If the product of all the unique positive divisors of $n$, a positive integer which is not a perfect cube, is $n^2$, then the product of all the unique positive divisors of $n^2$ is: (A) $n^3$ (B) $n^4$ (C) $n^6$ (D) $n^8$ (E) $n^9$
I tried using the formula: $\frac{n^{d(n)}}{2}$, where $d(n)$ is the number of factors and I got the answer as $n^7/2$.
On
Since the product of positive divisors of $n$ is $n^{\frac{d(n)}{2}}$. We want $$n^{\frac{d(n)}{2}}=n^2.$$ This means $d(n)=4$. So either $n=p^3$ or $n=pq$. The first possibility is ruled out since $n$ is not a cube. Thus $n=pq$, where $p$ and $q$ are distinct primes. Then $d(n^2)=(2+1)(2+1)=9$ and the product of divisors of $n^2$ is $$(n^2)^{\frac{d(n^2)}{2}}=(n^2)^{9/2}=n^9$$
Hint: First, note that for the above condition to hold, $n$ cannot be prime and hence can be considered as the product of two primes, $p$ and $q$.
Then, try to list out the factors of $n^2= p^2q^2$.