On the Prime Divisors of Odd Perfect Numbers

On the Prime Divisors of Odd Perfect Numbers Justin Sweeney Department of Mathematics Trinity College Hartford, CT justin.sweeney@trincoll.edu April 27, 2009 1

Contents 1 History of Perfect Numbers 5 2 Introduction 8 3 Necessary Information 9 3.1 Cyclotomic Polynomials and Roots of Unity............ 9 3.2 Expressing σ(n).......................... 10 3.2.1 Theorem 2.1........................ 10 3.2.2 Proof of Theorem 2.1.................... 10 3.3 The Divisors of the Cyclotomic Polynomial............ 10 3.3.1 Theorem 2.2........................ 10 3.3.2 Proof of Theorem 2.2.................... 11 3.3.3 Theorem 2.3........................ 14 3.3.4 Proof of Theorem 2.3.................... 14 3.3.5 Theorem 2.4........................ 14 3.3.6 Proof of Theorem 2.4.................... 14 3.4 Kanold s Theorem......................... 15 3.4.1 Theorem 2.5........................ 15 3.4.2 Proof of Theorem 2.5.................... 15 3.4.3 Theorem 2.6........................ 16 3.4.4 Proof of Theorem 2.6.................... 16 3.5 Factorization of N.......................... 16 3.5.1 Theorem 2.7........................ 16 3.5.2 Proof of Theorem 2.7.................... 16 4 Statement of Theorem from Jenkins 18 4.1 Jenkins Theorem.......................... 18 4.2 Proof of Theorem 3.1........................ 18 5 Acceptable Values of F r (p) 18 6 Inadmissable Small Primes 19 6.1 Proof that 1093 N......................... 19 6.2 Proof that 151 N......................... 22 7 Restrictions on Exponents in the Prime Power Decomposition of N 22 7.1 Lemma 6.1............................. 22 7.2 Proof of Lemma 6.1......................... 23 2

8 Four Sets 24 8.1 Proposition 7.1........................... 24 8.2 Proof of Proposition 7.1....................... 24 8.3 Proposition 7.2........................... 25 8.4 Proof of Proposition 7.2....................... 25 8.5 Proposition 7.3........................... 25 8.6 Proof of Proposition 7.3....................... 25 8.7 Proposition 7.4........................... 25 8.8 Proof of Proposition 7.4....................... 26 8.9 Computer Search for Number of Elements of S,T,U and V.... 26 9 Proof of Jenkins Theorem 26 10 Extension of Jenkins Theorem to 10 8 27 10.1 Proposition 2.1........................... 27 10.2 Acceptable Values of F r (p)..................... 28 10.3 Proof of Acceptable Values of F r (p)................ 28 10.3.1 Lemma 3.1......................... 28 10.3.2 Lemma 3.2......................... 28 10.4 Inadmissable Small Primes..................... 28 10.5 Restrictions on Exponents in the Prime Power Decomposition of N 29 10.6 Four Sets.............................. 29 10.6.1 Proof of (1) of Proposition 2.5............... 30 10.7 Proof that an Odd Perfect Number Has a Prime Factor Greater than 10 8.................................. 30 11 Computer Algorithms 31 11.1 Jenkins Algorithm.......................... 31 11.2 Proposition 4.1........................... 31 11.3 Goto and Ohno Algorithm..................... 32 12 Conclusion 32 A Table A: Acceptable Values of F r (p) for 3 p 10 7 and r 7. 33 B Table B: Acceptable Values of F r (p) for 3 p 10 7 and r 7. 37 3

Abstract This paper studies various results on the nature of the prime divisors of an odd perfect number. The paper will detail the history of the study of odd perfect numbers. This paper will enumerate the various results that have been found related to odd perfect number to this date. It will also provide a detailed commentary on results related to the prime divisors of an odd perfect number. In particular, this paper will closely examine the result found by Paul M. Jenkins in his thesis.[5] The thesis by Jenkins was submitted in September, 2000 and was published in January, 2003. Jenkins proved a theorem that an odd perfect number must have a prime divisor greater than 10 7. This paper will detail how this result was achieved and explain the background information necessary to comprehend the proof of this theorem. This paper will also explore the more recent result by Takeshi Goto and Yasuo Ohno, which proved that an odd perfect number must have a prime divisor greater than 10 8.[3] This paper was published in February, 2008. This paper will explain how Goto and Ohno were able to extend this bound and how their proof differed from the proof by Jenkins. 4

1 History of Perfect Numbers The study of perfect numbers has been in progress for as long as many other important mathematical fields. Although it is unknown when the study of perfect numbers first began, there is clear evidence of perfect numbers being studied as early as Pythagoras. Around 300 B.C., Euclid achieved the first documented result related to perfect numbers in his book Elements. In Proposition 36 of Book IX of the Elements, Euclid stated: If as many numbers as we please beginning from a unit be set out continuously in double proportion, until the sum of all becomes a prime, and if the sum multiplied into the last make some number, the product will be perfect.[6] This proposition essentially creates a sequence of numbers beginning with 1, where each number in the sequence is twice the preceeding number. The proposition states that when the sum of these numbers is prime, then the product of the sum and the last number in the sequence will be a perfect number. In modern terms, we can restate Proposition 36 using the fact that 1 + 2 + 4 +... + 2 k-1 = 2 k -1, so that the result reads: If, for some k >1, 2 k -1 is prime then 2 k-1 (2 k -1) is a perfect number.[8] As a separate note, prime numbers of the form 2 k -1 are commonly known as Mersenne primes. This is a major area of mathematical research on its own and there is currently a large-scale distributed search to find such numbers.[15] Around 100 A.D. Introductio Arithmetica was written by Nicomachus. In this text, he classified all numbers into three categories. These categories are superabundant numbers, deficient numbers, and perfect numbers. Superabundant numbers, now referred to simply as abundant numbers, have the property that their divisors sum to be greater than the twice the number, or notationally: σ(n)>2n.[10] Deficient numbers have the property that their divisors sum to be less than twice the number, or notationally: σ(n)<2n.[12] Perfect numbers have the property that their divisors sum to equal twice the number, or notationally, σ(n)=2n.[16] In this text, Nicomachus also stated five results concerning perfect numbers, but did not prove any of these results. The five results stated are: 1. The nth perfect number has n digits. 5

