Even Perfect Number except 6 is Congruent to 1 Modulo 9

Theorem
Let $n$ be an even perfect number, but not $6$.

Then:
 * $n \equiv 1 \pmod 9$

Proof
From Theorem of Even Perfect Numbers:


 * $n = 2^{p - 1} \left({2^p - 1}\right) = \dfrac {2^p \left({2^p - 1}\right)} 2$

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:
 * $n = \dfrac {\left({3 k + 2}\right) \left({3 k + 1}\right)} 2$

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:

and so the result does not hold.