On a Conjecture of Dris Regarding Odd Perfect Numbers

Transcription

1 O a Cojecture of Dris Regardig Odd Perfect Numbers Jose Araldo B. Dris Departmet of Mathematics, Far Easter Uiversity Maila, Philippies jadris@feu.edu.ph,josearaldobdris@gmail.com arxiv: v9 [math.nt] 4 Apr 2017 Abstract: Dris cojectured i [4] that the ieuality k < always holds, ifn = k 2 is a odd perfect umber give i Euleria form. I this ote, we show that either of the two coditios < k or / < σ()/ holds. This is achieved by first provig that / σ()/ k, where σ(x) is the sum of the divisors of x. Usig this aalysis, we show that the coditio < < k holds i four out of a total of six cases. By utilizig a separate aalysis, we show that the coditio < < 3 holds i four out of a total of five cases. We coclude with some ope problems related to Sorli s cojecture thatk = 1. Keywords: Odd perfect umber, abudacy idex, Sorli s cojecture. AMS Classificatio: 11A25. 1 Itroductio Let N = k 2 be a odd perfect umber give i Euleria form (i.e., is prime with k 1 (mod 4) adgcd(,) = 1). Therefore,. It follows that either < or <. Dris [3] proved that < implies Sorli s cojecture that k = 1 [6]. By the cotrapositive, k > 1 implies that <. Acuaah ad Koyagi [1] showed that all the prime factors r of N satisfy r < (3N) 1/3. I particular, ifk = 1, the < (3N) 1/3 = 3 < 3N = 3 2 = < 3. Therefore, regardless of the status of Sorli s cojecture, we kow that < 3 must be true. Notice that, if Acuaah ad Koyagi s estimate for the Euler prime (uder the assumptio k = 1) could be improved to < (2N) 1/3, the it would follow that < 2, ad this would hold ucoditioally. 1

2 Let σ(x) be the sum of the divisors of the positive itegerx. Let be the abudacy idex ofx. I(x) = σ(x)/x 2 Mai Results Here, we examie this problem. Determie the correct orderig for the followig uatities: Recall the followig results:, σ(), σ(k ) k, σ() ad 1 < I() = σ() I( k ) = σ(k ) σ() k < 2, < k = {k = 1 < }. I geeral, sice1 < k is deficiet (beig a proper factor of the perfect umbern = k 2 ), the we have ad I a similar vei, σ( k ) σ() k. σ(). Note that the followig implicatios are true: We wat to show that Suppose to the cotrary that < σ() σ() < Sice gcd(,) = 1 ad is prime, we have: = < 2, = <. σ() k. = σ() k. = σ() k N. 2

3 This meas that we get 1 = σ() k. But sice = +1is eve whileis odd, we the have: From the ieuality from which we obtai But the we fially have which is a cotradictio. Coseuetly, we obtai: 2 σ() σ() We ow cosider two separate cases: Case 1: 2 = σ() k. 2 σ() σ() < 1. = I() < 2 < 1 < 2 = σ() k σ(), σ() k. < σ() k Sicek 1, this implies that < σ() σ(). k Coseuetly, uder Case 1, we have the coditio: < σ(). From a previous remark, we kow that this implies < 2. Case 2: σ() < k Agai, sicek 1, this implies that σ() k < σ(k ). This implies that, uder Case 2, we have the coditio: σ() k < σ(k ). 3

4 But recall that we have the followig ieuality [3]: σ( k ) k = I( k ) < 3 2 < I() = σ(). Together, the last two ieualities imply that: < k. This implies that the bicoditioal k = 1 < is true. Now, suppose that σ() = σ(k ). Agai, sicegcd(,) = 1 ad is prime, we have σ() = σ(k ) N. It follows that But 1 σ() = σ(k ). σ( k ) k +1 2 (mod 4), sicek 1 (mod 4), whileis odd. Therefore, We claim that Assume that σ() 2 σ() = σ(k ) σ() Sice gcd(,) = 1 ad is prime, we have = σ(k ). = k > 1. = σ(k ). Coseuetly, we obtai (as before) σ() = σ(k ) N. 2 σ() = σ(k ) (siceσ( k ) ). From the ieuality 2 σ() 4

5 we get 2 σ() = I() < 2 from which we obtai < 1. This last ieuality implies that <. Now, from the ieualityi( k ) < 3 2 < I() (see [3]), we get: Note that from the followig ieuality: σ( k ) σ() < k. we get 2 σ(k ) σ() < 2 σ( k ) from which we obtai 1 < σ(k ) σ() < k. Thereupo, we get < k, which as we ve oted before, implies that the bicoditioal k = 1 < is true. But we have already obtaied <. Therefore, we kow that k > 1. A shorter way to prove the implicatio σ() = σ(k ) = k > 1 would be to ote that, i geeral, the coditio σ() is true. I particular, σ( k ) = σ() from which it follows that σ( k ). It follows from the last ieuatio thatk 1. Sice we kow that k 1, it follows thatk > 1. We ow summarize the results we have obtaied so far: σ() k. 5

