formofevenperfectnumbers

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