Note that $a+3-n=\frac{(a+1)^3}{a^2+b+3},$ and so $a^2+b+3 \mid (a+1)^3$. Since $v_p(a^2+b+3) \leq 2$ for all primes $p$, if $p \mid a^2+b+3$ then $p \mid a+1$ and so $v_p((a+1)^2) \geq 2 \geq v_p(a^2+b+3)$. Hence, $(a^2+b+3) \mid (a+1)^2$. Thus, the RHS above is a multiple of $a+1$, hence so is $a+3-n$. However $0<a+3-n <a+3$, and so $a+3-n=a+1$, implying that $n=2$. $n=2$ indeed works, since we may take $a=b=2$.

We have \begin{align*} n&=\frac{ab+3b+8}{a^+b+3}\\ &=\frac{a((a^2+b+3)-(a^2+3))+3((a^2+b+3)-(a^2+3))+8}{a^2+b+3}\\ &=\frac{(a+3)(a^2+b+3)-a^3-3a-3a^2-9+8}{a^+b+3}\\ &=a+3-\frac{(a+1)^3}{a^2+b+3}. \end{align*}For all $p\mid a+1$, we have $\nu_p((a+1)^3)\geq3$ and $\nu_p(a^2+b+3)\leq2$. Therefore, $\nu_p\left(\frac{(a+1)^6}{(a^2+b+3)^3}\right)\geq0$. If $p\nmid a+1$, then $a^2+b+3\mid (a+1)^3$ implies $p\nmid a^2+b+3$, so $\nu_p\left(\frac{(a+1)^6}{(a^2+b+3)^3}\right)=0$. Therefore, $\frac{(a+1)^6}{(a^2+b+3)^3}=\left(\frac{(a+1)^2}{a^2+b+3}\right)^3$ is an integer, so $a^2+b+3\mid(a+1)^2$. If $a^2+b+3\leq\frac12(a+1)^2$, then $2a^2+2b+6\leq a^2+2a+1$, which means $a^2-2a+5+2b=(a-1)^2+2b+4\leq0$, which is impossible. Therefore, $a^2+b+3=(a+1)^2$, so $n=a+3-\frac{(a+1)^3}{(a+1)^2}=a+3-(a+1)=2$. If $a=b=2$, then $a^2+b+3=9$ is not divisible by the cube of a prime, and $n=\frac{ab+3b+8}{a^2+b+3}=\frac{4+6+8}{4+2+3}=2$. This means that the only possible value of $n$ is $\boxed2$.

Nice one! Firstly, we must have $ab+3b + 8 \geq a^2 + b + 3$, so $b \geq \frac{a^2-5}{a+2} > a-3$ and hence $b\geq a-2$. Write $b=a+\ell-3$ for an integer $\ell \geq 1$ so that $n = \frac{a^2 + \ell a +3\ell - 1}{a^2 + a +\ell} = 1 + \frac{\ell(a+2) - (a+1)}{a^2 + a + \ell}$. Since the numerator is positive, we must have $\ell(a+2) - (a+1) \geq a^2 + a + \ell$ which is equivalent to $\ell \geq a+1$, so write $\ell = a + 1 + m$ for an integer $m\geq 0$. This gives $n = 1+\frac{(a+1)^2+m(a+2)}{(a+1)^2+m} = 2 + \frac{m(a+1)}{(a+1)^2+m} = 2 + \frac{km}{k^2+m}$ (where $k=a+1\geq 2$) and we require $k^2 + m$ to be not divisible by a cube of a prime. If $m=0$, then any cube-free $k\geq 2$ works and $n=2$. For $m>0$ we have $km \equiv -k^3 \pmod {m+k^2}$, so $m+k^2 \mid k^3$, but since $m+k^2$ is cube-free we must now have $m + k^2 \mid k^2$ and thus $m+k^2 \leq k^2$, contradicting $m>0$. Therefore the only possible $n$ is $2$.

