Determine all triples $(x, y, z)$ of positive integers for which

$$
x \mid (y+1), \quad y \mid (z+1) \quad \text{and} \quad z \mid (x+1)
$$

hold.

(Walther Janous)

Answer. There are ten triples that satisfy the three conditions. They are given by $(1,1,1),(1,1,2),(1,3,2)$, $(3,5,4)$ and their cyclic permutations.

Remark. For the sake of easier readability, we use the notation $a \mid b+c$ instead of $a \mid (b+c)$ in the following.
The equations $y+1=a x, z+1=b y$ and $x+1=c z$ with $a, b, c \in \mathbb{N}_{\geq 1}$ must hold.

Multiplication of these three equations yields $abcxyz = (x+1)(y+1)(z+1)$, so $abcxyz > xyz$, meaning $abc > 1$.

Therefore, at least one of the three factors $a, b$, or $c$ must be greater than 1. Since the three divisibility conditions are invariant under cyclic permutation, we may assume without loss of generality that $a \geq 2$. Thus, we have (since $b, c \geq 1$):

$$
y+1 \geq 2 x, z+1 \geq y \text{ and } x+1 \geq z.
$$

Adding these three inequalities gives $x+y+z+3 \geq 2 x+y+z$, i.e., $x \leq 3$.

We now consider the three possible cases.

- For $x=1$, $1 \mid y+1$ is satisfied, and it follows that $z \leq 1+1=2$ and thus $y \leq 2+1=3$.
- Let $y=1$. Then $1 \mid z+1$ is satisfied, and $z \mid 2$ gives $z=1$ or $z=2$.
- With $y=2$, it follows from $2 \mid z+1$ that $z=1$. Thus, $1 \mid x+1$ is also satisfied.
- For $y=3$, $3 \mid z+1$ implies $z=2$, and $2 \mid x+1$ is satisfied.
- For $x=2$, we have $y+1 \geq 4$, i.e., $y \geq 3$, and $2+1 \geq z$, i.e., $z \leq 3$, and $3+1 \geq y$, so $y=3$ or $y=4$. Since $2 \mid y+1$, only $y=3$ is possible. With $3 \mid z+1$, it follows that $z=2$. Then $2 \mid 2+1$ must hold, which is an obvious contradiction!
- For $x=3$, we get $y+1 \geq 6$, i.e., $y \geq 5$, and $3+1 \geq z$, i.e., $z \leq 4$, and $4+1 \geq y$, so $y=5$. This and $5 \mid z+1$ lead to $z=4$. Thus, all three divisibility conditions are satisfied.

Summary: The triples $(x, y, z)$ for which the three divisibility relations are satisfied are

$$
(1,1,1),(1,1,2),(1,3,2),(3,5,4)
$$

and the cyclic permutations thereof.