Product of Divisors/Proof 2

Theorem

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

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

Then:

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

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

Proof

\(\ds \map D n^2\)

\(=\)

\(\ds \paren {\prod_{d \mathop \divides n} d}^2\)

by definition

\(\ds \)

\(=\)

\(\ds \paren {\prod_{d \mathop \divides n} d} \paren {\prod_{d \mathop \divides n} \dfrac n d}\)

Sum Over Divisors Equals Sum Over Quotients

\(\ds \)

\(=\)

\(\ds \prod_{d \mathop \divides n} n\)

combining the two products

\(\ds \)

\(=\)

\(\ds n^{\map {\sigma_0} n}\)

Definition of Divisor Count Function

The result follows by taking the (positive) square root of both sides.

$\blacksquare$