## 1 Abstract

The following questions will be investigated throughout the course of this project:

2) All known perfect numbers are even. Are there any odd perfect numbers? To date, none have been found, and no one has proven that odd perfect numbers do not exist.

3) Define the sigma function to yield the sum of the proper divisors of a given number. For example: sigma(6) =1 + 2 + 3 = 6, sigma(15) = 1 + 3 + 5 = 9, sigma(12) = 1+ 2 + 3 + 4 + 6 = 16. If, given an initial number X, the sigma function is repeatedly applied to the result, a sequence of numbers is formed. For example: sigma(15) = 9, sigma(9) =4, sigma(4) =3, sigma(3) = 1. Thus, given the number 15, the resulting sequence is 15, 9, 4, 3, 1. I have grown accostomed to calling such a sequence a "factor sum sequence". Are there any infinitely long factor sum sequences? Explore the possibilities of applying factor sum sequences to cryptography.

## 2 A Quick Introduction To Terminology

A "perfect" number is a number whose divisors sum to twice itself. And easy examle is 6, whose proper divisors are 1, 2, 3, and 6, the sum of which is 1 + 2 + 3 + 6 = 12 = 2*6. Euclid was one of the first to study perfect numbers, although, due to the difficulty of finding such numbers, he only knew of the first four perfect numbers: 6, 28, 496, and 8128. The next three consecutive perfect numbers are: 33,550,336; 8,589,869,056; and 137,428,691,328. As you can see, perfect numbers are quite dispersed along the number line.

Number theorists have defined many different types of numbers, such as 'deficient', 'abundant', 'social', 'quasi-perfect', 'multi-perfect',  the list goes on. Each of these types are rich fields in themselves, so I will largely restrict myself to perfect numbers, while also calling on deficient and abundant numbers.

A 'deficient' number is, as one might guess, a number whose proper divisors sum to less than twice itself. Prime numbers and perfect squares are two examples of deficient numbers. On the other hand, an 'abundant' number is one whose proper divisors sum to greater than twice itself. The smallest abundant number is 12 (verify this yourself).

A unit fraction is any fraction with 1 in the numerator. It can be shown that the unit fractions whose denominators are the divisors of a perfect number will sum to two. For example: The divisors of the number 6 are: 6, 3, 2, and 1. Summing the reciprocals of these numbers yields 1/6 +1/3 + ½ + 1/1 = 2. This relationship holds with any perfect number, although the proof is not provided here. Along the same lines, the unit fractions of the divisors of a deficient number will sum to less than two (consider the number 9: 1/9 + 1/3 + 1/1 < 2) Likewise, the unit fractions of the divisors of an abundant number will sum to greater than two (consider the number 12: 1/12 + 1/6 + 1/4 + 1/3 + ½ + 1/1 = 7/3 > 2). This relationship between the unit fractions of the divisors of a number and the classification of the number as perfect, abundant, or deficient will prove useful when we try to find perfect numbers according to theory, rather than searching for them by trial and error.

## 3 How I Became Involved With Perfect Numbers

During my freshman year at the University of Arizona, my sister (Michele Gaberdiel, two years my senior) challenged me to show that all known perfect numbers are triangular numbers. I couldn't do it at first. The demonstration required a proof by the principle of mathematical induction, a technique I had not yet learned. I failed her challenge, but I suceeded in acquainting myself with a particular type of number that would fill my everyday thoughts for months: perfect numbers.

My knowledge of perfect numbers at the time was near non-existant, based solely on definition: a "perfect" number is an integer whose divisors sum to twice itself. When my sister challenged me, however, I began to investigate these numbers further. She gave me a short list of the first seven consecutive perfect numbers, and after examining the prime factorization of these numbers, I soon found a formula that could produce each perfect number on the list.

This is the formula: (2^(x-1))*(2^x - 1), where x is a natural number.

The first seven consecutive perfect numbers seen in this form are:
6 = (2)(3) = (2^(2-1))*(2^2 - 1)
28 = (4)(7) = (2^(3-1))*(2^3 - 1)
496 = (16)(31) = (2^(5-1))*(2^5 - 1)
8,128 = (64)(127) = (2^(7-1))*(2^7 - 1)
33,550,336 = (4096)(8191) = (2^(13-1))*(2^13 - 1)
8,589,869,056 = (65536)(131071) = (2^(17-1))*(2^17 - 1)
137,438,691,328 = (262144)(524287) = (2^(19-1))*(2^19 - 1)

The first thing I noticed about this formula was that only prime 'x' values yielded a perfect number. The following 'x' values all yield a perfect number in the preceding formula: x = 2, 3, 5, 7, 13, 17, and 19. It seemed at first glance that all prime numbers would yield a perfect number. When I noticed that 11 was not on the list, I naively assumed that (2^(11-1))*(2^11 - 1) would also be a perfect number. I quickly (too quickly) jotted down the number (2^(11-1))*(2^11 - 1) = 2096128, listed its factors
(Factors of 2096128: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2047, 4094, 8188, 16376, 32752,
65504, 131008, 262016, 524032, 1048064, 2096128)
and found the sum of the factors, which is 4192256 = (2)(2096128). Voila! It was perfect! I immediately called my sister and informed her I had found a new perfect number. Silent for a few moments as she calculated away, she proceded to inform me that 2047 is a composite number (2047 = 23*89) and so the factors of 2096128 sum to greater than 2*2096128. In short, I had jumped the gun and come to a false conclusion based on hasty calculations.

