More on the Nonexistence of Odd Perfect Numbers of a Certain Form
aa r X i v : . [ m a t h . N T ] D ec MORE ON THE NONEXISTENCE OF ODD PERFECTNUMBERS OF CERTAIN FORMS
PATRICK A. BROWN
Abstract.
Euler showed that if an odd perfect number exists, it must be ofthe form N = p α q β . . . q β k k , where p, q , . . . , q k are distinct odd primes, α , β , . . . , β k ∈ N , with p ≡ α ≡ N is not perfect, if 3 | N or 7 | N and each β i ≡ | N or 7 | N and showthat N is not perfect, simply, if each β i ≡ Introduction
We define σ ( N ) to be the sum of the positive divisors of N . We say N isperfect when σ ( N ) = 2 N . For example, σ (6) = 1 + 2 + 3 + 6 = 12 and σ (28) =1+2+4+7+14+28 = 56, making both 6 and 28 perfect. It is still an open questionas to whether or not there exists an infinite number of even perfect numbers or evena single example of an odd perfect number. Nevertheless, we let O be the set of allodd perfect numbers.Euler showed that if an odd perfect number exists, it must be of the form: N = p α q β . . . q β k k (1.1)where p, q , . . . , q k are distinct odd primes, α , β , . . . , β k ∈ N = { , , , . . . } , with p ≡ α ≡ p is often referred to as the special prime of N , and p α as the Eulerian component of N . Throughout this paper, when we say N ∈ O ,unless otherwise stated, we assume N has the form given in 1.1.Assuming β = · · · = β k = β , it has been shown for all fixed β ≤
14, except β = 9, that N cannot be odd perfect. Additional results include infinite congruenceclasses for β . For example, McDaniel proved in [7] that N is not perfect if each β i ≡ N is not perfect if 3 | N andeach β i ≡ β i ≡ Theorem 1.
Suppose N ∈ O and each β i ≡ , then gcd ( N,
21) = 1 and p ≡ . Using this result, and applying a different method of proof, we show
Date : November 2015.2010
Mathematics Subject Classification.
Key words and phrases.
Odd Perfect Numbers.Thanks go to Kevin G. Hare for endorsing me so that this paper could appear on ArXive.org.His endorsement goes only so far as to admit that this work might look like real mathematics.Any mistakes are the sole responsibility of the author.
Theorem 2.
Suppose N is as described in (1.1) and each β i ≡ , then N is not perfect. Theorem 2 subsumes a number of theorems like Theorem 1.2.
Preliminary Results
We start with some elementary properties of σ :(1) if q is prime, k ∈ N , then σ ( q k ) = 1 + q + q + . . . + q k .(2) σ is multiplicative. That is, if gcd( m, n ) = 1, then σ ( m ∗ n ) = σ ( m ) ∗ σ ( n ).For N ∈ O : σ ( N ) = σ ( p α ) σ ( q β ) . . . σ ( q β k k ) = 2 N (2.1)Which makes clear that any prime r dividing σ ( q β i i ) for some i , must also divide N . Additionally, α odd implies that for α = 2 a + 1 σ ( p α ) = 1 + p + · · · + p k + p a +1 = (1 + p ) + . . . + (1 + p ) p a = (1 + p )(1 + p + · · · + p a ) (2.2)Thus, ( p + 1) | σ ( p α ) and so any odd prime dividing ( p + 1) also divides N .We will make use of the following lemmas. The proof of the first can be foundin [8] and the second in [6]. Lemma 1. σ ( s f ) | σ ( s f +( f +1) m ) for all primes s and all m, f ∈ N . Let N ∈ O be written as in (1.1). Define γ i = 2 β i + 1 for 1 ≤ i ≤ k . Lemma 2.
Suppose d | γ i for each i, then d | N . Remark 1.
Observe the assumption in Theorem 2 that each β i ≡ isequivalent to assuming d = 5 in Lemma 2. Further note, with f = 4 and 2 β i ≡ m + 4 for some m ∈ N , Lemma1 implies σ ( s ) | σ ( s m +4 ) = σ ( s β i ) . (2.3)Though Lemma 1 is often discussed in the context of cyclotomic polynomials, wewon’t need results any more powerful than the above. These two ideas lie at thefoundation of every result similar to Theorem 1 and Theorem 2.3. σ -Chains Under the hypotheses of Theorem 2, suppose r is a prime that divides σ ( q i ).It follows immediately that r | σ ( q i ) | σ ( q β i i ) | N ⇒ r | N. (3.1)Thus, if r r cannot be the special prime in the primefactorization of N , so there must be some q i where r = q i . Remark 2.
