Find all integer-coefficient polynomials $P(x)$ such that there exists an infinite integer sequence $\left\{a_{n}\right\}$, where any two terms are distinct, and satisfies
$$
P\left(a_{1}\right)=0, P\left(a_{k+1}\right)=a_{k}(k=1,2, \cdots) .
$$
(2016, Peking University Mathematical Science Summer Camp Preliminary Contest)
First, we prove a lemma.
Lemma If $F(x)$ is a polynomial with integer coefficients, and $u, v$ are integers, then
$$
(u-v) \mid (F(u)-F(v)).
$$

Proof Let
$$
F(x)=b_{n} x^{n}+b_{n-1} x^{n-1}+\cdots+b_{1} x+b_{0}.
$$

Then $F(u)-F(v)$
$$
\begin{array}{l}
=b_{n}\left(u^{n}-v^{n}\right)+b_{n-1}\left(u^{n-1}-v^{n-1}\right)+ \\
\cdots+b_{1}(u-v).
\end{array}
$$

By the factorization formula, we know
$$
(u-v) \mid b_{k}\left(u^{k}-v^{k}\right) \quad (k=1,2, \cdots, n).
$$

Adding all terms together yields the conclusion of the lemma.
If $P(x)$ is a constant function, then from $P\left(a_{1}\right)=0$, we have
$$
P(x)=0.
$$

Therefore, for any $k \in \mathbf{Z}_{+}$, we have
$$
a_{k}=P\left(a_{k+1}\right)=0,
$$

which contradicts the condition that any two terms of the sequence $\left\{a_{n}\right\}$ are distinct.
Thus, $\operatorname{deg} P \geqslant 1$.
By the problem statement, we have
$$
P\left(a_{k+2}\right)=a_{k+1}, \quad P\left(a_{k+1}\right)=a_{k}.
$$

By the lemma, we get
$$
\left(a_{k+2}-a_{k+1}\right) \mid \left(P\left(a_{k+2}\right)-P\left(a_{k+1}\right)\right)
$$
$$
\begin{array}{l}
\Rightarrow \left(a_{k+2}-a_{k+1}\right) \mid \left(a_{k+1}-a_{k}\right) \\
\Rightarrow \left|a_{k+2}-a_{k+1}\right| \leqslant \left|a_{k+1}-a_{k}\right|
\end{array}
$$
$\Rightarrow \left\{\left|a_{k+1}-a_{k}\right|\right\}$ is a monotonically decreasing sequence of positive integers
$\Rightarrow$ there exists $M \in \mathbf{Z}_{+}$, such that when $k \geqslant M$, $\left|a_{k+1}-a_{k}\right|$ is a constant $c\left(c \in \mathbf{Z}_{+}\right)$.

When $k \geqslant M$, if $a_{k+1}-a_{k}$ is not a constant, then there must exist $k_{0} \geqslant M$, such that
$$
\left\{\begin{array}{l}
a_{k_{0}+1}-a_{k_{0}}=c, \\
a_{k_{0}+2}-a_{k_{0}+1}=-c
\end{array} \text { or } \left\{\begin{array}{l}
a_{k_{0}+1}-a_{k_{0}}=-c, \\
a_{k_{0}+2}-a_{k_{0}+1}=c
\end{array}\right.\right.
$$

Adding the two equations in each case, we get
$$
a_{k_{0}+2}-a_{k_{0}}=0,
$$

which contradicts the condition that any two terms of the sequence $\left\{a_{n}\right\}$ are distinct.
Thus, for any $k \geqslant M$, $a_{k+1}-a_{k}$ is a constant $C(|C|=c)$, i.e.,
$$
a_{k+1}-P\left(a_{k+1}\right)=C.
$$

Then the polynomial equation $x-P(x)=C$ has infinitely many distinct integer solutions $a_{k+1}, a_{k+2}, a_{k+3}, \cdots$
Hence, $x-P(x) \equiv C \Rightarrow P(x)=x-C$.
For any $C \in \mathbf{Z} \backslash\{0\}$, taking $a_{j}=j C(j \in \mathbf{Z}_{+})$, the sequence $\left\{a_{n}\right\}$ satisfies the condition.
In summary, all polynomials that satisfy the condition are $P(x)=x-C$ (where $C$ is a non-zero integer).