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.

Hint: The way to go is by prime factorisation. You need only consider the case up to four distinct prime factors (the rest should follow easily from your discussion). If you realise, all your found examples follow the pattern $pq^2$, where $p,q$ are primes.
