Divisibility of even perfect numbers

Question

I noticed that all even perfect numbers (looked at first 40) other than 6 are divisible by 4 with 0 remainder. Is there a reason for this that we know of?

Answer 1

In Euclid's Elements, it is shown that if $2^n-1$ is prime, then $2^{n-1}(2^n-1)$ is perfect.

Euler proved a converse. Every even perfect number is of the form $2^{p-1}(2^p-1)$, where $2^p-1$ is prime (and therefore $p$ is prime).

Euler's result in particular shows that every even perfect number greater than $6$ is divisible by $4$.

Answer 2

For even perfect number, Euler proved that all of them must be of the form (see http://en.wikipedia.org/wiki/Perfect_number): $\displaystyle\frac{p(p+1)}{2}$, where $p$ is a prime number and $p=2^m-1$ for some $m\in\mathbb{Z}_+$. So let $n$ be an even perfect number, we have \begin{eqnarray*} n=\frac{(2^m)\cdot(2^m-1)}{2}=2^{m-1}(2^m-1). \end{eqnarray*}

If $m=2$, $p=3$, we have $n=6$. If $m\geq 3$, we have $4\mid n$.