I soon realized that it was not only 'x' that needed to be prime, but also the value of (2^x -1). Months later, I read a proof of this and learned that number theorists traditionally label prime numbers of the form (2^x - 1) a "Mersenne Prime", in honor of a man who devoted himself to finding many primes of this form. More of Mersenne Primes later.

This, I must admit, was the extent to which I explored perfect numbers during my freshman year. It was during the following summer that my sister gave me a seemingly unrelated problem to solve while my family was driving from Arizona to Montana. The challenge was this:
For a given value 'Z', find the maximum number of distinct unit fractions
with denominators less than or equal to 'Z' that sum to 1.
For example, let Z = 20. Then the maximum number of distinct unit fractions with denominators less than or equal to 20 is six: 1/3 + Ό + 1/6 + 1/9 + 1/12 + 1/18 = 1.
Though this problem seemed utterly mind-numbing and useless, I devoted myself fully to this problem. What else had I to do on a cross country trip?

My first goal with this challenge was to break what my sister said was the class record: some one had found a little over fifty unit fractions summing to 1 with denominators less than or equal to 1000. I began by looking for a solution with nothing but chance helping me. I quickly found 13 distinct unit fractions summing to 1, then, after finding a neat trick, 32 distinct unit fractions. The trick I found was to start with the smallest possible unit fraction, ½, and to multiply it with the next biggest unit fraction, 1/3, to yield 1/6. These fractions are related according to the first being the sum of the latter two: ½ = 1/3 + 1/6. A general equation of this process is this: 1/q = 1/(q+1) + 1/(q(q+1)), which can be shown to be true for any natural number q through a little algebraic manipulation. Using this equation, I expressed each unit fraction as a some of two others:
½
1/3 + 1/6
(1/4 + 1/12) (1/7 + 1/42)
etcetera.
After applying this method repeatedly, I arrived at a list of 32 distinct unit fractions, which I then began to make larger by my first method: trial and error. Not soon after, I achieved a total of 68 distinct unit fractions summing to 1.
This record was sufficient to make me happy, so I turned my attention toward attempting to theoretically determine the maximum number of distinct unit fractions that sum to 1 with denominators less than or equal to Z. I did this by systematically solving this questions for Z = 1, 2, 3, 4, . 20, 21, ., etc., until an undeniable (and ultimately provable) pattern emerged:

Pattern 1: If all of the divisors of a perfect number are treated as denominators of unit fractions, then these unit fractions sum to exactly 2.
Consider the perfect number 6: 1/1 + ½ + 1/3 + 1/6 = 2
Pattern 2: If all of the divisors of a deficient number are treated as denominators of unit fractions, then these unit fractions sum to a number less than 2.
Consider the deficient number 5: 1/1 + 1/5 = 6/5 < 2
Pattern 3: If all of the divisors of an abundant number are treated as denominators of unit fractions, then these unit fractions sum to a number greater than 2.
Consider the abundant number 12: 1/1 + ½ + 1/3 + Ό + 1/6 + 1/12 = 7/3 > 2

PROOF OF PATTERN 1:
Let the divisors of N be listed in increasing order: 1, n1, n2, n3, , n N.
Property1: It is a general property of a list of divisors that the product of the first and last divisors equals N,
the product of the second and second to last divisors equals N,
the product of the third and third to last divisors equals N,
etcetera.
Since N is a perfect number, the sum of said divisors has the property that
1 + n1 + n2 + n3 +  + N = 2N
Divide this equation through by N to yield
1/N + n1/N + n2/N + n3/N +  + N/N = 2N/N
Now, simplify each term in this equation using Property 1
1/N +  + 1/n3 + 1/n2 + 1/n1 + 1 = 2
Thus, all of the divisors of an even perfect number N, if treated as denominators of unit fractions, will sum exactly to 2. QED

Similarly, one can show that Pattern 2 and Pattern 3 are also true.

As insignificant as this seems, and as infinitely more insignificant this entire question of unit fractions seems, this discovery was by far the most valuable and inspiring to my future work with perfect numbers. That summer, for the first time in months, I returned my thoughts to perfect numbers with fresh ideas. More than that, I found hope that there was more to this field than searching blindly for the next largest perfect number by mindlessly summing the divisors of huge numbers. What I had found in the above patterens was a tool I could use to study perfect numbers according to theory.

Here's an example: I want to know if there is a perfect number, N, that is the product of two prime numbers, a and b. Thus, when can N = a*b be perfect, if ever?

Use the fact that for a*b to be perfect, the unit fractions whose denominators are divisors of a*b must sum to 2.

