Perfect numbers in binary

We will prove a more general result about the binary form of all even perfect numbers.

We'll start with the Euler-Euclid result that all even perfect numbers are of the form

$2^{p-1}(2^{p}-1)$

(For this to yield a perfect number p must be prime and the second factor must also be prime, but these details are not needed to solve the problem.)

Writing this product out in binary gives

$(100\dots00)(111\dots111)$

where the first factor has $(p-1)$ zeros and the second factor consists of $(p)$ ones. Just as in our home base of ten, a multiplication like this is easy - we just append the zeros from the first factor to the second factor to get

$111\dots11100\dots00$

and so

$a-b=(p)-(p-1)=\boxed{1}$

The formula for a perfect number is $2^k(2^{k+1}-1)$ where $k+1$ is prime.

In its binary expansion, $2^{k+1}-1$ is a string of ' $1$ 's of length $k+1$ .

$2^{k}-1$ is a string of ' $0$ 's of length $k$ at the back of the string of ' $1$ 's.

Hence $a-b=k+1-k$ $=\boxed{1}$