Number Theory problem 2 by Dhaval Furia

$\begin{aligned} \frac {n^2+7n+12}{n^2-n-12} & = \frac {(n+3)(n+4)}{(n+3)(n-4)} = \frac {n+4}{n-4} = 1 + \frac 8{n-4} \end{aligned}$ . For $\dfrac {n^2+7n+12}{n^2-n-12}$ to be a positive integer, $8$ must be divisible by $n-4$ . And the largest $n$ is when $n-4=8$ $\implies n = \boxed{12}$ .