Perfect
Numbers - A Case Study
Perfect numbers
are those numbers that equal the sum of all their divisors including
1 and excluding the number itself.
Most numbers do
not fit this description. At the heart of every perfect number is a
Mersenne prime. All of the other divisors are either powers of 2 or
powers of 2 times the Mersenne prime.
Let's examine the
number 496 - one of the known perfect numbers. In order to demonstrate
that 496 is a perfect number, we must show that
496
= (the sum of all its divisors including 1 and excluding 496)
We might just start
by dividing and working out the divisors the long way. Or, we might
begin by noting that, in the notation that includes a Mersenne prime,
496
= 24 (25 - 1) = 24 x 31.
From this expression,
we may easily obtain the divisors of 496, namely:
1,
2, 22, 23, 24, 31, (2×31), (22×31),
(23×31), and (24×31).
The sum of all the
divisors excluding 496 is then
1
+ 2 + 22 + 23 + 24 + 31× (1 +
2 + 22+ 23)
In order to break
this mess down a bit, let us examine the partial sum
u
= 1 + 2 + 22 + 23 + 24
If we multiply by
2 (i.e., 2u = 2 + 22 + 23 + 24 + 25)
then subtract, we find
u
= 2u - u = 25 - 1 = 31
Thus, the sum of
the divisors becomes
=
31 + 31(24 - 1)
= 24×31
= 496
as we were to show.
To show that any perfect number may be broken down in this way, we let
nP
= 2c(2c+1 - 1)
be a perfect number
with (2c+1 -1) being the embedded Mersenne prime. Then the divisors
of nP are
1,
2, 22, ... , 2c, 2c+1 - 1, 2(2c+1
- 1), ..., 2c(2c+1 - 1)
The sum of all the
divisors excluding np is then
1
+ 2 + 22 + ... + 2c + (2c+1 - 1) (1
+ 2 + ... + 2c-1)
Again, we examine
the partial sum:
u
= 1 + 2 + 22 + ... + 2c.
We multiply by 2
(i.e., 2u = 2 + 22 + ... + 2c + 2c+1)
then subtract:
u
= 2u - u = 2c+1 - 1
The sum of the divisors
becomes:
=
(2c+1 - 1) + (2c+1 - 1)(2c - 1)
= 2c(2c+1 - 1)
as we were to show.