Product of Divisors/Proof 1

Proof
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 Counting 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 Counting 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:

Hence the result.