The answer is $n=2$ only. To show this is achievable, take $(a,b)=(2,2)$. We eliminate the variable $b$ using standard techniques: \begin{align*} a^2+b+3 &\mid ab+3b+8 \\ \implies a^2+b+3 &\mid (ab+3b+8) - (a+3)(a^2+b+3) \\ &= -(a+1)^3. \end{align*}Because of the cube-free condition, this implies \[ a^2+b+3 \mid (a+1)^2. \]Since $(a+1)^2 < 2(a^2+b+3)$, the divisibility only occurs when \[ a^2+b+3 = (a+1)^2 \implies b = 2a-2. \]Plugging that back into the original equation gives \[ n = \frac{(a+3)(2a-2)+8}{a^2+(2a-2)+3} = \frac{2a^2+4a+2}{a^2+2a+1} = 2. \]Remark: In fact, this lets us characterize all $(a,b)$. The set of solutions is $(a,2a-2)$ for any positive integer $a$ such that $a+1$ is squarefree (so that $a^2+b+3=(a+1)^2$ is cube-free).

It's similar to Canada MO 2019/2. (This is also Thailand TSTST 3.2.) The only solution is $n=2$. Consider $ab+3b+8$ modulo $a^2+b+3$. $$ab+3b+8\equiv a(-a^2-3)+3(-a^2-3)+8\equiv -(a+1)^3\pmod{a^2+b+3}$$Thus, $a^2+b+3\mid (a+1)^3$. Let $p$ be any prime factor of $a^2+b+3$. From the problem's condition, we have $v_p(a^2+b+3)=1,2$. Also, note that $$p\mid (a+1)^3\implies p^3\mid (a+1)^3.$$Let $a^2+b+3=p_1^{q_1}p_2^{q_2}\dots p_m^{q_m}$ where $p_1,p_2,\dots,p_m$ are different primes and $q_i\in\{1,2\}$. $$(p_1p_2\dots p_m)^3\mid (a+1)^3\implies (a+1)^3\geq (p_1p_2\dots p_m)^3\geq (a^2+b+3)^{\frac{3}{2}}$$Hence, $(a+1)^2\geq a^2+b+3$ and so $2a-2\geq b$. Claim: $n\neq 1$ Proof. Assume the contrary. We have $ab+3b+8=a^2+b+3.$ $$b=\frac{a^2-5}{a+2}\implies a+2\mid (a^2-5)-(a-2)(a+2)=-1$$Thus, $a+2=1,-1$ which is absurd. $$\frac{ab+3b+8}{a^2+b+3}=n\geq 2\implies ab+3b+8\geq 2a^2+2b+6$$Thus, $ab+b\geq 2a^2-2$ and so $b\geq 2a-2$. But we also know that $2a-2\geq b$. Hence, the equality holds which means that $n=2$. Note that $(a,b)=(2,2)$ works, and we are done.

Note that \[(a+3)(a^2+b+3)=a^3+3a^2+ab+3b+3a+9=(a+1)^3+ab+3b+8.\]Therefore, \[n=(a+3)-\frac{(a+1)^3}{a^2+b+3}.\]Suppose $\nu_p(a+1)=k$ for $k>0$ and prime $p.$ Then $\nu_p((a+1)^3)=3k.$ However, it is given that $p^3\nmid a^2+b+3,$ so $\nu_p(a^2+b+3)\le 2.$ In particular, $\nu_p(a^2+b+3)\le\tfrac{2}{3}\nu_p((a+1)^3).$ Repeating this for all $p$ dividing $a+1,$ we get the inequality \[a^2+b+3\le [(a+1)^3]^{2/3}=(a+1)^2.\]Plugging this into our expression for $n,$ \[n\le (a+3)-\frac{(a+1)^3}{(a+1)^2}=2.\]If $n=2,$ then $(a,b)=(2,2)$ works. If $n=1,$ then we must have $\tfrac{(a+1)^3}{a^2+b+3}=a+2.$ This implies $a+2\mid(a+1)^3,$ which is impossible as $\gcd(a+1,a+2)=1.$ In conclusion, the only solution is $\boxed{n=2}.$

The answer is $\boxed{n=2}$ only, which works by $(a,b) = (2,2)$. We now prove this is the only solution. We have $a^2+b+3\mid ab+3b+8$ implies that \begin{align*} a^2+b+3\mid (a+3)(a^2+b+3) - (ab+3b+8) \\ = (a+1)^3 \\ \end{align*} Since $a^2+b+3 $ is cube free, we have $a^2 + b+3 \mid (a+1)^2=a^2 + 2a+1$. Now, \[2(a^2+b+3) = 2a^2 + 2b + 6 > a^2 + 2a + 1\]So \[a^2+ b + 3 = a^2 + 2a + 1\implies b = 2a - 2\] So \begin{align*} n\\ =\frac{(a+3)(2a-2) + 8}{a^2 + 2a+1} \\ =\frac{2a^2 + 4a + 2}{a^2 + 2a+1} \\ =2, \end{align*}as desired.

