Question 61
Find all integers $n \ge 1$ for which there exists a pair of positive integers $(a, b)$ such that
$$
\frac{ab + 3b + 8}{a^2 + b + 3} = n
$$
and no cube of a prime divides $a^2 + b + 3$.
$$
\frac{ab + 3b + 8}{a^2 + b + 3} = n
$$
and no cube of a prime divides $a^2 + b + 3$.