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:

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

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} 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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