If $a>b+2$ $a^2+b+3|ab+3b+8$ $\rightarrow$ $a(a-b) \leq 2b+5$ absurd $a=b$ $\rightarrow$ $a^2+a+3|a^2+3a+8-a^2-a-3=2a+5$ $\rightarrow$ $a=b=1$ $a=b+1$ $\rightarrow$ $b^2+3b+4|b^2+4b+8$ no exist $b$ positive integer $a=b+2$ $\rightarrow$ $b^2+5b+3|b^2+5b+8$ no exist $b$ positive integer $\color{green} \rule{20 cm}{2pt}$ $\color{black} \clubsuit$ $\color{green} \rule{20 cm}{2pt}$ If $b\geq a$ claim$1:$$a+3-n=\frac{(a+1)^3}{a^2+b+3}$ $v_{p}(a^2+b+3)\leq 2$ for all $p$ primes If $p|a^2+b+3$ $\rightarrow$ $p|a+1$ $v_{p}((a+1)^2)\leq 2\leq v_{p}(a^2+b+3)$ $\rightarrow$ $(a+1)^2 \leq a^2+b+3$ $\rightarrow$ $a^2+b+3|(a+1)^2$ this is absurd because $b \leq a$

Note that $(ab+3b+8)=(a^2+b+3)(a+3)-(a+1)^3$ Which means $(a+3)-\frac{(a+1)^3}{a+b+3}=n$, but the problem says $a^2+b+3$ is cubefree Hence $a^2 + b+3 \mid (a+1)^2$. But clearly $a^2+b+3=(a+1)^2$, since $(a+1)^2 < 2(a^2+b+3)$. Hence we have $b = 2a-2$ Plugging into the original equation gives $n=2$

The answer is $n = 2$ only, which is satisfied by $a=b=2$. Rewrite the equation as $$(3+a) - n = \frac{(a+1)^3}{a^2 + b + 3}.$$Writing $a+1 = p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ for primes $p_i$ and exponents $\alpha_i \geq 1$ yields $(a+1)^3 = p_1^{3\alpha_1}p_2^{3\alpha_2}\cdots p_k^{3\alpha_k}$. Since $a^2+b+3$ is not divisible by the square of any prime, the exponents in its prime factorization must be at most $2 \leq 2\alpha_2$, so $a^2+b+3$ is a divisor of $p_1^{2\alpha_1}p_2^{2\alpha_2}\cdots p_k^{2\alpha_k} = (a+1)^2$. If $a^2+b+3=(a+1)^2$, then $$(3+a)-n = a+1\iff n = 2,$$which we saw works with $a=b=2$. Else, $a^2+b+3$ is less than or equal to $\frac{(a+1)^2}{2}$, so $$(3+a) - n = \frac{(a+1)^3}{a^2+b+3} \geq \frac{(a+1)^3}{\frac{(a+1)^2}{2}} = 2(a+1)\iff 1-a\geq n\implies 1 > n,$$a contradiction. Thus, we must have $a^2+b+3 = (a+1)^2\implies n = 2$, as desired.

