Determine all integers $n \geqslant 1$ for which there exists a pair of positive integers $(a, b)$ such that no cube of a prime divides $a^{2}+b+3$ and  
$$ \frac{a b+3 b+8}{a^{2}+b+3}=n $$
As $b \equiv -a^{2}-3\left(\bmod a^{2}+b+3\right)$, the numerator of the given fraction satisfies 
$$ 
a b+3 b+8 \equiv a\left(-a^{2}-3\right)+3\left(-a^{2}-3\right)+8 \equiv -(a+1)^{3} \quad\left(\bmod a^{2}+b+3\right) 
$$ 
As $a^{2}+b+3$ is not divisible by $p^{3}$ for any prime $p$, if $a^{2}+b+3$ divides $(a+1)^{3}$ then it does also divide $(a+1)^{2}$. Since 
$$ 
0<(a+1)^{2}<2\left(a^{2}+b+3\right) 
$$ 
we conclude $(a+1)^{2}=a^{2}+b+3$. This yields $b=2(a-1)$ and $n=2$. The choice $(a, b)=(2,2)$ with $a^{2}+b+3=9$ shows that $n=2$ indeed is a solution.