Product of Proper Divisors

Theorem

Let $n$ be an integer such that $n \ge 1$.

Let $\map P n$ denote the product of the proper divisors of $n$.

Then:

$\map P n = n^{\map {\sigma_0} n / 2 - 1}$

where $\map {\sigma_0} n$ denotes the divisor count Function of $n$.

Let $\map D n$ denote the product of all the divisors of $n$.

$\map D n = n^{\map {\sigma_0} n / 2}$

The proper divisors of $n$ are defined as being the divisors of $n$ excluding $n$ itself.

Thus:

$\map P n = \dfrac {\map D n} n = \dfrac {n^{\map {\sigma_0} n / 2} } n = n^{\map {\sigma_0} n / 2 - 1}$

$\blacksquare$

The product of the proper divisors of $12$ is $144 = 12^2$.

The product of the proper divisors of $20$ is $400 = 20^2$.