Even Perfect Number is Hexagonal

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Theorem

All perfect numbers which are even are hexagonal.


Proof

Let $a$ be an even perfect number.

From the Theorem of Even Perfect Numbers, $a$ is in the form $2^{p - 1} \paren {2^p - 1}$ where $2^p - 1$ is prime.

Thus:

\(\ds a\) \(=\) \(\ds \paren {2^p - 1} 2^{p - 1}\)
\(\ds \) \(=\) \(\ds 2^{p - 1} \paren {2 \times 2^{p - 1} - 1}\)
\(\ds \) \(=\) \(\ds n \paren {2 n - 1}\) where $n = 2^{p - 1}$

The result follows from Closed Form for Hexagonal Numbers.

$\blacksquare$


Sources

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