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.

If p is prime
$N= p^5$ -  all divisors product is $p \ p^2 p^3  p^4  p^5 = p^{15}  = N^3$