The divisors of a*b are: a*b, a, b, and 1.
Thus, show: 1/1 + 1/a + 1/b + 1/a*b = 2
Simplify this by finding a common denominator
a*b/a*b + b/a*b + a/a*b + 1/a*b = 2
(a*b + b + a + 1)/a*b = 2
a*b + b + a + 1 = 2a*b
1 + a + b =a*b
Now solve for a:
a - a*b = -b - 1
a(1 - b)= -1*(1+b)
a = (b + 1) / (b - 1)
since b is prime, b+1 is even, of the form 2k for some integer k
also, b-1 is even, of the form 2k - 2
Thus, a = (2k)/(2k - 2)
a = k/(k-1)
The expression k/(k-1) will only yield a prime number 'a' if the denominator is equal to 1.
Thus k - 1 = 1, or k = 2.
So a = 2/(2-1) = 2.
Recall that (b+1) = 2k. Thus (b+1) = 4, or b = 3.

In conclusion, if a = 2, and b =3, then N = 6 is a perfect number, as is well known.
Further, N = 6 is the only perfect number that is the product of two primes.

Note: It is true that the preceding example could have been solved by simply observing that for a number N = a*b to be perfect, the sum of the divisors of a*b must sum to 2a*b. In other words, the requirement that 1 + b + a + a*b = 2a*b is obvious without knowing that a seemingly obscure sum of unit fractions equals 2. However, in other cases, knowing the link between unit fractions and perfect numbers can prove very useful.

## 4 My Attention Turns To Odd Perfect Numbers

1) What is the next largest even perfect number? As of 1999, there were 38 known.
2) Is there an odd perfect number? No one has found one, but no one has proven that they done't exist.

As to the first question, much is already known about even perfect numbers, most of which will be covered in this report. It is known that all even perfect number are of the form (2^(x-1))*(2^x - 1) where (2^x - 1) is a Mersenne prime. Thus, the problem of finding even perfect numbers is reduced to the problem of finding prime numbers of the form (2^x - 1) where 'x' is a natural number. In the main stream of the mathematical community, this problem is tackled with faster and faster computers and more and more efficient factoring algorithms. I neither own a fast computer nor am I familiar with factoring algorithms, so leave question #1 to the community while I turn my attention to question #2: are there any odd perfect numbers?

At first glance, there does not appear to be any odd perfect numbers. My initial thought was to use a computer program to simply look for an odd perfect number in the range from 1 - 1000. None were found, but in the world of mathematics, absence of evidence is not evidence of absence. Instead of continuing to look through larger and larger numbers for an odd perfect number, I took a systemmatic approach. I did this be eliminating certain types of odd numbers that simply could not be perfect. My lofty ambition told me that if I could eliminate every type of odd number as a candidate for an odd perfect number, I would have solved the problem.

These are the few of the 'types' of odd numbers I considered first:
1 primes
2 primes raised to any power
3 the product of two primes
4 the product of any number of primes
5 the product of any number of primes, each raised to any power n

I will give here a proof for why the first 3 types of odd numbers cannot be perfect.

1) Prime numbers cannot be perfect. By definition, a number N is perfect if the sum of its divisors is 2N. For any prime number P, it's divisors are P and 1. The sum of these divisors is (P+1), which is always less than 2P.
2) A prime number, P, raised to any power, n, cannot be perfect. The divisors of a number P^n are:
P^n, P^(n-1), P^(n-2),  P^2, P^1, and 1. I must show that
P^n + P^(n-1) + P^(n-2) + + P^2 + P^1 + 1 < 2P^n for all P and all n.
Regroup the left side of the inequality
P^n + P^(n-1) + P^(n-2) + + P^2 + P^1 + 1 = P^n + [P^(n-1) + P^(n-2) + + P^2 + P^1 + 1]
But the list of numbers in brackets is simply a geometric sequence whose sum is [P^n - 1] / [P - 1] (this is the formula for the sum of a generic geometric sequence).
Thus,
P^n + P^(n-1) + P^(n-2) + + P^2 + P^1 + 1 = P^n + [P^n - 1] / [P - 1]
But [P^n - 1] / [P - 1] < P^n for all P and all n, so
P^n + P^(n-1) + P^(n-2) + + P^2 + P^1 + 1 = P^n + [P^n - 1] / [P - 1] < P^n + P^n = 2P^n
Therefore, a prime P raised to any power 'n' cannot be perfect.
3) The product of two primes, p and q, cannot be perfect. The divisors of the number p*q are:
pq, p, q, and 1. I must show that 1 + p + q + pq < 2pq for all prime numbers p and q.
It is known that for all distinct primes, p and q, greater than or equal to 3,
(q - 1)*(p - 1 ) > 2
Simplifying this expression yields qp - p - q + 1 > 2.
Rewriting this expression yields qp > p + q + 1
Adding pq to each side of the inequality yields 2pq > pq + p + q + 1.
Thus, the product of any two primes cannot be perfect.

The above mentioned 4th and 5th type of odd numbers could not be so easily dismissed as candidates for odd perfect numbers. Literal hours upon hours were spent looking for ways to solve, for example, that the product of three primes could not be perfect. This would require showing:
(p + 1)*(q+1)*(t+1) < 2pqt for all odd primes p, q, and t. Unfortunately, I could find no way to show this.

Another way to tackle the preceding types of odd numbers is to use a proof by mathematical induction.

