, 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
$$\begin{aligned} (n+l)(n+l+1) &= \left(n + \frac{n(n+1)}{m}\right) \left(n + 1 + \frac{n(n+1)}{m}\right) \\ &= \frac{l(m+n+1)(m+n)}{m^2}. \end{aligned}$$
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 $(m, n) \neq (n, l)$. Thus starting with the pair $(1, 2)$, we can generate new pair $(2, 6)$ and the process may be continued indefinitely.