a|b c-1, \quad b| c a-1, \quad c \mid a b-1
$$

## 1st solution
Obviously, $a, b, c$ are pairwise divisors, because, for example, every common divisor of $a$ and $b$ is also a divisor of 1 because of $a \mid b c-1$ . Since $b c-1$ is divisible by $a$, $a b+b c+c a-1$ must also be divisible by $a$, and for reasons of symmetry the same applies to $b$ and $c$. Because of the divisor strangeness of $a, b$ and $c$, we therefore obtain

$$
a b c \mid a b+b c+c a-1
$$

In particular, the left-hand side is not greater than the right-hand side, i.e. after division by $a b c$

$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a b c} \geq 1
$$

We can now assume $a \leq b \leq c$ and distinguish three cases:

(i) For $a \geq 3$ the left-hand side is less than 1, contradiction.

(ii) Let $a=2$. For $b \geq 4$ the left-hand side is less than 1, therefore $b=2.3$. Only $b=3$ leads to the solution $c=5$.

(iii) For $a=1$ we obtain $c \mid b-1$ from the original conditions. Because of $c \geq b$ it follows that $b=1$.

It is easy to see that in the cases $(a, b, c)=(2,3,5)$ and $a=b=1$ the conditions are fulfilled, so the triples $(a, b, c)$ we are looking for are the permutations of

$$
(2,3,5) \quad \text { and } \quad(1,1, n), \quad n \geq 1
$$