2. All perfect numbers are even. 3. All perfect numbers end in 6 and 8 alternately. 4. All perfect numbers are of the form: 2 k-1 (2 k -1) for some k>1, where 2 k -1 is prime. That is, Euclids formula will produce all perfect numbers. 5. There are infinitely many perfect numbers. At this point, we know that statements 1 and 3 are false, but statements 2, 4, and 5 are still open to be proved or disproved. Obviously, statement 2 is of the most concern of this thesis. Nicomachus assertions were taken as fact for a significant period of time and it wasnt really until the 1500s that his assertions began to be questioned. In 1536, Hudalrichus Regius published the work Utriusque Arithmetices in which he provided the factorization 2 1 1 1 = 2047 = 23 89. This showed that there exists at least one prime p such that 2 p-1 (2 p -1) is not a perfect number. Also, in this work, Regius found that 2 13-1 is a prime number and thus, 2 12 (2 13-1) = 33550336 is a perfect number. This was a significant result because this was the fifth perfect number, but the number consisted of 8 digits. This showed that Nicomachus assertion that the nth perfect number has n digits was incorrect. In 1604, Cataldi created a table of the first 132 primes. He used this list of primes to find the sixth perfect number, 2 16 (2 17-1) = 8589869056. This result was important because it showed that Nicomachus assertion that perfect numbers ended in 6 and 8 alternately was incorrect because the fifth and sixth perfect numbers both ended in 6. Cataldi also presented the seventh perfect number, 2 18 (2 19-1). Over the next few centuries, various correspondences related to perfect numbers took place between mathematicians such as Fermat, Mersenne, Descartes and others. Fermat was the next mathematician to make significant progress. In a letter to Mersenne, Fermat created the following structure: 1 2 3 4 5 6 7 8 9 10 11 12 13... 1 3 7 15 31 63 127 255 511 1023 2047 4095 8191... As can be seen, the top row of numbers is simply the sequence of all natural numbers. The bottom row of numbers is a progression beginning at 1, where each number is one more than twice the previous number. That is, a i+1 = 2a i + 1. 6

Now, Fermat called the bottom row of numbers the radicals of perfect numbers. This is because whenever a number in the bottom row is prime, that number produces a perfect number. Now the numbers in the top row were called the exponents of the radicals of perfect numbers. Using this structure, Fermat stated the following three results: 1. When the exponent of a radical number is composite, its radical is also composite. 2. When the exponent of a radical number is a prime number, its radical minus one is divisible by twice the exponent. As an example, 7 is the exponent of 127, and 7 is prime. Thus, 127-1 = 126 is a multiple of 2 7 = 14. 3. When the exponent of a radical number is a prime number, its radical cannot be divisible by any other prime except those that are twice the exponent plus one. Mersenne, possibly using some of Fermat s results, went on to claim that 2 p -1 is prime for p=2, 3, 5, 7, 13, 17, 19, 31, 67, 127, 257, and for not other value of p less than 257. This result was, impressively, almost entirely correct, with only 5 mistakes being found by later mathematicians. Euler was the next mathematician to make significant progress on the topic of perfect numbers. Euler was able to prove that every even perfect number is of the form 2 p-1 (2 p -1), thus proving the converse of Euclid s assertion. Euler also proved Descartes assertion that there are no odd perfect numbers, unless they are composed of a single prime number multiplied by a square whose root is composed of several other prime numbers.[8] In fact, Euler proved that any odd perfect number must have the form: (4n + 1) 4k+1 b 2 where 4n+1 is prime. Over the course of the next few centuries, a few results were made regarding perfect numbers, particularly the finding of more even perfect numbers. In that time, no odd perfect numbers were found, but significant study took place regarding odd perfect numbers. To this date, no odd perfect numbers have been found, but a significant search has taken place. In 1957, Kanold proved that no odd perfect numbers exist up 7

to 10 20. In 1973, Tuckerman extended this bound to 10 36. Later in 1973, Hagis extended this bound to 10 50. Then, in 1989 and 1991, Brent et al. extended this bound to 10 160 and 10 300 respectively. Currently, work is being done to prove that no odd perfect numbers exist up to 10 500.[4] Also, significant studies have been done regarding the number of prime factors of an odd perfect number. Chein and Hagis showed that an odd perfect number must have at least 8 distinct prime factors. Nielsen extended this result to show that an odd perfect number must have at least 9 distinct prime factors. A significant amount of work has also been done to study the nature of these prime factors. In 1944, Kanold proved that an odd perfect number must be divisible by a prime number exceeding 60. Hagis and McDaniel improved this bound to 10 4 and then 10 5. Hagis and Cohen improved this bound to 10 6. Jenkins then showed that an odd perfect number must be divisible by a prime exceeding 10 7. Finally, Goto and Ohno extended this bound to 10 8. The most recent two results, by Jenkins and Goto and Ohno, is the work that is of the most concern to this thesis and will be discussed extensively.[3] 2 Introduction A perfect number is any number N such that σ(n) = 2N. That is, the sum of the divisors of N add up to be twice the number N. To this date, all known perfect numbers are even. All of the even perfect numbers that have been found have been of the form N=2 p-1 (2 p -1), where p is prime and 2 p -1 is a Mersenne prime. There have been numerous papers written on the existence of odd perfect numbers and a number of conditions have been laid out which an odd perfect number would have to meet if such a number exists. Many of these will be discussed in the following chapter detailing the history of perfect numbers. This paper deals specifically with the condition that the largest prime divisor of an odd perfect number must be greater than some value x. The fundamental theorem of number theory provides us with a unique prime factorization of every number. Every natural number n (>1) can be expressed as the product of primes (prime factors) in the form N = p 1 p 2... p r, (r 1) 8

There is only one such expression as a product, if the order of the factors is not taken into consideration.[7] Also, we have that if we let p 1, p 2,..., p v denote all distinct prime factors of N, we may then express N in the form N = u i=1 p a i i (1) [7] We shall use this expression for N throughout this paper. Numerous papers have studied this factorization and have proven claims regarding the magnitude of the prime factors. The two papers that will be looked at specifically in this paper are Odd Perfect Numbers Have a Prime Factor Exceeding 10 7, by Paul M. Jenkins, and Odd Perfect Numbers Have a Prime Factor Exceeding 10 8, by Takeshi Goto and Yasuo Ohno. For the sake of this paper, we will let N represent an odd perfect number, and we will let p, q, and r be primes. Also, we will let the dth cyclotomic polynomial be denoted by F d, so that: F p (x) = 1 + x + x 2 +... + x p 1 (2) Also, if two numbers p and m are relatively prime, we will let (p,m) represent the order of p modulo m. 3 Necessary Information 3.1 Cyclotomic Polynomials and Roots of Unity This section will provide background information regarding cyclotomic polynomials and roots of unity. This will provide the background information for a better understanding of the proofs contained in the following sections. First, we must discuss Euler s ϕ function. If we have two integers a and b, then we say that a and b are relatively prime and (a, b) = 1 if a and b have no common divisor greater than 1. Now, Euler s ϕ function, also know as the totient function, is defined on a natural number n as: ϕ(n) is the number of natural numbers n which are relatively prime to n.[17] We also have, from [7] a mathematical equation to express ϕ(n). 9

