2
$\begingroup$

For example:
$6$ has this property since proper divisors of $6$ are: $2$ and 3.

From this thread: What does the product of all proper divisors equal to?

My attempt was:
If n = p_1^{a_1} \times p_2^{a_2} \times ... \times p_k^{a_k}
Then
n = n^{\frac{\tau(n)}{2} - 1}$. Where $\tau(n) = (a_1 + 1) \times (a_2 + 1) \times ... \times (a_k + 1)

So is it good enough to stop here, or we can express n$ in a better formula?

  • 0
    @Ross Millikan: Opp! My bad. Thanks for pointing out.2011-02-28

2 Answers 2

5

$n = n^{\tau(n)/2 - 1}$ implies $\tau(n)=4$ and so $n$ is a product of two primes or the cube of a prime.

  • 0
    Nice thought! Thanks for this clever hint.2011-02-28
4

Any number of the form $n = p_1 \times p_2$ where $p_1,p_2$ are primes.

  • 1
    You've missed $n=p^3$.2011-02-28