Even Perfect Number except 6 is Congruent to 1 Modulo 9
Jump to navigation
Jump to search
Theorem
Let $n$ be an even perfect number, but not $6$.
Then:
- $n \equiv 1 \pmod 9$
Proof
From Theorem of Even Perfect Numbers:
- $n = 2^{p - 1} \paren {2^p - 1} = \dfrac {2^p \paren {2^p - 1} } 2$
where $p$ is prime.
From Odd Power of 2 is Congruent to 2 Modulo 3:
- $2^p \equiv 2 \pmod 3$
for odd $p$.
Thus:
\(\ds n\) | \(=\) | \(\ds \dfrac {\paren {3 k + 2} \paren {3 k + 1} } 2\) | for some $k \in \Z_{>0}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {9 k^2 + 9 k + 2} 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 9 \frac {k \paren {k + 1} } 2 + 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 9 T_k + 1\) | Closed Form for Triangular Numbers |
So $n$ is $1$ more than $9$ times the $k$th triangular number for some $k$.
That is:
- $n \equiv 1 \pmod 9$
When $n = 6$ the situation is different.
We have:
\(\ds n\) | \(=\) | \(\ds \dfrac {2^2 \paren {2^2 - 1} } 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {3 k + 1} \paren {3 k} } 2\) | where $k = 1$ | |||||||||||
\(\ds \) | \(\equiv\) | \(\ds 0 \pmod 3\) |
and so the result does not hold.
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $28$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $28$