The ninety-first problem: Given a positive integer $n$, find all real-coefficient polynomials $f(x)$ such that $\operatorname{deg} f=2 n+1$, and $(x-$ $1)^{n} \mid (f(x)+1), (x+1)^{n} \mid (f(x)-1)$.
The 91st problem:
Solution: Let $f(x)=\sum_{i==0}^{2 n+1} a_{i} x^{i}$, and let $g(x)=f(x)-\left(a_{2 n+1} x+a_{2 n}\right)(x+1)^{n}(x-1)^{n}$. Then $\deg g \leq 2 n-1$, and $(x-1)^{n}\left|(g(x)+1),(x+1)^{n}\right|(g(x)-1)$. This is equivalent to $x=1$ being an $n$-fold root of $g(x)+1$, and $x=-1$ being an $n$-fold root of $g(x)-1$. That is, it is equivalent to $g(1)+1=0, g(-1)-1=0$, and $(x-1)^{n-1} \mid g(x), (x+1)^{n-1} \mid g(x)$. Noting that $\deg g \leq 2 n-2$, we have
$$
\begin{array}{l}
g^{\prime}(x)=c(x-1)^{n-1}(x+1)^{n-1}=c\left(x^{2}-1\right)^{n-1} \\
=c \sum_{i=0}^{n-1} C_{n-1}^{i}(-1)^{n-1-i} x^{2 i}
\end{array}
$$

Thus,
$$
g(x)=\int c \sum_{i=0}^{n-1} C_{n-1}^{i}(-1)^{n-1-i} x^{2 i}=c \sum_{i=0}^{n-1} \frac{C_{n-1}^{i}(-1)^{n-1-1}}{2 i+1} \cdot x^{2 i+1}+d
$$

Here, $c$ and $d$ are real numbers to be determined.
Substituting $g(1)+1=0, g(-1)-1=0$, we get
$$
\left\{\begin{array}{c}
c \sum_{i=0}^{n-1} \frac{C_{n-1}^{i}(-1)^{n-1-i}}{2 i+1} + d + 1 = 0 \\
- c \sum_{i=0}^{n-1} \frac{C_{n-1}^{i}(-1)^{n-1-i}}{2 i+1} + d - 1 = 0
\end{array} \Rightarrow \left\{\begin{array}{c}
c=-\frac{1}{\sum_{i=0}^{n-1} C_{n-1}^{i}(-1)^{n-1-i}} \\
d=0
\end{array}\right.\right.
$$

Thus, we have $g(x)=-\frac{1}{\sum_{i=0}^{n-1} \frac{C_{n-1}^{i}(-1)^{n-1-i}}{2 i+1}} \cdot \sum_{i=0}^{n-1} \frac{C_{n-1}^{i}(-1)^{n-1-i}}{2 i+1} \cdot x^{2 i+1}$. Therefore,
$$
f(x)=(a x+b)(x+1)^{n}(x-1)^{n}-\frac{1}{\sum_{i=0}^{n-1} \frac{C_{n-1}^{i}(-1)^{n-1-i}}{2 i+1}} \cdot \sum_{i=0}^{n-1} \frac{C_{n-1}^{i}(-1)^{n-1-i}}{2 i+1} \cdot x^{2 i+1}
$$

Here, $a, b \in \mathbb{R}$, and $a \neq 0$.
Note: Further, we can prove that $\sum_{i=0}^{n-1} \frac{C_{n-1}^{i}(-1)^{n-1-i}}{2 i+1}=\frac{2^{2 n-2}((n-1)!)^{2}(-1)^{n-1}}{(2 n-1)!}$.
To do this, we only need to prove:
$$
\sum_{i=0}^{n} \frac{C_{n}^{i}(-1)^{n-i}}{2 i+1}=\frac{(-1)^{n} 2^{2 n}(n!)^{2}}{(2 n+1)!}
$$

In fact, taking the function $f(x)=\frac{1}{1+2 x}$, then $f(i)=\frac{1}{1+2 i}$, so
$$
\sum_{i=0}^{n} \frac{C_{n}^{i}(-1)^{n-i}}{2 i+1}=\sum_{i=0}^{n} C_{n}^{i}(-1)^{n-i} f(i)=\Delta^{n} f(0)
$$

And by induction, it is easy to know that $\Delta^{n} f(x)=\frac{(-1)^{n} \cdot(2 n)!}{(2 x+1)(2 x+3)(2 x+5) \ldots(2 x+2 n+1)}$, so
$$
\Delta^{n} f(0)=\frac{(-1)^{n} \cdot(2 n)!!}{(2 n+1)!!}=\frac{(-1)^{n} 2^{2 n}(n!)^{2}}{(2 n+1)!}
$$

Thus, (1) holds.