Consider the number N = (p^n)*(q^m) where 'p' and 'q' are distinct odd primes and 'n' and 'm' are natural numbers. The sum of the divisors of N can be written as (1 + p +  + p^n)*(1 + q +  + q^m), which must equal 2*(p^n)*(q^m) if N is to be perfect. In order to show that this cannot happen, consider the divisors of N as denominators of unit fractions, and show their sum is always less than 2.
In other words, show it is impossible that (1 + 1/p +  1/p^n)*(1 + 1/q +  + 1/q^m) = 2
We want (1 + 1/p +  1/p^n)*(1 + 1/q +  + 1/q^m) to be as large as possible, so allow n and m to tend toward infinity.
In such a case, the sum of an infinite geometric series is 1/(1-r) where r is the common ratio.
Thus, as n and am approach infinity
(1 + 1/p +  1/p^n)*(1 + 1/q +  + 1/q^m) approaches [1/(1 - 1/p)]*[1/(1 - 1/q)] = [p/(p - 1)]*[q/(q - 1)]
Now, show that for any odd primes p and q, [p/(p - 1)]*[q/(q - 1)] < 2.

Base Case
Let p = 3 and q = 5.
We want (1 + 1/3 +  1/3^n)*(1 + 1/5 +  + 1/5^m) to be as large as possible, so allow n and m to tend toward infinity.
Thus, as n and m approach infinity
(1 + 1/3 +  1/3^n)*(1 + 1/5 +  + 1/5^m) approaches [1/(1 - 1/3)]*[1/(1 - 1/5)] = 1.875

Inductive Hypothesis
Let p and q be defined such that (1 + 1/p +  1/p^n)*(1 + 1/q +  + 1/q^m) < 2.

Let (p + h) and (q + g) be defined such that h and g are numbers such that (p + h) and (q + g) are prime numbers greater than p and q, respectively.
Through algebraic manipulation, [p/(p - 1)]*[q/(q - 1)] < 2
Can be shown to be equivalent to the expression: -1 + p + q < pq/2
After substituting the values of (p + h) and (q + g) for p and q, respectively,
The expression can be simplified to [-1 + p + q] + g + h < [pq/2] +(hq + pg + gh)/2
The terms in brackets are simply the inductive hypothesis, so they can be removed from the inequality to yield g + h < hq/2 + pg/2 + gh/2
Since q and p are each odd primes, they are each greater than or equal to 3.
Thus, g + h < 1.5h + 1.5g + gh/2 <= hq/2 + pg/2 + gh/2.
Therefore, it is true that [p/(p - 1)]*[q/(q - 1)] < 2 for all odd primes p and q.

The preceding result implies that there are no odd perfect numbers of the form N = (p^n)*(q^m)
where 'p' and 'q' are distinct odd primes and 'n' and 'm' are natural numbers. In other words, an odd perfect number must have at least three distinct prime factors.

Constructing proofs for when a particular type of odd number cannot be perfect is fine and dandy, but at some point, my method for tackling the odd perfect number problem had to change. After writing the preceding proofs, I had eliminated 4 special cases of odd numbers that couldn't be perfect, but there were still an infinite number of special cases waiting. Realizing this, I knew I needed a fresh perspective, so I went to the library and began researching with books instead of my calculator.

## 5 The Sigma and Tau Functions

Many number theory books define two incredibly useful functions - the sigma and tau - before delving into the field of perfect numbers and related topics.

THE SIGMA FUNCTION
The sigma function, for a number N, yields the sum of all divisors of N. To reiterate,
When sigma(N) < 2N, N is a deficient number.
When sigma(N) = 2N, N is a perfect number.
When sigma(N) > 2N, Ni is an abundant number.

Here are some examples:
sigma(6) = 6 + 3 + 2 + 1 = 12 = 2*6 Thus, 6 is a perfect number.
sigma(9) = 9 + 3 + 1 = 13 < 2*9 Thus, 9 is a deficient number.
sigma(12) = 12 + 6 + 4 + 3 + 2 + 1 = 28 > 2*12 Thus, 12 is an abundant number.

Fortunately, sigma is a nice function to work with because you don't have to actually list the divisors of N in order to find sigma(N). The following three properties of the sigma function make it simple to find sigma(N) for any natural number N.

-For a prime number P, sigma(P) = P + 1
By definition, the only divisors of a prime number are itself, P, and 1.
-For a prime number P raised to any power n, sigma(P^n) = [P^(n+1) - 1] / [P - 1]
This follows from the fact that the divisors of P^n are the terms in the geometric sequence
1, P^1, , P^(n-1), P^(n)
-For relatively prime numbers A and B, sigma(A*B) = [sigma(A)]*[sigma(B)]
A "multiplicative function" has the property that the function of a product A*B is equal to the product of the function of A and the function of B. Thus, sigma is a multiplicative function.

Thus, to find sigma(N) for any natural number N, simply rewrite N in it's prime factored form and apply the previously stated properties of the sigma function. For example:
Sigma(36) = sigma(4*9) = [sigma(2^2)]*[sigma(3^2)] = ([2^(2+1) - 1]/[2-1])*([3^3 - 1]/[3 - 1]) = 91
Indeed, the sum of the divisors of 36 is 1 + 2 + 3 + 4 + 6 + 9 + 12 + 18 + 36 = 91

THE TAU FUNCTION
The tau function, for a number N, yields the total number of divisors of N.

