[Number Theory]
Problem: 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

$$
\frac{a b+3 b+8}{a^{2}+b+3}=n
$$

Solution: 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.

Final answer: 2