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Solving the Odd Perfect Number Problem: Some Old and I I Αi ≤ 3 Pi New Approaches”, M

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Solving the Odd Perfect Number Problem: Some Old and I I Αi ≤ 3 Pi New Approaches”, M Solving the Odd Perfect Number Problem: Some New Approaches Jose Arnaldo B. Dris Master of Science in Mathematics Graduate (2008) De La Salle University, 2401 Taft Avenue, 1004 Manila, Philippines [email protected] Abstract—A conjecture predicting an injective and surjec- k 2 σ(p ) σ(m ) tive mapping X = ,Y = between OPNs pk m2 σ(m2) k 2 k µ = N = p m (with Euler factor p ) and rational points on the 1 m2 hyperbolic arc XY = 2 with 1 <X< 1.25 < 1.6 <Y < 2 and 2.85 < X + Y < 3, is disproved. We will show that if an σ(m2) k 2 2 µ2 = OPN N has the form above, then p < 3 m . We then give a pk somewhat weaker corollary to this last result (m2 −pk ≥ 8) and give possible improvements along these lines. We will also attempt Proof: We refer the interested reader to the author’s to prove a conjectured improvement to pk < m by observing master’s thesis completed in August of 2008 [2]. k k σ(p ) σ(p ) σ(m) 57 11 that =6 1 and =6 k in all cases. Lastly, we m m p Lemma II.2. 20 <ρ1 + µ1 < 3 < 3 ρ2 + µ2 r ≤ αi also prove the following generalization: If N = Y pi is the Proof: Again, the interested reader is referred to [2], i=1 where it is shown that ”any further improvement to the lower α N i 2 bound of 57 for ρ µ would be equivalent to showing canonical factorization of an OPN N, then σ(pi ) ≤ α 20 1 + 1 3 pi i 2 2−r 1 2 r−1 that there are no OPNs of the form 5m which would be for all i. This gives rise to the inequality N ≤ ( 3 ) 3 , a very major result. Likewise, any further improvement on which is true for all r, where r = ω(N) is the number of distinct prime factors of N. the upper bound of 3 would have similar implications for all Index Terms—Odd perfect number, Euler factor, inequalities, arbitrarily large primes and thus would be a very major result.” OPN components, non-injective and non-surjective mapping (These assertions, which are originally Joshua Zelinsky’s, were readily verified by the author using Mathematica.) I. INTRODUCTION σ(pk) σ(m) Lemma II.3. Define ρ3 = and µ3 = . Then A perfect number is a positive integer N such that the m pk ρ µ , and the following statements hold: sum of all the positive divisors of N equals 2N, denoted by 3 = 3 6 k σ(N)=2N. The question of the existence of odd perfect • If ρ3 < 1, then p <m. numbers (OPNs) is one of the longest unsolved problems of • Suppose that 1 <ρ3. 4 k √ number theory. This paper gives some new and nontraditional – If ρ3 <µ3, then 5 m<p < 2m. k attempts and approaches to solving the Odd Perfect Number – If µ3 <ρ3, then m<p . (OPN) Problem. arXiv:1206.1548v2 [math.NT] 15 Oct 2013 Proof: The interested reader is again referred to [2]. The σ(pk) σ(m) Hereinafter, we shall let N = pkm2 denote an OPN with crucial part of the argument is in showing that < . pk m Euler (prime-power) factor pk (with p k 1 (mod 4) and gcd(p,m)=1), assuming at least one≡ such≡ number exists. III. MAIN RESULTS II. SOME PREPARATORY LEMMAS First, we prove that a conjectured one-to-one correspon- The following lemmas will be very useful later on: dence is actually both not surjective and not injective. Conjecture III.1. For each N = pkm2 an OPN with Lemma II.1. 300 0 <ρ 2 < 1 <ρ < 5 < 8 <µ < 2 < 3 µ where: N > 10 , there corresponds exactly one ordered pair of 2 ≤ 3 1 4 5 1 ≤ 2 σ(pk) σ(m2) rational numbers , lying in the region σ(pk) pk m2 k 2 ρ1 = k σ(p ) 5 8 σ(m ) p 1 < < , < < 2, and pk 4 5 m2 k σ(pk) 57 σ(p ) σ(m2) ρ = < + < 3, and vice-versa. 