To make $9^{n+2}$ a composite number, we need to ensure that it has divisors other than 1 and itself.

We know that: $9^{n+2} = (3^2)^{n+2} = 3^{2(n+2)}$

This expression is a power of 3 for any integer $n$, which implies that for any $n$, $9^{n+2}$ is a composite number. A composite number has at least one divisor other than 1 and itself, and powers of 3 (other than $3^1$) always have divisors other