Here are some examples:
Tau(6) = 4 The divisors of 6 are: 6, 3, 2, 1
Tau(9) = 3 The divisors of 9 are: 9, 3, 1
Tau(12) = 6 The divisors of 12 are: 12, 6, 4, 3, 2, 1

There is no easy way to relate tau(N) with N to determine if N is a perfect number as in the way sigma(N) = 2N implies N is a perfect number. However, it will later be shown that for a perfect number of the form (2^x - 1)*(2^(x -1)), tau((2^x - 1)*(2^(x -1))) = 2x.

Like the sigma function, you don't need to list the divisors of N in order to calculate tau(N). The following three properties of the tau function make it simple ti calculate tau(N) for any natural number N.

-For a prime number P, tau(P) = 2
Again, by definition, any prime number only has two divisors: itself and one.
-For a prime number P raised to any power n, tau(P^n) = n + 1
This follows from observing the that the geometric sequence -- 1, P^1, , P^(n-1), P^(n) -
Has 'n' terms with P as a factor and one term that is simply 1.
-For relatively prime numbers A and B, tau(A*B) = [tau(A)]*[tau(B)]
Tau is also a multiplicative function.

Thus, to find tau(N) for any natural number N, simply rewrite N in it's prime factored form and apply the previously stated properties of the tau function. For example:
Tau(36) = tau(4*9) = [tau(2^2)]*[tau(3^2)] = (2+1)*(2+1) = 9
Indeed, 36 has nine divisors: 36, 18, 12, 9, 6, 4, 3, 2, and 1.

## 6 Applying The Sigma And Tau Function To The Study Of Even Perfect Numbers

It was previously conjectured that numbers of the form (2^(x-1))*(2^x - 1) are perfect numbers if (2^x - 1) is prime. A stronger form of this statement will be prove here with the aid of the sigma function.

**Note: The names of the following "Theorems" are not technical or commonly used. I simply made up
the names to suit the subject matter.

EVEN PERFECT NUMBER THEOREM
An even number is perfect if and only if it is of the form (2^(x - 1))*(2^x - 1) where (2^x - 1) is prime.

PART 1 (The following proof is similar to the proof Euler gave to the same effect)
First, I will prove that an even number is perfect if it is of the form (2^(x - 1))*(2^x - 1) where (2^x - 1) is prime.

Let 2n be an even perfect number.
By definition, sigma(2n) = 2(2n).
Let y be the highest power of 2 in the number 2n, and let q by the odd remainder of 2n divided by 2^y.
Thus, 2n = (2^y)*q.
Furthermore, using the properties of the sigma function, sigma(q*2^y) = [sigma(q)]*[sigma(2^y)], which
can be simplified to [sigma(q)]*[(2^(y+1) - 1)/(2-1)] = [sigma(q)]*[2^(y+1) - 1]
Equating the various solutions of sigma(2n) yields: 2(2n) = 2(q*2^y) = [sigma(q)]*[2^(y+1) - 1]
Or, q*2^(y+1) = [sigma(q)]*[2^(y+1) - 1]
Rewrite the (2^(y+1)) term on the left as (2^(y+1) - 1 + 1) to yield
q*(2^(y+1) - 1 + 1) = [sigma(q)]*[2^(y+1) - 1]
Rewrite again as q*(2^(y+1) - 1) + q = [sigma(q)]*[2^(y+1) - 1]
Solve for [sigma(q)] by dividing through by [2^(y+1) - 1] to yield
q + q/[2^(y+1) - 1] = sigma(q)
Since q is an odd integer, and sigma(q) is also an integer, [2^(y+1) - 1] must divide q.
Let q = [2^(y+1) - 1]*k for some integer k.
Then q + ([2^(y+1) - 1]*k)/ [2^(y+1) - 1] = sigma(q)
Simplifying this expression yields q + k = sigma(q)
Since q and k both divide q, they must constitute all of the divisors of q.
Only prime numbers have two divisors. Thus, q is a prime number and k = 1.
Since we let q = [2^(y+1) - 1]*k for some integer k, and k = 1, then q = [2^(y+1) - 1]
We stated originally that 2n is our even perfect number and that 2n = (2^y)*q.
Substitute in our value of q to yield 2n = (2^y)*[2^(y+1) - 1]
Finally, let x = y + 1.
Then 2n = (2^(x - 1))*(2^x - 1).
Therefore, an even number is perfect if it is of the form (2^(x - 1))*(2^x - 1) where (2^x - 1) is prime.QED

PART 2
Second, I will prove that a number of the form (2^(x - 1))*(2^x - 1) where (2^x - 1) is prime
is a perfect number.

