W&M ScholarWorks Dissertations, Theses, and Masters Projects Theses, Dissertations, & Master Projects 1965 On Perfect Numbers David Thomas Eastham College of William & Mary - Arts & Sciences Follow this and additional works at: https://scholarworks.wm.edu/etd Part of the Mathematics Commons Recommended Citation Eastham, David Thomas, "On Perfect Numbers" (1965). Dissertations, Theses, and Masters Projects. Paper 1539624586. https://dx.doi.org/doi:10.21220/s2-thss-sr14 This Thesis is brought to you for free and open access by the Theses, Dissertations, & Master Projects at W&M ScholarWorks. It has been accepted for inclusion in Dissertations, Theses, and Masters Projects by an authorized administrator of W&M ScholarWorks. For more information, please contact [email protected]. ON PERFECT NUMBERS A Thesis Presented to The Faculty of the Department of Mathematics The College of William and Mary in Virginia In Partial Fulfillment Of the Requirements for the Degree of Master of Arts By David Thomas Eastham August 196? APPROVAL SHEET This thesis is submitted in partial fulfillment of the requirements for the degree of Master of Arts Author Approved, August 1965: Thomas L* Reynolds, Ph.Ds Hugh B. Easier, M.A. LutherT. Conner, M.A. i i ACKNOWLEDGMENT The writer wishes to express his grateful appreciation to Professor Thomas L. Reynolds for his personal guidance and assistance throughout the compilation of this manuscript* The author is also indebted to Professor Luther T* Conner * and Professor Hugh B. Easier for their helpful comments and corrections. He also wishes to thank his wife, Kathleen, * for the typing of this manuscript* i i i TABLE OF CONTENTS Page ACKNOWLEDGMENT........................... iii ABSTRACT ......... ...... v INTRODUCTION . ............ 2 Chapter I. EVEN PERFECT NUMBERS . ...........' . 3 II. ODD PERFECT NUMBERS . ............................ 9 III. TECHNIQUES OF PROOF ........... 2? BIBLIOGRAPHY ...... .............. 35 iv ABSTRACT A perfect number is a number the sum of whose divisors is equal to twice the number. Even numbers that are perfect have been known for centuries. However, the existence or non-existence of odd numbers that are perfect has never been established. In this paper, some of the properties of perfect numbers are discussed together with conditions for their existence. Chapter I lists the main result concerning even perfect numbers together with some minor conclusions concerning their properties. In Chapter II some of the basic principles are set forth concerning the question of odd perfect numbers. This is accompanied by a brief historical review of the progress in various areas. In order to illustrate the means employed in the attack on odd perfect.numbers, Chapter III selects the most comprehensive and most important of a variety of methods and explains their structure. In the light of the evidence presented, it would seem unlikely that any odd perfect number exists. This, however has not yet been proved. v INTRODUCTION The aliquot parts of a number are the divisors which are less than the number* If a given natural number is equal to the sum of its aliquot parts it is termed perfect* which is equivalent to saying that the sum. of its divisors is equal to twice the number. If this aliquot sum is less than the given number it is termed deficient. If the sum of the •divisors is greater than twice the given number, it is termed abundant. The quality of such a number that equals the sum of its parts has for hundreds of years been associated with perfection, and hence its name. The establishment of a simple criterion by which one could determine whether or not a given natural number is perfect has been a goal sought by mathematicians since the days of the early Greeks» We will discuss, during the next few chapters, the results which have been established concerning such numbers and the methods employed in their establishment. 