If N is a natural number with the different prime factors p 1, p 2,..., p r, then ϕ(n) = N(1 1 p 1 )(1 1 p 2 )... (1 1 p r ) Now, we will discuss roots of unity. The nth roots of unity are roots of the form e 2πik/n of the cyclotomic equation x n = 1. The roots of unity are also known as de Moivre numbers.[9] We will let ε k represent the kth nth root of unity. Now, we also note that +1 is always an nth root of unity and -1 is an nth root of unity only if n is even.[9] We have from [7] that the nth roots of unity have the form: ε m = cos 2πm n + isin2πm, (m = 0, 1, 2,..., n 1). n Now, if ε m denotes one of the nth roots of unity and we have that ε 0 m, ε 1 m, ε 2 m,..., ε n 1 m are distinct, then we have that ε m is a primitive root of unity and the numbers ε 0 m, ε 1 m, ε 2 m,..., ε n 1 m represent all of the nth roots of unity. We also have from [7] that: A necesary and sufficient condition for ε m to be a primitive nth root of unity is that the integer m be prime to n. Finally we must describe the cyclotomic polynomial. The cyclotomic polynomial of index n is the polynomial F n (x) = a (x ε a), with the product taken over all primitive nth roots of unity. Due to the fact that the product is taken over the primitive nth roots of unity, we have that the polynomial has degree ϕ(n). Now we have the necessary background information to have sufficient understanding of the proofs in the following sections. 3.2 Expressing σ(n) 3.2.1 Theorem 2.1 σ(p a ) = F d (p) d (a+1),d>1 10

3.2.2 Proof of Theorem 2.1 First of all, we know from (1) that N = p a 1 1 pa 2 2... pau u. Also, we know that σ is a multiplicative function.[13] Thus, since all of the terms are unique primes and are thus relatively prime, we are able to write: σ(n) = σ(p a 1 1 )σ(pa 2 2 )... σ(pau u ) (3) Therefore, now we must obtain an expression for σ(p a ). The divisors of p a are: 1, p, p 2,..., p a. Thus we are able to write the equation [13]: σ(p a ) = 1 + p + p 2 +... + p a = (pa+1 1) (p 1) (4) Now, from [2] we have the factorization x n 1 = d n F d (x). (5) So, if we let x = p, and n = a+1, we can use this equation to adjust the equation for σ given in (4). σ(p a ) = d (a+1) F d(p) p 1 Now, we have that F 1 (x) = x 1, so F 1 (p) = p 1.[7] Thus we can cancel the F 1 (p) from the numerator with (p 1) in the denominator, to get the final expression, σ(p a ) = d (a+1),d>1 F d(p). (6) 3.3 The Divisors of the Cyclotomic Polynomial Now that we have reduced the expression for σ(p a ) to consist solely of the cyclotomic polynomial, we must consider the divisors of the cyclotomic polynomial. We will consider the cyclotomic polynomial F n (p). Also, we will suppose there exists some prime q which does not divide n. 11

3.3.1 Theorem 2.2 It is true that q F m (p) if and only if m = q b h(p; q). If b>0, then q F m (p). If b=0, then q 1 (mod m). 3.3.2 Proof of Theorem 2.2 First, consider the case where b = 0, then q 1 (mod n), so n (q 1). By Fermats litle theorem, a q 1 1 (mod q) for all a 0 (mod q).[14] Thus, (x q 1 1) = (x a), a 0 (mod q) Now, from (5), we have that (x q 1 1) = d (q 1) F d(x). Then, since n (q 1), F n (x) (x q 1 1). So, F n (x) (x a), a 0 (mod q) Now, if x is a primitive nth root of 1 (mod q), then F n (x) 0 (mod q) and if F n (x) 0 (mod q) has solutions, then the solutions must be primitive nth roots of 1 (mod q). Since a q 1 1 (mod q) for all a 0 (mod q), then by Fermat s theorem, n (q 1) and q 1 (mod n). Thus, q F n (p) if and only if n = h(p,q). Now, we will need to prove that if q F n (x), then F n (x) is divisible by exactly the same power of q as x n 1. Theorem 94 of [7] states that: If q is a prime which does not divide n, we have: 1. The necessary and sufficient condition for the congruence F n (x) 0 (mod q) (7) to be solvable is that q 1 (mod n). 2. If q 1 (mod n), the solutions of (7) are the numbers which belong to the exponent n modulo q. Thus the number of incongruent solutions modulo q is ϕ(n). 12

3. If x is a solution of congruence (7), the number F n (x) is divisible by exactly the same power of q as x n 1. The proof of this theorem is as follows. First we know that F n (0) = 1 and thus it must be that for any solution x of congruence (7), x 0 (mod q). We must introduce the notation v = (x n p 1 p 2...pv F n (x) = 0 1 1). Then, we have the identity, 2... 3... (8) Now we can use this identity to say that if q F n (x), then at least one of the factors in the numerator on the right-hand side of (8) is divisible by q. From this, we know that x n 1 is divisible by q. We will assume that the solution x belongs to the exponent µ modulo q, the number µ must be a divisor of n. We will further suppose that n µ > 1, and we will let p 1, p 2,..., p m be the distinct prime factors of n µ. If q divides the number x n d 1, where d is a product of different prime factors of n, then the number n d must be a multiple of µ and thus, must also be a multiple of d. Therefore, every prime divisor of d must belong to the set of primes p 1, p 2,..., p m. Now we will suppose that x µ 1 is divisible by q s for some integer s and by no higher power of q. Then, x µ = 1 + tq s, where t is not divisible by q. We will now raise each side of this equation to the kth power. From this, we get, x kµ = 1 + ktq s + q 2s t 1 = 1 + t 2 q s, where t 1 and t 2 are both integers. If k is not divisible by q, then t 2 cannot be divisible by q. Therefore, we have that x kµ 1 is divisible by the same power of q as x µ 1. From above, we have that n d = kµ. In this equation, k must be an integer not divisible by q because we know n is not divisible by q. From all of the above, we may conclude that the product 0 2 4... is divisible by the power of q where the exponent of q is s(1+( ) m 2 + ( m ) 4 +...). Also, the product 1 3... is divisible by the power of q whose exponent is s( ( ) ( m ) Q Q 1 + m +...). Then the fraction Q0 2... 3... must be divisible by 1 Q3 q s(1+ ( m 2 )+( m 4 )+...). This equation is equivalent to q q s( ( m 1 )+( s(1+(m m 2 )+( m 4 )+...) s(( m 1 )+( m 3 )+...) 3 )+...) which is equivalent to q s(1 (m 1 )+( m 2 ) ( m 3 )+...) = (1 1) m = 0. Thus, if n µ > 1, then F n (x) is not divisible by q. If µ = n, then F n (x) is divisible by the same power of q as x n 1. Therefore we have that theorem 94 is proved. Having proved 13

