Over 2300 years ago Euclid
proved that If 2k-1 is a prime
number (it would be a Mersenne
prime), then 2k-1(2k-1) is a perfect
number. A few hundred years ago Euler
proved the converse (that every even perfect number has this form).
It is still unknown if there are any odd perfect numbers (but if there are,
they are large and have many prime factors).
If 2k-1 is a prime number, then 2k-1(2k-1)
is a perfect number and every even perfect number has this form.
Proof: Suppose first that p = 2k-1
is a prime number, and set n = 2k-1(2k-1).
To show n is perfect we need only show sigma(n)
= 2n. Since sigma is multiplicative
and sigma(p) = p+1 = 2k, we know
sigma(n) = sigma(2k-1).sigma(p)
= (2k-1)2k = 2n.
This shows that n is a perfect number.
On the other hand, suppose n is any even perfect number
and write n as 2k-1m where m is an odd
integer and k>2. Again sigma is multiplicative so