Exploration on odd perfect numbers
Shatin Pui Ying College
(Registration Number : 51082)
Teacher
Mr. Lai Chi Keung
Team Members
Man Kong Fard,
Yip Shun,
Wong Tsz Ching,
Li Ling Chit,
Leung Ka Wai
A report submitted to the Scientific Committee of the Hang Lung Mathematics Award,
2006
19 Aug 2006
Exploration on odd perfect numbers p.1
Abstract
The aim of our project is to investigate the existence of odd perfect numbers. In our
investigation, we obtained three main results.
First of all, we analyzed the properties of hypothetical odd perfect numbers and
showed that any odd perfect number W must exist in the form of (4k + 1) 4l + 1 M 2 ,
where (4k + 1, M ) = 1 , k , M Î N and 4k + 1 is prime . Using this result, we further
studied the unit digits of 4k + 1 and M2 and succeeded to rule out some of their
possibilities.
Secondly, we explored the properties of the function s (n) , which represents the
sum of all factors of n. We found out that if c = a ×b , and b is a positive integer other
than 1, then
s (c ) s ( a )
>
. By this property, we proved that multiples of a perfect
c
a
number would never be a perfect number.
After all, we made use of the above property of s (n) to further our study on
multiply perfect numbers. Surprisingly, we had found that 2 times of an odd perfect
number is a 3-perfect number and concluded that if the number of 3-perfect number is
finite, then the number of odd perfect number is also finite.
Exploration on odd perfect numbers p.2
First Stage
First of all, let’s define what a perfect number is. By the definition in [3], a
natural number Ω is said to be perfect if and only if the sum s (W) of all the proper
factors of Ω is equal to itself, i.e. s (W) = 2W. Obviously, when A and B are relatively
prime,
s ( AB) = s ( A)s ( B) .
Let N = p1a 1 p2a 2 ... pna n , where p1 , p2 ,..., pn are distinct prime numbers,
then it is well-known that
s ( N ) = (1 + p1 + p12 + ...... + p1a1 )(1 + p2 + p22 + ...... + p2a 2 )......(1 + pn + pn2 + ...... + pna n ) …… (*)
We now focus on finding the general form of an odd perfect number and its properties.
Theorem 1:
If Ω is an odd perfect number, then W= (4k + 1) 4l + 1 M 2 for some M , k , l Î N
satisfying both (4k + 1, M ) = 1 and 4k + 1 is prime.
Proof:
Let Ω= p1a 1 p2a 2 ... pna n , where p1 , p2 ...... pn are distinct prime numbers.
As Ω is odd, 2Ω is a multiple of 2 but not a multiple of 4.
By (*), we obtained the following equation:
(1 +
p1 + p12 + ... + p1a1 )(1 + p2 + p22 + ... + p2a 2 )...(1 + pn + pn2 + ... + pna n ) = 2W
WLOG, we assume that (1 + p1 + p12 + ...... + p1a1 ) is the only even factor, then
(1 + p2 + p22 + ...... + p2a 2 ), (1 + p3 + p32 + ...... + p3a 3 ),......(1 + pn + pn2 + ...... + pna n ) are
odd numbers. Consequently,  2 ,  3 ,...... m are all even and  1 is odd.
Now a 1 is either a 1 = 4l + 3 or a 1 = 4l + 1
If a 1 = 4l + 3 . then
(1 + p1 + p12 + ...... + p1a 1 ) = (1 + p1 ) (1 + p12 ) (1 + p14 + p18 + ...... + p14l ) ,
even
even
which is a multiple of 4.
This violates with the fact that 2Ω is a multiple of 2 but not a multiple of 4.
So  1 must be in a form of 4l + 1 .
Exploration on odd perfect numbers p.3
Similarly, p1 is either in the form 4k + 3 or 4k + 1 .
If p1 = 4k + 3 = 4(k + 1) - 1 , then
s ( p1a1 )
= (1 + p1 + p12 + ...... + p1a1 )
= (1 + [4(k + 1) - 1] + [4(k + 1) - 1]2 + ...... + [4( k + 1) - 1)]4 n+ 1
= 4 N + 1- 1 + 1- 1...... + 1- 1
= 4N
Since N is an integer, so s ( p1a 1 ) is a multiple of 4 which violates with the fact that 2Ω
is a multiple of 2 but not a multiple of 4.
So p1 = 4k + 1 and p1a 1 = (4k + 1) 4l + 1 .
As  2 , 3 ,...... n are all even numbers, p2a 2 ... pna n is a perfect square.