By definition, if N is a perfect number, sigma(N) = 2N
Thus, sigma((2^(x - 1))*(2^x - 1)) should be equal to 2*(2^(x - 1))*(2^x - 1)
(provided, of course, that (2^x - 1) is prime)
Using the properties of the sigma function, find sigma((2^(x - 1))*(2^x - 1))
Since (2^x - 1) is a prime number that is never equal to 2, (2^x - 1) is relatively prime to (2^(x-1)).
Thus, sigma((2^(x - 1))*(2^x - 1)) = [sigma(2^(x - 1))]*[sigma(2^x - 1)]

Since (2^x - 1) is prime, sigma(2^x - 1) = (2^x - 1 + 1) = (2^x)
Also, sigma(2^(x - 1)) = [2^x - 1]/[2-1] = (2^x - 1)

Thus, [sigma(2^(x - 1))]*[sigma(2^x - 1)] = (2^x - 1)*(2^x)
This final expression can be rewritten as 2*(2^(x - 1))*(2^x - 1)

Indeed, when (2^x - 1) is a prime number, sigma((2^(x - 1))*(2^x - 1)) = 2*(2^(x - 1))*(2^x - 1)
Therefore, a number of the form (2^(x - 1))*(2^x - 1) where (2^x - 1) is prime is a perfect number. QED

THEOREM CONCERNING TAU(N) WHEN 'N' IS AN EVEN PERFECT NUMBER:
If N is an even perfect number, then tau(N) = 2x

Let N be an even perfect number.
Thus N = (2^(x-1))*(2^x - 1) for some value 'x' where (2^x - 1) is prime.
Using the properties of the tau function:
tau((2^(x-1))*(2^x - 1)) = [tau(2^(x - 1))]*[tau(2^x - 1)] = [(x - 1) + 1]*(2) = 2x
Therefore, if N is an even perfect number, then tau(N) = 2x

## 7 The Search For Mersenne Primes

Now that we know that an even number is perfect if and only if it is of the form (2^(x - 1))*(2^x - 1) where (2^x - 1) is prime, how do we go about selecting an appropriate value of x from the infinity of natural numbers? As previously stated, a prime of the form (2^x - 1) is commonly referred to as a Mersenne prime. What restrictions can we place on 'x' to make (2^x - 1) prime?

FIRST MERSENNE PRIME THEOREM
If 'x' is composite, then (2^x - 1) is also composite.

Let 'x' be a composit number expressed as x = a*b where 'a' and 'b' are any natural numbers greater than 1
Then (2^x - 1) = [2^(ab) - 1]
In general, a number of the form [X^(ab) - 1]
can be rewritten as [X^a - 1]*[X^(ab - a) + X^(ab - 2a) +  + X^(ab - ba)]
Thus, if (2^x - 1) = [2^(ab) - 1]
And [2^(ab) - 1] = [2^a - 1]*[2^(ab - a) + 2^(ab - 2a) +  + 2^(ab - ba)],
Then (2^x - 1) is composite. QED

Nothing can be said, however, if 'x' is prime. It is sometimes the case that when 'x' is prime, (2^x -1) is prime:
Let x = 2. (2^2 - 1) = (4 - 1) = 3
And it is sometimes the case that when 'x' is prime, (2^x - 1) is composite:
Let x = 11. (2^11 - 1) = 2047 = 23*89

As previously stated, primes of the form (2^x - 1) are called Mersenne primes. According to the previous theorem, looking for Mersenne primes means looking for values of (2^x - 1) when x is prime. For any given prime 'x', (2^x - 1) could be prime or not, and all you can do to check is actually factor the number. This is where fast computers and highly efficient factoring algorithm come into the picture, and I and my lack of experience with said tools exit.

Note: As of yet, it is still not known whether there are an infinite number of Mersenne primes.

Using the sigma and tau functions, much can be determined about what characteristics an odd perfect number would have (if such numbers actually exist).

Note: In the proofs that follow, 'D' always represents the hypothetical odd perfect number.

PROPERTY 1: EVEN NUMBER OF DIVISORS

Suppose D is an odd perfect number.
By definition, sigma(D) = 2D.
Thus, the sum of the divisors of D is an even number.
Each individual divisor of D is odd, so there must be an even number of divisors.

PROPERTY 2: AT LEAST ONE ODD EXPONENT IN THE PRIME FACTORIZATION OF 'D'

Suppose D is an odd perfect number and let D = (a^x1)*(b^x2)*(c^x3)**(n^xn).
By definition, tau(D) = (x1 + 1)*(x2 + 1) * (x3 + 1) **(xn + 1).
But an odd perfect number has an even number of divisors.
Thus, tau(D) = (x1 + 1)*(x2 + 1) * (x3 + 1) **(xn + 1) = 2k for some integer k.
For (x1 + 1)*(x2 + 1) * (x3 + 1) **(xn + 1) to be even, at least one of the quentities (xi + 1)
must also be even (here, 1 <= i <= n).
If (xi + 1) is even, then (xi +1) = 2q for some integer q. So, xi = 2q -1.
Therefore, at least one of the exponents in the prime factorization of D is odd.

SideNote: A similar proof shows that an even perfect number, N, must also have at least one odd exponent in its prime factorization. Thus, in general, a square number can never be perfect.

PROPERTY 3: AT MOST ONE ODD EXPONENT IN THE PRIME FACTORIZATION OF 'D'

Supose D is an odd perfect number and let D = (a^x1)*(b^x2)**(n^xn).
By definition, sigma(D) = 2D
Or, sigma((a^x1)*(b^x2)**(n^xn)) = 2*(a^x1)*(b^x2)**(n^xn)
Also, sigma((a^x1)*(b^x2)**(n^xn)) = [sigma(a^x1)]*[sigma(b^x2)]**[sigma(n^xn)]
Thus, in the solution of [sigma(a^x1)]*[sigma(b^x2)]**[sigma(n^xn)],
exactly one of [sigma(i^xi)] can be even (here, 1 <= i <= n).
Since each term in [sigma(i^xi)] = 1 + i + i^2 + i^xi
is odd, there must be an even number of divisors.
So tau(i^xi) = 2k for some integer k. As before, xi must therefore be odd.
Therefore, there is at most one odd exponent in the prime factorization of 'D'.

