perfect number

perfect number

An

positive integer

n

is called

perfect

if it is the sum of all

positive

divisors

of

n

less than

n

itself. It is not known if there are any odd perfect numbers, but all even perfect numbers have been classified according to the following lemma:

Lemma 1

.

An

even number

is perfect if and only if it equals

2

k

-

1

⁢

(

2

k

-

1

)

for some

integer

k

>

1

and

2

k

-

1

is prime.

Proof.

Let

σ

denote the

sum of divisors function

. Recall that this

function

is

multiplicative

.

Necessity

: Let

p

=

2

k

-

1

be prime and

n

=

2

k

-

1

⁢

p

. We have that

σ

⁢

(

n

)

=

σ

⁢

(

2

k

-

1

⁢

p

)

=

σ

⁢

(

2

k

-

1

)

⁢

σ

⁢

(

p

)

=

(

2

k

-

1

)

⁢

(

p

+

1

)

=

(

2

k

-

1

)

⁢

2

k

=

2

⁢

n

,

which shows that

n

is perfect.

Sufficiency: Assume

n

is an even perfect number. Write

n

=

2

k

-

1

⁢

m

for some odd

m

and some

k

>

1

. Then we have

gcd

⁡

(

2

k

-

1

,

m

)

=

1

. Thus,

σ

⁢

(

n

)

=

σ

⁢

(

2

k

-

1

⁢

m

)

=

σ

⁢

(

2

k

-

1

)

⁢

σ

⁢

(

m

)

=

(

2

k

-

1

)

⁢

σ

⁢

(

m

)

.

Since

n

is perfect,

σ

⁢

(

n

)

=

2

⁢

n

by definition. Therefore,

σ

⁢

(

n

)

=

2

⁢

n

=

2

k

⁢

m

. Piecing together the two

formulas

for

σ

⁢

(

n

)

yields

2

k

⁢

m

=

(

2

k

-

1

)

⁢

σ

⁢

(

m

)

.

Thus,

(

2

k

-

1

)

∣

2

k

m

, which forces

(

2

k

-

1

)

∣

m

. Write

m

=

(

2

k

-

1

)

⁢

M

. Note that

1

≤

M

<

m

. From above, we have:

2

k

⁢

m

=

(

2

k

-

1

)

⁢

σ

⁢

(

m

)

2

k

⁢

(

2

k

-

1

)

⁢

M

=

(

2

k

-

1

)

⁢

σ

⁢

(

m

)

2

k

⁢

M

=

σ

⁢

(

m

)

Since

m

∣

m

by definition of divides (

http://planetmath.org/Divides

) and

M

∣

m

by

assumption

, we have

2

k

⁢

M

=

σ

⁢

(

m

)

≥

m

+

M

=

2

k

⁢

M

,

which forces

σ

⁢

(

m

)

=

m

+

M

. Therefore,

m

has only two positive divisors,

m

and

M

. Hence,

m

must be prime,

M

=

1

, and

m

=

(

2

k

-

1

)

⁢

M

=

2

k

-

1

, from which the result follows.
∎

The lemma can be used to produce examples of (even)

perfect numbers

:

•

If

k

=

2

, then

2

k

-

1

=

2

2

-

1

=

3

, which is prime. According to the lemma,

2

k

-

1

⁢

(

2

k

-

1

)

=

2

2

-

1

⋅

3

=

6

is perfect. Indeed,

1

+

2

+

3

=

6

.

•

If

k

=

3

, then

2

k

-

1

=

2

3

-

1

=

7

, which is prime. According to the lemma,

2

k

-

1

⁢

(

2

k

-

1

)

=

2

3

-

1

⋅

7

=

28

is perfect. Indeed,

1

+

2

+

4

+

7

+

14

=

28

.

•

If

k

=

5

, then

2

k

-

1

=

2

5

-

1

=

31

, which is prime. According to the lemma,

2

k

-

1

⁢

(

2

k

-

1

)

=

2

5

-

1

⋅

31

=

496

is perfect. Indeed,

1

+

2

+

4

+

8

+

16

+

31

+

62

+

124

+

248

=

496

.

Note that

k

=

4

yields that

2

k

-

1

=

2

4

-

1

=

15

, which is not prime.

The

sequence

of known perfect numbers appears in the OEIS as sequence

http://www.research.att.com/ njas/sequences/?q=A000396

A000396.

Title

perfect number

Canonical name

PerfectNumber

Date of creation

2013-03-22 11:45:29

Last modified on

2013-03-22 11:45:29

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

24

Author

Wkbj79 (1863)

Entry type

Definition

Classification

msc 11A05

Classification

msc 20D99

Classification

msc 20D06

Classification

msc 18-00

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