We proceed in several steps. Step 1: $(a+1)^3$ is divisible by $a^2+b+3$ Proof: Simply note that: $$ a^2+b+3 \mid 3(a^2+b+3)-ab-3b-8 =3a^2-ab +1$$On the other hand: $$ a^2+b+3 \mid a(a^2+b+3) +(3a^2-ab+1) = a^3+3a^2+3a+1 =(a+1)^3 $$as desired. Step 2: $(a+1)^2$ is divisible by $a^2+b+3$ Proof: Observe that for any prime $p$ dividing $a^2+b+3$, we have $\nu_p(a^2+b+3) \le 2$. On the other hand $p \mid (a+1)^3 \implies p \mid a+1$. Therefore $\nu_p(a+1) \ge 1 \implies \nu_p((a+1)^2) =2\nu_p(a+1) \ge 2$. This implies that $\nu_p((a+1)^2) \ge \nu_p(a^2+b+3)$. Repeating this argument for every prime $p$ dividing $a^2+b+3$ yields to $a^2+b+3 \mid (a+1)^2$ as needed. Step 3: $(a+1)^2 = a^2+b+3$. Proof: Suppose that $(a+1)^2 = k(a^2+b+3)$ where $k \ge 2$. We rewrite equation as follows: $$ ka^2+kb+3k = a^2+2a+1$$Note that $a^2+1 \ge 2a $, therefore: $$ a^2+2a+1 = (k-1)a^2+a^2 +1 +kb +3k >a^2 +2a + 1 $$which is obviosly a condradiction. Now we finish the problem as follows. Just note that: $$ a^2+2a+1 = a^2+b+3 \implies b =2a-2 $$Moreover: $$ n =\frac{ab+3b+8}{a^2+b+3} = \frac{a(2a-2)+3(2a-2)+8}{a^2-2a+1} =\frac{2(a^2-2a+1)}{a^2-2a+1}=2 $$For $n=2$, $a=2$, $b=2$ works, since $a^2+b+3 = 9$, which is not divisible by a cube.

We claim the answer is $n=2$ only, achieved with $(a,b)=(2,2)$. This can be shown to work trivially. Note that $$0 \equiv ab+3b+8 \equiv (a+3)(-a^2-3)+8 \equiv (a+1)^3 \pmod{a^2+b+3}.$$Thus $a^2+b+3|(a+1)^3$. Claim: $a^2+b+3|(a+1)^2$. Proof: Assume not. Then for some prime $p$, $2v_p(a+1)<v_p(a^2+b+3) \le 3v_p(a+1)$. This obviously implies that $v_p(a+1) \ge 1$. We have that the condition gives $v_p(a^2+b+3) \le 2$, so we also have that $2v_p(a+1)<v_p(a^2+b+3) \le 2$, so $v_p(a+1) \le 1$, a contradiction so we have proven the claim. Assume $a^2+b+3 \neq (a+1)^2$. This would mean $2(a^2+4) \le 2(a^2+b+3) \le (a+1)^2$, so $(a-1)^2+6 \le 0$, an obvious contradiction. Thus we must have $a^2+b+3=(a+1)^2 \implies b=2a-2$. This gives that $$n=\frac{(a+3)(2a-2)+8}{a^2+2a+1}=\frac{2a^2+4a+2}{a^2+2a+1}=2,$$so this means that $n=2$ is the sole solution and we are done.

Write $$n = a + 3 - \frac{(a+1)^3}{a^2 + b + 3} \implies a^2 + b + 3 | (a+1)^3$$Claim : $a^2 + b + 3 | (a+1)^2$ Proof : Suppose for any arbitrary prime $p$ , $p | a^2 + b + 3 \implies p|a+1 \implies 2v_p(a+1) \geq 2 \geq v_p(a^2 + b + 3)$ from the problem hypothesis . Thus $n - 3 - a \equiv n - 2 \equiv 0$(mod $a+1$) $\implies n = (a+1)k + 2$ for some positive integer $k$ , but note that $(a+1)k + 2 < a + 3 \implies k = 0$. Therefore $n = 2$ is the only answer .( Take $(a,b) = (2,2)$).

Writing as partial fractions: $$n=\frac{-(a+1)^3}{a^2+b+3} +a+3 \implies a^2+b+3\mid -(a+1)^3$$but because $a^2+b+3$ isn't divisible by cube of primes $a^2+b+3\mid (a+1)^2.$ This forces $(a+1)^2=a^2+b+3$ as $2a^2+6+2b> (a+1)^2.$ Solving we get $b=2a-2$ and finally $n=2.$

Nice and easy NT, appropriate for this position. Note that $b \equiv -a^2-3 (mod a^2+b+3)$, so plugging into the divisibility gives $a^2+b+3 | (a+1)^3$. Due to condition (1) we get $a^2+b+3 | (a+1)^2$, and if $a^2+b+3\geq \frac{(a+1)^2}{2}$ one easily gets contradiction. Thus $a^2+b+3=(a+1)^2$ and $b=2a-2$. Plugging into (2) one obtains $2a(a+3)=(a+1)^2n$ and it's easy to see $a+1$ can't be divisible neither by an odd prime $p$, nor by $4$, so now it's easy to finish.

