Question 46 Find all 3-tuples of positive integers $(a, b, c)$, such that for infinitely many positive integers $n$, we have $(a+$ $n) \mid (b+c \cdot n!)$

保留源文本的换行和格式，翻译结果如下：

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Question 46 Find all 3-tuples of positive integers $(a, b, c)$, such that for infinitely many positive integers $n$, we have $(a+$ $n) \mid (b+c \cdot n!)$
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Question 46:
Solution: Let $S=\left\{n\left|n \in Z^{+},(a+n)\right|(b+c \cdot n!)\right\}$, then $S$ is an infinite set.
Take a positive integer $\mathrm{M}=\max \{\mathrm{a}, \mathrm{b}-\mathrm{a}, 2 \sqrt{\mathrm{a}+1}+2\}$, then for any positive integer $\mathrm{n}>M$, we have $\frac{\mathrm{n}+\mathrm{a}}{2}b$, and $\mathrm{n}>2 \sqrt{\mathrm{a}+\mathrm{n}}$.
Let $S^{\prime}=\{\mathrm{n} \mid \mathrm{n} \in S, \mathrm{n}>M\}$, then $\mathrm{S}^{\prime}$ is an infinite set.
We first prove a lemma: If $n \in S^{\prime}$, then $n+a$ is a prime number.
Proof of the lemma: If there exists $n \in S^{\prime}$ such that $n+a$ is a composite number. Let $p$ be the smallest prime factor of $n+a$, and $n+a=\mathrm{ps}$, then $\mathrm{s} \geq \mathrm{p}$.

If $\mathrm{p}=\mathrm{s}$, then $\mathrm{p}\mathrm{b}$, a contradiction.

If $\mathrm{p}b$, a contradiction.
In summary, the lemma is proved.
Now, let's return to the original problem. For any $n \in S^{\prime}$, since $n+a$ is a prime number, according to Wilson's Theorem, we have:
$$
\begin{array}{l}
(n+a-1)!\equiv-1(\bmod n+a) \\
\Rightarrow n!\cdot(n+1)(n+2) \ldots(n+a-1) \equiv-1(\bmod n+a) \\
\Rightarrow n!\cdot[-(a-1)][-(a-2)] \ldots(-1) \equiv-1(\bmod n+a) \\
\Rightarrow n!\cdot(-1)^{a-1}(a-1)!\equiv-1(\bmod n+a) \\
\Rightarrow n!\cdot(-1)^{a}(a-1)!\equiv 1(\bmod n+a) \\
\Rightarrow n!\equiv\left[(-1)^{a}(a-1)!\right]^{-1}(\bmod n+a) \\
\text { Therefore, }
\end{array}
$$

Therefore,
$$
\begin{array}{l}
0 \equiv b+c \cdot n!\equiv b+c \cdot\left[(-1)^{a}(a-1)!\right]^{-1}(\bmod n+a) \\
\Rightarrow 0 \equiv b \cdot(-1)^{a}(a-1)!+c(\bmod n+a)
\end{array}
$$

By the arbitrariness of $\mathrm{n}+\mathrm{a}$, we know $\mathrm{b} \cdot(-1)^{\mathrm{a}}(\mathrm{a}-1)!+\mathrm{c}=0$, i.e., $\mathrm{c}=\mathrm{b} \cdot(-1)^{\mathrm{a}-1}(\mathrm{a}-1)!$. Noting that $a, b, c$ are all positive integers, hence $a$ is an odd number.

On the other hand, if $(a, b, c)=(2 k+1, t, t(2 k)!)$, where $k$ is any natural number, and $t$ is any positive integer. For any prime number $p$ greater than $a$, take $n=p-a$, then
$$
b+c \cdot n!=t+t(2 k)!\cdot(p-2 k-1)!\equiv t+t(p-1)!\equiv 0(\bmod p)
$$

Thus, $(a+n) \mid(b+c \cdot n!)$.
In summary, the solutions satisfying the conditions are $(a, b, c)=(2 k+1, t, t(2 k)!)$, where $k$ is any natural number, and $t$ is any positive integer.
Note: It is easy to prove: For sufficiently large positive integers $n$, if $(a+n) \mid(b+c \cdot n!)$, then $a+n$ must be a prime number.
In fact, if $a+n$ is a composite number, let the smallest prime factor of $a+n$ be $p$, then $p \leq \frac{a+n}{p} \leq \frac{a+n}{2} \leq n$ (when $n \geq$ $a$), hence $p|n!\Rightarrow p| b ; \frac{a+n}{p}\left|n!\Rightarrow \frac{a+n}{p}\right| b$. Therefore, $\left.p \cdot \frac{a+n}{p} \right\rvert\, b^{2} \Rightarrow a+n \leq b^{2} \Rightarrow n \leq b^{2}-a$. In other words, when $n>\max \left\{a, b^{2}-a\right\}$, if $(a+n) \mid(b+c \cdot n!)$, then $a+n$ must be a prime number.