Product of Divisors

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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 1

We have by definition that:

$\map D n = \ds \prod_{d \mathop \divides n} d$

Also by definition, $\map {\sigma_0} n$ is the number of divisors of $n$.


Suppose $n$ is not a square number.

Let $p \divides n$, where $\divides$ denotes divisibility.

Then:

$\exists q \divides n : p q = n$

Thus the divisors of $n$ come in pairs whose product is $n$.

From Divisor Count Function is Odd Iff Argument is Square, $\map {\sigma_0} n$ is even.

Thus $\dfrac {\map {\sigma_0} n} 2$ is an integer.

Thus there are exactly $\dfrac {\map {\sigma_0} n} 2$ pairs of divisors of $n$ whose product is $n$.

Thus the product of the divisors of $n$ is:

$\ds \prod_{d \mathop \divides n} d = n^{\map {\sigma_0} n / 2}$


Now suppose $n$ is square such that $n = r^2$.

Then from Divisor Count Function is Odd Iff Argument is Square, $\map {\sigma_0} n$ is odd.

Hence the number of divisors of $n$ not including $r$ is $\map {\sigma_0} n - 1$.

As before, these exist in pairs whose product is $n$.

Thus:

\(\ds \prod_{d \mathop \divides n} d\) \(=\) \(\ds r \times n^{\paren {\map {\sigma_0} n - 1} / 2}\)
\(\ds \) \(=\) \(\ds r \times n^{\map {\sigma_0} n / 2 - 1 / 2}\)
\(\ds \) \(=\) \(\ds n^{\map {\sigma_0} n / 2 - 1 / 2 + 1 / 2}\) as $r = n^{1/2}$
\(\ds \) \(=\) \(\ds n^{\map {\sigma_0} n / 2}\)

Hence the result.

$\blacksquare$


Proof 2

\(\ds \map D n^2\) \(=\) \(\ds \paren {\prod_{d \mathop \divides n} d}^2\) 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.

$\blacksquare$