Applying Theorem 1, we can reach this same conclusion so long as r . Additionally, since we know N , (2.2) implies r . ONEXISTENCE OF OPN 3
For the rest of this paper, we assume our special prime p satisfies these two condi-tions.Applying the same reasoning to a prime, say r , such that r | σ ( r ), so long as r does not satisfy the conditions of Remark 2, then we may conclude r | N . Con-tinuing this process, we may construct a chain { r , r , r , . . . } of primes satisfying r i +1 | σ ( r i ) whereby each r i +1 must divide N .Traditionally, these chains have been used to accumulate enough primes to showthat σ ( N ) >
2, thus contradicting σ ( N ) = 2. We will be using them to create acontradiction on a different internal structure of N . For the moment, we will usethem to show the following Proposition 1. If N ∈ O satisfies the hypotheses of Theorem 2, each prime in thecollection T = { , , , , , , , , , , } divides N .Proof. By Remark 1, we know 5 | N , which we use to construct our first two chains: { , }{ , , , , , , } Once we know 211 | N , we may construct the chain: { , , , , , , , } and after 191 | N , { , , }{ , , , , , , , , , }{ , , , , , , , } A quick scan through each chain shows that none of the primes are congruent to1 (mod 12), except 181 which is also 6 (mod 7). Thus, none of the primes can bethe special prime, and since each element of T appears, we are done. (cid:3) Proposition 2.
For the same N , each prime in T divides N at least times.Proof. We skip most of the details and simply point out that11 divides σ (5 ) , σ (31 ) , σ (71 ) , σ (191 ), and σ (311 )31 divides σ (101 ) , σ (281 ) , σ (1031 ) , σ (1151 ), and σ (1721 )and similarly, every element of T can be shown to appear in at least 5 σ -chainsof various primes under 10000, except 181, which requires us to go to as high as18521. (cid:3) For the rest of the paper we expand the definition of T to include all of theprimes demonstrated through σ -chains to divide N . Suppose now that it has beendemonstrated { , } ⊂ T . Consider the following two chains: { , , }{ , , } Observe that 61 , , and 2521 each satisfy the conditions of Remark 2 and arecandidates for being the special prime. Neither chain has any primes in commonexcept for the last element, 61. At most, one of 241 or 2521 may be the specialprime. By virtue of this fact, one of the two chains shows 61 ∈ T . PATRICK A. BROWN A Closer Look at the Form of N In 1980, Ewell [4] demonstrated the following:
Theorem 3.
Let N = p α q β . . . q β k k be odd perfect. Set α = 4 ǫ +1 and q i = 2 π i − .Assume N , then (1) p ≡ , (2) ǫ ≡ or − , (3) π i β i ≡ or − for each i , (4) The number of elements in the set { π β , . . . , π k β k } for which π i β i ≡ − is even when ǫ ≡ and odd when ǫ ≡ − . Applying our assumption β i ≡ N , after some relabeling, may be written as N = p α s Y i =1 q β i +4 i t Y j =1 r γ j +4 j u Y k =1 s δ k +24 k (4.1)Where p, q i , r j , and s k are primes with p ≡ q i ≡ − r j , s k ≡ β , γ , δ , and α ∈ N with α ≡ N is a bit cumbersome, but it does clearly demonstrate one usefulpiece of information. Remark 3.