Lets look at the equation $ n = \frac{ab + 3b + 8}{a^2 + b + 3}$ First things first, It's a N1, so this must be resolvable by simple ideas, like divisibility manipulation. Notice that $a^2+ b + 8 $must divide $ab + 3b + 8 $, so it must divide $(ab + 3b + 8) - (a+3)*(a^2 + b + 3)$. So, we have that $(a^2 + b + 3) | - ( a^3 + 1 ) $ . Now, lets look at the $Vp$, since $Vp(ab + 3b + 8) \leq 2 $ , we conclude that, if there is a prime dividing $a^2 + b + 3$, so the prime divide $(a + 1)^3 $ and it's Vp is at least 3, so, $a^2 + b + 3$ divides $(a+1)^2$ . Since the number $\frac{(a+1)^3)}{a^2 + b + 3)}$, the number must be a multiple of (a+1). Since the RHS is a +3 - n (by divisibility, manipulations) we have that $a +3 - n \geq a + 1$. So, $ 2 \geq n $. Since (a,b,n) being (2, 2, 2) is a solution, we can prove that n = 1 results in a contradiction. Let n = 1, so $a^2 + b + 3 = ab + 3b + 8$, making some algebric manipulations trying to isolate b, we get that $\frac{a^2 - 5}{2-a} = b$, since for all a (you can check) this number is negative and b is a positive integer, n = 1 isn't a solution. So, the unique solution is (2,2,2).

The condition is equivalent $\frac{(a+1)^3}{a^2+b+3}$ being an integer and since $a^2+b+3$ is not divisible by the cube of any prime, this implies that $\frac{(a+1)^2}{a^2+b+3}$ is an integer, or that $a^2+b+3\mid 2a-b-2$. Suppose that $2a\neq b+2$. Then, $a^2+b+3\le 2a-b-2$ or $a^2+b+3\le b+2-2a$. Now note that the second inequality only holds when $a=-1$, meaning that $a^2+b+3\le 2a-b-2$. But now this gives $0\le a^2-2a+1\le -2b-4$ which is also a contradiction. Thus $2a=b+2$ which gives $\boxed{n=2}$. We are done. $\blacksquare$

Cute one! I enjoyed this one. Solution: We will show that $(a,b) = (a, 2a-2)$ for $a>1$ where $a$ is such that there exists no prime $p$ such that $p^2 \mid a+1$. As a matter of fact, we show that $n=2$ is the only solution. For verification of answer, just put $b = 2a-2$ to get \[n = \frac{ab+3b+8}{a^2+b+3} = \frac{2(a+1)^2}{(a+1)^2} = 2\]and the problem condition of ``$a^2+b+3$ is not divisible by the cube of any prime" is also satisfied. Some easy computations give \begin{align*} a^2+b+3 &\mid ab+ 3b +3 -3(a^2+b+3) \\ a^2+b+3 &\mid 3a^2+1-ab \\ a^2+b+3 &\mid 3a^2+1-ab + a(a^2+b+3) \\ a^2+b+3 &\mid (a+1)^3 \end{align*}As $a^2+b+3$ is not divisible by cube of any prime, we get \begin{align*} a^2+b+3 &\mid (a+1)^2 \\ a^2+b+3 &\mid a^2+2a+1 - (a^2+b+3) \\ a^2+b+3 &\mid 2a-b-2 \end{align*}Now $a^2+b+3 \le |2a-b-2|$ for $2a-b-2 \ne 0$. Note that if $2a-b-2 = 0$ then we get the solution we claimed. We now have to consider two cases. Case 1: $2a-b-2 > 0$. The inequality we get is \[a^2+b+3 \le 2a -b -2 \iff (a-1)^2 +2b +4 \le 0\]which is just impossible. Case 2: $2a-b-2 < 0$. The inequality we get this time will be \[a^2 + b +3 \le b+2-2a \iff (a+1)^2 \le 0\]which is again impossible. As we have exhausted all other possibilities, we have shown that the claimed solutions are the only ones. $\blacksquare$

We have $a+3-n = \frac{(a+1)^3}{a^2+b+3}$, so $a^2+b+3 \mid (a+1)^3$. Since $a^2+b+3$ is cubefree, we get that $a^2+b+3 \mid (a+1)^2$. But $2(a^2+b+3) > 2(a^2 + 1) \ge 2(a+1)^2$, so $a^2+b+3 = (a+1)^2$. This means $b = 2a-2$, and $n = 2$. Indeed, $n=2$ works because the tuple $(a,b) = (2,2)$ does the job.