this theorem, we will suppose q n, so n = q a n 1, where n 1 is not divisible by q. If q F n (x) then q (x n 1) and therefore, x n 1 (mod q). So, by Fermat s Little Theorem, x q x (mod q), so x n (x qa ) n 1 x n 1 1 (mod q). Thus, q x n 1 1, so by theorem 94 of [7], q F n1 (x) and n 1 = h(p,q). This proves that if q F n (x), then F n (x) is divisible by exactly the same power of q as x n 1. Now, from identity (8) above, we are able to see that F np (x) = F n(x p ) F n (x), (9) as long as p is a prime which does not divide n. Now from this we can see that F n1 q a = Fn 1 (xqa ) F n1 (x qa 1 ), since q is a prime which does not divide n 1. Now, we already know that n 1 q a = n, so we have that F n (x) = Fn 1 (xqa ). Now, by theorem 94 F n1 (x qa 1 ) from [7], F n1 (x qa ) is divisible by the same power of q as x qa n 1 1. Also, we have that F n1 (x qa ) = (x qa 1 n 1 1)(x (q 1)qa 1 n 1 + x (q 2)qa 1 n 1 +... + x qa 1 n 1 + 1). Now, if we take this equation modulo q, we have that, F n (x) x (q 1)qa 1 n 1 + x (q 2)qa 1 n 1 +... + x qa 1 n 1 + 1 (mod q). Since we have that n 1 = h(p, q), each of the q terms in the expression is equivalent to 1 (mod q). Thus, F n (x) (1 + 1 +... + 1) = q (mod q). Therefore, we have that q F n (x). Thus, we have that q F n (x) if and only if n = q b h(p, q). Now, we must consider theorem 95 from [7], which states: Suppose that q is a prime factor of n, and let n = q a n 1, where n 1 is not divisible by q. Then we have: 1. The necessary and sufficient condition for the congruence F n (x) 0 (mod q) to be solvable is that q 1 (mod n 1 ). 2. If q 1 (mod n 1 ), the solutions of the congruence above are the 14

numbers which belong to the exponent n 1 modulo q. Thus, the number of incongruent solutions modulo q is ϕ(n 1 ). 3. If x is a solution of the congruence above, the number F n (x) is divisible by q and not by q 2, provided that n > 2. We will now prove the above theorem. Suppose that n is not a power of 2. Then for q=2 we must have n 1 > 2. We have that if p divides n, then F np (x) = F n (x p ). (10) So, from (9) and (10) above, we get F n (x) = F n 1 (x qa ) F n1 (x qa 1 ). (11) If F n (x) is divisible by q, then F n1 (x qa ) is also divisible by q. So, by theorem 94 from above, x qa must belong to the exponent n 1 modulo q. Thus, x must belong to the exponent n 1 modulo q. For q=2, this implies that n 1 = 1. Then, since we assumed n is not a power of 2, q must be odd. Now, we know that F n (1) = p, when n is a power of the prime p. First we will suppose x = ±1. If x=1, then n 1 = 1 and F n (1) = q. If x=-1, then n 1 = 2 and F n ( 1) = F 1 n(1) = q. Now, 2 suppose x ±1. If x belongs to the exponent n 1, then F n1 (x qa ) is divisible by q. Then, according to part 3 of theorem 94, F n1 (x qa ) is divisible by the same power of q as (x qa ) n 1 1 = x n 1 0. In this case, F n1 (x qa 1 is also divisible by the same power of q as (x qa 1 ) n 1 1 = x n q 1 0. Suppose that x n q 1 is divisible by q s and not by q s+1. Then, x n q = 1 + tq s, where t is not divisible by q. Now, if we raise each side of this equation to the qth power, we get, x n = 1 + qtq s + ( ) q t 2 q 2s +... = 1 + t 1 q s+1, 2 where t 1 is not divisible by q since q > 2. Then, we have that if F n1 (x qa 1 ) is divisible by q s, then F n1 (x qa ) is divisible by q s+1. From (11) it follows that F n (x) is divisible by q and not by q 2. Finally, when n is a power of 2, we have F n (x) = x 2m + 1, where m = n 4. Thus, F n(x) is never divisible by 4. Thus we 15

have theorem 95 for all values of n. Combining the above results, we have Theorem 2.2. 3.3.3 Theorem 2.3 Let r be a prime number. If q F r (p) then either r=q and p 1 (mod q), so that q F r (p), or q 1 (mod r). 3.3.4 Proof of Theorem 2.3 First we let r be a prime number. Also, let q F r (p). Then from theorem 2.2, we have that r = q b h(p, q). If b=0, then from theorem 2.2, we have that q 1 (mod r). Otherwise, b > 0, but q cannot divide r since r is prime, so r = q and p 1 (mod q). In this case, q F r (p). Thus, we have theorem 2.3. 3.3.5 Theorem 2.4 If q=3 or 5 and m > 1 is odd, then q F m (p) (and q F m (p)) if and only if m = q b and p 1 (mod q). 3.3.6 Proof of Theorem 2.4 First we will make a note that if p 2 (mod 3), then h(p,3) = 2. If p 4 (mod 5), then h(p,5) = 2. If p 2 or 3 (mod 5), then h(p,5)=4. Let q=3 or 5 and let m > 1 be odd. Also, let m = q b and p 1 (mod q). Then, h(p,q)=1, so m = q b h(p, q), so by theorem 2.2, q F m (p). Now, let q F m (p), then by theorem 2.2, m = q b h(p, q), and since m is odd, q b and h(p,q) must be odd. If q=3 or 5, then h(p,q) is only odd if p 1 (mod q). Therefore, p 1 (mod q), so h(p,q)=1 and m = q b. Thus, we have theorem 2.4. 3.4 Kanold s Theorem From [7] we know that F m (x) = (x ɛ 1 )(x ɛ 2 )... (x ɛ ϕ(m) ), where ɛ i is a primitive nth root of unity, ɛ i = 1. 16