From Proposition 2, it is immediate that | N at least 14 times, | N at least 14 times, | N at least 24 times, and so on, according to each prime’sresidue (mod 6). The Sum of the Reciprocal of the Primes of N We now consider [1]. Cohen demonstrates the following for odd perfect numberssuch that 5 | N and 3 N : X q | N q < . Divisors LowerBound U pperBound N, | N . . N, N . . | N, | N . . N, N . . N , it isnatural we try to improve on this bound for our particular case. Cohen starts hisproof off by demonstrating a stronger form of the following inequality, for 0 < x ≤ x + x > exp( x ) (5.1) ONEXISTENCE OF OPN 5
We assume (5.1) and let N ∈ O . Thus,2 N = σ ( N ) = (1 + p + p + . . . + p α ) k Y i =1 (1 + q i + q i + . . . + q β i i )Dividing through by N and utilizing α ≥ ≥ (1 + 1 p ) k Y i =1 (1 + 1 q i + 1 q i + . . . + 1 q β i i )We separate the q i based on whether or not they appear in T . As per Remark3, we let γ i be 14 or 24 depending on whether q i ≡ − q i ≡ q i ∈ T . And finally, we truncate for the primes not in T .2 > (1 + 1 p ) Y q i ∈ T (1 + 1 q i + 1 q i + . . . + 1 q γ i i ) Y q i T (1 + 1 q i + 1 q i )Apply (5.1) to the components of q i T and take the log,ln 2 > ln (1 + 1 p ) + X q i ∈ T ln (1 + 1 q i + 1 q i + . . . + 1 q γ i i ) + X q i T q i (5.2)Let ∆ = X q i ∈ T ln (1 + 1 q i + 1 q i + . . . + 1 q γ i i ) − q i We substitute ∆ into (5.2), add p to both sides, and rearrange,ln 2 − ∆ − ln (1 + 1 p ) + 1 p > p + k X i =1 q i (5.3)We now observe the right hand side of the inequality is the sum of the reciprocalof the primes that divide N . Because ∆ is a straight forward calculation, it wouldseem our upper bound is singularly dependent upon p , however, we can still do alittle better by adding a few more primes to T . Remark (2) implies p = 37 , , p ≥
109 are the only options for the special prime. It will be convenient for usto refer to γ q for the exponent of a particular prime q ∈ T as we consider each casein turn.(1) p = 37, then 19 | ( p + 1) (and thus N ). Additionally, we know 61 is notthe special prime and since we have seen two chains (with appropriateassumptions) that demonstrate 61 | N we can put 19 , ∈ T with γ = γ = 4.(2) p = 61, then 31 | ( p + 1), which is already in T .(3) p = 73, then 37 | ( p +1), and again, we can put 37 , ∈ T with γ = γ = 4.(4) p ≥ ∈ T with γ = 4.When p ≥
109 we must change our bound slightly. By virtue of the Maclaurinseries of ln (1 + x ), ln (1 + x ) > x − x for small x . Now, (5.3) may be written asln 2 − ∆ + 12 p > p + k X i =1 q i After computing the different incarnations of ∆ for each case, we get the followingupper bounds for the right hand side of (5.3):
PATRICK A. BROWN
U pperBoundp = 37 . p = 61 . p = 73 . p ≥ . the upper bound for our N .This may seem like a lot of work to lower the previous upper bound by lessthan .013. However, when one considers optimal solutions, by which we mean,every allowable prime in T without concern for special primes. It turns out that T must contain 100,369 primes; 5 and every 1 (mod 5) prime less than 5,826,451.This improved bound requires 5 and every 1 (mod 5) prime less than 2,647,111 ora mere 48,250 primes. Of course, the primes that appear in σ -chains is far fromoptimal and about 20% of them are special primes, which as we have seen, requireextra attention. 6. Programming Methodology
It should be noted that σ -chains as a term is a bit misleading. A more appropriateterm would be σ -trees. Of course, presenting such information in tree form becomesunwieldly. Though as chains get progressively longer, they too become unwieldly.Considering we had to construct chains to include over 960,000 primes, even aftertaking great pains to reduce the upper bound in Section 5, the approximately 10,000pages of data is likely why Theorem 2 has not been demonstrated sooner.In an effort to simplify the process of constructing chains, we started with chainsof length 2: { Known Seed Prime } → {
New Prime } Chains resulting in a potential special prime or a composite too large for Math-ematica to conveniently factor were set aside . The new primes were put into a“seed” list. The next ordered pair, or 2-chain, was created using the smallest num-ber from the seed list. This is not a new idea. The advantage is always workingwith the smallest numbers from the full σ -tree, thus avoiding the need to factorlarge numbers to continue a chain. More importantly, keeping the data as uni-form 2-chains made programming chain manipulations much easier to automateand separate between two computers when convenient.Breaking the upper bound without including candidates for the special prime,though technically feasible, would probably take years. The 2-chains with specialprimes, previously set aside, were extended to 3-chains where the 3rd element wasalso a special prime, as follows { Seed Prime } → {
Special Prime } → {
Special Prime } This collection of 3-chains was then sorted by the last element. The first twoappearances of a special prime in the 3rd element were paired together and usedto confirm that prime was an element of T . After removing chains with duplicate The chains involving large composites were never used. We were able to generate enoughprimes to prove Theorem 2 without factoring any of these numbers.