2
3
And so we can conclude that W= (4k + 1)4l + 1 M 2 , where M  p 2 2 p3 2 ...p n
n
2
Clearly (4k + 1, M ) = 1 and 4k + 1 is prime.
(Q.E.D.)
Throughout this paper, we denote an odd perfect number by
W= (4k + 1) 4l + 1 M 2 = p1a1 p2a 2 ... pna n ,
where p1 = 4k + 1 and a 1 = 4l + 1
We now explore on the properties of the unit digits of p1 and M 2 using our Main
Theorem 1.
The unit digit of p1 can be 1, 3, 5, 7 or 9, and the unit digit of M 2 can be 1, 5 or 9.
So there are totally 15 combinations that are shown in the following table.
p1 º 1(mod10)
p1 º 3(mod10)
p1 º 5(mod10)
p1 º 7(mod10)
p1 º 9(mod10)
M 2 º 1(mod10)
1, 1
3, 1
5, 1
7, 1
9, 1
M º 5(mod10)
1, 5
3, 5
5, 5
7, 5
9, 5
M º 9(mod10)
1, 9
3, 9
5, 9
7, 9
9, 9
Unit digits
2
2
Exploration on odd perfect numbers p.4
Corollary 2
The unit digits of p1 and M 2 cannot be 5 and 5, 9 and 1, or 9 and 9.
Proof:
The only prime number having unit digit of 5 is 5.
It is trivial that p1 and M 2 cannot both have unit digits of 5 because they are
relatively prime.
When the unit digit of p1 is 9, then p1 º - 1(mod10) .
And
1 + p1 + p12 + ...... + p14 n+ 1
º 1 + (- 1) + (- 1) 2 + ...... + (- 1) 4 n+ 1 (mod10)
º 1- 1 + 1- 1 + ... + 1- 1
(mod10)
º 0
(mod10)
Therefore Ω is divisible by 5.
That implies M is a multiple of 5, and so the unit digit of M 2 cannot be 1 or 9.
(Q.E.D.)
Corollary 3
If the unit digits of p1 and M 2 are respectively 3 and 5, 5 and 1, 5 and 9, or 7 and 5,
then the unit digit of pi must be 1 for some i = 2,3,..., n .
Proof:
Suppose that the unit digits of p1 and M 2 are respectively 3 and 5, 5 and 1, 5 and 9,
or 7 and 5. Then W is a multiple of 5 in all four cases and by
2  ()  (p11 )(p 22 )...(p n n ) ,
we obtained 5 | s ( pi ) for some i = 1, 2,..,n.
We now first study whether 5 | s ( p1a1 ) .
As p1  3, 5 or 7 (mod 10), we shall prove that s ( p1a1 ) is not divisible by 5.
When p1  3 (mod10) , s ( p1a1 )
Exploration on odd perfect numbers p.5
= s ( pi4l + 1 )
= 1 + pi + pi2 + ...... + pi4l + 1
º (1 + 3 + 9 + 7) + ... + (1 + 3 + 9 + 7) + (1 + 3) (mod10)
º 4
(mod10)
When p1  5 (mod10) , s ( p1a1 )
º 1 + 5 + 25 + ...... + 54l + 1 (mod 10)
º 1
(mod 10)
When p1  7 (mod10) , s ( p1a1 )
= s ( p14l + 1 )
= 1 + p1 + p12 + ...... + p14l + 1
º (1 + 7 + 9 + 3) + ... + (1 + 7 + 9 + 3) + 1 + 7 (mod 10)
º 8
(mod 10)
Conclusively, s ( p1a1 ) is not divisible by 5 for p1  3,5 or 7 (mod10)
and therefore
5 | s ( pi ) for some i = 2, 3,..,n.
Now consider i = 2, 3, …, n.
For any pi º 5(mod10)
s ( p1a 1 )
a1
º 1 + 5 + 25 + ...... + 5 (mod 10)
º 1 + 5 + 5 + ...... + 5
(mod 10)
º 1 or 6
(mod 10)
s ( pi a i ) is never a multiple of 5.
For any pi º 3(mod10)
s ( pia i )
= s ( pi2b )
= 1 + pi + pi2 + ...... + pi2b
ïì (1 + 3 + 9 + 7) + ... + (1 + 3 + 9 + 7) + (1 + 3 + 9) (mod 10) if b is odd
º ïí
ïïî (1 + 3 + 9 + 7) + ... + (1 + 3 + 9 + 7) + 1
(mod 10) if b is even
ìï 3
(mod 10) if b is odd
º ïí
ïïî 1
(mod 10) if b is even
s ( pi a i ) is never a multiple of 5.