3.4.1 Theorem 2.5 For an integer x, if r F m (x) and m = p a 1 1 pa 2 2... pa k k 3, with p 1 < p 2 <... < p k, then: If r = p k then p k 1 (mod p a 1 1 pa 2 2... pa k 1 k 1 ). If p k 1 (mod p a 1 1 pa 2 2... pa k 1 k 1 ) then r 1 (mod m) and p2 k F m(x). 3.4.2 Proof of Theorem 2.5 From Theorem 2.2, if r F m (x) then either r m or r 1 (mod m). Now, we have that x m 1 x 1 = xm 1 + x m 2 +... + x + 1 = F m (x) Let m l = m/p l for l = 1, 2,..., k. Then d m,d 1,m F d (x). (1+x+x 2 +...+x m l 1 )(1+x m l +x 2m l +...+x (p l 1)m l ) = x m 1 +x m 2 +...+x+1. Let a l = (1 + x + x 2 +... + x m l 1 ) and a l = (1 + x + x 2 +... + x m l 1 ) and b l = (1 + x m l + x 2m l +... + x (p l 1)m l ). Since a l = F ml (x) d m l,d 1,m F d(x), it must be true that a l d m,d 1,m F d(x). But, a l b l = F m (x) d m,d 1,m F d(x), so F m (x) b l. Now, suppose r l b l, where r l is a prime. Then x m = x m lp l 1 (mod r l ), since r l b l and a l b l (x m 1). Suppose that x m l 1 (mod r l ). Then, b l (mod r l ) = 1 + x m l + x 2m l +... + x (p l 1)m l 1 + 1 +... + 1 = p l (mod r l ). But since r l b l, b l 0 (mod r l ), so r l = p l since both are prime. In this case, x m l = 1 + sp l for some s. Thus, b l = 1 + x m l + x 2m l +... + x (p l 1)m l = 1 + (1 + sp l ) + (1 + sp l ) 2 +... + (1 + sp l ) p l 1 = p l + sp l + 2sp l +... + (p l 1)sp l + Ap 2 l = p l + sp l ( p l(p l 1) 2 ) + Ap 2 l p l (mod p 2 l ). Thus, p 2 l b l. The other case is that x m l 1 (mod r l ). We know that x m = x m lp l 1 (mod r l ) and that x rl 1 1 (mod r l ) by Fermat s Little Theorem. Let t = gcd(r l 1, m l p l ). Then, x t 1 (mod r l ). 17

Let n l = m/p a l l. Then (x n lp a l 1 l ) p l 1 (mod r l ), but x n lp k l 1 (mod r l ) for k < a l. Thus, the order of x n l (mod r l ) is p a l l. By Lagrange s Theorem, the order of an element divides the order of a group.[1] For this case, we will be considering the multiplicative group with r l 1 elements, so p a l l (r l 1). Thus, r l 1 (mod p a l l ) for l = 1... k. Thus, if r is a prime divisor of F m (x), then for all l either r = p l or r 1 (mod p a l l ). But r = p l for at most one l, so r = p d and r 1 (mod p a l l ) for l d. If d k then we have p a k k (r 1) and (r 1) < pa k k, a contradiction that gives us d=k. It is thus shown that if r = p k then p k 1 (mod p a 1 1 pa 2 2... pa k 1 k 1 ). If p k 1 (mod p a 1 1 pa 2 2... pa k 1 k 1 ), then by contrapositive it is true that p k r, and that r 1 (mod p a l l ) for all l, so r 1 (mod m). Since F m (x) b l andp 2 l b l, it must be true that p 2 l F m (x). Thus we have Theorem 2.5. 3.4.3 Theorem 2.6 If p is an odd prime and m 3, then F m (p) has at least one prime factor q such that q 1 (mod m). 3.4.4 Proof of Theorem 2.6 If F m (x) has no prime divisors congruent to 1 (mod m), p k is the only divisor of F m (x). If this is the case, then p 2 k F m(x), then p k = F m (x). Suppose that m 3 and x 3. Since x 3, x ɛ i > 2, and F m (p) > 2 ϕ(m). Now, since ϕ is multiplicative, ϕ(m) = ϕ(p 1 )ϕ(p 2 )... ϕ(p k ) ϕ(p k ) = (p k 1), and 2 ϕ(m) 2 pk 1 p k. From this we have that F m (p) > p k. Therefore, we have that p k cannot be the only divisor of F m (x). Thus, we have Theorem 2.6. 3.5 Factorization of N 3.5.1 Theorem 2.7 2N = u i=0 σ(p a i i ) = u i=0 d (a i +1),d>1 F d (p i ) 18

3.5.2 Proof of Theorem 2.7 First, assume p is odd. We know from [11] that if p is an odd prime, then F p (x) = x p 1 x 1 = xp 1 + x p 2 +... + x + 1. Thus, we have that F d (p) = 1 + p + p 2 +... + p d 1. From this we know that F d (p) has d terms. Also, since p is odd and the product of two odd numbers is odd, we have that each of the d terms is odd. Thus, if we assume d is an even number, then we have the sum of an even number of odd numbers. Since the sum of an even number of odd numbers is even, then F d (p) is even. If we assume d is odd, then we have the sum of an odd number of odd numbers. Since the sum of an odd number of odd numbers is odd, then F d (p) is odd. Since we are considering odd perfect numbers, we know that N is odd and σ(n) = 2N. Also, we know from (1) that N = p a 0 0 pa 1 1... pau u. Also, from (3) we know that σ(n) = 2N = σ(p a 0 0 )σ(pa 1 1 )... σ(pau u ). Now, we also know from theorem 2.1 that σ(p a ) = d (a+1),d>1 F d(p). Thus, we have that 2N = u i=0 σ(pa i i ) = u i=0 d (a i +1),d>1 F d(p i ), as required. Thus, we have Theorem 2.7. Since σ(n) = 2N, we know that 2 σ(n). Since we have that σ(n) = u i=0 σ(pa i i ), then in order for this to be true, we must have that 2 σ(pa i i ) for at least one i. We will use a simple proof to show that we have the previous fact for exactly one i. Consider a number s such that s = a 1 a 2 a k. Also, let 2 s, so 2 s, but 4 s. Now, let 2 a 1, 2 a 2,..., 2 a j for some j < k. Then, we know that a 1 = 2 x and a 2 = 2 y for some x and y. Thus, we have that s = 2x 2y a 3 a k = 4 xy a 3 a k and thus 4 s, which is a contradiction. Thus, 2 a j for exactly one a j. This proof translates to our problem to show that 2 σ(p a i i ) for exactly one i. Thus, we have that there is exactly one even F d (p). We stated earlier that F d (p) is even if d is even. Thus, for one F d (p), 2 d. Now, for F d (p) to be included in the factorization of N, d (a i +1). Now since 2 d for one d, it must be true that 2 (a i +1) and thus 2 a i for exactly one a i. We can assume without loss of generality that this a i is a 0, which is the exponent for the prime p 0, which we will call the special prime. We will now show that a 0 1 (mod 4). We will proceed by contradiction. First, assume a 0 2 (mod 4). Then, we have that a 0 must be even, which is a 19