2 m2 20 pk m2 8 Proof: (Note that this is actually a refutation of the a<b x equals Cx + o(x) where C .” ([3], [1]) σ(x) σ(pk) ≤ ≥ 147 conjecture.) First, we note that the equation = (Note that C here is the same as the (natural) density of x pk friendly integers.) Given Erdos’ result then, this means that has the sole solution x = pk since prime powers are 8 k 2 eventually, as m , there will be at least 10150 σ(p ) σ(m ) 2 147 solitary. Next, let X = and Y = . It is → ∞ 2 2 pk m2 σ(m1 ) σ(m2 ) solutions m ,m to , a number which is straightforward to observe that, since the abundancy index ( 1 2) 2 = 2 m1 m2 is an arithmetic function, then for each N = pkm2 an OPN obviously greater than 1. This finding, though nonconstructive, (with N > 10300), there corresponds exactly one ordered still proves that the mapping defined in the backward direction pair of rational numbers (X, Y ) lying in the hyperbolic arc of the conjecture is not injective. XY = 2 bounded as follows: 1 <X< 1.25, 1.6 <Y < 2, Next, we show how the components pk and m2 of an OPN and 2.85 <X + Y < 3. (Note that these bounds are the same are related. ones obtained in Lemma II.1 and Lemma II.2.) 2 Theorem III.1. pk < m2 3 We now disprove the backward direction of the conjecture. k Proof: This theorem follows from the inequalities ρ1 > 1 σ(p ) 2 We do this by showing that the mapping X = k and and ρ . p 2 ≤ 3 σ(m2) A somewhat weaker result than Theorem III.1 is the fol- Y = is neither surjective nor injective in the specified m2 lowing: region. (X, Y ) is not surjective. We prove this claim by producing a Theorem III.2. m2 pk 8 − ≥ rational point (X0, Y0) lying in the specified region, and which pk σ(pq) Proof: From Theorem III.1, we know that m2 pk > . satisfies X0 = where p and q are primes satisfying − 2 pq So in particular, we are sure that m2 pk > 0. But m odd 5 <p<q. Notice that implies that m2 1 (mod 4), and we− also know that pk 1 (p + 1)(q + 1) 1 1 8 12 96 (mod 4). Thus, ≡m2 pk 0 (mod 4), which is equivalent≡ 1 <X0 = = 1+ 1+ = pq p q ≤ 7 11 77 to saying that 4 (m2− pk≡). Since m2 pk > 0, then this | − − 96 2 implies that m2 pk 4. Suppose m2 pk =4. where < 1.2468 < 1.25. Now, by setting Y0 = , the − ≥ − 77 X0 other two inequalities for Y0 and X0 +Y0 would follow. Thus, Then pk = m2 4 = (m + 2)(m 2). Hence, we have − − − we now have a rational point (X0, Y0) in the specified region, the simultaneous equations pk x = m +2 and px = m 2 k σ(p ) where k 2x +1. Consequently, we have pk−x px =− 4, and which, by a brief inspection, satisfies X0 = for 6 pk which implies≥ that px(pk−2x 1) = 4 where k −2x is odd. all primes p and positive integers k (since prime powers are Since (p 1) (py 1) y 1−, this last equation− implies that solitary). Consequently, the mapping defined in the backward (p 1) 4−. Likewise,| − the∀ congruence≥ p 1 (mod 4) implies direction of the conjecture is not surjective. that− 4 (|p 1). These two divisibility≡ relations imply that p 1=4| ,− or p =5. Hence, 5x(5k−2x 1)=4. Since 5 does Remark. Since the mapping is not onto, there are rational not− divide 4, x = 0 and thus 5k 1=4− , which means that points in the specified region which do not correspond to any k =1. − OPN. Therefore, p = 5, k = 1, x = 0. Consequently, (X, Y ) is not injective. It suffices to construct two distinct pk−x =51−0 =5= m +2 and px =50 =1= m 2. Either OPNs N = p k1 m 2 and N = p k2 m 2 that correspond to 1 1 1 2 2 2 way, we have m =3. − the same rational point (X, Y ). Since it cannot be the case k1 k2 2 2 that p1 = p2 , m1 = m2 , we consider the scenario k 2 k1 k6 2 2 2 All of these computations imply that N = p m = p1 = p2 , m1 = m2 . Thus, we want to produce a pair 1 2 6 σ(m 2) σ(m 2) 5 3 = 45 is an OPN. But this contradicts the fact that m ,m satisfying 1 2 . (A computer check σ(N) 26 ( 1 2) 2 = 2 m1 m2 = < 2 (i.e., N = 45 is deficient). by a foreign professor using Maple produced no examples for N 15 this equation in the range 1 m1 < m2 300000. But Thus, m2 pk 8. ≤ ≤ − ≥ then again, in pure mathematics, absence of evidence is not Lastly, we prove the following generalization to the evidence of absence.) Now, from the inequalities pk <m2 and 2 2 N inequality σ(pk) m2 = . N = pkm2 > 10300, we get m2 > 10150. A nonconstructive ≤ 3 3 pk σ(m 2) σ(m 2) approach to finding a solution to 1 2 would 2 = 2 ω(N) m1 m2 150 2 2 αi then be to consider 10 < m1 < m2 and Erdos’ result Theorem III.3.
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