Exploration on odd perfect numbers p.6
For any pi º 7(mod10)
s ( pia i )
= s ( pi2b )
= 1 + pi + pi2 + ...... + pi2b
ìï (1 + 7 + 9 + 3) + ... + (1 + 7 + 9 + 3) + (1 + 7 + 9) (mod 10) if b is odd
º ïí
ïïî (1 + 7 + 9 + 3) + ... + (1 + 7 + 9 + 3) + 1
(mod 10) if b is even
ìï 7
(mod 10) if b is odd
º ïí
ïïî 1
(mod 10) if b is even
s ( pi a i ) is never a multiple of 5.
For any pi º 9(mod10)
s ( pia i )
= s ( pi2b )
= 1 + pi + pi2 + ...... + pi2b
º (1 + 9) + ... + (1 + 9) + 1 (mod 10)
º 1
(mod 10)
s ( pi a i ) is never a multiple of 5.
There is only one possibility left i.e. pi º 1(mod10) for some i=2, 3, ...,n
(Q.E.D.)
The Corollaries 2 and 3 are summarized as below:
p1 º 1(mod10) p1 º 3(mod10) p1 º 5(mod10)
Unit digits
p1 º 7(mod10)
p1 º 9(mod10)
M º 1(mod10)
1, 1
3, 1
5, 1 (**)
7, 1
9, 1 (not exist)
M º 5(mod10)
1, 5
3, 5 (**)
5, 5 (not exist)
7, 5 (**)
9, 5
M 2 º 9(mod10)
1, 9
3, 9
5, 9 (**)
7, 9
9, 9 (not exist)
2
2
Out of 15 possibilities, we certainly rule out 3 of them i.e. (5, 5), (9, 1), and (9, 9).
For the other four cases (3, 5), (5, 1), (7, 5) and (5, 9), we obtained the necessary condition for
being odd perfect is pi º 1(mod10) for some i = 2, 3, ...,n .
The remaining cases are very difficult. So we shifted our attention to the multiples of odd
perfect numbers and stepped into our second stage in which we had found some interesting
properties of odd perfect numbers other than their unit digits.
Exploration on odd perfect numbers p.7
Second Stage
In our first stage, we focused on the unit digits of the components p1 and M2 of odd
perfect numbers, hoping to get some contradictions to the existence of odd perfect
numbers. We had thought for the remaining cases for a long time without any progress,
then we began to consider the multiples of odd perfect numbers. We had found some
interesting results
Lemma 4:
For any positive integers a and c,
If c = a ×b for some positive integer b greater than 1, then
s (c ) s ( a )
>
.
c
a
Proof:
a
a
a
Let a = p1 1 p2 2 ... pn n
s (a)
1
1
1
1
1
1
= (1 +
+ 2 + ... + a1 )...(1 +
+ 2 + ... + a n )
a
p1 p1
p1
pn pn
pn
And so
Now a is multiplied by a number b = p1b1 p2b 2 ... pnb n .k1w1 k2w2 ...knwn ,
where k1 , k2 ...kn are positive integers relatively prime to a .
Let c = a ×b , then
s (c )
= s (a ´ p1b1 p2b 2 ... pnb n .k1w1 k2w2 ...knwn )
= (1 + p1 + p12 + ... + p1a 1 + b1 )(1 + p2 + p22 + ... + p2a 2 + b 2 )...(1 + pn + pn2 + ... + pna 2 + b 2 ) ×
(1 + k1 + k12 + ... + k1w1 )...(1 + kn + kn2 + ... + knw1 )
Then
s (c )
c
s (a ´ p1b1 p2b 2 ... pnb n ´ k1w1 k2w2 ...knwn )
=
a ´ p1b1 p2b 2 ... pnb n ´ k1w1 k2w2 ...knwn
1
1
1
1
1
1
1
1
+ 2 + ... + a 1 + ... + a 1 + b1 )...(1 +
+ 2 + ... + a n + ... + a 1 + b1 ) ×
p1 p1
p1
p1
pn p n
pn
pn
= (1 +
(1 +
> (1 +
=
1
1
1
1
1
1
+ 2 + ... w1 )...(1 +
+ 2 + ... wn )
k1 k1
k1
kn kn
kn
1
1
1
1
1
1
+ 2 + ... + a1 )...(1 +
+ 2 + ... + a n )
p1 p1
p1
pn pn
pn
s (a )
a
(Q.E.D.)
