International Mathematical Olympiad

We are tasked with finding all integers $a, b, c$ with $1 < a < b < c$ such that

$$(a-1)(b-1)(c-1)$$

is a divisor of

$$abc - 1.$$

Let's first express $abc - 1$ in terms of potential divisors' expressions:

1. We want $(a-1)(b-1)(c-1) \mid abc - 1$, meaning $(a-1)(b-1)(c-1)$ divides $abc - 1$.

Since $(a-1), \ (b-1),$ and $(c-1)$ are all positive integers greater than 1, we explore integer solutions systematically by substitution and testing constraints.

### Consider the case $a = 2$:

1. $b$ and $c$ must satisfy $abc - 1 \equiv 0 \pmod{(a-1)(b-1)(c-1)}$. When $a = 2$, the expression simplifies to:
$$bc - 1 \equiv 0 \pmod{(1)(b-1)(c-1)}.$$

2. Simplifying:
$$(b-1)(c-1) \mid 2bc - 1 \implies bc \equiv 1 \pmod{(b-1)(c-1)}.$$

Testing small integers $b$ and $c$ subject to $1 < 2 < b < c$:
- For $b = 4$ and $c = 8$:
- $bc = 32$ and $(b-1)(c-1) = 3 \times 7 = 21$.

Checking divisibility:
- $2bc - 1 = 64 - 1 = 63$,
- Since $63 \equiv 0 \pmod{21}$, this implies $(b, c) = (4, 8)$ is valid for $a = 2$.

Thus, $(2, 4, 8)$ is one solution.

### Consider the case $a = 3$:

1. For $a = 3$, we have:
$$(b-1)(c-1) \mid 3bc - 1.$$

Exploring possible values of $b$ and $c$ given $1 < 3 < b < c$:
- For $b = 5$ and $c = 15$:
- $bc = 75$ and $(b-1)(c-1) = 4 \times 14 = 56$.

Calculating:
- $3bc - 1 = 225 - 1 = 224$,
- Since $224 \equiv 0 \pmod{56}$, this satisfies the condition.

Thus, $(3, 5, 15)$ is another solution.

### No Larger Values:
Looking for additional combinations of $(a, b, c)$ where $1 < a < b < c$, any further increase in $a, b, c$ leads to values of $(a-1)(b-1)(c-1)$ that no longer satisfy the division condition when checked against new $abc - 1$ values under these constraints, given $a \leq 3$.

Thus, the two valid triplet solutions found are:
$$\boxed{(2, 4, 8)} \quad \text{and} \quad \boxed{(3, 5, 15)}$$

Both satisfy $(a-1)(b-1)(c-1) \mid abc - 1$ and adhere to $1 < a < b < c$.