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perfect number∗ Wkbj79† 2013-03-21 12:24:34 An positive integer n is called perfect if it is the sum of all positive divisors of n less than n itself. It is not known if there are any odd perfect numbers, but all even perfect numbers have been classified according to the following lemma: Lemma 1. An even number is perfect if and only if it equals 2k−1 (2k − 1) for some integer k > 1 and 2k − 1 is prime. Proof. Let σ denote the sum of divisors function. Recall that this function is multiplicative. Necessity: Let p = 2k − 1 be prime and n = 2k−1 p. We have that σ(n) = σ(2k−1 p) = σ(2k−1 )σ(p) = (2k − 1)(p + 1) = (2k − 1)2k = 2n, which shows that n is perfect. Sufficiency: Assume n is an even perfect number. Write n = 2k−1 m for some odd m and some k > 1. Then we have gcd(2k−1 , m) = 1. Thus, σ(n) = σ(2k−1 m) = σ(2k−1 )σ(m) = (2k − 1)σ(m). Since n is perfect, σ(n) = 2n by definition. Therefore, σ(n) = 2n = 2k m. Piecing together the two formulas for σ(n) yields 2k m = (2k − 1)σ(m). ∗ hPerfectNumberi created: h2013-03-21i by: hWkbj79i version: h30206i Privacy setting: h1i hDefinitioni h11A05i h20D99i h20D06i h18-00i † This text is available under the Creative Commons Attribution/Share-Alike License 3.0. You can reuse this document or portions thereof only if you do so under terms that are compatible with the CC-BY-SA license. 1 Thus, (2k − 1) | 2k m, which forces (2k − 1) | m. Write m = (2k − 1)M . Note that 1 ≤ M < m. From above, we have: 2k m = (2k − 1)σ(m) 2 (2 − 1)M = (2k − 1)σ(m) k k 2k M = σ(m) Since m | m by definition of divides and M | m by assumption, we have 2k M = σ(m) ≥ m + M = 2k M, which forces σ(m) = m + M . Therefore, m has only two positive divisors, m and M . Hence, m must be prime, M = 1, and m = (2k − 1)M = 2k − 1, from which the result follows. The lemma can be used to produce examples of (even) perfect numbers: • If k = 2, then 2k − 1 = 22 − 1 = 3, which is prime. According to the lemma, 2k−1 (2k − 1) = 22−1 · 3 = 6 is perfect. Indeed, 1 + 2 + 3 = 6. • If k = 3, then 2k −1 = 23 −1 = 7, which is prime. According to the lemma, 2k−1 (2k − 1) = 23−1 · 7 = 28 is perfect. Indeed, 1 + 2 + 4 + 7 + 14 = 28. • If k = 5, then 2k − 1 = 25 − 1 = 31, which is prime. According to the lemma, 2k−1 (2k − 1) = 25−1 · 31 = 496 is perfect. Indeed, 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 = 496. Note that k = 4 yields that 2k − 1 = 24 − 1 = 15, which is not prime. The sequence of known perfect numbers appears in the OEIS as sequence A000396. 2