The Abundancy index of divisors of odd perfect numbers Part III

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1 Notes on Number Theory Discrete Mathematics Print ISSN , Online ISSN Vol 23, 2017, No 3, The Abundancy index of divisors of odd perfect numbers Part III Jose Arnaldo B Dris Department of Mathematics Physics, Far Eastern University Nicanor Reyes Street, Sampaloc, Manila, Philippines jadris@feueduph, josearnaldobdris@gmailcom Received: 9 July 2017 Accepted: 16 October 2017 Abstract: Dris gave numerical bounds for the sum of the abundancy indices of k n 2, where k n 2 is an odd perfect number, in his master s thesis In this note, we show that improving the limits for this sum is euivalent to obtaining nontrivial bounds for the Euler prime Keywords: Odd perfect number, Abundancy index, Euler prime AMS Classification: 11A25 1 Introduction Let σn denote the sum of the divisors of the natural number N If σn = 2N N is odd, then N is called an odd perfect number Denote the abundancy index I of N as IN = σn/n Euler proved that every odd perfect number N has to have the form N = k n 2 where is prime called the Euler prime of N satisfying k 1 mod 4 gcd, n = 1 Dris proved the following lemmas in [2]: Lemma 11 If N = k n 2 is an odd perfect number with Euler prime, then we have the bounds L < I k + In 2 U, where L = U = = 3 2 =

2 Lemma 12 If N = k n 2 is an odd perfect number with Euler prime, then we have the numerical bounds 20 < Ik + In 2 < 3, with the further result that they are best-possible In this note, we will prove the following results: Theorem 11 If N = k n 2 is an odd perfect number with Euler prime, then 97 if only if I k + In Theorem 12 If N = k n 2 is an odd perfect number with Euler prime, then > 5 if only if I k + In 2 > All of the proofs given in this note are elementary 2 The proofs of Lemma 11 Lemma 12 Let N = k n 2 be an odd perfect number with Euler prime First, since N is perfect I is weakly multiplicative, we have the following euation ineualities: IN = I k n 2 = I k In 2 = 2 In 2 = 2 I k + 1 = I I k = σk k 2 1 = k k = k+1 1 k 1 < < In 2 = 2 I k k+1 k 1 = 1 Notice that I k < / 1 5/4 since prime with 1 mod 4 implies that 5, so that we have In 2 = 2/I k > 8/5 > 5/4 > I k Here is another way to prove I k < In 2 So from before, we have I k < 2 1 < In 2 Since is an integer, 1 2 1, 1 because otherwise 2 = 2 1 which implies that contradicting / 1 is rational 1 = 2 54

3 Now, we want to prove that I k < In 2 It suffices to show that < Suppose to the contrary that This implies that 2 1 < < 2 which further means that < 2 or < 0 Since 5, = = 7, which is a contradiction This then gives an alternative albeit longer proof for I k < In 2 To summarize what we have so far: 1 < + 1 I k < 1 < 5 4 < 8 5 < 2 1 < In < 2 Note that, when k = 1, we have the slightly stronger bounds: I k = I = + 1 = < 5 3 In2 = 2 I k = 2 I We want to show first that L < I k + In 2 U, where Consider the product L = U = I k In 2 1 = 3 2 = This product is negative since I k < / 1 < In 2 Therefore, = I k In 2 + < I k + In 2 from which we obtain Finally, we have L = < Ik + In = = 3 2, 55

4 as desired Next, consider the product I k + 1 In This product is nonnegative since + 1/ I k < In 2 Conseuently, we have = I k In I k + In 2 from which we obtain Finally, we have U = I k + In 2 = = , as desired This proves Lemma 11 We follow the discussion in [1] for our proof of Lemma 12 here By using a method similar to that used to prove Lemma 11, we can show that 20 < Ik + In 2 < 3 We now prove that these bounds are best-possible It suffices to get the minimum value for L the maximum value for U in the interval [5,, or if either one cannot be obtained, the greatest lower bound for L the least upper bound for U for the same interval would likewise be useful for our purposes here From basic calculus, we get the first derivatives of L U determine their signs in the interval [5, : L = > U = > 0 which means that L, U are increasing functions of on the interval [5, Hence, L attains its minimum value on that interval at L5 = /20, while U has no maximum value on the same interval, but has a least upper bound of This confirms our earlier findings that lim U = 3 20 < Ik + In 2 < 3 with the further result that such bounds are best-possible This proves Lemma 12 Note that when k = 1 we have the slightly stronger bound 43/15 I k + In 2 56

5 3 The proofs of Theorem 11 Theorem 12 Let N = k n 2 be an odd perfect number with Euler prime First, we want to show that 97 if only if I k + In Suppose that L < I k + In This implies that 100 L = 3 2 < Since 5, we have < 2 2 < This implies that < 0 < from which we obtain 97 since is prime 1 mod 4 Does the converse hold? That is, if N = k n 2 is an odd perfect number with Euler prime, does 97 imply that I k + In 2 299? 100 To this end, assume that 97, suppose to the contrary that This means that so that we obtain Since 5, we get This implies that < Ik + In 2 U U = > , > > > 0 > contradicting 97 This proves Theorem 11 We now show that > 5 if only if I k + In 2 > 43 15

6 Suppose that U I k + In 2 > 43/15 This means that U = > 43 15, which implies that Since 5, we have 2 15 > > > > 0, from which we obtain > 5 Now assume that > 5 We want to prove that I k + In 2 > 43/15 Suppose to the contrary that I k + In 2 43/15 Since is prime 1 mod 4, this implies that 13 This implies that we have the bounds I k < < < In2, from which we obtain I k contradicting 2 + This proves Theorem = I k In In < < Ik + In 2 < 0 I k + In 2 > I k + In = I k + In 2 4 Generalization of Theorem 11 It is possible to prove the following generalization of Theorem 11: Theorem 41 Let N = k n 2 be an odd perfect number with Euler prime, let ɛ satisfy 0 < ɛ < Then I k + In 2 3 ɛ if only if < ɛ ɛ 2 6ɛ + 1 2ɛ 58

7 Here, we only prove one direction of Theorem 41 We show that if I k + In 2 3 ɛ, then < ɛ ɛ 2 6ɛ + 1 2ɛ Since then we have from which we get Finally we obtain 3 2 = L < Ik + In 2 3 ɛ, 0 < ɛ < 2 ɛ 2 ɛ < 2 ɛ 2 ɛ < 0 < ɛ ɛ 2 6ɛ + 1 2ɛ Notice how Theorem 11 becomes a special case of Theorem 41 for ɛ = Concluding remarks The results in this paper were motivated by the remarks of Joshua Zelinsky on a post of the author in the Math Forum at Drexel in 2005 Showing the truth of the improvements outlined in this paper remains an open problem Observe that, from Lemma 11, when Lx Ux are viewed as functions on the domain D = R \ { 1, 0, 1}, then Lx + 1 = Ux U2 = U3 = L3 = 17 6 < 20 Acknowledgments The author would like to thank Joshua Zelinsky for providing helpful ideas, the anonymous referees for feedback towards improving the overall style presentation of this manuscript References [1] Dris, J A B 2008 Solving the Odd Perfect Number Problem: Some Old New Approaches, M Sc Math thesis, De La Salle University, Manila, Philippines [2] Dris, J A B 2012 The abundancy index of divisors of odd perfect numbers, J Integer Se, 15, Article