The answer is n=2, with (a,2a-2). Subtracting a+3 multiples of the LHS from the right, we get $a^2+b+3\mid(a+1)^3$. However, $$p \mid a^2+b+3\implies p \mid a+1\implies v_p((a+1)^2) \ge 2 \geq v_p(a^2+b+3),$$so we can get reduce RHS to (a+1)^2. Since 2LHS>RHS, we must have RHS=LHS so $a^2+b+3=(a+1)^2\implies b=2a-2\implies n=2$.

Claim: $b + 3 = 2a + 1$ Proof. Note that \[ (a + 3)(a^2 + b + 3) = a^3 + ab + 3a + 3a^2 + 3b + 9 = (a + 1)^3 + ab + 3b + 8 \]so it is equivalent to have that $a^2 + b + 3 \mid (a + 1)^3$. Since $a^2 + b + 3$ is not divisible by any cube of a prime, it follows that $a^2 + b + 3 \mid a^2 + 2a + 1$. Since $a^2 < a^2 + b + 3 \le (a + 1)^2$, it must follow that $b + 3 = 2a + 1$ or that $b = 2a - 2$. $\blacksquare$ Then, the expression simplifies as \[ n = \frac{2a^2 + 4a + 2}{a^2+2a+1}. \]so $n = 2$ is the only such integer.

Note that \begin{align*} a^2+b+3 &\mid ab+3b+8\\ a^2+b+3 & \mid (a+3)(a^2+b+3)-(ab+3b+8)\\ a^2+b+3 &\mid (a+1)^3 \end{align*}Now, since $a^2+b+3$ is not divisible by the cube of any prime, we must infact have \[a^2+b+3 \mid (a+1)^2\]Thus, $b+3 \leq 2a+1$. Now, we have \[a^2+b+3 \mid 2a-b-2\]If $2a-b-2\neq 0$ we require, \[2a-b-2\geq a^2+b+3 \implies (a-1)^2\leq -2(b+1)<0\]which is clearly impossible. Thus, the only possibility is when $b=2a-2$. Plugging this into the given fraction we obtain that then $n=2$ which is also clearly not divisible by a cube of any prime.

The idea is that the only way for the numerator to get big relative to the denominator is for $b$ to get big; thus we try to eliminate $b$ from the dividend. Write $$b+(a^2+3) \mid (a+3)b + (a^2+3)(a+3) - (a+3)b - 8 = (a^2+3)(a+3) - 8 = (a+1)^3.$$By the given condition, it follows that $b+a^2+3 \mid (a+1)^2$, which implies $b \leq 2a-2$. However, notice that $$ab+3b + 8 > 2(a^2+b+3) \iff b > 2a-2.$$It thus follows that $n \leq 2$. When $n=1$, the equation solves to $$b = \frac{a^2 - 5}{a+2} = a-2 + \frac 1{a+2}$$which has no positive integer solutions. $n=2$ obviously works with $b=2a-2$.

The answer is only $n=2$. The problem condition implies that $a^2 + b + 3 \mid ab + 3b + 8$. So we get that, \begin{align*} a^2 + b + 3 &\mid ab+3b+8 \equiv (ab+3b+8) - 3(a^2 + b+3)\\ &= ab - 1 - a^2 \equiv (ab - 1 - a^2) - a(a^2 + b + 3) \\ &= -(a+1)^3 \equiv (a+1)^3 .\end{align*} This means that $a^2 + b + 3 \mid (a+1)^3$. But now due to the condition that $a^2 + b + 3$ is not divisible by the cube of any prime, we can restrict the condition to $a^2 + b + 3 \mid (a+1)^2$ since $\nu_{p}(a^2 + b+3)\le 2$ and that $2\nu_{p}(a+1) \ge 2$ anyways. Thus $a^2 + b + 3 \mid (a+1)^2 = a^2 + 2a + 1 \equiv (a^2 + 2a + 1) - (a^2 + b+ 3) = 2a -b -2$. Now a simple bound shows that $a^2 + b + 3 \ge \max\{2a, b+2\}$ for all $a\ge 3$. For now we assume $a\ge 3$. Then this forces $2a = b+2 \implies b = 2a - 2$. Now substituting this into the problem statement, we get $n=\dfrac{2(a+1)^2}{(a+1)^2} = 2$ for square-free $a+1$. This gives the possibility for $n=2$. Otherwise, if $a = 2$, then $b \in \left\{9,27\right\}$ which gives $(a,b)=(2,9)$ or $(2,20)$; none of which work. Now for the other case when $a=1$, we get $b=4$ which gives $(a,b) = (1,4)$ which does not work either due to the cube condition.

