Even Perfect Number is Triangular
Jump to navigation
Jump to search
Theorem
All perfect numbers which are even are triangular.
Proof 1
Let $a$ be an even perfect number.
From the Theorem of Even Perfect Numbers, $a$ is in the form $2^{p - 1} \left({2^p - 1}\right)$ where $2^p - 1$ is prime.
Thus:
\(\ds a\) | \(=\) | \(\ds \left({2^p - 1}\right) 2^{p - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \left({2^p - 1}\right) \frac {2^p} 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {n \left({n + 1}\right)} 2\) | where $n = 2^p - 1$ |
The result follows from Closed Form for Triangular Numbers.
$\blacksquare$
Proof 2
Follows from:
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $28$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $28$