Austrian Mathematical Olympiad

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

Without loss of generality, let $x$ be the smallest of the three numbers (or one of the smallest), i.e. $x\le y$ and $x\le z$. From $z\mid x+1$ we obtain $x\le z\le x+1$. Thus we have to consider two cases.

Case 1. Let $z=x$. Then $z=x\mid x+1$ leads to $x=z=1$ and $y\mid z+1=2$. Therefore $y=1$ or $y=2$, and we get the two solutions $(1,1,1)$ and $(1,2,1)$.

Case 2. Let $z=x+1$. Then the two conditions $x\mid y+1$ and $y\mid x+2$ must be fulfilled. In particular, we obtain $x\le y+1$ and $y\le x+2$. This yields $x-1\le y\le x+2$ and we have to examine the following cases for $y$.

Case 2a. Let $0<y=x$. The conditions $x\mid x+1$ and $x\mid x+2$ can only hold simultaneously for $x=1$, giving the solution $(1,1,2)$.

Case 2b. Let $y=x+1$. Then the two conditions are $x\mid x+2$ and $x+1\mid x+2$. They cannot hold simultaneously.

* Case 2c. Let $y=x+2$. The condition $y=x+2\mid x+2=z+1$ is trivially fulfilled. The requirement $x\mid y+1=x+3$ can only hold for $x\mid 3$. And, indeed, for either $x=1$ or $x=3$ the condition is fulfilled and we obtain the solutions $(1,3,2)$ and $(3,5,4)$.

Summing up, the triples $(1,1,1)$, $(1,2,1)$, $(1,1,2)$, $(1,3,2)$ and $(3,5,4)$ fulfill all three conditions. As each of the three numbers can be the minimum, every cyclic permutation of these triples is a solution as well.