This property of an odd perfect number puts some closure to an issue I could never quite answer using other techniques. Previously, I had asked if the product of three distinct odd primes could be perfect, i.e.: if a, b, and c are distinct primes, can a*b*c be perfect? There is a proof that the product of two distinct primes can not be perfect, but try as I might, I could not find a proof that the product of three, or four, or five, or any arbitrary amount of primes could not be perfect. Property 3 clearly states that one and only one of the primes in the prime factorization of an odd perfect number can be odd. Thus, numbers that are the product of odd primes (all to the first power) cannot be perfect. This implies the truth of a whole slew of inequalities:
(Let the following variables all represent odd primes to the first power)
(a + 1) < 2a
(a+1)*(b+1) < 2ab
(a+1)*(b+1)*(c+1) < 2abc
etcetera
Each of the preceding inequalities can be rewritten as
1 < a
1+a+b < ab
1+a+b+c+ab+bc+ac < abc
etcetera
This means that, when considering a set of k distinct odd primes, all possible product combinations of said primes will sum to less than the product of the k primes.

PROPERTY 4: IN THE PRIME FACTORIZATION OF 'D', THERE MUST EXIST A PRIME LESS THAN OR EQUAL TO i^xi WHERE 'i' IS THE PRIME WITH ODD EXPONENT 'xi'.

It has been shown that sigma(p^n) < 2p^n for all prime numbers p and natural numbers n.
So, for the prime - i^xi - with odd exponent in the prime factorization of 'D'
there must exist a z = q^m such that sigma(i^xi) = 2z < 2i^xi
Thus, z < p^n.
Therefore, 'D' contains in it's prime factorization a prime number that is less than or equal to the prime
factor with odd exponent.
Further, since sigma(I^xi) = 1 + i + i^2 +  + i^xi = 2z, i cannot divide 2z.

PROPERTY 5: AT LEAST 3 PRIME FACTORS IN 'D'

This result was proved in Section 4 of this report.

PROPERTY 6: FOR THE PRIME 'p' WITH ODD EXPONENT '2x + 1'
'p' IS CONGRUENT TO ONE MODULO FOUR

Consider sigma(p^(2x + 1)) where 'p' is the prime in 'D' with the odd exponenet.
sigma(p^(2x + 1)) = [p^(2x + 1 + 1) - 1] / [p - 1] = [p^(2x + 2) - 1] / [p - 1]
As disusses previously, since 2x +2 is composite, [p^(2(x+1)) - 1] can be factored like so:
[p^(2(x+1)) - 1] = (p^2 - 1)*[p^(2x + 2 - 2) + p^(2x + 2 - 4) +  + 1]
Thus, sigma(p^(2x+1)) = [(p^2 - 1)*[p^(2x + 2 - 2) + p^(2x + 2 - 4) +  + 1]] / [p - 1]

Lemma: 8 divides (p^2 - 1) for all odd primes 'p'.
Let p = 2k +1 for some integer k.
Then (2k+1)^2 - 1 = 4k^2 + 4k + 1 - 1 = 4k^2 +4k = 4(k)(k+1)
If k is an even integer, then k = 2q for some integer q, and (k)(k+1) = (2q)(2q +1) = 2(2q^2 + q)
Thus 4(k)(k+1) = 4[2(2q^2 + q)] = 8(2q^2 +q).
If k is an odd integer, then k = 2q + 1 for some integer q, and (k)(k+1) = (2q+1)(2q + 1 + 1) = 2(q+1)(2q+1)
Thus 4(k)(k+1) = 4[2(q+1)(2q+1)] = 8(q+1)(2q+1).
Indeed, in any case, 8 divides 4(k)(k+1) = (2k+1)^2 - 1 = (p^2 - 1) for some odd integer p = 2k+1

According to the preceding Lemma, 8 always divides the numberator of
sigma(p^(2x+1)) = [(p^2 - 1)*[p^(2x + 2 - 2) + p^(2x + 2 - 4) +  + 1]] / [p - 1]
If sigma(p^(2x+1)) is to have only one 2 in its prime factorization, the denominator of sigma(p^(2x+1))
must at least be divisible by 4.
So, p - 1 = 4k for some integer k. Or, solving for p, p = 1 + 4k

Therefore, in the prime factorization of 'D', the prime 'p' with odd exponent '2x + 1'
is congruent to one modulo four.

IN SUM

The existence of an odd perfect number has yet to be proved or disproved. But if an odd perfect number does exist, it will have each the preceding properties. In addition, it will be no less than 10^300.

(This estimate was quoted from page 44 of the book "Unsolved Problems in Number Theory", written by Richard K. Guy, published in 1994 by Springer-Verlag)

## 9 Curiousities Relating to Perfect Numbers and Related Topics

In this section, I will peruse many of the interesting tid-bits about perfect numbers and related topics.

