Product of Factors of Perfect Number
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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
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $28$