Product of Factors of Perfect Number

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $P$ be the perfect number $2^{n - 1} \paren {2^n - 1}$.

Then:

$\ds \prod_{d \mathop \divides P} d = P^n$


Proof

The factors of $P$ are:

$1, 2, 4, \dots, 2^{n - 1}, 2^n - 1, 2 \paren {2^n - 1}, \dots, 2^{n - 1} \paren {2^n - 1}$

Therefore their product is:

\(\ds \prod_{d \mathop \divides P} d\) \(=\) \(\ds \paren {\prod_{i \mathop = 0}^{n - 1} 2^i} \paren {\prod_{i \mathop = 0}^{n - 1} 2^i \paren {2^n - 1} }\)
\(\ds \) \(=\) \(\ds \paren {\prod_{i \mathop = 0}^{n - 1} 2^i}^2 \paren {2^n - 1}^n\)
\(\ds \) \(=\) \(\ds \paren {2^\frac {n \paren {n - 1} } 2}^2 \paren {2^n - 1}^n\)
\(\ds \) \(=\) \(\ds \paren {2^{n - 1} }^n \paren {2^n - 1}^n\)
\(\ds \) \(=\) \(\ds P^n\)

$\blacksquare$


Sources