Prime and Perfect Numbers 0.3 Infinitude of prime numbers 0.3.1 Euclid s proof Euclid IX.20 demonstrates the infinitude of prime numbers. 1 The prime numbers or primes are the numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29,... (1) which cannot be resolved into smaller factors. 2... We have to prove that there are infinitely many primes, i.e., that the series (1) never comes to an end. Let us suppose that it does, and that 2, 3, 5,..., P is the complete series (so that P is the largest prime); and let us, on this hypothesis, consider the number Q defined by the formula Q = 2 3 5 P + 1. It is plain that Q is not divisible by any of 2, 3, 5,..., P ; for it leaves the remainder 1 when divided by any one of these numbers. But, if not itself prime, it is divisible by some prime, and therefore there is a prime (which may be Q itself) greater than any of them. This contradicts our hypothesis, that there is no prime greater than P ; and therefore this hypothesis is false. 0.3.2 Euler s proof Suppose, for a contradiction, that there be only finitely many many primes p 1, p 2,...,p N. By the Fundamental Theorem of Arithmetic, 1 + 1 2 + 1 3 + 1 4 + + 1 n + 1 The presentation is taken from G. H. Hardy, A Mathematician s Apology, pp. 93 94. 2 There are technical reasons for not counting 1 as a prime.
8 CONTENTS = = ( 1 + 1 + 1 ) ( p 1 p 2 + 1 + 1 + 1 ) ( 1 p 2 p 2 + 1 + 1 2 1 1 1 p 1 1 1 1 p 2 1 1 1 p N, p N + 1 contradicting the divergence of the harmonic series 1 + 1 2 + 1 3 + + 1 n +. 0.4 Some important theorems on prime numbers p 2 N ) + Theorem 0.1 (Tchebycheff). 3 For every integer k 2, there exists a prime number p in the range k < p < 2k. Theorem 0.2 (Dirichlet (1805-1859)). If a and b are relatively prime, the arithmetical progression an + b contains infinitely many prime numbers. Theorem 0.3 (The Prime Number Theorem). 4 If π(x) denotes the number of primes between 1 and x, then π(x) as x. x log x 0.4.1 Some unsolved problems about prime numbers 5 1. (Goldbach conjecture) 6 Every even number > 6 is a sum of two distinct odd primes. 2. (Twin prime conjecture) There are infinitely many pairs of prime numbers of the form p and p + 2. 7 3. Does the sequence n 2 + 1 contain infinitely many primes? 0.5 Perfect numbers A number is perfect if it is equal to the sum of all its divisors, including 1 but excluding the number itself. The first five even perfect numbers are 6, 28, 496, 8128, 33550336. For each positive integer n, let σ(n) denote the sum of all positive divisors of n, including 1 and n itself. A number n is a perfect number if and only if σ(n) = 2n. 3 This is also known as the Bertrand postulate. It was conjectured by J.L.F. Bertrand (1822 1900) in 1845, and was proved by P.L. Tchebycheff (1821 1894) in 1850). 4 The prime number theorem was conjectured by A.M. Legendre (1752 1833) and C.F Gauss (1777 1855) around 1800, first proved at the end of the 19th century by J. Hadamard (1865 1963) and de la Vallée Poussin (1866 1962) using deep analytical methods, but by elementary (but difficult) methods by Paul Erdös (1913 1996) and A. Selberg in 1949. 5 For more unsolved problems about prime numbers, see Chapter 1 of R.K. Guy. Unsolved Problems in Number Theory, (2nd ed.), Springer Verlag, 1994. 6 Proposed to Euler in 1742 by C. Goldbach (1690 1764). 7 As of 1990, the largest known pair of twin primes are 1706595 2 11235 ± 1.
