Problem 1160

By comparing the coefficients of each power of $x$ on both sides of the equation, we have
$$\left\{\begin{array}{l} b_{0}+b_{1}+\cdots+b_{n}=a_{0}+a_{1}+\cdots+a_{n}, \\ \sum_{0 \leqslant i \leqslant j \leqslant n} b_{i} b_{j}=\sum_{0 \leqslant i \leqslant j \leqslant n} a_{i} a_{j}, \\ \cdots \cdots \\ \sum_{i=0}^{n} \frac{b_{0} \cdots b_{n}}{b_{i}}=\sum_{i=0}^{n} \frac{a_{0} \cdots a_{n}}{a_{i}}, \\ (-1)^{n+1}\left(b_{0} \cdots b_{n}-a_{0} \cdots a_{n}\right)=c, \end{array}\right.$$

Thus, by Newton's formulas, we have
$$\left\{\begin{array}{l} b_{0}+b_{1}+\cdots+b_{n}=a_{0}+a_{1}+\cdots+a_{n}, \\ b_{0}^{2}+b_{1}^{2}+\cdots b_{n}^{2}=a_{0}^{2}+a_{1}^{2}+\cdots+a_{n}^{2}, \\ \cdots \cdots \\ b_{0}^{n}+b_{1}^{n}+\cdots b_{n}^{n}=a_{0}^{n}+a_{1}^{n}+\cdots+a_{n}^{n}, \\ b_{0}^{n+1}+b_{1}^{n+1}+\cdots+b_{n}^{n+1}=a_{0}^{n+1}+a_{1}^{n+1}+\cdots+a_{n}^{n+1}-(n+1) c . \end{array}\right.$$

Consider the polynomial $f(x)=(x-1)(x-2) \cdots(x-(n+1)) \stackrel{\text { set }}{=} x^{n+1}+d_{1} x^{n}+\cdots+d_{n+1}$.
$$\begin{aligned} \sum_{i=0}^{n} f\left(b_{i}\right) & =\sum_{i=0}^{n}\left(b_{i}-1\right)\left(b_{i}-2\right) \cdots\left(b_{i}-(n+1)\right)=\sum_{i=0}^{n}\left(b_{i}^{n+1}+d_{1} b_{i}^{n}+\cdots+d_{n+1}\right) \\ & =\left(\sum_{i=0}^{n} a_{i}^{n+1}\right)-(n+1) c+\sum_{i=0}^{n}\left(d_{1} a_{i}^{n}+\cdots+d_{n+1}\right) \\ & =\left(\sum_{i=0}^{n}\left(a_{i}-1\right)\left(a_{i}-2\right) \cdots\left(a_{i}-(n+1)\right)\right)(n+1) c \end{aligned}$$

Each term in the sums on both sides of the above equation is the product of $n+1$ consecutive integers, which is a multiple of $(n+1)!$. Therefore,
$$(n+1)!\mid \sum_{i=0}^{n}\left(b_{i}-1\right) \cdots\left(b_{i}-(n+1)\right)-\sum_{i=0}^{n}\left(a_{i}-1\right) \cdots\left(a_{i}-(n+1)\right),$$

which implies $(n+1)!|(n+1) \Leftrightarrow n!| c$.