Download Perfect Numbers and Perfect Squares

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 = n 2 where n = p1 1 , p 2 2 , ..., pc c , (m) = 
a
( n 2 ) = (( p1 1 , p 2 2 , ..., pc c ) 2 ) = ( p1
a
a
 2* m = 2 n 2 = 2 * p1
a
2 a1
, p2
2 a2
2 a1
, p2
, ..., pc
2 ac
2 a2
, ..., pc
2 ac
) =  ( p1
a
2 a1
)  ( p2
a
2 a2
) … ( pc
2 ac
).
.
Definitions:
Perfect Square: A number m is said to be a perfect square if it can be written as n 2 where n =
a
a
a
p1 1 , p 2 2 , ..., pc c , so m = p1
2 a1
, p2
2 a2
, ..., pc
2 ac
. For example, 225 = ( 32 ) * ( 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 + … + p a ) (1 + q + q 2 + … + q b ) = 2 *
( p a ) * ( q b ) . For example,  (28) =  ( ( 23 ) * ( 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  ( p 2 ) = 1 + p + p 2  2 * p 2  p 2 + p 2 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 )
In general, ( ( p a )
2
2
) = 2 6 = (1 + 2  2 3 + … + 2 6 ) = 127  2* 2 6 = 128.
) = ( 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 2a ) =
p *  ( p 2 a 1 )  1
2
( p 2a ) 
p *  ( p a 1 )  1
p *  ( p a 1 )

2
2
( p 2a ) 
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 ( p a ) .
Let's assume that ( p a ) is a perfect number. This means that  ( p a ) = 2 * ( p a ).
But
 ( p a ) = p*  ( p a  1 ) + 1. (This is proved in the last page.)
2 * ( p a ) = p*  ( p a  1 )+ 1.
( pa ) = (
( pa ) =
p *  ( p a 1 )  1
)
2
p *  ( p a 1 )
p *  ( p a 1 )
when p = 2 and ( p a ) 
when p is a prime number other than 2.
2
2
So, ( p a ) 
p *  ( p a 1 )
p *  ( p a 1 ) 1
 ). This contradicts the previously stated fact
and ( p a )  (
2
2
2
that the sum of the proper divisors of ( p a ),  ( p a  1 ) is equal to ( p a ).
Therefore,
Property 4: ( p a ) 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 + p 2 ) (1 + q + q 2 ) .
If p = 2, the only even prime number, 1 + p + p 2 = 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 + p 2 = 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 + p 2 ) (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
a
a
A perfect square m can be written as n 2 where n = p1 1 , p 2 2 , ..., pc c , so m =
p1
2 a1
, p2
2 a2
, ..., pc
2 ac
.
(m) =  ( n 2 ) = (( p1 1 , p 2 2 , ..., pc c ) 2 )
= ( p1
2 a1
=  ( p1
2 a1
, p2
2 a2
a
a
, ..., pc
2 ac
)  ( p2
2 a2
a
)
) … ( pc
= ( 1  p1  p 21  ...  p1
1
2 a1
2 ac
)
) ( 1  p 2  p 2 2  ...  p 2
1
2 a2
) ( 1  pc  p 2 c  ...  pc
1
We will prove it by contradiction. Let m be a perfect number. (m) = 2 * p1
2 a1
, p2
2 a2
2 ac
).
, ..., pc
2 ac
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 pi and ai .
When p = 2, ( 1  p  p 2  ...  p
2a
) = 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
2a
) = 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
2a
) = 1 + even = odd.
Hence, we can conclude that whether p is an even number or odd number, (m) =
( 1  p1  p 21  ...  p1
1
2 a1
) ( 1  p 2  p 2 2  ...  p 2
1
2 a2
) … ( 1  pc  p 2 c  ...  pc
1
2 ac
) is the product of
odd numbers and so (m) is an odd number. This contradicts the previously stated fact that (m) = 2
* p1
2 a1
, p2
2 a2
, ..., pc
2 ac
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 + p2 + … + pa )
p* p a  1 + 1 = p ( 1 + p + p 2 + … + p a  1 ) + 1
= ( p + p2 + … + pa ) + 1 = ( 1 + p + p2 + … + pa )
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".