contradiction with the above statement that 2 a 0. The same is true for a 0 0 (mod 4). Thus, we have that a 0 2 (mod 4) and a 0 0 (mod 4). Now, assume a 0 3 (mod 4). Then, we have that a 0 + 1 4 (mod 4), so 4 (a 0 + 1). Then we would have that F 4 (p 0 ) N. Since p is odd, we have that F 4 (p 0 ) = p 3 0 +p2 0 +p 0 +1. Now, since p 0 is odd, we have that either p 0 1 (mod 4) or p 0 3 (mod 4). If p 0 1 (mod 4), then p 2 0 1 (mod 4) and p3 0 1 (mod 4). Then, F 4(p 0 ) = p 3 0 +p2 0 +p 0 +1 1+1+1+1 (mod 4) 4 (mod 4) 0 (mod 4). If p 0 3 (mod 4), then p 2 0 1 (mod 4) and p3 0 3 (mod 4). Then, F 4(p 0 ) = p 3 0 + p2 0 + p 0 + 1 3+1+3+1 (mod 4) 8 (mod 4) 0 (mod 4). Thus, F 4 (p 0 ) 0 (mod 4). Then, we would have that 4 F 4 (p 0 ) and since F 4 (p 0 ) N, 4 N. Thus, we would have that N is even, which is a contradiction. Therefore, we have that a 0 1 (mod 4). Now, we will show that p 0 3 (mod 4). It is obvious as we have shown above that if p 0 2 (mod 4) or p 0 0 (mod 4), then p 0 is even and thus we would have N being even. Thus, assume p 0 3 (mod 4). Then we can see that p b 0 1 (mod 4) whenever b is even and p b 0 3 (mod 4) whenever b is odd. So, we can see that if we took consecutive pairs of terms of the cyclotomic polynomial, F a0 +1(p 0 ), one term would be equivelent to 1 (mod 4) and one would be equivelent to 3 (mod 4) and thus these terms would sum to 0 (mod 4). Now, since there must be an even number of terms in F a0 +1(p 0 ), then we have that all of the terms sum to 0 (mod 4) and thus 4 F a0 +1(p 0 ) and therefore, 4 N, which is a contradiction. Therefore we must have that p 0 1 (mod 4). 4 Statement of Theorem from Jenkins 4.1 Jenkins Theorem If N is odd and perfect, then N has a prime factor greater than 10 7. 4.2 Proof of Theorem 3.1 The proof that will follow is a proof by contradiction. We will assume that p i < 10 7 for every p i in equation (1). We know, from Theorem 2.7, that the set of p i in (1) is identical to the set of odd prime factors of the cyclotomic polynomial F d (p i ). So, it follows that all prime factors of each F d (p i ) must be less than 10 7. In particular, 20

we have that if we let r be a prime divisor of a i +1, then every prime factor of F r (p) must be less than 10 7. Thus, we will focus upon the cyclotomic polynomial F r (p) and the set of prime factors of said polynomial in the following proof. We will lay out all of the necessary lemmas and propositions in order to prove this theorema and will culminate in the proof of the theorem. 5 Acceptable Values of F r (p) For this proof it is obvious that we must define F r (p) to be acceptable if every prime divisor of F r (p) is less than 10 7. We will first define a limit on the value of r. Since we know that p is odd, then from equation 2.6, we know that F r (p) has at least one prime factor q 1 (mod r). Now this means q must be equal to 1, r+1, 2r+1, 3r+1,..., since those are the only values that will be congruent to 1 (mod r). Now, 1 is not a prime number, so q cannot be equal to 1. Since r is prime and thus odd, r+1 is an even number, so q cannot be equal to r+1. Thus we have that q 2r + 1. Now, q must be less than 10 7, so 2r + 1 < 10 7. Thus, we have that 2r < 10 7 1 = 9999999, so r < (10 7 1)/2 = 4999999.5. Since we are working only with integers, we have that r < 5000000. Now, we will only define p to be an inadmissable prime if F r (p) is unacceptable, as defined above, for every prime r, such that r < 5000000. Also, we only need to consider r = 2 in the case that it is possible that p is the special prime for N. In order to determine which pairs of p and r create acceptable values of F r (p), a computer program must be written to perform a search. The computer program found that F r (p) is acceptable only for the pairs of p and r listed in Table A, at the end of this paper. 6 Inadmissable Small Primes In this section of the paper, we will prove the following lemma: Lemma 5.1. Let X be the set of primes {3,5,7,11,13,17,19,23,29,31,37,43,61,71, 113,127,131,151,197,211,239,281,1093} If p X, then p N. 21

The following proof will consider each prime in the set X one by one and prove, by contradiction, that p N. For this proof, it is convenient to consider the primes in the order below: 1093,151,31,127,19,11,7,23,31,37,43,61, 13,3,5,29,43,17,71,113,197,211,239,281. As stated above each for each p we will use a proof by contradiction. That is, we will assume p N and then use Table A to find all acceptable values of F r (p) for r 7. We will factor F 3 (p) and F 5 (p) by hand to check if those cyclotomic polynomials are acceptable. We will factor F 2 (p) only if p could be the special prime, meaning p 1 (mod 4) and no other prime has already been assumed to be the special prime. If a prime p is assumed to be the special prime, then we will denote it as p*. From Theorem 2.7, we know that for at least one acceptable F r (p), F r (p) 2N. Also, we know that each odd prime divisor of this specific F r (p) must divide N. Thus, we select an odd prime divisor q of F r (p) for each acceptable F r (p) and we find the acceptable values of F r (q). This procedure is repeated until a contradiction has been found for all acceptable values of F r (p), thus proving that p N. The proofs to show that 1093 N and 151 N are shown below. The proofs for the rest of the primes in X can be found in the appendix of [5]. 6.1 Proof that 1093 N Assume 1093 N. Then, we have that one value of F r (1093) must divide N. Below we list the values of F r (1093) found in Table A as well as for r = 2,3,5. F 2 (1093) = 2 547 F 3 (1093) = 3 398581 F 5 (1093) = 11 31 4189129561 We can see that F 5 (1093) is unacceptable since 4189129561 > 10 7. Thus, we can see that either 547 N or 398581 N. First, assume 547 N. Then, we have that 2 1093, so 1093 is the special prime. Now, one value of F r (547) must divide N. Below the values of F r (547) from Table A and for r = 3,5 are listed. 22

F 3 (547) = 3 163 613 F 5 (547) = 431 208097431 We can see that F 5 (547) is unacceptable since 208097431 > 10 7. Thus, we can see that 613 N. Then, we have that one value of F r (613) must divide N. Below the values of F r (613) from Table A and for r = 3, 5 are listed. F 3 (613) = 3 7 17923 F 5 (613) = 131 20161 53551 Thus, we have that either 17923 N or 20161 N. Assume that 17923 N. Then, one value of F r (17923) must divide N. Below the values of F r (17923) from Table A and for r = 3,5 are listed. F 3 (17923) = 3 13 31 265717 F 5 (17923) = 11 27308381 343540871 We can see that F 5 (17923) is unacceptable since 343540871 > 10 7. Thus, we have that 265717 N. Therefore, one value of F r (265717) must divide N. Below we list the values of F r (265717) from Table A and for r = 3,5. F 3 (265717) = 3 7 3362180467 F 5 (265717) = 31 41 101 38833995162778271 We can see that F 3 (265717) is unacceptable since 3362180467 > 10 7. Also, F 5 (265717) is unacceptable since 38833995162778271 > 10 7. Thus, neither of the values of F r (265717) are acceptable, so 265717 is inadmissable. Thus, 17923 N, so it must be that 20161 N. Then, one value of F r (20161) must divide N. Below the values of F r (20161) from Table A and for r = 3,5 are listed. F 3 (20161) = 3 135495361 F 5 (20161) = 5 11 1801 1667989905611 We can see that F 3 (20161) is unacceptable since 135495361 > 10 7. Also, F 5 (20161) is unacceptable since 1667989905611 > 10 7. Thus, we have that 20161 is inadmissable and thus neither 17923 nor 20161 divide N, so 613 N. Thus, we have 23

