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Perfection: a Brief Introduction Perfection: A brief introduction Paul Pollack University of Georgia June 4, 2020 1 of 30 For example, 5 is deficient (s(5) = 1, and similarly for any prime), 12 is abundant (s(12) = 1 + 2 + 3 + 4 + 6 = 16), and 6 is perfect (s(6) = 1 + 2 + 3 = 6). Three kinds of natural numbers P Let s(n) = djn;d<n d be the sum of the proper divisors of n. Abundant: s(n) > n. Deficient: s(n) < n. Perfect: s(n) = n. 2 of 30 Three kinds of natural numbers P Let s(n) = djn;d<n d be the sum of the proper divisors of n. Abundant: s(n) > n. Deficient: s(n) < n. Perfect: s(n) = n. For example, 5 is deficient (s(5) = 1, and similarly for any prime), 12 is abundant (s(12) = 1 + 2 + 3 + 4 + 6 = 16), and 6 is perfect (s(6) = 1 + 2 + 3 = 6). 2 of 30 You can see a lot just by looking Abundants: 12, 18, 20, 24, 30, 36, 40, 42, 48, 54, 56, 60, 66, 70, 72, 78, 80, 84, 88, 90, 96, 100, 102, . Deficients: 1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 19, 21, 22, 23, 25, 26, 27, . Perfects: 6, 28, 496, 8128, 33550336, 8589869056, 137438691328, 2305843008139952128, 2658455991569831744654692615953842176, .... 3 of 30 Theorem • The limiting proportion of abundant numbers exists, is ≈ 24.76%. • The limiting proportion of deficient numbers exists, is ≈75.24%. • The limiting proportion of perfect numbers exists, is exactly 0%. \Limiting proportion" = asymptotic density: Count up to x, divide by x, send x ! 1. Nicomachus: No rule needed to generate abundants or deficients. Just as . ugly and vile things abound, so superabundant and deficient numbers are plentiful and can be found without a rule. 4 of 30 • The limiting proportion of deficient numbers exists, is ≈75.24%. • The limiting proportion of perfect numbers exists, is exactly 0%. \Limiting proportion" = asymptotic density: Count up to x, divide by x, send x ! 1. Nicomachus: No rule needed to generate abundants or deficients. Just as . ugly and vile things abound, so superabundant and deficient numbers are plentiful and can be found without a rule. Theorem • The limiting proportion of abundant numbers exists, is ≈ 24.76%. 4 of 30 • The limiting proportion of perfect numbers exists, is exactly 0%. \Limiting proportion" = asymptotic density: Count up to x, divide by x, send x ! 1. Nicomachus: No rule needed to generate abundants or deficients. Just as . ugly and vile things abound, so superabundant and deficient numbers are plentiful and can be found without a rule. Theorem • The limiting proportion of abundant numbers exists, is ≈ 24.76%. • The limiting proportion of deficient numbers exists, is ≈75.24%. 4 of 30 Nicomachus: No rule needed to generate abundants or deficients. Just as . ugly and vile things abound, so superabundant and deficient numbers are plentiful and can be found without a rule. Theorem • The limiting proportion of abundant numbers exists, is ≈ 24.76%. • The limiting proportion of deficient numbers exists, is ≈75.24%. • The limiting proportion of perfect numbers exists, is exactly 0%. \Limiting proportion" = asymptotic density: Count up to x, divide by x, send x ! 1. 4 of 30 In this talk, we focus our attention on perfect numbers, which are arguably the most interesting of the three kinds. The existence of these densities, as well as the fact that the perfect numbers have density 0, is a consequence of a general theorem proved by Davenport in early 1930s. (See book!) Calculating the densities of the abundant (or deficient) numbers is another story, for another time. 5 of 30 The existence of these densities, as well as the fact that the perfect numbers have density 0, is a consequence of a general theorem proved by Davenport in early 1930s. (See book!) Calculating the densities of the abundant (or deficient) numbers is another story, for another time. In this talk, we focus our attention on perfect numbers, which are arguably the most interesting of the three kinds. 5 of 30 Main theorems Theorem (Euclid { Euler) If 2n − 1 is prime, then N := 2n−1(2n − 1) is perfect. Conversely, if N is an even perfect number, then N has this form. But what about odd perfect numbers? 