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!