**Mathematics in Depth**

Though Stubblefield wrote and
published numerous works, we are going to focus on his one

paper entitled *Lower bounds for odd perfect numbers (beyond the googol)*.
The purpose of

this paper is to “provide a newer method by which, when a natural number
M is given, we can

either (a) find an odd perfect number less than M or (b) determine that an
odd perfect number

less than M does not exist (lower, 211).”

**What are Natual Numbers?**

Natural numbers are the numbers that one would
“naturally use for counting: 1,2,3,4…

(natural).” A positive integer n is said to be a perfect number if n
is equal to the sum of all of its

positive divisors, excluding n itself. For example, let n = 6. The divisors
of 6, not including 6,

are 1,2,3. Hence 1+2+3 = 6, and so 6 is a perfect number. Now let n = 12.
The divisors of 12,

not including 12, are 1,2,3,4,6. Hence 1+2+3+4+6 = 16. Since 16 does not equal
12, then 12

is not a perfect number.

**What are Prime Numbers?**

Prime numbers are those that have divisors
of only 1 and itself. This means that the sum of

all its positive divisors, excluding itself is always going to equal 1 for
all n that are prime. For

example, let n = 3. Notice 3 is prime, so its only positive divisor, not including
3, is 1. Since this

is going to be the case for all prime numbers, prime numbers can never be
perfect.

**What Stubblefield Shows in His Paper:**

Using the idea of perfect numbers, Stubblefield
shows that if a square has a side that has a

measure equal to an odd perfect number, then the area of that particular square
is “beyond the

googol.” He claims in his paper that it is sufficient to let M = 10^50
in order to show that the

odd perfect number, that is the length of one of the sides of the square,
is greater than 10^50.

To show that this claim is adequate, let X = the length of one side of a square.
Hence, the area

of the square is X^2, which we will assume is greater than googol, or 10^100
(lower, 211).

area = X^2 > 10^100

odd perfect number = X = sqrtX^2 > sqrt10^100

odd perfect number = X = X >10^50

odd perfect number > 10^50

Through the above calculations, we can see that
Stubblefield’s claim is indeed correct. If the

length of a side of a square is an odd perfect number than the length of the
side is greater than

10^50, which in turn means that the area of the square is “beyond the
googol (lower, 211).”

**Proof:**

Another idea that Stubblefield used to help
him in his paper, was the fact that “Every positive

integer n>1 can be expressed as a product of primes; this representation
is unique, apart from

the order in which the factors occur (Burton, 54).” To show that this
idea is indeed a fact, we

will go through a proof of it. n can be one of two things, a prime number
or a composite number.

If n is a prime number, then there is nothing further to prove. If n happens
to be a composite

number, then an integer d exists that satisfies d|n and 1<d<n. Out of
all the integers contained in

d, choose p1 to be the smallest. Hence p1 must be prime. If not, it would
also have a divisor q

with 1<q<p1, but then q|p1 and p1|n imply that q|n. This is a contradiction
to the choice of p1

as the smallest positive divisor, not equal to 1, of n (Burton, 54).

Therefore, we may write n=p1n1, where p is prime
and 1<n1<n. We will have our

representation if n1 is a prime number. In the opposing case, the argument
is repeated to come

up with a second prime number p2 such that n1=p2n2. In other words,

n=p1p2n1, 1<n2<n1

Again, if n is a prime number, then we have nothing else to
prove. Otherwise, we keep going by

writing n2=p3n3, with p3 a prime:

n=p1p2p3n3 1<n2<n1

The decreasing sequence n>n1>n2>…>1 cannot
continue forever, so that after a finite number

of steps nk-1 is a prime pk. This leads to the prime factorization n=p1p2…pk
(Burton. 54).