6 of 30 Is there a simple formula for odd perfect numbers, like for even perfect numbers? Probably not. Theorem (Dickson, 1913) For each positive integer k, there are only finitely many odd perfect numbers with ≤ k distinct prime factors. 7 of 30 The rest of this talk is devoted to giving more or less complete proofs for these three theorems. Dickson's theorem does not rule out the existence of odd perfect numbers. It is even compatible with their being infinitely many of them. But we can at least prove that it cannot be \too big" of an infinity. Theorem (Hornfeck) p There are ≤ x odd perfect numbers ≤ x, for all x ≥ 1. 8 of 30 Dickson's theorem does not rule out the existence of odd perfect numbers. It is even compatible with their being infinitely many of them. But we can at least prove that it cannot be \too big" of an infinity. Theorem (Hornfeck) p There are ≤ x odd perfect numbers ≤ x, for all x ≥ 1. The rest of this talk is devoted to giving more or less complete proofs for these three theorems. 8 of 30 = 2n · (1 + 2 + 22 + ··· + 2n−1) = 2n · (2n − 1) = 2N: So N is perfect. Euclid{Euler Observe: N perfect () σ(N) = 2N. Euclid's half of the theorem is \easy", given modern notations and notions. If 2n − 1 is prime, and N := (2n − 1)2n−1; then σ(N) = σ(2n − 1)σ(2n−1) 9 of 30 = 2n · (2n − 1) = 2N: So N is perfect. Euclid{Euler Observe: N perfect () σ(N) = 2N. Euclid's half of the theorem is \easy", given modern notations and notions. If 2n − 1 is prime, and N := (2n − 1)2n−1; then σ(N) = σ(2n − 1)σ(2n−1) = 2n · (1 + 2 + 22 + ··· + 2n−1) 9 of 30 Euclid{Euler Observe: N perfect () σ(N) = 2N. Euclid's half of the theorem is \easy", given modern notations and notions. If 2n − 1 is prime, and N := (2n − 1)2n−1; then σ(N) = σ(2n − 1)σ(2n−1) = 2n · (1 + 2 + 22 + ··· + 2n−1) = 2n · (2n − 1) = 2N: So N is perfect. 9 of 30 σ(N) Define h(N) = N . So N is perfect () h(N) = 2. Lemma The function h(N) is multiplicative. It is also \multiplicatively strictly increasing": If a j b, then h(a) ≤ h(b), with equality only if a = b. Proof. Multiplicativity: Inherited from σ(N) and N. Multiplicatively strictly increasing: Follows from the identity X 1 h(N) = : d djN 10 of 30 By the last proposition, σ(N) σ(m) m + 1 ≥ ≥ : N m m Equality holds throughout. We need σ(m) = m + 1, so m is prime. And we need h(N) = h(m), so N = m. Lemma m+1 Let m be an integer > 1. If h(N) = m , then m is prime, and N = m. Proof. σ(N) m+1 Write N = m . RHS is in lowest terms. Thus, m j N. 11 of 30 Equality holds throughout. We need σ(m) = m + 1, so m is prime. And we need h(N) = h(m), so N = m. Lemma m+1 Let m be an integer > 1. If h(N) = m , then m is prime, and N = m. Proof. σ(N) m+1 Write N = m . RHS is in lowest terms. Thus, m j N. By the last proposition, σ(N) σ(m) m + 1 ≥ ≥ : N m m 11 of 30 Lemma m+1 Let m be an integer > 1. If h(N) = m , then m is prime, and N = m. Proof. σ(N) m+1 Write N = m . RHS is in lowest terms. Thus, m j N. By the last proposition, σ(N) σ(m) m + 1 ≥ ≥ : N m m Equality holds throughout. We need σ(m) = m + 1, so m is prime. And we need h(N) = h(m), so N = m. 11 of 30 σ(Q) 2k+1 m+1 Rearrange: h(Q) = Q = 2k+1−1 = m where m = 2k+1 − 1: By lemma: m = 2k+1 − 1 is prime, and Q = m = 2k+1 − 1. So N = 2k (2k+1 − 1). Proof that if N is even perfect, N has Euler's form. Write N = 2k Q, where Q is odd. Starting from σ(N) = 2N, get 2k+1Q = σ(2k )σ(Q) = (2k+1 − 1)σ(Q): 12 of 30 Proof that if N is even perfect, N has Euler's form. Write N = 2k Q, where Q is odd. Starting from σ(N) = 2N, get 2k+1Q = σ(2k )σ(Q) = (2k+1 − 1)σ(Q): σ(Q) 2k+1 m+1 Rearrange: h(Q) = Q = 2k+1−1 = m where m = 2k+1 − 1: By lemma: m = 2k+1 − 1 is prime, and Q = m = 2k+1 − 1. So N = 2k (2k+1 − 1). 12 of 30 We will give a supernatural proof of this theorem, due (essentially) to HN Shapiro. Dickson's finiteness theorem Theorem (Dickson, 1913) For each positive integer k, there are only finitely many odd perfect numbers with ≤ k distinct prime factors. 13 of 30 Dickson's finiteness theorem Theorem (Dickson, 1913) For each positive integer k, there are only finitely many odd perfect numbers with ≤ k distinct prime factors. We will give a supernatural proof of this theorem, due (essentially) to HN Shapiro.
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