>0$ for all $x \in \mathbb{R}$. This means $P(x)$ is strictly increasing (and hence bijective). Define $f: \mathbb{R} \to \mathbb{Z}$ as follow: Let $f(1)=1, f(2)=2, \ldots, f(n)=n$. For $x \notin \{1,2,\ldots,n \}, x \in \mathbb{R}$, let $f(x)=1$. In this way, the range of $f$ has exactly $n$ elements. Now let's verify the condition. If $x \in \{1,2,\ldots,n \}$, then $P(x)=f(x)=x$, so $f(P(x))=f(x)=x=P(x)=P(f(x))$. If $x \notin \{1,2,\ldots,n \}$, because $P(x)$ is bijective, $P(x) \notin \{P(1),P(2),\ldots,P(n) \}=\{1,2,\ldots,n \}$, so $f(P(x))=1=P(1)=P(f(x))$ as well.