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 Stubblefields 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).
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,
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:
The decreasing sequence n>n1>n2>
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
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 Noahs 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).
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 Stubblefields 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).