What is the largest positive integer such that $\frac{n^2 + 7n + 12}{n^2 - n - 12}$ is also a positive integer?

We need to find the largest positive integer $n$ such that $\frac{n^2 + 7n + 12}{n^2 - n - 12}$ is also a positive integer.

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Let us start by factoring both parts of the fraction.

For the numerator $n^2 + 7n + 12$, we need two numbers that multiply to 12 and add to 7. Those are 3 and 4.

So: $n^2 + 7n + 12 = (n + 3)(n + 4)$

For the denominator $n^2 - n - 12$, we need two numbers that multiply to -12 and add to -1. Those are -4 and 3.

So: $n^2 - n - 12 = (n - 4)(n + 3)$

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Now we can rewrite the original expression:

$\frac{n^2 + 7n + 12}{n^2 - n - 12} = \frac{(n + 3)(n + 4)}{(n - 4)(n + 3)}$

Since $(n + 3)$ appears in both numerator and denominator, we can cancel it out (provided $n \geq -3$):

$\frac{(n + 3)(n + 4)}{(n - 4)(n + 3)} = \frac{n + 4}{n - 4}$

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We can rewrite this fraction by performing polynomial division:

$\frac{n + 4}{n - 4} = \frac{(n - 4) + 8}{n - 4} = 1 + \frac{8}{n - 4}$

Key insight: For this expression to be a positive integer, $\frac{8}{n - 4}$ must be a non-negative integer. This means $(n - 4)$ must be a positive divisor of 8.

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The positive divisors of 8 are: 1, 2, 4, and 8.

So we need:

$n - 4 = 1 \rightarrow n = 5$

$n - 4 = 2 \rightarrow n = 6$

$n - 4 = 4 \rightarrow n = 8$

$n - 4 = 8 \rightarrow n = 12$

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Let us check each value:

$n = 5$: $\frac{5 + 4}{5 - 4} = \frac{9}{1} = 9$

$n = 6$: $\frac{6 + 4}{6 - 4} = \frac{10}{2} = 5$

$n = 8$: $\frac{8 + 4}{8 - 4} = \frac{12}{4} = 3$

$n = 12$: $\frac{12 + 4}{12 - 4} = \frac{16}{8} = 2$

All values give positive integers as required.

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We also need to ensure that $n \geq -3$ (which would make the original denominator zero after factoring) and $n \geq 4$ (which would make the simplified denominator zero). Since all our solutions are positive and different from these values, they're all valid.

Therefore, the largest positive integer is $n = 12$.