that 547 N. Therefore we must have that 398581 N. Since 547 N, 1093 is no longer the special prime. Thus, one value of F r (398581) must divide N. Below the values of F r (398581) from Table A and for r = 2,3,5 are listed. F 2 (398581) = 2 17 19 617 F 3 (398581) = 3 1621 32668561 F 5 (398581) = 5 1866871 2703853428809791 We can see that F 3 (398581) is unacceptable since 32668561 > 10 7. Also, F 5 (398581) is unacceptable since 2703853428809791 > 10 7. Thus, we have that 617 N which means that 398581 is the special prime. One value of F r (617) must divide N. Below the values of F r (617) from Table A and for r = 3,5 are listed. F 3 (617) = 97 3931 F 5 (617) = 145159381141 We can see that F 5 (617) is unacceptable since 145159381141 > 10 7. Thus, we have that 3931 N. One value of F r (3931) must divide N. Below the values of F r (3931) from Table A and for r = 3,5 are listed. F 3 (3931) = 3 7 31 23743 F 5 (3931) = 5 11 4342701505151 We can see that F 5 (3931) is unacceptable since 4342701505151 > 10 7. Thus, we have that 23743 N. One value of F r (23743) must divide N. Below the values of F r (23743) from Table A and for r = 3,5 are listed. F 3 (23743) = 3 37 5078863 F 5 (23743) = 11 2 251 10464092501131 We can see that F 5 (23743) is unacceptable since 10464092501131 > 10 7. Thus, 5078863 N. One value of F r (5078863) must divide N. Below the values of F r (5078863) from Table A and for r = 3,5 are listed. 24

F 3 (5078863) = 3 19 2341 193311109 F 5 (5078863) = 11 2 10831 507705831495587075591 We can see that F 3 (5078863) is unacceptable since 193311109 > 10 7. Also, F 5 (5078863) is unacceptable since 507705831495587075591 > 10 7. Thus, 5078863 is inadmissable. Since 5078863 is inadmissable, 23743 is inadmissable, and thus 3931 is inadmissable, and finally 617 is inadmissable. Since 617 is inadmissable, 398581 is inadmissable, so since we already showed that 547 is inadmissable, we have that 1093 N, as required. 6.2 Proof that 151 N Assume 151 N. Then, we have that one value of F r (151) must divide N. Below we list the values of F r (151) found in Table A as well as for r = 2,3,5. F 3 (151) = 3 7 1093 F 5 (151) = 5 104670301 We can see that F 3 (151) is unacceptable since we proved that 1093 N above. Also, F 5 (151) is unacceptable since 104670301 > 10 7. Thus, since no value of F r (151) divides N, we have that 151 N, as required. Using this method, we are able to eliminate all of the primes in the set X and thus we have Lemma 5.1. Knowing that if p X, then p N, we are able to eliminate all entries in Table A, where p X. Thus, we get a new table, which we will call Table B, of the remaining possible combinations of p and r for which F r (p) could divide N. Table B can be found in the appendix. 7 Restrictions on Exponents in the Prime Power Decomposition of N We will now prove the following lemma which will restrict the possibilities for exponents in the prime power decomposition of N. In order to do this, we will use Theorem 2.7, which shows that 2N = u i=0 σ(pa i i ) = u i=0 d (a i +1),d>1 F d(p i ). Also, we know that we have generated a table of all possible F r (p) which could divide N for r >= 7. Thus we know that the only other possible F r (p) which could 25

divide N must have r = 2, r = 3, or r = 5, with r = 2 only if p is the special prime. Using this information we will attempt to prove the following lemma. 7.1 Lemma 6.1 If p a N and p is not the special prime p 0, then a + 1 = 3 b 5 c where (b + c) > 0. If p a 0 0 N, then a 0 + 1 = 2 3 b 5 c where (b + c) 0. 7.2 Proof of Lemma 6.1 In order to prove this lemma, we must show that none of the values of F r (p) that appear in Table B are acceptable. Suppose that p a N and r (a + 1) with r > 5. Then, we know from Theorem 2.7 that F r (p) must appear in Table A, and thus F r (p) N. Now, from Lemma 5.1, we have that F r (p) must appear in Table B, since we eliminated all F r (p) from Table A except those that appear in Table B. From Table B, we have that r = 7 and p {67, 607, 619, 653, 1063, 1453, 2503, 4289, 5953, 9103, 9397, 10889, 12917, 19441, 63587, 109793, 113287, 191693, 6450307, 7144363}. Now, we will proceed to eliminate all values of p from this set by using Lemma 5.1 to show that they are inadmissable. In the following section we say a prime p is inadmissable if F r (p) is unacceptable, as defined in section 5, for every prime r, such that r < 5000000. Since we have that Table B contains all acceptable F r (p) for r 7, then we know that p is inadmissable if F r (p) does not appear in Table B, F 3 (p) is unacceptable, and F 2 (p) is unacceptable, where we only consider F 2 (p) if p could be the special prime. From Table B, if p = 67, then 175897 N. Since F r (175897) is acceptable only for r = 2 and r = 3, with F 2 (175897) = 2 37 2377 and F 3 (175897) = 3 3121 3304489. Since, from Lemma 5.1 we have that 3 N and 37 N, it follows that p 67. Similarly, if p = 173, then 3144079 N. We have that only F 3 (3144079) = 3 13 67 13267 285151 is acceptable, but 3 N, so p 173. 607. If p = 607, then 2210419 N. We have that 2210419 is inadmissable, so p If p = 619, then 4453751 N. We have that only F 3 (4453751) = 13 522061 2922721 is acceptable, but 13 N, so p 619. 26

