, 2017

Solution:

Firstly note that $\frac{(a+b)(a+b+1)}{a b}=2+\frac{a^{2}+b^{2}+a+b}{a b}$. Let $c$ be this fraction so that $(a+b)(a+b+1)=a b(c+2)$ for some integers $a, b, c$. Suppose $(a, b)$ with $a \geq b$ is a solution for some $c$. Consider the quadratic
$$x^{2}-(b c-1) x+b^{2}+b=0$$
It has one root $a$, and the other root is therefore $b c-a-1$. Furthermore the other root can also be expressed as $\frac{b^{2}+b}{a} \leq \frac{b^{2}+b}{b+1}=b$, so that $0<b c-a-1 \leq b$. In particular, $(b, b c-a-1)$ is a solution as well.
Thus all solutions $(a, b)$ reduce to a solution where $a=b$, at which point $c=2+\frac{2}{a}$. Since $a, c$ are positive integers we thus have $a=1,2$, and so $c=3,4$.
Through this jumping process, we iteratively find the solutions for $c=3$ :
$$(2,2) \rightarrow(2,3) \rightarrow(3,6) \rightarrow(6,14) \rightarrow(14,35) \rightarrow(35,90)$$
and $c=4$ :
$$(1,2) \rightarrow(2,6) \rightarrow(6,21) \rightarrow(21,77)$$
so that the desired pair is $(35,90)$.