noice Trivially $a^2 + b + 3 \equiv 0 \mod{a^2 + b + 3} \implies b \equiv -a^2 - 3 \mod{a^2 + b + 3}$. Since $ab + 3b + 8 \equiv 0\mod{a^2 + b + 3}$, we have $-a^3 - 3a^2 - 3a - 1 = -(a + 1)^3 \equiv 0\mod{a^2 + b + 3}$. From the prime statement given, we must have $a^2 + b + 3 \mid (a + 1)^2$. Since $(a + 1)^2 < 2(a^2 + b + 3)$ and it's positive, we must have $a^2 + 2a + 1 = a^2 + b + 3 \implies 2a + 1 = b + 3 \implies b = 2a - 2$. Plugging this back, we have $\frac{2a^2 + 4a + 2}{a^2 + 2a + 1} = 2 = n$, which means $n = 2$. $(a, b) = (2, 2)$ confirms this result.

noice Trivially $a^2 + b + 3 \equiv 0 \mod{a^2 + b + 3} \implies b \equiv -a^2 - 3 \mod{a^2 + b + 3}$. Since $ab + 3b + 8 \equiv 0\mod{a^2 + b + 3}$, we have $-a^3 - 3a^2 - 3a - 1 = -(a + 1)^3 \equiv 0\mod{a^2 + b + 3}$. From the prime statement given, we must have $a^2 + b + 3 \mid (a + 1)^2$. Since $(a + 1)^2 < 2(a^2 + b + 3)$ and it's positive, we must have $a^2 + 2a + 1 = a^2 + b + 3 \implies 2a + 1 = b + 3 \implies b = 2a - 2$. Plugging this back, we have $\frac{2a^2 + 4a + 2}{a^2 + 2a + 1} = 2 = n$, which means $n = 2$. $(a, b) = (2, 2)$ confirms this result.

We claim the only possible value is $\boxed{n = 2}$. Rewrite the divisibility as, \begin{align*} a^2 + b + 3 &\mid ab + 3b + 8\\ \iff a^2 + b + 3 &\mid a^3 - 3b + 3a - 8\\ \iff a^2 + b + 3 &\mid a^3 + 3a^2 + 3a + 1\\ \iff a^2 + b + 3 &\mid (a+1)^3 \end{align*}Now since $a^2 + b + 3$ is cube-free we know that $a^2 + b + 3 \mid (a+1)^2$. Then we have, \begin{align*} a^2 + b + 3 &\mid a^2 + 2a + 1\\ \iff a^2 + b + 3 &\mid 2a - b - 2 \end{align*}Now assume $2a - b - 2 \neq 0$. Then we must have, \begin{align*} a^2 + b + 3 \leq |2a - b - 2| \end{align*}Then either, \begin{align*} a^2 + 2b + 5 \leq 2a \end{align*}which is absurd as $a^2 + 1 \geq 2a \iff (a - 1)^2 \geq 0$ for all $a$. The other case is, \begin{align*} (a+1)^2 &\leq 0 \end{align*}which is absurd for positive $a$. Thus equality holds and we always have $2a = b + 2 \iff 2a - 2 = b$. Then plugging this into our original equation, \begin{align*} n &= \frac{2a^2 + 4a + 2}{a^2 + 2a + 1}\\ &= 2 \end{align*}This can be verified for say $(a, b) = (2, 2)$ so we are done. $\square$

