A positive integer $n$ has the property that $n^2$ has exactly $9$ positive divisors. If $n$ is a prime power, how many positive divisors does $n$ have?

For a prime power $n = p^a$, the number of positive divisors of $n$ is $a + 1$, and the number of positive divisors of $n^2 = p^{2a}$ is $2a + 1$. Set $2a + 1 = 9 \Rightarrow a = 4$. Therefore $n = p^4$ has $a + 1 = 5$ positive divisors. Check with $p = 2$: $n = 16$ has divisors $\{1, 2, 4, 8, 16\}$ — five of them. And $n^2 = 256 = 2^8$ has divisors $\{1, 2, 4, 8, 16, 32, 64, 128, 256\}$ — nine, as required. Trap A ($3$) would correspond to $n^2$ having 5 divisors, not 9.