For the second part of our proof, the uniqueness
of the prime factorization, we will let the

integer n be represented as a product of primes in two ways:

n=p1p2…pr=q1q2…qs, r<=s

with p and q being primes, and written in increasing magnitude so that p1<=p2<=…

<=pr,q1<=q2<=…<=qs. Since p1|q1q2…qs, we know that p1=qk
for some k, but then

p1>=q1. Based on similar reasoning, we can say q1>=p1, hence p1=q1.
When we cancel out

the common factor, we get p2p3…pr=q2q3…qs. We then repeat this process
to get p2=q2

and this turns out to be p3p4…pr=q3q4…qs. This process is continued.
If the inequality r<s

holds, we will arrive at 1=qr+1qr+2…qs which is ridiculous, since each
q1>1. Therefore r=s

and p1=q1,p2=q2,…,pr=qr. This makes the two factorizations of n identical,
as desired

(Burton, 55).

**The History of a "Perfect Number":**

The name “perfect number” was given
by the Pythagoreans, who often attributed mystical

qualities to numbers. For centuries, philosophers were not interested in the
mathematical

properties of perfect numbers, but were more concerned with their mystical
or religious

significance. Saint Augustine used the idea of perfect numbers when talking
about the creation

of the world. He said that God chose to create the world in 6 days instead
of 1, because the

perfection of his work is symbolized by the perfect number 6. Another argument
of the

perfection of the universe was by the early critics of the Old Testament.
They claimed that the

number 28, also perfect, represents the perfection of the universe since 28
is the number of days

it takes the moon to circle the earth. Alcuin of York, an 8th century theologian,
observed that

the whole human race came from the 8 souls on Noah’s Ark and that this
second Creation is

not as perfect as the first, 8 being an imperfect number. The ancient Greeks
only knew of 4

perfect numbers. P1 = 6, P2 = 28, P3 = 496, P4 = 8128. The Greeks claimed
that these

numbers were formed in an “orderly” fashion, one among the ones,
one among the tens, one

among hundreds, and one among the thousands. Based on this small amount of
evidence, two

things were inferred:

1.the nth perfect number Pn contains exactly
n digits

2.the even perfect numbers end, alternately,
in 6 and 8.

Both of these claims were wrong. A perfect
number with five digits does not exist. The next

perfect number is P5 = 33,550,336. While P5 does end in a 6, the next perfect
number P6 =

8,589,869,056 also ends in a six, not an 8 as thought by the Greeks (Burton,
250-254).

Since almost the beginning of mathematical time,
there has been a problem of determining

the general form of all perfect numbers. Thanks to Euclid and Euler, we now
know the format

for even perfect numbers. If 2k-1 is prime (k>1), then n = 2k-1 (2k-1)
is perfect and every

even perfect number is of this form. Now that the form of even perfect numbers
is known, the

problem of finding even perfect numbers is reduced to the search for prime
numbers in the form

2k-1. One of the difficulties in finding additional perfect numbers is the
unavailability of tables of

primes. This raises the question of whether there are infinitely many prime
numbers that are of

the form 2P-1, with p being prime (Burton, 251-254).

**Conclusion:**

Looking at how large the sixth perfect number
already is, one can begin to see just how rare

perfect numbers really are. Though we do know they are rare, it is not yet
known whether there

are infinitely many, or finitely many perfect numbers. This of course would
depend on the

answer to the above question of whether there are infinitely many prime numbers
that are of the

form 2P-1, with p being prime (Burton, 252-254). Though perfect numbers are
very rare,

number theorists today find them useful for things such as string theory (string).

The question of the existence of odd perfect
numbers has yet to be answered. For years

people have been trying to answer this question. There has not been a single
odd perfect

number discovered yet, but there is also not any proof saying that they do
not exist. We are

expected to have an answer to the question of their existence sometime in
the near future.

Stubblefield is one of the many mathematicians
that have contributed in trying to determine if

odd perfect numbers exist. In order to try and prove or disprove the existence
of odd perfect

numbers, mathematicians have resulted to the use lower bounds. The idea behind
this approach

is to find a number n and prove that there does not exist any odd perfect
numbers that are less

than n. Hence, n being the lower bound. In Stubblefield’s paper, he chose
n=1050. He goes on

to prove that indeed there are no odd perfect numbers that are less that 1050.
As a result, he

had found a new lower bound for odd perfect numbers, 1050 (lower, 211-222).