0.6 Mersenne primes 9 Theorem 0.4 (Euclid IX.36). If as many numbers as we please beginning from an unit be set out continuously in double proportion, until the sum of all becomes prime, and if the sum multiplied into the last make some number, the product will be perfect. Modern formulation: Suppose M k := 1 + 2 + 2 2 + 2 k 1 = 2 k 1 is prime. Then 2 k 1 (2 k 1) is a perfect number. (Note that such a perfect number is even). Proof. σ(2 k 1 (2 k 1)) = σ(2 k 1 )σ(2 k 1) = (2 k 1) 2 k = 2 2 k 1 (2 k 1). Theorem 0.5 (Euler). Every even perfect number is of the form described in Eucl.IX.36: If n is an even perfect number, then n = 2 k 1 (2 k 1) for some integer k and M k = 2 k 1 is prime. Proof. Write n = 2 k 1 q, q odd. Since n is perfect, 2 k q = 2n = σ(n) = σ(2 k 1 )σ(q) = (2 k 1)σ(q). From this, σ(q) = q + q 2 k 1. Since σ(q) is an integer, 2k 1 must be a divisor of q. Indeed, 2 k 1 = q, and σ(q) = q + 1. This means that q = 2 k 1 is a prime. Remark. It is not known if an odd perfect number exists. 0.6 Mersenne primes The number M k := 2 k 1 is called the kth Mersenne number. If M k is prime, then k must be a prime. (Exercise). For example, M 3 = 7, M 5 = 31 and M 7 = 127 are primes, but M 11 = 2 1 1 1 = 2047 is not a prime. (Exercise. Factorize it). 0.6.1 Primality tests for M k Theorem 0.6 (Fermat). If k is a prime, then every prime factor of M k = 2 k 1 must be of the form 2kr + 1 for some integer r. Remark. In the early 20-th century, F.N.Cole, Professor of Mathematics in Columbia University, spent the sundays of three consecutive years on the factorization of M 67, and obtained M 67 = 193707721 761838257287. Theorem 0.7 (Lucas - Lehmer). Let (v k ) be the sequence defined recursively by v i+1 = v 2 i 2, v 2 = 4. For a prime number k, the Mersenne number M k = 2 k 1 is prime if and only if M k divides v k. Remark. It is not known if there are infinitely many Mersenne primes, (equivalently even perfect numbers).
10 CONTENTS 0.6.2 Records of Mersenne primes k Year Discoverer k Year Discoverer 2 Ancient 3 Ancient 5 Ancient 7 Ancient 13 Ancient 17 1588 P.A.Cataldi 19 1588 P.A.Cataldi 31 1750 L.Euler 61 1883 I.M.Pervushin 89 1911 R.E.Powers 107 1913 E.Fauquembergue 127 1876 E.Lucas 521 1952 R.M.Robinson 607 1952 R.M.Robinson 1279 1952 R.M.Robinson 2203 1952 R.M.Robinson 2281 1952 R.M.Robinson 3217 1957 H.Riesel 4253 1961 A.Hurwitz 4423 1961 A.Hurwitz 9689 1963 D.B.Gillies 9941 1963 D.B.Gillies 11213 1963 D.B.Gillies 19937 1971 B.Tuckerman 21701 1978 C.Noll, L.Nickel 23209 1979 C.Noll 44497 1979 H.Nelson, D.Slowinski 86243 1982 D.Slowinski 110503 1988 W.N.Colquitt, L.Welsch 132049 1983 D.Slowinski 216091 1985 D.Slowinski 756839 1992 D.Slowinski,P.Gage 859433 1993 D.Slowinski 1257787 1996 Slowinski and Gage 1398269 1996 Armengaud, Woltman et al. 2976221 1997 Spence, Woltman, et.al. 3021377 1998 Clarkson et. al 6972593 1999 Hajratwala et. al 13466917 2001 Cameron, Woltman, 20996011 2003 Michael Shafer 24036583 2004 Findlay 25964951 2005 Nowak 30402457 2005 Cooper, Boone et al 32582657 2006 Cooper, Boone et al 37156667 9/8/2008 43112609 8/8/2008 The most recently discovered Mersenne primes M 37156667 and M 43112609 have about 11.1 million and 12.9 million digits and are the largest known primes. 0.7 Fermat numbers The number F k := 2 2k + 1 is called the kth Fermat number. Fermat (1601-1665) observed that F 0 = 3, F 1 = 5, F 2 = 17, F 3 = 257, F 4 = 65537 are all primes, and conjectured that all F k are primes. This was later refuted by Euler (1707-1773) who found the factorization No other Fermat primes are known. F 5 = 2 32 + 1 = 4294967297 = 641 6700417. Remarks. (1) Every factor of F k is of the form 2 k+2 a + 1 for some integer a. (2) F. G. Eisenstein (1823 1852) conjectured that 2 2 +1, 2 22 +1, 2 222 +1,..., are primes. However, it was found out that 2 222 + 1 = F 16 is composite (J. L. Selfridge). Relation with geometry. Gauss (1777 1855) discovered that it is possible to construct (using only ruler and compass) a regular polygon of n sides if n is the product of a power of 2 and distinct Fermat primes. He stated that the converse is also true. This was later proved by P. L. Wantzel (1814 1848). For Gauss construction of the regular 17 gon.
0.7 Fermat numbers 11 Exercise (a) q is a prime number if and only if σ(q) = 1 + q. (b) σ(2 k 1 ) = 2 k 1. Determine if M 13 = 8191 and M 23 = 8388607 are prime numbers. Use the Lucas - Lehmer test for (re-)confirm that M 5 = 31 and M 7 = 127 are primes, but that M 11 = 2047 is not.