Gazeta Matematica 6-7-8/2017, Problem 27391

Problem:
Find all triples of consecutive integers such that the sum of all the possible quotients between two numbers is an integer.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Let $n-1,n,n+1$ be three consecutive integers. We have to find all the integers $n$ such that $$\dfrac{n-1}{n}+\dfrac{n-1}{n+1}+\dfrac{n}{n-1}+\dfrac{n}{n+1}+\dfrac{n+1}{n-1}+\dfrac{n+1}{n}=N$$ is an integer. Observe that
$$\begin{eqnarray*} N&=&2+\dfrac{2n+1}{n-1}+\dfrac{2n-1}{n+1}\\&=&6+\dfrac{3}{n-1}-\dfrac{3}{n+1}\\&=&6+\dfrac{6}{n^2-1}. \end{eqnarray*}$$ Hence $N$ is an integer if and only if $6/(n^2-1)$ is an integer, i.e. if and only if $(n^2-1)|6$. Therefore, $n^2-1 \in \{\pm 1, \pm 2, \pm 3, \pm 6\}$. By an easy check we get $n=\pm 2$. We conclude that the only triples which satisfy the conditions are $(-3,-2,-1)$ and $(1,2,3)$.