6 The followig ieuatios are trivial: σ( k ) σ() k ad Also, ote that σ(). σ(k ) σ() σ(). k 3 Syopsis We ow list all the possible orderigs for: {, σ(), σ(k ) k, σ() } A: B: C: D: E: F: σ(k ) < σ() σ() k < σ() < σ(k ) k < σ() < σ() k σ() < σ(k ) σ() < k < σ() σ(k ) σ() < k < σ(k ) < σ() σ() = σ() < k = σ(k ) Note that, uder cases B, C, D, E ad F, we have the ieuality σ() < σ(k ) k which implies that < k. Furthermore, ote that, uder cases A, B, C, D ad E, we have the coditio < σ(). 6

7 Lastly, otice that, uder cases B, C, D ad E, we actually have the ieualities < < k sicek > 1 i each of these cases. For aother case-to-case aalysis, we first prove the followig claim: Claim: Either/ < 2 orσ()/ < 2. Sice / σ()/ ad I() = (/) (σ()/) < 2, by symmetry, it suffices to cosider the case: / < σ()/. Suppose to the cotrary that 2 < / < σ()/. This leads to 2 < (/) (σ()/) = I() < 2, which is a cotradictio. (The case σ()/ < / is treated similarly.) This establishes the claim. Now, there are four cases to cosider: Case I: / < 2 < σ()/ There are two further subcases here: Sub-Case I-1: / < 1 < 2 < σ()/ Uder this subcase, <. Sub-Case I-2: 1 < / < 2 < σ()/ < 2 Uder this subcase, < < 2. Case II: σ()/ < 2 < / < Uder this case, < < 3. Case III: 1 < / < σ()/ < 2 Uder this case, < < 2. Case IV:1 < σ()/ < / < 2 7

8 Uder this case, < < 2. It therefore remais to improve the upper boud for uder Case II. 4 Some Additioal Improvemets Let N = k 2 be a odd perfect umber give i Euleria form. We wat to show that Suppose to the cotrary that σ() σ() σ(k ). = σ(k ). Note that this is oly possible i Cases B ad D above. Uder these cases, we have < < k. I particular,k > 1. This implies that σ() = σ( k ) which further meas that σ() ad σ( k ) (sicegcd(,) = 1). Coseuetly, σ() = σ(k ) = x N. If the immediately precedig euatio is true, the ( x = σ( k ) ) ) 1 (σ() = x) = σ( x σ(k ) = x. There are o solutios to this last euatio for k > 1 ad k 1 (mod 4), for particular values of x. We haveσ(x) X, where the ieuality is strict forx > 1; hece ( ) 1 σ x σ(k ) 1 x σ(k ) 1 x k, so we would eed k 1 x 2. But fork > 1 ad k 1 (mod 4), we havek 5, so that k , sice is prime with 1 (mod 4). Hece the origial system of euatios will oly have a solutio whe x = x 25. Coseuetly, we obtai 25 σ(k ) 8 = σ().

9 Multiplyig both sides of the ieuality ad euatio by σ()/ k, we get 25 σ() σ() σ() = σ(k ) k k σ() = I( k ) < 2. k Multiplyig both sides of the ieuality ad euatio by/, we get ad 25 σ(k ) = σ() = I() < 2. Therefore, we haveσ()/ k < 2/25 ad / < 2/25, from which it follows that σ() < 2 k 25 < 25 σ(k ) < 2 25 (This sectio is curretly a work i progress.) σ() < Ope Problems I the paper [5], a heuristic motivatig the pursuit of a proof for the cojectures of Sorli (k = 1) ad Dris ( k < ) is preseted. I particular, it seems fruitful to try to establish the followig predictio: Cojecture A: k = 1 = < Sicek > 1 = < is true, the truth of this cojecture would imply <. Note that the truth of Cojecture A together with Sorli s cojecture will imply Dris s cojecture. I a recet preprit, Brow [2] claims a proof for CojectureA. Brow also shows that k < holds i may cases. Ackowledgemets The authors would like to thak the aoymous referees for their valuable feedback ad suggestios which helped i improvig the presetatio ad style of this mauscript. Refereces [1] Acuaah, P., S. Koyagi, O prime factors of odd perfect umbers, It. J. Number Theory, 08 (2012), 1537, doi: [2] Brow, P. A., A partial proof of a cojecture of Dris, preprit, 9

10 [3] Dris, J. A. B., The abudacy idex of divisors of odd perfect umbers, J. Iteger Se., 15 (Sep. 2012), Article , ISSN [4] Dris, J. A. B., Solvig the odd perfect umber problem: some old ad ew approaches, M. Sc. thesis, De La Salle Uiversity, Maila, Philippies, 2008, [5] Dris, J. A. B., Euclid-Euler heuristics for (odd) perfect umbers, preprit, [6] Sorli, R. M., Algorithms i the study of multiperfect ad odd perfect umbers, Ph. D. Thesis, Uiversity of Techology, Sydey, 2003, 10