2023 IMO Shortlist N4

Let $a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n}$ be $2 n$ positive integers such that the $n+1$ products $$\begin{gathered} a_{1} a_{2} a_{3} \cdots a_{n} \\ b_{1} a_{2} a_{3} \cdots a_{n} \\ b_{1} b_{2} a_{3} \cdots a_{n} \\ \vdots \\ b_{1} b_{2} b_{3} \cdots b_{n} \end{gathered}$$ form a strictly increasing arithmetic progression in that order. Determine the smallest positive integer that could be the common difference of such an arithmetic progression. (Canada) Answer: The smallest common difference is $n!$.

设 $a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n}$ 为 $2 n$ 个正整数，使得 $n+1$ 乘积 $$\begin{gathered} a_{1} a_{2} a_{3} \cdots a_{n} \\ b_{1} a_{2} a_{3} \cdots a_{n} \\ b_{1} b_{2} a_{3} \cdots a_{n} \\ \vdots \\ b_{1} b_{2} b_{3} \cdots b_{n} \end{gathered}$$ 按此顺序形成严格递增的算术级数。确定可能是此类算术级数的公差的最小正整数。 (加拿大)答案：最小公差是$n!$。