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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 = [pic]where n = [pic], ((m) = ( ([pic]) = ((([pic])[pic]) = (([pic]) = ( ([pic]) ( ([pic]) …( ([pic]) .

← 2* m = 2[pic]= 2 * [pic].

Definitions:

Perfect Square: A number m is said to be a perfect square if it can be written as [pic] where n = [pic], so m = [pic]. For example, 225 = [pic]= [pic] = [pic]; 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 = [pic], ( (n) = (1 + p +[pic]+ … + [pic]) (1 + q +[pic]+ … + [pic]) = 2 * [pic]. For example, ( (28) = ( ([pic]) = (1 + 2 [pic] (1 + 3 [pic] = (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 ( ([pic]) = 1 + p + [pic] ( 2 * [pic] ([pic] + [pic] 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) = ( ([pic]) =[pic] = (1 + 2 [pic]+ … +[pic]) = 127 ( 2*[pic] = 128.

In general, (([pic]) = (([pic] ) ( 2[pic] or ([pic]) ( ( ([pic] ).

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 (([pic] ) must be equal to ([pic]) .

Lets assume that ([pic]) is a perfect number. This means that (([pic] ) = 2*[pic] . But (([pic] )= p* ( ([pic] )+ 1. (This is proved in the last page.)

2 * ([pic]) = p* ( ([pic] )+ 1.

([pic]) = [pic]

([pic]) ( [pic] ( [pic]

([pic]) ( [pic]

Therefore, [pic]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 (([pic] ) must be equal to ([pic]) .

Let's assume that ([pic]) is a perfect number. This means that ( ([pic]) = 2 * ([pic]).

But ( ([pic]) = p* ( ([pic]) + 1. (This is proved in the last page.)

2 * ([pic]) = p* ( ([pic] )+ 1.

([pic]) = ([pic] )

([pic]) = [pic] when p = 2 and ([pic]) ( [pic] when p is a prime number other than 2. So, ([pic]) ( [pic] and ([pic]) ( ([pic] ). This contradicts the previously stated fact that the sum of the proper divisors of ([pic]), ( ([pic] ) is equal to ([pic]).

Therefore,

Property 4: ([pic]) 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 = [pic][pic], (( n ) = ( ([pic][pic]) ( 2 *[pic][pic] .

We will prove it by contradiction.

Lets' assume that ( ([pic][pic]) = 2 *[pic][pic]. Therefore, ((n) = ( ([pic][pic]) is a even number. In addition,

(( n ) = ( ([pic][pic]) = ( ([pic]) . ( ([pic]) = (1 + p + [pic] ) (1 + q + [pic] ) .

If p = 2, the only even prime number, 1 + p + [pic] = 1 + even + even = odd and similarly, 1 + q + [pic] = 1 + even + even = odd.

When p is a prime number other than 2, 1 + p + [pic] = 1 + odd + odd = 1 + even = odd. The same truth holds for 1 + q + [pic]. Therefore we can conclude that whether the prime numbers are even or odd, (1 + p + [pic] ) (1 + q + [pic] ) = odd * odd = odd which contradicts the previously stated fact that (( n ) = ( ([pic][pic]) 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 [pic] where n = [pic], so m = [pic].

((m) = ( ([pic]) = ((([pic])[pic])

= (([pic])

= ( ([pic]) ( ([pic]) …( ([pic])

= ([pic] ) ([pic] ) ([pic] ).

We will prove it by contradiction. Let m be a perfect number. ((m) = 2 *[pic]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 [pic].

When p = 2, ([pic] ) = 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, ([pic] ) = 1 + odd + odd + …+ odd + odd. While disregarding 1, there are even power 2a, an evenly paired odd number gives an even number,

([pic] ) = 1 + even = odd.

Hence, we can conclude that whether p is an even number or odd number, ((m) = ([pic] ) ([pic] ) … ([pic] ) is the product of odd numbers and so ((m) is an odd number. This contradicts the previously stated fact that ((m) = 2 *[pic]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: ( ([pic]) = p* ( ([pic] )+ 1

Proof: ([pic] = ( 1 + p +[pic]+ … + [pic])

p* [pic] + 1 = p ( 1 + p +[pic]+ … + [pic]) + 1

= ( p +[pic]+ … + [pic]) + 1 = ( 1 + p +[pic]+ … + [pic])

Therefore, ( ([pic]) = p* ( ([pic] )+ 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".

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