The hundredth problem, determine all integer-coefficient polynomials $P(x)$, such that for all real numbers $s, t$, if $P(s), P(t)$ are both integers, then $P(st)$ is also an integer.



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The hundredth problem, determine all integer-coefficient polynomials $P(x)$, such that for all real numbers $s, t$, if $P(s), P(t)$ are both integers, then $P(st)$ is also an integer.
The 100th problem,
Solution: When $\operatorname{deg} P=0$, it clearly satisfies the condition. Now assume $\operatorname{deg} P=n \geq 1$.
Notice that if $P(x)$ is a solution, then for any integer $k$, $P(x)+k$ and $-P(x)+k$ are also solutions. Therefore, we can assume that the coefficients of $P(x)$ are positive, and $P(0)=0$.
Let $P(x)=\sum_{i=1}^{n} a_{i} x^{i}\left(a_{n}>0\right)$.
Choose a prime $p>\sum_{i=1}^{n}\left|a_{i}\right|$, since the leading coefficient of $P(x)$ is positive, there exists a real number $t>1$ such that $P(t)=p$.
Let $f(x)=(P(x)-p, P(2 x)-P(2 t))$ be the greatest common divisor of the monic polynomials, then $(x-t) \mid f(x)$.
Since $P(t)=p \in Z, P(2) \in Z$, hence $P(2 t) \in Z$. Therefore, $f(x)$ is a non-constant polynomial with rational coefficients. If $\operatorname{deg} f(x) \neq n$, set
$$
P(x)-p=f(x) \cdot g(x)
$$

Then $g(x)$ is a non-constant polynomial. According to Gauss's lemma:
$$
P(x)-p=f_{1}(x) \cdot g_{1}(x)
$$

Here $f_{1}, g_{1}$ are the constant multiples of the integer coefficient polynomials of $f, g$.
Notice that the constant term of $P(x)-p$ is $-p$, so one of $f_{1}, g_{1}$ must have a constant term of $\pm 1$. Without loss of generality, assume the constant term of $f_{1}$ is $\pm 1$, then by Vieta's theorem, $f_{1}$ must have a root $r$ with a modulus no greater than 1, and thus $r$ is also a root of $P(x)-p$. But
$$
|P(r)-p| \geq p-|P(r)| \geq p-\sum_{i=1}^{n}\left|a_{i}\right||r|^{i} \geq p-\sum_{i=1}^{n}\left|a_{i}\right|>0
$$

This leads to a contradiction. Therefore, $\operatorname{deg} f(x)=n$.
Notice that $\operatorname{deg}(P(x)-p)=\operatorname{deg}(P(2 x)-P(2 t))=n$, so $(P(x)-p) \mid(P(2 x)-P(2 t))$. Comparing the leading coefficients, we have:
$$
P(2 x)-P(2 t)=2^{n}(P(x)-p)
$$

Comparing the coefficients of both sides of (*), we find that $P(x)=a_{n} x^{n}$, and $P(2 t)=2^{n} p$.
Let $a=b=a_{n} \frac{-1}{n}$, then $P(a)=P(b)=1$, according to the condition, $P\left(a^{2}\right)=\frac{1}{a_{n}} \in Z \Rightarrow a_{n}=1$. Therefore, $P(x)=x^{n}$, which clearly satisfies the condition.
In summary, the required $P(x)= \pm x^{n}+k$, where $n \in N, k \in Z$.