A Diophantine equation Let $a,b$ be integers, $a>b\ge 1$, 
$a^2(a+1)$ be divisible by $b$, and 
$3a^2$ be divisible by $b$.
Let us consider the following expression:
$\frac {1+3a+3a^2+a^2(a-b+1)/b}
{1+\frac{3}{a}+\frac{3}{a^2}+\frac{b}{a^2(a-b+1)}}$.
This fraction is always integer if $b=1$. For $b>1$, I know only one pair $a,b$ such that 
the fraction is integer, namely, $a=15,b=9$ (I checked all pairs with $a<10^4$).
Can anyone prove that there are no other pairs $a,b$ such that the fraction is integer?
This question is related to the algebraic graph theory.
Thanks for any comments or hints!

If we substitute $f= \frac{a-b+1}{ab}$ and subtract $a^3$ the question becomes; when is this an integer:
$$\frac{a^3 \times (f-1) \times (a^3 + \frac{1}{f})}{(a+1)^3 + \frac{1}{f} -1}$$
Note that $b=1$ implies that $f=1$ and this quotient is zero.
Using Maple I found that $a=14$, $b=12$ is a second example which satisfies your conditions.
(sorry, not a full answer, I'm new here and not sure how to post)
