2.57 A quirky mathematician has a ladder with $n$ steps, and he climbs up and down the ladder, each time ascending $a$ steps or descending $b$ steps. Here, $a, b$ are fixed positive integers.

If he can start from the ground, climb to the very top step of the ladder, and then return to the ground, find the minimum value of $n$ (expressed in terms of $a, b$) and prove it.
[Solution] The minimum value is $a+b-(a, b)$, where $(a, b)$ denotes the greatest common divisor of $a$ and $b$.

Since the level the mathematician reaches each time is a multiple of $(a, b)$, we can ignore other levels, treating $(a, b)$ as the first level, and using $\frac{a}{(a, b)}, \frac{b}{(a, b)}$ to replace $a$ and $b$. Therefore, we only need to consider the case where $(a, b)=1$.
(1) $n=a+b-1$.

In this case, since $(a, b)=1$, then $a, 2a, 3a, (b-1)a, ba \pmod{b}$ are distinct residues, forming a complete residue system modulo $b$.

If the mathematician is at level $r$ and $r+a > n$, then it must be that $r > b$. The mathematician can descend several times until reaching level $n$. Thus, he is forever confined to this residue class and cannot move to the other residue classes (especially the $ba$ class), leading to a contradiction!

In summary, when $(a, b)=1$, the minimum value is $n=a+b-1$.