Htun Soe
Math 300: Conjecture and Proof
Prof. David Housman
11.28.2001
Perfect Numbers and Perfect Squares
Let n, m, a, b, c, d be integers and p, q, r, s be prime numbers.
Conjecture: Perfect squares are not prefect numbers.
In mathematical form, for any perfect number m =
2 n where n = p
1 a
1 , p
2 a
2 , ..., p c a c ,

(m) =

( n
2
) =

(( p
1 a
1 p a
, 2
2
, ..., p c a c )
2
) =

( p
1
2 a
1 , p
2
2 a
2 , ..., p c
2 a c ) =

( p
1
2 a
1 )

( p
2
2 a
2
) … 
( p c
2 a c ) .

2* m = 2
2 n = 2 * p
1
2 a
1 , p
2
2 a
2 , ..., p c
2 a c .
Definitions:
Perfect Square: A number m is said to be a perfect square if it can be written as
2 n where n = p
1 a
1 , p
2 a
2 , ..., p c a c , so m = p
1
2 a
1 , p
2
2 a
2 , ..., p c
2 a c . For example, 225 = ( 3
2
) * ( 5
2
) = ( 3 * 5 )
2
= ( 15 )
2
; 225 is therefore a perfect square.
Perfect Number: A number n is said to be perfect number if

(n) = 2n. In other words, n is perfect if the sum of its proper divisors (those divisors that are strictly less than n) is equal to n. For example, 28 is a perfect number because

(28) = 1 + 2 + 4 + 7 + 14 + 28 = 56 = 2 * 28. Or because the sum of the proper divisors of 28 is 1 + 2 + 4 + 7 + 14 = 28. 27 is not a perfect number because

(27) = 1 + 3+ 9
+ 27 = 40

2 * 27. In other words, because the sum of the proper divisors of 27 is 1 + 3 + 9

27.
In addition, if n = ( p a
) * ( q b
) ,

(n) = (1 + p + p
2 + … + a p ) (1 + q + q
2 + … + b q ) = 2 *
( p a
) * ( q b
) . For example,

(28) =

( ( 2
3
) * ( 3 ) ) = (1 + 2

2
2 
2
3
) (1 + 3

3

3
2
) =
(1 + 2 + 4 + 8) (1 + 3) = 56 = 2 * 28. Therefore, 28 is a perfect number.
Proof:
Step 1: A prime number is not a perfect number.
The sum of the divisors of a prime number p (i.e. 1 and itself) is not equal to 2*p. In mathematical form,

(p) = (1 + p)

2* p

p + p unless p = 1. But in definition of a prime number, we take 1 as non-prime number.
Property 1: A prime number is not a perfect number.

Step 2: Let's consider the case of the perfect squares of a single prime number: 4 and 9.

(4) = 1 + 2 + 4

2 * 4

4 + 4

8

(9) = 1 + 3+ 9

2 * 9

9 + 9

18
Generally, we can conclude that

(
2 p ) = 1 + p + p
2 
2 * p
2  2 p +
2 p because the underlined parts, (1 + p)

p* p when p is a prime number.
Property 2: The perfect square of a prime number is not a perfect number.
Step 3: Let's consider the case of a perfect square number that is the square of a power of a prime number.

(64) =

( ( 2
3
)
2 ) = 2
6 = (1 + 2

2
3 + … +
2
6 ) = 127

2* 6
2 = 128.
In general,

( ( p a
)
2 ) =

( p
2 a )

2 p
2 a or ( p
2 a )
 
( p
2 a

1 ).
We will prove it by contradiction too.
For a number in order to be a perfect number, the sum of its proper divisors (those divisors that are strictly less than n) is equal to n. This in turn means that

( p
2 a

1
) must be equal to ( p
2 a
) .
Lets assume that ( p
2 a
) is a perfect number. This means that

( p
2 a
) = 2* p
2 a
. But

( p
2 a
)= p*

( p
2 a

1
)+ 1. (This is proved in the last page.)
2 * ( p
2 a
) = p*

( p
2 a

1
)+ 1.
( p
2 a
) = p *

( p
2 a

1
)

1
2
( p
2 a
)
 p *

( p a

1
)

2 p *

( p a

1
)

1
2
( p
2 a
)
 p *

( p
2 a

1
)

1
2
Therefore, p
2 a is not a perfect number.
Property 3: A perfect square number that is the square of a power of a prime number is not a perfect number.
Step: 4: Is the power of a prime number a perfect number?
We will prove it by contradiction.

