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Number theory Difficulty 4.7 AIME Prove it Taiwan

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

Solution

The only integer with property is $n = 2$ .
As $b \equiv -a^2 - 3 \pmod{a^2 + b + 3}$ , the numerator of the given fraction satisfies
$ab + 3b + 8 \equiv a(-a^2 - 3) + 3(-a^2 - 3) + 8 \equiv -(a + 1)^3 \pmod{a^2 + b + 3}$
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
$0 < (a+1)^2 < 2(a^2 + b + 3),$
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 $a^2 + b + 4 = 9$ shows that $n = 2$ indeed is a solution.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.