Even Perfect Number except 6 is Congruent to 1 Modulo 9

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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

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