Determine all integers $n \geq 1$ for which there exists a pair $(a, b)$ of positive integers with the following properties:
i) No third power of a prime divides $a^{2}+b+3$.
ii) It holds that $\frac{a b+3 b+8}{a^{2}+b+3}=n$.

Answer: The only integer with these properties is $n=2$.
Let $p$ be any prime factor of $a^{2}+b+3$ (such a prime exists because $a^{2}+b+3 \geq 5$). Then $b \equiv -a^{2}-3 \bmod p$. Due to the integrality of $n$, $p$ must also be a prime factor of $a b+3 b+8$. Therefore, $0 \equiv a b+3 b+8 \equiv a\left(-a^{2}-3\right)+3\left(-a^{2}-3\right)+8 \equiv -a^{3}-3 a^{2}-3 a-1 \equiv -(a+1)^{3} \bmod p$, so $p$ must also be a prime factor of $a+1$. Because of $i$), $p$ can occur in $a^{2}+b+3$ at most quadratically, i.e., at most twice as often as in $a+1$. Since this holds for every prime factor, we get $a^{2}+b+3 \leq (a+1)^{2}$. If a prime factor $p$ occurred in $a^{2}+b+3$ at most simply, we would have $a^{2}+b+3 \leq \frac{(a+1)^{2}}{p} \leq \frac{(a+1)^{2}}{2}$, in contradiction to $\frac{(a+1)^{2}}{2} \leq a^{2}+1 < a^{2}+4 \leq a^{2}+b+3$.
Therefore, $a^{2}+b+3 = (a+1)^{2}$ must hold. Consequently, $b = 2a - 2$ and thus $a b + 3 b + 8 = a(2a - 2) + 3(2a - 2) + 8 = 2a^{2} + 4a + 2 = 2(a+1)^{2}$.
Hence, $n = 2$ is the only possible solution. The example $(a, b) = (2, 2)$ with $a^{2} + b + 3 = 9$ shows that $n = 2$ is indeed a solution. ㅁ