math 409/409g history of mathematics

Math 409/409GHistory of Mathematics

Perfect Numbers

What’s a perfect number?

A positive integer is a perfect number if it is equal to the sum of all its positive divisors except itself.

For example, 6 is a perfect number since the positive divisors of 6 are 1, 2, 3, and 6 and 6 1 + 2 + 3 (the sum of all positive divisors of 6 except 6).

In fact, P1 6 is the smallest perfect number. The next in line is P2 28. (The positive divisors of 28 are 1, 2, 4, 7, 14, and 28 and 28 1 + 2 + 4 + 7 + 14.) The third and fourth perfect numbers are P3 496 and P4 8128.

As another example, no prime number p can be a perfect number since the only divisors of p are 1 and p and p ≠ 1.

Why study perfect numbers?

Ancient philosophers thought that perfect numbers had mystical and religious significance: God created the world in 6 days and rested on the seventh; it takes 28 days for the moon to circle the earth.

We, of course, study perfect numbers for the sheer beauty of the mathematics.

How do you find the positive divisors of a number?

Look at the prime factorization of the number.

The positive divisors of n are of the form

where

31 21 2 3

rkk k krn p p p p

31 21 2 3

raa a ard p p p p

0 1,2,3, , . for i ia k i r

For example, 28 22 ·7. So the positive divisors of 28 are of the form d 2a ·7b where a 0, 1, or 2 and b 0 or 1. Since there are 3 choices for a and 2 for b, 28 has 3·2 6 positive divisors.

With b 0 we get the divisors 20, 21, and 22. And with b 1 we get 20 ·7, 21 ·7, and 22 ·7. So the six positive divisors of 28 are 1, 2, 4, 7, 14, and 28.

Finding all positive divisors of a number can be an exhausting task.

For example, if n 25 ·33 ·72 ·11, then when determining the positive divisors of n we have: 6 choices for the exponent of two, 4 for the exponent of three,

3 for the exponent of seven, and 2 for the exponent of eleven.

So n has 6·4·3·2 144 positive divisors!

To determine if n 25 ·33 ·72 ·11 is a perfect number we would have to find the 143 divisors other than n itself and then add them up.

There’s got to be a better way! We need a formula for the sum of these divisors.

The sigma function

If n is a positive integer, then σ(n) is defined to be the sum of all the positive divisors of n.

Examples:

The positive divisors of 6 are 1, 2, 3, and 6. So σ(6) 1 + 2 + 3 + 6 12.

The positive divisors of 28 are 1, 2, 4, 7, 14, and 28. So σ(28) 56, the sum of all these positive divisors.

Since σ(n) is the sum of all the positive divisors of n, σ(n) – n is the sum of all positive divisors except n. So n is a perfect number when σ(n) – n n. That is

n is a perfect number if σ(n) 2n.

Examples: 6 and 28 are perfect numbers since σ(6) 12 2·6 and σ(28) 56 2·28.

Theorem:

Example:

1 21 2

rk k krn p p p

1 21 1 11 2

1 2

1 1 1( ) .

1 1 1

rk k kr

r

p p pn

p p p

3 22 1 2 1 7 1

28 2 7 (28) 56.2 1 7 1

Proof that

For i 1, 2, 3, …, r, let

Ex. For n 28 22 ·71,

P1 1 + 2 + 22 and P2 1 + 7.

Consider the product P1P2P3···Pr .

1 2

1 2

1 1 11 2

1 21 2

1 1 1( ) .

1 1 1

r

r

k k kk k k r

rr

p p pn p p p n

p p p

2 31 .iki i i i iP p p p p

Ex. For n 28,

P1P2 (1 + 2 + 4)(1 + 7)

1 + 2 + 4 + 7 + 14 + 28.

Each positive divisor of n appears exactly once in the expansion of this product, so σ(n) P1P2P3···Pr .

But is a

geometric series. Thus

2 31 iki i i i iP p p p p

1 1.

1

iki

ii

pP

p

We now have our desired formula:

1 2

1 2

1 1 11 2

1 2

( )

1 1 1.