2 ON PERFECT NUMBERS CHAPTER I EVEN PERFECT NUMBERS Euclid [6 ]^ gave a sufficient condition for an even number to be perfect with: Theorem 1: If p and 2P- 1 are primes, then N = (2P~-*-) (2P- l) is perfect and even. Proof; Obviously, (2P“^)(2P- 1 ) is divisible by 1 , 2 ,------ 2P~1 , 2P- 1 , 2 (2P- 1 ),...., (2P~2 )(2P- l) but not by any other number less than (2P“*^)(2P- 1 )* The sum of these divisors is (2^^)(2P- 1 ). Clearly N is even. Euler [6 ] later proved that this condition was also necessary. Hence, Theorem 2; If N is even and perfect, then N = (2P"^)(2P- 1 ), where p and 2P- 1 are primes. Proof: We will denote V ^ byO^CN). Letting a | isn N = 2 \ , where (2k ,,j.) = 1 , we have G“(N) = 2N =(r(2kn) =G'(2k )(r(fi)» giving (A) 2k+1- lg(u) = 2k+l 2-1 . I1’ — . The numbers in brackets refer to the bibliography. , 3 b But 2k+1-l is odd and must be a factor of ja. So H = (2k+1-l)V or (2k + 1-lXT(p.) = 2k+ 1[(2k+ 1-l)v], whence cr(|j.) = (2k + 1 )v. But also G*((i.) = cr[(2k + 1-i)v]. Now, either V = 1 or V > 1 . Assume V > 1. The divisors of (2k+1-l)V are: (2k+1-l)V, 2 k+1-l, V, 1 , ..... Their sum, S, is: S = (2k+1)V - V + 2k+1-l + V + 1 + ..... = (2k+1)V + (2k+1 + .......+.) =G'[(2k+1-l)V] = CTCp.) which from a previous equation is equal to (2k+1)V, or (2k+1)V = (2k+1)V + (2k+1 + k*t*l which is impossible, therefore V = 1, and |x = 2 -1 . From (A), G~(p.) = 2k+1 5 or CT(2k+1- 1) = 2k+1. Now, the divisors of 2* - l are: 2 k+ X i *i Adding, we get CT(2k+1- I) = 2k+-L- l + l + (extra terms) = 2k+^ + (extra terms), k+1 thus (extra terms) = 0 * So 2 - l must be prime® To show that k + 1 is a prime we assume k + 1 = a«-b® Then 2k+1- l = 2a#b- 1 = (2a )b- 1 ts xb- 1 ® This xb- 1 is divisible by x — 1 = 2a- 1 , but Xb- 1 = 2k+1- 1 was prime, hence a contradiction and k + X is prime® Primes of the form p = 1 + 2 + 22 +••••+ 2^“^ = 2b- 1 are called Mersenne primes and since the time of Euler the search for even perfect numbers has been reduced to the search for Mersenne primes® Due to the difficulty of ascertaining whether or not a number, of the form 2^- 1 is a prime, few correct lists of Mersenne primes were published until the advent of the modern computers. Even now the only even perfect numbers known are those corresponding to p = 2 , 3 , 6 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 228x, 3217, ‘+253, and Mf23 [l]. It is still not known whether there is a finite number of even perfect number s• Some of the properties of even perfect numbers follow: Theorem 3 : Every multiple of an even perfect number • is abundant. Proof: Let N = (2^”*^*) (2^-1) be perfect* . Then NK = (2p_1 )(2p-l)K and £ K(2P-1 )(2P-1 ) d|N d<N = 2(2P-1)K + 22 (2P-1)K +.+ 2p-2 (2p-l)K + N + K + 2K +.+ (2p-1)K = (2P-1) (2P’"1-1)K + N + (2P-1)K Let (2p_l)(2P-1)K - (2p-l)(2p-1-l)K +N + K(2P-1) = (2P-1 )(2P_1-1 )K + (2P_1 )(2P-1 ) + (2p-l)K, hence, (2P-1)K = (2P_1-1 )K + 2P_1 +' K = K>2p-1-K + 2P“1 +K _ (2P_1)K + 2P“1 . 7 This holds only for p = 1, which does not make N perfect. Otherwise NK is obviously abundant. Theorem *+: Every divisor of an even perfect number is deficient. Proof; Let N = (2P-1)(2P_1) be perfect. The conclusion is immediate since, if p is prime, then (a) £ d = 1 d|p d<p and (b) Y 2k = 1 + 2 + 22 + *••• + 2k_1 = 2k- 1 . i —l - d|2k d<2 Theorem 5: If (2p"b(2P- 1) is perfect, then (2p-1 )(2p-l) = l(mod 9 ). Proof: We shall denote the ultimate sum of the digits of a number by S(N). Example: S(398) = 3 + 9 + 8 = 20* 2+0=2 and S(398) = 2 Brooke [*+] showed that S[ (2n~'L) (2>n- l)] forms the sequence 1, 3* 1> 9> 1* 8 , 1 , 3> • •• etc., for n > 3* However, if (2n“^)(2n- 1 ) is to be perfect, n must be prime and those elements occupying the positions in the sequence generated when n was even would be eliminated. Hence the only numbers to be considered are those such that S(N) = 1. The result follows from the easily proven fact that if S(N) = l(mod 9), then N = 1 (mod 9 ). Theorem 6 : Every even perfect number is triangular* 8 Proof: Triangular numbers are of the form 1/2 [n(n + 1 )], n > 0 , Given an even perfect number it is of the form (2P-1)(2P_1) by Theorem 2 and (2p-l)(2p"1 ) = 1/2 [2P(2P-1 )]. Theorem 7: For any integer of the form 2n“^(2n-l), 2n”l<2n-l for every positive integer n >1. In particular, this holds for any even perfect number N = 2P~‘*‘(2P-1) where p is a prime. Proof: If n = 2, 2 < 3. Assume that. 2k“l < 2k-l for some integer k > 2 . Then 2k < 2k+1-2 or 2k < 2k+1-l and the induction is complete* An alternate result may be found in [l]* Theorem 7*: Six is the only square free even perfect number• Proof: In Theorem 2 we proved that for N = (2P“^ )(2P-1) to be perfect, 2P~1 and p had to be prime.