Saint Mary's College of California Department of Mathematics and Computer Science On Perfect Numbers Advisors: Dr. James Sauerberg Author: Dr. Kathryn Porter Arturo Garcia Reader: Dr. Kristen Beck May 17, 2016 Abstract A perfect number is a number that is equal to the sum of its divisors excluding itself. Theorems proven by Euclid and Euler show that finding a new Mersenne prime is equivalent to the discovery of a new even perfect number. Even perfect numbers have a systematic way of being found and have various ways in which they can be characterized, but the same is not true for odd perfect numbers. In fact, the existence of an odd perfect number has not been proven or disproven. This paper discusses the categorizations of even perfect numbers and possible characteristics of odd perfect numbers. 1 Introduction A perfect number is a positive integer that is equal to the sum of its positive divisors excluding itself. The ancient Greeks were well versed in the existence of perfect numbers and held those that they were aware of (the first four) in high regard. Perfect numbers were thought to have mystical healing properties and were even used to illustrate the existence of God. Examples of perfect numbers include 6; 28; 496; and 8128 because: 6 = 1 + 2 + 3 28 = 1 + 2 + 4 + 7 + 14 496 = 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248: 1 Such examples could entice the notion that perfect numbers are found fre- quently and are relatively close to one another, but this could not possibly be further from the truth. As of the writing of this paper, there are 49 known perfect numbers, with the newest one having been discovered in early January 2016. This newest perfect number number is 2(74;207;280)(274;207;281 − 1) and is 44; 677; 235 digits long. Large primes and perfect numbers such as these are found by many people who cooperate together as part of the Great Internet Mersenne Prime Search (GIMPS). While finding new perfect num- bers is certainly exciting, mathematicians are more concerned with finding a systematic and reliable method of finding perfect numbers. Perfect numbers have continued to be of interest for thousands of years because they have been studied by some of the brightest mathematicians in history, but have yet to reveal the entirety of their nature. This paper seeks to explain the progress made in the understanding of perfect numbers and to discuss why perfect numbers continue to be one of the world's oldest unsolved puzzles. 2 Euclid's Perfect Number Theorem Born in Alexandria around 330 B.C, the great mathematician Euclid was instrumental in the advances made in the study of perfect numbers. In Book 2 IX Proposition 36 of his famous work, Elements, he states the following: \If as many numbers as we please beginning from a unit are set out continuously in double proportion until the sum of all becomes prime, and if the sum multiplied into the last makes some number, then the product is perfect." The result of Euclid's studies of perfect numbers is Euclid's Perfect Number Theorem. Euclid's Theorem is widely considered to be the first step mankind took to understanding the nature of perfect numbers [9]. Theorem 1 (Euclid's Perfect Number Theorem). If 2p−1 is a prime number, then 2p−1(2p − 1) is a perfect number. Proof. Let p be an integer such that 2p − 1 is a prime number. We aim to show that 2p−1(2p − 1) is a perfect number. Now let q = 2p − 1, so that 2p−1(2p − 1) = 2p−1q. We can write out the divisors of 2p−1 as follows: 1; 2; 4; 8; 16;:::; 2p−1. Then, we can write out the other divisors of 2p−1q as follows: q; 2q; 4q; 8q; 16q; : : : ; 2p−2q. We proceed by adding 1; 2; 4; 8; 16;:::; 2p−1 first. Recall that xn − 1 1 + x + x2 + x3 + ··· + xn−1 = : x − 1 In this case, n = p and x = 2 so 2p − 1 1 + 2 + 4 + 8 + 16 + ··· + 2p−1 = = 2p − 1 = q (1) 2 − 1 Next, we can use the same formula to find q +2q +4q +8q +16q +···+2p−2q. 3 It is clear that q + 2q + 4q + 8q + 16q + ··· + 2p−2q = q(1 + 21 + 22 + 23 + ··· + 2p−2) so now x = 2 and n = p − 1. This gives us: 2p−1 − 1 q(1 + 21 + 22 + 23 + ··· + 2p−2) = q( ) 2 − 1 = q(2p−1 − 1) (2) Finally, we can sum all the divisors of 2p−1q by combining (1) and (2): 1 + 2 + 4 + 8 + 16 + ··· + 2p−1 + q + 2q + 4q + 8q + 16q + ::: 2p−2q = q + q(2p−1 − 1) = q + q(2p−1) − q = 2p−1q = 2p−1(2p − 1) So we have shown that if q = 2p −1 is prime, then 2p−1(2p −1) is perfect. 