1 1 1

r

r

k k kr

r

n PP P

p p p

p p p

Example: Is n 25 ·33 ·72 ·11 a perfect number?

Solution:

So n is not a perfect number.

6 4 3 2

5 4

2 1 3 1 7 1 11 1( )

2 1 3 1 7 1 11 1

2 3 5 7 19 2

n

n

Can we generate perfect numbers?

Consider the first four perfect numbers

They are each of the form of 2k ·p where p is a prime number.

11

22

43

64

6 2 3

28 2 7

496 2 31

8128 2 127

P

P

P

P

Will any number of the form 2k ·p where p is a prime number be a perfect number?

No. Consider 44 22 ·11.

So 44 is not a perfect number.

3 22 1 11 1(44) 84 2 44.

2 1 11 1

What condition must be placed on the prime p and the exponent k of 2?

1

2

4

1

2

3

46

6 2

28 2

4

3

7

396 1

1

2

8128 2 27

P

P

P

P

We now see that the first four perfect numbers are of the form 2k(2k + 1 – 1).

Is every number of this form a perfect number?

1 1

2 2

4 4

1

2

3

2

3

5

74

6 6

6 2

28 2

4

2 (2 1)

2 (2 1)

2 (2 1)

3

7

31

2

96

127 (2 1)

2

8128 2

P

P

P

P

Consider n 28(29 – 1) 28 ·7·73 130,816.

So n is not perfect, and thus not all numbers of the form 2k(2k + 1 – 1) are perfect numbers.

Which numbers of the form 2k(2k + 1 – 1) are perfect numbers?

9 2 22 1 7 1 73 1( ) 302,512 2

2 1 7 1 73 1n n

The pattern we found for the first four perfect numbers was that they were of the form 2k(2k + 1 – 1) where 2k + 1 – 1 is prime.

1 1

2 2

4 4

1

2

3

2

3

5

74

6 6

6 2

28 2

4

2 (2 1)

2 (2 1)

2 (2 1)

3

7

31

2

96

127 (2 1)

2

8128 2

P

P

P

P

In the last example of a number that was not perfect, n 28(29 – 1) where 29 – 1 7·73 is not prime.

So will the numbers of the form

2k(2k + 1 – 1) where 2k + 1 – 1 is prime

generate perfect numbers?

Yes, as shown in the next theorem.

Euclid’s Theorem: If 2k + 1 – 1 is prime, then 2k(2k + 1 – 1) is perfect.

Pf: Let p 2k + 1 – 1 be prime and set n 2k

·p.

Then

But p 2k + 1 – 1, and thus p + 1 2k + 1.

So n is perfect since

11

22 1 1( ) .

2 1 1(2 1 ( 1))k

k

pp

pn

1( ) 2 2( .2 ) 2k k p npn

So Euclid’s theorem will generate even perfect numbers. Actually, it can be shown that all even perfect numbers are of the form 2k(2k + 1 – 1) where 2k + 1 – 1 is prime.

So Euclid’s theorem will generate all even perfect numbers. But generating these even perfect numbers not as easy as it looks since for each k, you have to determine if 2k + 1 – 1 is prime.

If we look at the first four perfect numbers for inspiration to help us determine when 2k + 1 – 1 is prime, we suspect that k + 1 must be prime.

1 1

2 2

4 4

1

2

3

2

3

5

64

6 7

6

28

4

2 3 2 (2 1)

2 7 2 (2 1)

2 31 2 (2 1)

2 1

96

27 2 (812 28 1)

P

P

P

P

Although I will not prove it here, it has been shown that if 2k + 1 – 1 is prime, then k + 1 must be prime.

Putting this together with Euclid’s theorem shows that the every even perfect numbers is of the form 2p - 1(2p – 1) where p is prime.

But we are still not guaranteed that every number of the form 2p - 1(2p – 1) where p is prime will be a perfect number.

Ex: p 11 is prime, but 211 – 1 23·89 is not prime. So n 210(211 – 1) is not a perfect number.

The proof of this example is left as an exercise.

This ends the lesson on

Perfect Numbers