For a number in order to be a perfect number, the sum of its proper divisors (those divisors that are strictly less than n) is equal to n. This in turn means that

( p a

1
) must be equal to ( a p ) .
Let's assume that ( p a
) is a perfect number. This means that

( p a
) = 2 * ( a p ).
But

( p a
) = p*

( p a

1
) + 1. (This is proved in the last page.)
2 * ( p a
) = p*

( p a

1
)+ 1.
( a p ) = ( p *

( p a

1
)

1
)
2
( a p ) = p *

( p a

1
)
2
when p = 2 and ( p a
)
 p *

( p a

1
)
2
when p is a prime number other than 2.
So, ( p a
)
 p *

( p a

1
)
2
and ( p a
)

( p *

( p a

1
)
2

1
2
). This contradicts the previously stated fact that the sum of the proper divisors of ( p a
),

( p a

1
) is equal to ( a p ).
Therefore,
Property 4: ( a p ) is not a perfect number.
Step: 5: Let's consider the case of the perfect square of number that is a product of two prime numbers.

(2*3)^2 = (1 + 2 + 4) (1 + 3+ 9 )= 91

2* 36
In general, if n = p
2 q
2
,

( n ) =

( p
2 q
2
)

2 * p
2 q
2
.
We will prove it by contradiction.
Lets' assume that

( p
2 q
2
) = 2 * p
2 q
2
. Therefore,

(n) =

( p
2 q
2
) is a even number. In addition,

( n ) =

( p
2 q
2
) =

( p
2
) .

( q
2
) = (1 + p +
2 p ) (1 + q + q
2
) .
If p = 2, the only even prime number, 1 + p +
2 p = 1 + even + even = odd and similarly, 1 + q + q
2
=
1 + even + even = odd.
When p is a prime number other than 2, 1 + p +
2 p = 1 + odd + odd = 1 + even = odd. The same truth holds for 1 + q + q
2
. Therefore we can conclude that whether the prime numbers are even or odd, (1 + p +
2 p ) (1 + q + q
2
) = odd * odd = odd which contradicts the previously stated fact that

( n ) =

( p
2 q
2
) is an even number.
Hence, n is not a perfect number.
Property 5: The perfect square of number that is a product of two prime numbers is not a perfect number.

Step 6: Now we will consider the most general case of all perfect squares.
A perfect square m can be written as
2 n where n = p
1 a
1 , p
2 a
2 , ..., p c a c , so m = p
1
2 a
1 , p
2
2 a
2 , ..., p c
2 a c .

(m) =

( n
2
) =

(( p
1 a
1 p a
, 2
2
, ..., p c a c )
2
)
=

( p
1
2 a
1 , p
2
2 a
2 , ..., p c
2 a c )
=

( p
1
2 a
1 )

( p
2
2 a
2
) … 
( p c
2 a c )
= ( 1
 p
1
1  p
2
1

...
 p
1
2 a
1 ) ( 1
 p
2
1  p
2
2

...
 p
2
2 a
2 ) ( 1
 p c
1  p
2 c

...
 p c
2 a c ).
We will prove it by contradiction. Let m be a perfect number.

(m) = 2 * p
1
2 a
1 , p
2
2 a
2 , ..., p c
2 a c is an even number.
For a prime number p of an even power 2a, the number of its divisors = 2a +1, which implies that when we disregard 1, there will be 2a even number of divisors. Without losing the generality, we will make p and a represent all p i and a i
.
When p = 2, ( 1
 p
 p
2 
...
 p
2 a
) = 1 + even + even + … + even = odd. So the product of the prime divisors of m is the product of odd numbers so m is odd.
When p is a prime number other than 2, ( 1
 p
 p
2 
...
 p
2 a
) = 1 + odd + odd + …+ odd + odd.
While disregarding 1, there are even power 2a, an evenly paired odd number gives an even number,
( 1
 p
 p
2 
...
 p
2 a
) = 1 + even = odd.
Hence, we can conclude that whether p is an even number or odd number,

(m) =
( 1
 p
1
1  p
2
1

...
 p
1
2 a
1 ) ( 1
 p
2
1  p
2
2

...
 p
2
2 a
2 ) … ( 1
 p c
1  p
2 c

...
 p c
2 a c ) is the product of odd numbers and so

(m) is an odd number. This contradicts the previously stated fact that

(m) = 2
* p
1
2 a
1 , p
2
2 a
2 , ..., p c
2 a c is an even number.
So, m is not a perfect number.
Now, we can safely conclude that perfect squares are not perfect numbers.

Attached to Step 4.
Conjecture:

( p a
) = p*

( p a

1
)+ 1
Proof:

( p a
) = ( 1 + p + p
2 + … + a p ) p* p a

1
+ 1 = p ( 1 + p + p
2 + … + a

1 p ) + 1
= ( p + p
2 + … + p a ) + 1 = ( 1 + p + p
2 + … + p a )
Therefore,

( p a
) = p*

( p a

1
)+ 1
Recognition: This paper is indebted to a problem proposed by Carl Pomerance, University of Georgia. in page 6036 of June-Dec 1975 "Mathematical Monthly" and another problem proposed by P. Richard
Herr, University Park, Pennsylvania in page 1026 of 1974 "Mathematical Monthly".