Indija TS 2008

If we take $m=1$, $n=2$, we see that $\frac{(m+n)(m+n+1)}{mn}=6$. (Or we can start with $m=2$, $n=3$ as well.)

Suppose we have some pair $(m,n)$ of positive integers such that $m<n$ and $$\frac{(m+n)(m+n+1)}{mn}=k$$ is an integer. This may be written in the form $$nk=m+2n+1+\frac{n(n+1)}{m}$$ Thus $\frac{n(n+1)}{m}$ is also an integer, say equal to $l$. Observe that $$\begin{array}{rl}(n+l)(n+l+1)& =\left(n+\frac{n(n+1)}{m}\right)\left(n+1+\frac{n(n+1)}{m}\right)\\ & =\frac{l(m+n+1)(m+n)}{m^2}.\end{array}$$ Thus $$\frac{(n+l)(n+l+1)}{nl}=\frac{(m+n)(m+n+1)}{mn}=k.$$ Moreover $lm=n(n+1)>n^2>nm$ showing that $l>n$. Hence $(m,n)\ne (n,l)$. Thus starting with the pair $(1,2)$, we can generate new pair $(2,6)$ and the process may be continued indefinitely.