Htun Soe - Goshen College
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".
................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.
Related download
- solution to stat 350 exam 2 review questions spring 2015
- statistics 8 chapters 7 to 10 sample
- wordnet a lexical database for english george a miller
- esl diagnostic esld
- chapter 2 the properties of electromagnetic radiation
- words to use instead of said
- 100 synonyms for said kimberly dana
- over 300 ways to say said
- session 4 graphical solution to lp
- review chapter 5 and 6
Related searches
- college board college code lookup
- college board average college cost
- college ceeb codes college board
- college board college search
- college board college search tool
- college board college search engine
- goshen county wyoming property search
- college board college application waiver
- college football college football recruiting ranks
- college search engine college board
- college board college application checklist
- college scholarships for college 2021