If p = 653, then 706763 N. We have that only F 3 (706763) = 31 37 631 751 919 is acceptable, but 31 N, so p 653. If p = 1063, then 1371511 N. We have that only F 3 (1371511) = 3 1579 1759 225751 is acceptable, but 3 N, so p 1063. If p = 1453, then 2219491 N. We have that only F 3 (2219491) = 3 421 487 8008933 is acceptable, but 3 N, so p 1453. If p = 2503, then 3591869 N, but 3591869 is inadmissable, so p 2503. If p = 4289, then 1538951 N, but 1538951 is inadmissable, so p 4289. If p = 5953, then 1591927 N, but 1591927 is inadmissable, so p 5953. If p = 9103, then 1338331 N, but 1338331 is inadmissable, so p 9103. If p = 9397, then 1551383 N, but 1551383 is inadmissable, so p 9397. If p = 10889, then 686057 N. We have that only F 2 (686057) = 2 3 114343 is acceptable, but 3 N, so p 10889. If p = 12917, then 5194337 N. We have that only F 2 (5194337) = 2 3 181 4783 is acceptable, but 3 N, so p 12917. If p = 19441, then 2196979 N, but 2916979 is inadmissable, so p 19441. If p = 63587, then 6079823 N, but 6079823 is inadmissable, so p 63587. If p = 109793, then 7258639 N, but 7258639 is inadmissable, so p 109793. If p = 113287, then 5980619 N, but 5980619 is inadmissable, so p 113287. If p = 191693, then 208279 N, but 208279 is inadmissable, so p 191693. If p = 6450307, then 901279 N, but 901279 is inadmissable, so p 6450307. If p = 7144363, then 171823 N, but 171823 is inadmissable, so p 7144363. Thus, we have that we can elminate all entries from Table B. Therefore, we know that if p is not the special prime, then for F r (p) N, it must be that r = 3, or r = 5. So, since r (a + 1), a + 1 = 3 b 5 c, where (b + c) > 0. Now, if p is the special prime p 0, then for F r (p) N, it must be that r = 2, r = 3, or r = 5. Thus, since r (a 0 + 1), a 0 + 1 = 2 3 b 5 c, where (b + c) 0. Thus, we have Lemma 6.1. 8 Four Sets Let S = {47, 53, 59...} be the set of all primes p such that p 1 (mod 3), p 1 (mod 5) and 37 < p < 10 7. Let T = {61, 151, 181,...} be the set of all primes p 27

such that p 1 (mod 15) and 37 < p < 10 7. Let U = {73, 79, 103,...} be the set of all primes p such that p 1 (mod 3), p 1 (mod 5), F 5 (p) has a prime factor greater than 10 7, and 37 < p < 10 7. Let V = {3221, 3251, 3491,...} be the set of all primes p such that p 1 (mod 5), p 1(mod 3), F 3 (p) has a prime factor greater than 10 7, and 37 < p < 10 7. Note that S,T,U, and V are pairwise disjoint. 8.1 Proposition 7.1 The number N is divisible by at most one element of S. If there is such an element s, then s p 0 and s 47. 8.2 Proof of Proposition 7.1 Let p N and p F d (p i ) where d 2. Then, we have that d (a i + 1) and by Lemma 6.1, either 3 d or 5 d. Then, by Theorem 2.4, either p 1 (mod 3) or p 1 (mod 5), so p S. Now, assume p i S and p a i i N and p i F 2 (p 0 ). Then, we have that p a i i F 2(p 0 ). Assume, towards a contradiction, that two elements of S were divisors of F 2 (p 0 ). Then, F 2 (p 0 ) = p 0 + 1 2 47 2 53 2 = 12410162. This is a contradiction since p 0 < 10 7. Thus, we have that at most one element of S can divide F 2 (p 0 ). Thus, we have shown that no element of S can divide F d (p i ) for any d 2. Also, we have shown that only one element of S can divide F 2 (p 0 ), so we have that only one element of S can divide N. Now, we will prove that s p 0. We will assume towards a contradiction that p 0 S. The, we have that p 0 2 (mod 3) and thus, 3 (p 0 + 1) = F 2 (p 0 ), so 3 N, which is a contradiction of Lemma 5.1. Thus, p 0 S. Thus, we have proved Proposition 7.1. 8.3 Proposition 7.2 The number N is divisible by at most one element of T. If there is such an element, it is p 0 and thus p 0 61. 28

8.4 Proof of Proposition 7.2 We will prove that if p T then p = p 0. Suppose, towards a contradiction, that p i T and p i p 0. If p a i i N, then, by Lemma 6.1, either 3 (a i + 1) or 5 (a i + 1). So, by Theorem 2.4, either F 3 (p i ) N or F 5 (p i ) N. Thus, either 3 N or 5 N. In either case, we have a contradiction with Lemma 5.1. Thus, if p i T then p i = p 0, so we have Proposition 7.2. 8.5 Proposition 7.3 The number N is divisible by at most one element of U. If there is such an element it is p 0 and then p 0 73. 8.6 Proof of Proposition 7.3 We will prove that if p U, then p = p 0. Suppose, towards a contradiction, that p i U and p i p 0. If p a i i N, then by Lemma 6.1, either 3 (a i + 1) or 5 (a i + 1). If 3 (a i + 1), then F 3 (p i ) N and 3 N, which is a contradiction with Lemma 5.1. If 5 (a i + 1), then F 5 (p i ) N and then N has a factor greater than 10 7, which is a contradiction. Thus, p i N, so we have, by contradiction, that p i = p 0. Thus, we have Proposition 7.3. 8.7 Proposition 7.4 The number N is not divisible by any element of V. 8.8 Proof of Proposition 7.4 We will prove that no element of V divides N. Assume, towards a contradiction, that p i V and p i N. Since p i 1 (mod 3), then p i 2 (mod 3) or p i 0 (mod 3). Now, if p i 0 (mod 3), then 3 p i and thus 3 N, which is a contradiction with Lemma 5.1. Thus, p i 2 (mod 3), and so 3 (p i + 1) = F 2 (p i ). Now, F 2 (p 0 ) N and by Lemma 5.1, 3 N, so p i p 0. Let p a i i N, then by Lemma 6.1, either 3 (a i + 1) or 5 (a i + 1). If 5 (a i + 1), then F 5 (p i ) N and 5 N, which contradicts Lemma 5.1. If 3 (a i +1), then F 3 (p i ) N and N has a factor greater than 10 7, which is a contradiction. Thus, we have Proposition 7.4. 29

8.9 Computer Search for Number of Elements of S,T,U and V A computer search showed that S has 249278 elements and that S = p S p > 1.733190914437589931. (12) p 1 A computer search showed that T has 83002 elements and that T = p T p > 1.1791835683407662159. (13) p 1 A computer search showed that U has 694 elements and that U = p p 1 > p U, p < 20000 p p 1 p U > 1.239225225. (14) A computer search showed that V has 57 elements less than 20000, and that V = p V p p 1 > p V,p<20000 9 Proof of Jenkins Theorem There are 664567 primes p such that 37 < p < 10 7, and If p a N, then P = 41 p<10 7 p > 1.006054597. (15) p 1 p < 4.269448664996309337. (16) p 1 1 < σ(p a )/p a = (p a+1 1)/(p a (p 1)) < p/(p 1). Now, since σ is a multiplicative function, we have that, σ(n) N = σ(p 0)σ(p 1 )... < p 0 p 1... u i=0 p i p i 1 30