, 2022

Solution:

Let $p$ be any prime factor of $a^{2}+b+3$ (such a factor exists because $a^{2}+b+3 \geq 5$). Then $b \equiv -a^{2}-3 \bmod p$. Because $n$ must be an integer, $p$ must also be a prime factor of $a b+3 b+8$. Thus $0 \equiv a b+3 b+8 \equiv a\left(-a^{2}-3\right)+3\left(-a^{2}-3\right)+8 \equiv -a^{3}-3 a^{2}-3 a-1 \equiv -(a+1)^{3} \bmod p$, so $p$ must also be a prime factor of $a+1$. By i), $p$ can occur in $a^{2}+b+3$ at most quadratically, hence at most twice as often as in $a+1$. Since this holds for every prime factor, we obtain $a^{2}+b+3 \leq (a+1)^{2}$.

If, on the other hand, a prime factor $p$ occurred in $a^{2}+b+3$ at most once, then we would have $a^{2}+b+3 \leq \frac{(a+1)^{2}}{p} \leq \frac{(a+1)^{2}}{2}$, contradicting $\frac{(a+1)^{2}}{2} \leq a^{2}+1 < a^{2}+4 \leq a^{2}+b+3$.

Hence we must have $a^{2}+b+3=(a+1)^{2}$. Consequently $b=2 a-2$ and thus $a b+3 b+8=a(2 a-2)+3(2 a-2)+8=2 a^{2}+4 a+2=2(a+1)^{2}$.

Therefore $n=2$ 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

Solution:

If we solve condition ii) for $b$, we obtain $b=\frac{n a^{2}+3 n-8}{a+3-n}$.

Case 1: $n=a+3$. Here, since $b \in \mathbb{Z}^{+}$, the numerator must also vanish, which after substitution gives $(a+1)^{3}=0$, i.e. $a=-1$ - a contradiction to $a \in \mathbb{Z}^{+}$.

Case 2: $n \neq a+3$. From (1) it follows that
$$a^{2}+b+3=a^{2}+\frac{n a^{2}+3 n-8}{a+3-n}+3=\frac{a^{3}+3 a^{2}-n a^{2}+n a^{2}+3 n-8+3 a+9-3 n}{a+3-n}=\frac{(a+1)^{3}}{a+3-n}.$$

Now we denote by $v_{p}(z)$ the multiplicity with which a prime factor $p$ occurs in the number $z \in \mathbb{Z}^{+}$. If there existed a prime factor $p$ with $v_{p}(a+1) \geq v_{p}(a+3-n)+1$, then by (2) we would have $v_{p}\left(a^{2}+b+3\right) \geq 2 v_{p}(a+3-n)+3 \geq 3$, contradicting i). Hence $v_{p}(a+1) \leq v_{p}(a+3-n)$, and thus $a+1$ divides $a+3-n$. From (2) it also follows that $a+3-n>0$ and hence $a+1 \leq a+3-n$, so $n \leq 2$.

For $n=1$ we would need $a+1 \mid a+2$, which holds for no $a \in \mathbb{Z}^{+}$. Hence only $n=2$ can hold, which is confirmed by the example from the first solution. ㅁ