2

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?

Community
  • 1
roxrook
  • 12,399

2 Answers2

6

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

lhf
  • 221,500
4

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