Exploration on odd perfect numbers p.8
Theorem 2:
A multiple of a perfect number cannot be a perfect number.
Proof:
Let N be a perfect number and M = kN for some natural number k ³ 2 .
By Lemma 4, we get
s (M ) s ( N )
>
= 2
M
N
\ M is not a perfect number.
(Q.E.D.)
After proving that multiples of a perfect numbers are not perfect, we were just
wondering what they are. One day, we felt and believed that they should have some
properties that others don’t have. We began to explore such properties in the Third
Stage.
Third Stage
In the Third Stage, we explored the relationships between perfect numbers and
their multiples. Finally we succeeded to relate odd perfect numbers and 3- perfect
numbers and generate higher multiply perfect numbers by lower ones. As in [1],
multiply perfect numbers are defined as below :
For a given natural number k, a number n is called k-perfect (or k-multiply perfect)
if and only if the sum of all positive divisors σ(n)of n is equal to kn, i.e σ(n)=kn.
Lemma 5
If W is an odd perfect number, then 2 W is a 3-perfect number.
Proof:
Since (2, W) =1, therefore s (2W) = s (2)s (W) = 3(2W)
So 2 W is a 3-perfect number.
(Q.E.D.)
Theorem 3
If the number of 3-perfect numbers is finite, then the number of odd perfect numbers is
also finite.
Proof:
By Lemma 5, if there are infinitely many odd perfect numbers, then there are also
infinitely many 3-perfect numbers can be generated. The theorem is proved by
contrapositive.
(Q.E.D.)
Exploration on odd perfect numbers p.9
Similarly, we could generate higher perfect numbers by lower ones which are
relatively prime to each other.
Theorem 4
If there are n perfect numbers that are relatively prime to each other, then there exists a
2 n -perfect number.
Proof:
Let W1 , W2 ,......Wn be n perfect numbers that are relatively prime to each other
and M= WW
1 2 ......Wn .
n
n
Then (M) = s (WW
1 2 ......Wn ) = s (W
1 )s (W2 )......s (Wn ) = 2 WW
1 2 ......Wn = 2 M
Thus M is a 2 n -perfect number.
(Q.E.D.)
Exploration on odd perfect numbers p.10
Summary and Conclusions
After a long-term study, we were all satisfied with our fruitful outcomes, even
though it was not perfect. However, they were all come from our sweat and effort. Our
main goal is to find the general form of a hypothetical odd perfect number and
eliminating those which cannot be odd perfect numbers. The following are the results of
our investigation.
In the First Stage, we were glad to find that the general form of a hypothetical odd
number W is (4k + 1)4l + 1 M 2 which is very beautiful.
Using this form, modulo and
divisibility, we have been more familiar with their unit digits and proved that the unit
digits of pi and M 2 cannot be 5 and 5, 9 and 1 or 9 and 9. Furthermore, if the unit digit
of pi is not 1 for all i = 2,3,..., n then the unit digits of p i and M 2 are never 3 and
5, 5 and 1, 5 and 9, and 7 and 5.
After investigation their general form, we started to consider the multiples of odd
perfect numbers and found a beautiful inequality on the function (n) i.e.
s (c ) s ( a )
>
,
c
a
by which we proved that any multiple of a perfect number cannot be perfect.
In the Third Stage, we continued our study by making use of the multiply perfect
numbers. We found that if W is an odd perfect number, then 2 W is a 3-perfect number.
By this result, the number of odd perfect numbers is bounded by that of 3-perfect
numbers. We were all excited about this result since we may prove the non-existence
of odd perfect number by proving the number of 3-perfect number is finite.
Additionally, if there are n perfect numbers that are relatively prime to each other, there
must a 2 n -perfect number can be generated.
After all, we hope that our investigation can have a little contribution to the
advancement of Mathematics. In the coming future, we will still continue to explore the
world of Mathematics!
Exploration on odd perfect numbers p.11
Reference
Websites
[1]
[2]
[3]
[4]
[5]
[6]
[7]
http://en.wikipedia.org/wiki/Multiply_perfect_number
http://mathworld.wolfram.com/PerfectNumber.html
http://mathworld.wolfram.com/OddPerfectNumber.html
http://mathworld.wolfram.com/MultiperfectNumber.html
http://mathworld.wolfram.com/AlmostPerfectNumber.html
http://www-history.mcs.st-andrews.ac.uk/HistTopics/Perfect_numbers.html
http://www.math.niu.edu/~rusin/known-math/99/multperf
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Exploration on odd perfect numbers p.12