More about even perfect numbers

Knowing that every even perfect number is of the form

  \[ n=2^ kp=2^ k(2^{k+1}-1) \]    

makes it easy to list all its factors, and this enables us to prove some interesting results. Since $p$ is a prime, the factors of $n=2^ kp$ must either be a power of $2$ (up to $2^ k$) or a product of a power of $2$ and $p$. Listing these factors in a table we get:

Factors of $2^ k$Other factors
$2^ k$$2^ kp$

For example, the factors of $n=28=2^4 \times 7$ are:

Factors of $2^4$Other factors
$2$$2\times 7=14$
$2^2$$2^2\times 7=28$

We claimed that the product of the factors of an even perfect number $n=2^ kp$ is always equal to $n^{k+1}$. For $n=24$ this is clearly true, since

  \[  1\times 2 \times 4 \times 7 \times 14 \times 28=21952=28^3, \]    

as required.

For a general proof, note that the product of the factors in the first column of the table above is

  \begin{equation} 1 \times 2 \times 2^2 ...\times 2^ k = 2^{1+2+...+k}.\end{equation}   (1)

As we already noted in the main article, the sum of the first $k$ integers is

  \[ 1+2+..+k = k(k+1)/2, \]    

so (1) is equal to

  \[ 2^{1+2+...+k}=2^{k(k+1)/2}. \]    

Note that every factor in the second column of the table above is $p$ times the corresponding factor in the first table. Since there are $k+1$ entries in each column, the product of all the factors in the second column is therefore

  \[ p^{k+1} 2^{k(k+1)/2}. \]    

The product of all the factors is

  \[ 2^{k(k+1)/2}p^ k 2^{k(k+1)/2}=p^{k+1}2 ^{k(k+1)}=(2^ k p)^{k+1}=n^{k+1}. \]    

This proves our result.

A similar result involves adding up the reciprocals of all the factors of an even perfect number. Taking our example of $n=28$ again we see that

  \[ \frac{1}{1}+\frac{1}{2}+\frac{1}{4}+\frac{1}{7}+\frac{1}{14}+\frac{1}{28}=2. \]    

In fact the sum is equal to $2$ for every perfect number! As one might expect, the proof proceeds in the same way as above, replacing addition with multiplication. We leave it to you as an exercise.

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