ONEXISTENCE OF OPN 7 terminus (keeping one chain from the confirmed primes), the 3-chains were extendedto 4-chains, again, only extending in the cases where a special prime appeared asthe 4th element (that had not already been confirmed in the previous step). Thelast element of each 4-chain was then compared to the 3rd and 4th element of eachof the other 4-chains. Any matches were used to confirm the corresponding specialprime was in T . This process is not optimal. Among other considerations, we areallowed to have non-special primes appear in these extended chains, however, abroader approach turned out to be unnecessary.The calculations needed to create the primes included in our set T , required 2laptops a little over 3 weeks to perform. The previous month had been spent tryingto work with chains of arbitrary length. As aforementioned this becomes quitecumbersome. Once this 2-chain method was implemented, the previous month’swork (in some sense) was replicated in about 4 days.7. Proof Methodology
As was previously mentioned, the typical method of proof for theorems likeTheorem 1 and 2 is to accumulate enough primes to show σ ( N ) > N . In thecontext of optimal solutions, this bound can be easier to surpass than the sum ofthe reciprocal of primes. Taking every allowable prime to be in T without concernfor special primes, and assuming each prime divides N four times, we achieve σ ( N ) > N after 47335 primes; 5 and every 1 (mod 5) prime less than 2592521. Amarginal improvement compared to our methodology. Though without any specialknowledge regarding the number of times an arbitrary prime divides N , the sumof the reciprocal of primes may be an easier bound to surpass. Regardless, whennearly a million primes are needed for either method of proof, the choice largelybecomes a matter of taste.With regards to section 4. If one compares the improvement to the upper boundfrom increasing the number of times a prime divides N from 4 to 14 (or 24), theimprovements are minor. This begs the question, “Why use this in our proof?”The answer is simple. Since we are showcasing a different methodology for thesetypes of proofs, we hope something here may spur innovation in someone else bybroadening the view of the problem.8. Data Summary
Occasionally, Mathematica would hang trying to show a number to be provablyprime. This is why Module 17 has about half as much data as the other modules.The first time this happened, we started the next module, but after this instance,we skipped the seed prime, and restarted the module with the next number in theseed list.Despite setting the recursion level to infinity, Mathematica did not seem to likeusing our algorithm after about 24,000 new seeds cycled through. To remedy this,we stopped after 20,000 iterations and called it a data module . We have 30 datamodules for the non-special primes using this algorithm. A proper remedy, namelytaking the local variables within the program and making them global, solved thisproblem and allowed us to use as many as 50,000 seeds without issue in Modules32-34.Due to a lapse in programming, the primes that were 1 (mod 12) and also 6(mod 7) were originally collected with the special primes. After the oversight was
PATRICK A. BROWN noticed, we separated them out and called them Module 31. The new primes wereused as seeds for chains in Modules 32-34.The filters resulting from Modules 1-18 were used to filter Module 20. As aresult, 6305 primes were duplicated between Modules 19 and 20. The two data setswere combined for simplicity of filtering out duplicates.Data SummaryModule
References [1] G. L. Cohen. On odd perfect numbers.
Fibonacci Quarterly , 16:523–527, 1978.[2] Graeme L. Cohen. On odd perfect numbers (ii), multiperfect numbers and quasiperfect num-bers.
Journal of the Australian Mathematical Society , 29(3):369–384, 1980.
ONEXISTENCE OF OPN 9 [3] Ronald Evans and Jonathan Pearlman. Nonexistence of odd perfect numbers of a certain form.
Fibonacci Quarterly , 45(2):122–127, 2007.[4] John A. Ewell. On the multiplicative structure of odd perfect numbers.
Journal of NumberTheory , 12:339–342, 1980.[5] D. E. Iannucii and R. M. Sorli. On the total number of prime factors of an odd perfect number.
Mathematics of Computation , 72(244):2077–2084, 2003.[6] H. J. Kanold. Untersuchungen uber ungerade vollkommene zahlen.
J. Reine Angew , 183:98–109, 1941.[7] W. McDaniel. The non-existence of odd perfect numbers of a certain form.
Archiv der Math ,21:52–53, 1970.[8] Jonathan Pearlman. Necessary conditions for the non-existence of odd perfect numbers. Mas-ter’s thesis, University of California, San Diego, 2005.
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