ALL KNOWN PERFECT NUMBERS ARE TRIANGULAR

This is the first challenge my sister gave me. In hingsight, the proof is fairly simple.

Definition: a triangular number is any number of the form (1/2)*(k)*(k+1) where k is a natural number.
All known perfect numbers are of the form (2^x - 1)*(2^(x - 1)) where 'x' is a natural number.
This expression can be rewritten as (1/2)*(2^x)*(2^x - 1)
Let 2^x = k + 1.
Thus, all known perfect numbers are of the form (1/2)*(k + 1)*(k)
Therefore, all known perfect numbers are triangular numbers.

THERE ARE ODD ABUNDANT NUMBERS!!!

This fact struck me as quite amazing. For a long time, I made the naοve assumption that if odd numbers could not be perfect, then certainly odd numbers could not be abundant. It seems almost too obvious that if the divisors of an odd number 'd' cannot even sum to 2d, then the divisors of an odd number certainly couldn't sum to greater than twice itself. Alas, my assumptions were wrong. A few weeks ago I stumbled upon not one, but infinitely many odd abundant numbers.
Here are two odd abundant numbers: 945. sigma(945) = 1920 > 2*945 = 1890.
2205. sigma(2205) = 4446 > 2*2205 = 4410
Lemma: There exist infinitely many multiples of an abundant number that are also abundant.
Define 'A' to be an abundant number. Thus, sigma(A) >2A, or, sigma(A) = 2A + L
where L is the difference between sigma(A) and 2A.
Also define 'k' to be any prime number greater than the largest prime in A.
The Lemma can be restated thus: sigma(A*k) > 2A*k for all defined A and k.
Since A and k are relatively prime, the multiplicative property of the sigma function can be used to simplify
sigma(A*k) = [sigma(A)]*[sigma(k)] = [2A + L]*[k + 1] = 2Ak + Lk + 2A + L > 2Ak
Due to the infinitude of primes, there are infinitely many multiples of an abundant number that are also abundant. QED

Thus, having found one odd abundant number, the preceding lemma asserts that there are infinitely many odd abundant numbers.

FACTOR SUM SEQUENCES

Define the function SIGMA(N) = sigma(N) - N. Thus, SIGMA(N) calculates the sum of the divisors of N excluding the number N itself. For example: SIGMA(6) = 6.

Given an initial number X, if the SIGMA function is repeatedly applied to the result, a sequence of numbers is formed. For example: SIGMA(15) = 9, SIGMA (9) =4, SIGMA (4) =3, SIGMA (3) = 1. Thus, given the number 15, the resulting sequence is 15, 9, 4, 3, 1. I have grown accostomed to calling such a sequence seeded by the number X a "factor sum sequence" of X, or, FSS(X)

Much like any natural number can either be classified as deficient, perfect, or abundant, I have decided to classify factor sum sequences as either degenerate, stable, or generating

In a degenerate FSS, each term is smaller than the previous term.
Example: FSS(15) = 15, 9, 4, 3, 1 (as above)
A stable FSS ends with a perfect number.
Example: FSS(95) = 95, 25, 6
In a generating FSS, at some point in the sequence, a term must be larger than the previous term. A generating FSS can eventually degenerate or stabilize.
Example: a generating FSS that eventually degenerates: FSS(24) = 24, 36, 55, 17, 1
Example: a generating FSS that eventually stabilizes: FSS(608) = 608, 652, 496.

I used a computer program to print out the FSS(X) for every X between 1 and 1000. A large majority of all FSS's were either purely degenerate or degenerated eventually after briefly generating a few terms. Some FSS's actually repeated themselves, as in: FSS(220) = 220, 284, 220, 284,  The terms in such repeating FSS's are better known as social numbers, a topic I will introduce in the next section.

There was one type of FSS I could not find, and could not prove existed: an infinitely long generating FSS. Are there any infinitely long factor sum sequences? To date, no one has found one. However, according to Richard K. Guy in his book "Unsolved Problems In Number Theory" on page 60, one Mitchell Dickerman found the FSS with the most number of terms: FSS(1248) has 1075 terms. To calculate this FSS, Dickerman had to find the sum of the divisors of a 58 digit number.

SOCIAL NUMBERS

The best way I can define a social number is to simply give an example: 220 and 284 are social numbers.
The proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110; their sum is 284.
The proper divisors of 284 are 1, 2, 4, 71, 142; their sum is 220.
More specifically, 220 and 284 constitute an 'amicable pair', one type of social number. In his book Perfect Numbers, Richard Shoemaker defines social numbers "as being a chain of numbers in which each number is equal to the sum of all the proper divisors of the preceding number, the last being considered as preceding the first number of the chain" (page 27). 220 and 284 are therefore a 'chain' of two numbers. Here is an example of a chain of five: 14,288 --> 15,472 --> 14, 536 --> 14,264 --> 12, 496 --> 14,288.

There are many, many, many known chains of social numbers. Thus, I will only record here one more, my favorite, a chain of 28: the first number in the chain is 14,316. I'll leave it to you as a healthy exercise to find the remaining 27 numbers in the chain.

## 10 A Link To A List Of The Perfect Numbers Found To Date

The following site has the most comprehensive and informative list of perfect numbers that I have found:
http://www.utm.edu/research/primes/mersenne.shtml