Find all numbers such that "Product of all divisors=cube of number". While solving some old olympiad problems I came across this one. As I m stuck at it, so I m here.
The problem is: Find all positive integers $N$ such that the product of all the positive divisors of N is equal to $N^3$.
Since I was not able to solve this one mathematically hence I tried Hit and trial method to find the pattern and then work upon it. I got that:
12 has divisors 1,2,3,4,6,12 product of all of which give 1728($12^3$).Similarly 18,20,28 also follow the same case. I noticed that all of them have 4 factors, but I don't think it can take me any further (I also think that a perfect power(such as $2^3$)will not follow the case).
After all of  my efforts I m on U guys. Need help. Any Mathematical formulation or suggestion is heartily welcome. Thanks.

Note that $\displaystyle\left(\prod_{d \mid n} d\right)^2
=\prod_{d \mid n} d \prod_{d \mid n} \frac{n}{d}
=n^{\tau(n)}
$.
Therefore, we seek $n$ such that $n^{\tau(n)}=n^6$, that is, $n=1$ or $\tau(n)=6$.
Write $n=\prod_p p^{e_p}$. Then $\tau(n)=\prod_p (1+e_p)$. There aren't many possibilities if this is to be $6$ because each possibility corresponds to a factorisation of $6$:


*

*$6=6$ gives $n=p^5$.

*$6=2\cdot 3$ gives $n=pq^2$.