We claim the answer is $n=2$, which is possible through the ordered pair $(2,2)$. We now show it is the only one. By using long division, we can say: \[n=a+3-\frac{(a+1)^3}{b+a^3+3}.\] In order for $n$ to be an integer: \[3v_p(a+1)\geq v_p(a^3+b+3).\]Note that the RHS is always less than or equal to $2$, by the given condition. As the LHS takes only multiples of $3$, we can say: \[2v_p(a+1)\geq v_p(a^3+b+3),\]\[\frac{a^2+2a+1}{b+a^3+3}\in \mathbf{Z}\]if $n$ is an integer. This means that: \begin{align*} n&=a+3-(a+1)\cdot\frac{a^2+2a+1}{b+a^3+3}\\ &=a+3-(a+1)\cdot\left(1+\frac{2a-b-2}{a^2+b+3}\right)\\ &=2-(a+1)\cdot\frac{2a-b-2}{a^2+b+3}.\\ \end{align*} Consider: \[\frac{2a-b-2}{a^2+b+3},\]which clearly has to be an integer. The denominator is clearly positive, by the given condition. Therefore: \[|2a-b-2|\geq a^2+b+3,\text{ or}\]\[2a-b-2=0.\] If: \[|2a-b-2|\geq a^2+b+3,\]then either: \[(a-1)^2\leq -2(b+2)\iff b\leq -2,\]\[(a+1)^2 \leq 0\iff a=-1,\]both of which clearl dont work. If $2a-b-2=0$, then $n=2$, so $n$ can only be $2$, as desired $\blacksquare$.

Euclid tells us $a^2+b+3 \mid ab+3b+8$ implies the LHS also divides \[3(a^2+b+3)-(ab+3b+8)=ab-3a^2-1, \quad a(a^2+b+3)-(ab-2a^2-1) = (a+1)^3\]\[\implies a^2+b+3 \mid a^2+2a+1 \implies a^2+b+3 \mid b-2a+2.\] Since casework tells us $|a^2+b+3| < |b-2a+2|$, we must have \[b-2a+2 = 0 \implies b=2a-2 \implies \boxed{n=2}. \quad \blacksquare\]

We want $(b+(a^2+3))|((a+3)b+8)$ so $(b+(a^2+3))|((a^2+3)(a+3)-8)$ so $(b+(a^2+3))|(a+1)^3$. This means that for every prime $p|(a+1)$, $v_p(b+(a^2+3))\le 2$ and for every prime $p\not|(a+1)$, $p\not|(b+(a^2+3))$. Note that $b+(a^2+3)\ge (a+1)^2$ since dividing $(a+1)^2$ by any $p|(a+1)$ will make it too small. We claim that $v_p(a+1)\le 1$ for all $p$. FTSOC, let $v_p(a+1)\ge 2$ for some $p$. Then, $v_p((a+1)^2)\ge 4$ and $v_q((a+1)^2)\ge 2$ for all other $q|(a+1)$. Since $b+(a^2+3)\ge (a+1)^2$, this would imply that $b+(a^2+3)$ is divisible by the cube of some prime, which is not allowed. Therefore, $v_p(a+1)\le 1$ for all $p$. Since $b+(a^2+3)$ cannot be divisible by the cube of some prime, $b+(a^2+3)=(a+1)^2$, so $b=2a-2$. Plugging this back in gives $$n=\frac{(a+3)(2a-2)+8}{a^2+2a-2+3}=\frac{2a^2+4a+2}{a^2+2a+1}=2$$so $\boxed{n=2}$ is the only possible value of $n$.

https://drive.google.com/file/d/1DO7sC6soXC3vcoAUqdFWipjvZJWBOSzD/view?usp=drivesdk it's my soln.

why did this make the shortlist??? I claim the answer is $2$, achieved by $(a,b) = (2,2)$. We can write $\frac{ab + 3b + 8}{a^2 + b + 3} = a + 3 - \frac{(a + 1)^3}{a^2 + b +3}$. If $a^2 + b + 3 > (a + 1)^2$, this forces that $a^2 + b + 3$ divides the cube of some prime (observe if it did not divide the cube of some prime, it would be at most the product of the squares of the prime factors of $a + 1$ which is at most $(a + 1)^2$), so $(a^2 +b + 3) \le (a + 1)^2$. Since $\frac{(a + 1)^3}{a^2 + b + 3} \ge a + 1 $. Observe this cannot be greater than $a + 2$ since it would render the original expression as nonpositive, but it cannot be $a + 2$ since $(a + 1)^3$ cannot possibly have $a + 2$ as a divisor, so this is exactly $a + 1$, giving the only possible answer as $a + 3 -a - 1 = 2$.

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