Form of Even Perfect Numbers
If 2**k - 1 is prime (k>1), then n=(2**(k-1))(2**k - 1) is perfect and every even perfect number is of this form. 2**k - 1 is known as a Mersenne prime, and k is prime.
Proof:
Let p=2**k - 1
sigma(n)=sigma((2**(k-1))p)
=sigma(2**(k-1))sigma(p)
=(2**k - 1)(p+1)
=(2**k - 1)2**k
=2n
Conversely:
n=2**(k-1)m [n even perfect number][m odd][k>=2]
(2**k)m
=2n
=sigma(n)
=sigma((2**(k-1))m)
=sigma(2**(k-1))sigma(m)
=(2**k - 1)sigma(m)
Therefore m=(2**k - 1)M,
sigma(m)=(2**k)M
Since m and M are both divisors of m:
(2**k)M=sigma(m)>=m+M=(2**k)M
i.e. sigma(m)=m+M, m prime, M=1
LEMMA 1
If a**k - 1 is prime (a>0, k>=2), then a=2 and k is also prime
Proof:
a**k - 1=(a-1)(a**(k-1) + a**(k-2) + … + a + 1)
=> a-1 = 1, i.e. a=2
Suppose k=rs:
a**k - 1=(a**r)**s - 1
=(a**r - 1)(a**(r(s-1)) + a**(r(s-2)) + … + a**r + 1)
which would not be prime