if the product of all unique positive divisors of n a and p

Question

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: $n^{d(n)/2}$, where $d(n)$ is the number of factors and I got the answer as $n^{7/2}$.

please provide full solution

Answer 1

Since $n^2=n^{d(n)/2}$, we have $d(n)=4$. Then $n$ is the product of two different primes or the cube of a prime, a possibility excluded in the problem. Let then $n=pq$.

Now, $d(n^2)=d(p^2q^2)=3\cdot 3=9$, so the product of the divisors of $n^2$ is $$(n^2)^{d(n^2)/2}=n^9$$

Answer 2

Since $n^{d(n)/2}=n^2$ we have $d(n)=4=4\cdot1=2\cdot 2$. But $n$ is not a cube, so $n=pq$ for some primes $p\neq q$. Therefore $n^2=p^2q^2$ has $3\cdot 3=9$ positive divisors. Finally, the product of all positive divisors of $n^2$ is $(n^2)^{9/2}=n^9$.