3 Euler's Perfect Number Theorem Naturally, the question that arises after proving Euclid's Perfect Number Theorem is whether it describes all perfect numbers. That is to say, are 4 all perfect numbers of the form 2p−1(2p − 1)? The next significant step in the answering of this question and the understanding of perfect numbers was made two thousand years after Euclid's results by Swiss mathematician Leonhard Euler. Euler proved that Euclid's formula for perfect numbers holds true for all even perfect numbers. Theorem 2 (Euler's Perfect Number Theorem). If n is an even perfect number, then it is of the form n = 2p−1(2p − 1) where p is some prime and 2p − 1 is a Mersenne prime. A Mersenne prime is a prime of the form 2p − 1. Euler's theorem is indicates that there is a one-to-one relationship between Mersenne primes and even perfect numbers, so it is of significant importance that we prove this theorem. 3.1 The Sigma Function Before Euler's Perfect Number Theorem can be proved, it is imperative that we define and study the properties of what is known as the sum of divisors function. We define the sum of divisors function σ as the following: σ(n) = the sum of all unique divisors of n including 1 and n: 5 Examples include σ(4) = 1 + 2 + 4 = 7 σ(6) = 1 + 2 + 3 + 6 = 12 σ(18) = 1 + 2 + 3 + 6 + 9 + 18 = 39 Note that n is perfect when σ(n) = 2n. If we are to prove Euler's theorem, then naturally we must show that σ(2p−1(2p − 1)) = 2(2p−1(2p − 1)). Before we address such a proof we must prove the following: Theorem 3 (Properties of the Sigma Function). (a) If p is a prime and k ≥ 1, then pk+1 − 1 σ(pk) = 1 + p + p2 + ··· + pk = : p − 1 (b) If gcd(m; n) = 1, then σ(mn) = σ(m)σ(n): We begin by proving statement (a). Proof. If p is prime, then it is divisible only by 1 and itself. It follows that the divisors of pk are of the form pi for all i such that i ≤ k. That is to say, we can write σ(pk) = 1 + p + p2 + ··· + pk−1 + pk. Note that this sum can be 6 n written as having the form P arj. We begin by defining S such that j=0 n X S = arj j=0 and proceed by multiplying S by r: n X rS = r arj j=0 n X = arj+1 j=0 n+1 X = ark k=1 n X = ark + (arn+1 − a) k=0 = S + (arn+1 − a): Solving for S gives us that arn+1 − a S = r − 1 k pk+1−1 But in our case a = 1 and r = p, thus σ(p ) = p−1 To prove property (b), we must first prove the following lemma [8]. Lemma 1. Suppose a and b are relatively prime integers with fai : 1 ≤ i ≤ sg and fbj : 1 ≤ j ≤ tg being all the divisors of a and b respectively. Then 7 S = faibj : 1 ≤ i ≤ s; 1 ≤ j ≤ tg are all the divisors of ab. Proof. It is clear that each element of the form aibj is a divisor of ab. So it follows that S contains only divisors of ab. We aim to show that all of the divisors of ab are present in S. 0 d Suppose d j ab and let D = gcd(a; b) and d = D 2 Z. Because D j a, D = aj 0 0 0 for some j. It follows that d = Dd = ajd . We must show d divides b. 0 0 a Because d j ab, dk = ab for some integer k, so Dd k = ab if d k = D b. a b a 0 Because gcd(a; b) = D, we have that gcd( d ; D ) = 1, so gcd( d ; d ) = 1. Thus, 0 0 by Euclid's Lemma d j b so d = bi for some i as desired. Now we can proceed with proving property (b). Proof. Using the same notation as above, let fai : 1 ≤ i ≤ sg and fbi : 1 ≤ j ≤ tg be the divisors of a and b respectively. Then we have σ(a)σ(b) = P P P P P ( i ai)( j bj) = i j aibj = djab d = σ(ab). 3.2 Proving Euler's Perfect Number Theorem Now we use our newly found knowledge of σ(n) to prove Euler's theorem. Recall that Euler's Perfect Number Theorem states the following: if n is an even perfect number, then it is of the form n = 2p−1(2p − 1) where 2p − 1 is a Mersenne prime [11].