Determine all triplets $(a, b, c)$ of integers greater than or equal to 2 such that $a b c-1$ is divisible by $(a-1)(b-1)(c-1)$.
This is an International Olympiad exercise (1992/1), which does not present theoretical difficulty but requires a methodical approach. Let $q$ be the quotient: $\frac{a b c-1}{(a-1)(b-1)(c-1)}$, and assume $a \leq b \leq c$. For each value of $a$, we will determine the possible values of $q$, and for each value of $(a, q)$, we will have to solve a simple equation in $(b, c)$.

First, note that $q>1$, because $a b c-1>a b c-b c=(a-1) b c>(a-1)(b-1)(c-1)$. On the other hand, since $a \leq b \leq c, \frac{a}{a-1} \geq \frac{b}{b-1} \geq \frac{c}{c-1}$, so $q<\frac{a b c}{(a-1)(b-1)(c-1)} \leq\left(\frac{a}{a-1}\right)^{3}$. In particular, if $a \geq 5,1<q<\frac{125}{64}<2$, but there is no integer strictly between 1 and 2, so there is no solution for $a \geq 5$.

For $a=4, q$ can take the value 2, but this is impossible because $a b c-1$ is odd if $a$ is even.

For $a=3, q$ can take the values 2 and 3, leading to two equations to solve: $3 b c-1=2 \times 2(b-1)(c-1)$, which expands to: $b c-4 b-4 c+5=0$ or $(b-4)(c-4)=11$, giving the first solution: $(a, b, c)=(3,5,15)$. Or: $3 b c-1=3 \times 2(b-1)(c-1)$, which is impossible because $3 b c-1$ cannot be divisible by 3.

In general, the equation: $p b c-q b-q c+r=0$ factors as: $(p b-q)(p c-q)=q^{2}-r p$, so it suffices to find all the factors of $q^{2}-r p$ to solve it.

For $a=2, q$ can theoretically take the values $2,3, \ldots 7$, because $\left(\frac{2}{1}\right)^{3}=8$, but since $a b c-1$ is then odd, only the odd values 3, 5, and 7 are possible. This leads to three equations: $2 b c-1=7(b-1)(c-1)$, which can be written as: $5 b c-7 b-7 c+8=0$, which factors to $(5 b-7)(5 c-7)=9$, leading to a second solution: $(a, b, c)=(2,2,2)$. Or: $2 b c-1=5(b-1)(c-1)$, which can be written as: $3 b c-5 b-5 c+6=0$ or $(3 b-5)(3 c-5)=7$, giving a new solution: $(a, b, c)=(2,2,4)$. Or finally: $2 b c-1=3(b-1)(c-1)$, which is: $b c-3 b-3 c+4=0$ so $(b-3)(c-3)=5$, giving a fourth and final solution: $(a, b, c)=(2,4,8)$.