Let $f(x)$ be a polynomial with integer coefficients, and $f(x)=1$ has an integer root. We agree to denote the set of all such $f$ as $F$.

For any given integer $k>1$, find the smallest integer $m(k)>1$, which guarantees the existence of $f \in F$, such that
$$
f(x)=m(k)
$$

has exactly $k$ distinct integer roots.
(Zhang Zhusheng, provided)
Suppose $f \in F$ such that $f(x)=m(k)$ has exactly $k$ distinct integer roots, denoted as $\beta_{1}, \beta_{2}, \cdots, \beta_{k}$. Then there exists an integer polynomial $g(x)$ such that
$$
\begin{array}{l}
f(x)-m(k) \\
=\left(x-\beta_{1}\right)\left(x-\beta_{2}\right) \cdots\left(x-\beta_{k}\right) g(x) .
\end{array}
$$

Since $f \in F$, there exists an integer $\alpha$ such that $f(\alpha)=1$. Substituting $\alpha$ into the above factorization and taking the absolute value on both sides, we get
$$
\begin{array}{l}
m(k)-1 \\
=\left|\alpha-\beta_{1}\right| \cdot\left|\alpha-\beta_{2}\right| \cdots \cdot\left|\alpha-\beta_{k}\right| \cdot|g(\alpha)| .
\end{array}
$$

By the problem's conditions, $\alpha-\beta_{1}, \alpha-\beta_{2}, \cdots, \alpha-\beta_{k}$ are distinct integers, and since $m(k)>1$, they are all non-zero. To ensure that $m(k)$ is minimized, it is clear that $|g(\alpha)|=1$, and $\alpha-\beta_{1}, \alpha-\beta_{2}, \cdots, \alpha-\beta_{k}$ should take the smallest $k$ non-zero integers, i.e., they should be chosen sequentially from $\pm 1, \pm 2, \cdots$.
We will discuss the value of $m(k)$ for different cases of $k$.
When $k$ is even, $\alpha-\beta_{1}, \alpha-\beta_{2}, \cdots, \alpha-\beta_{k}$ should take $\pm 1, \pm 2, \cdots, \pm \frac{k}{2}$, with $\frac{k}{2}$ negative numbers. Considering the original factorization, $g(\alpha)$ must equal $(-1)^{\frac{k}{2}+1}$. Thus,
$$
m(k)=\left(\left(\frac{k}{2}\right)!\right)^{2}+1 .
$$

The corresponding $f$ can be taken as
$$
f(x)=(-1)^{\frac{k}{2}+1} \prod_{i=1}^{\frac{k}{2}}\left(x^{2}-i^{2}\right)+\left(\left(\frac{k}{2}\right)!\right)^{2}+1 \text {. }
$$

Similarly, when $k$ is odd, $\alpha-\beta_{1}, \alpha-\beta_{2}, \cdots, \alpha-\beta_{k}$ should take $\pm 1, \pm 2, \cdots, \pm \frac{k-1}{2}, \frac{k+1}{2}$, and $g(\alpha)$ equals $(-1)^{\frac{k-1}{2}+1}$. Thus,
$$
m(k)=\left(\frac{k-1}{2}\right)!\left(\frac{k+1}{2}\right)!+1 .
$$

The corresponding $f$ can be taken as
$$
\begin{aligned}
f(x)= & (-1)^{\frac{k+1}{2}} \prod_{i=1}^{\frac{k-1}{2}}\left(x^{2}-i^{2}\right)\left(x+\frac{k+1}{2}\right) \\
& +\left(\frac{k-1}{2}\right)!\left(\frac{k+1}{2}\right)!+1 .
\end{aligned}
$$