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Upper Bounds on the Sum of Principal Divisors of an Integer

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Upper Bounds on the Sum of Principal Divisors of an Integer 190 MATHEMATICSMAGAZINE Upper Bounds on the Sum of PrincipalDivisors of an Integer ROGER B. EGGLETON Illinois State University Normal, IL61790 [email protected] WILLIAM P. GALVIN School of Mathematical and Physical Sciences University of Newcastle Callaghan, NSW 2308 Australia A prime-poweris any integerof the form p", where p is a primeand a is a positive integer.Two prime-powersare independentif they are powers of differentprimes. The Fundamental Theorem of Arithmetic amounts to the assertion that every positive integerN is the productof a uniqueset of independentprime-powers, which we call theprincipal divisors of N. Forexample, 3 and4 arethe principaldivisors of 12, while 2, 5 and 9 are the principaldivisors of 90. The case N = 1 fits this description,using the conventionthat an emptyset has productequal to 1 (andsum equalto 0). In a recentinvited lecture, Brian Alspach noted [1]:Any odd integerN > 15 thatis not a prime-power is greater than twice the sum of its principal divisors. For instance, 21 is morethan twice 3 plus 7, and 35 is almostthree times 5 plus 7, but 15 falls just shortof twice 3 plus 5. Alspachasked for a "nice"(elegant and satisfying) proof of this observation,which he used in his lectureto provea resultabout cyclic decomposition of graphs. Respondingto the challenge,we proveAlspach's observation by a veryelementary argument.You, the reader,must be the judge of whetherour proof qualifiesas "nice." We also show how the same line of reasoningleads to several strongeryet equally elegantupper bounds on sums of principaldivisors. Perhaps surprisingly, our meth- ods will not focus on propertiesof integers.Rather, we considerproperties of finite sequencesof positivereal numbers,and use a classicalelementary inequality between the productand sum of any such sequence.But first,let us put Alspach'sobservation in its numbertheoretic context. Aliquotparts and principaldivisors The positivedivisors of a positiveinteger have fascinatedhuman minds for millennia. The divisorsof an integerN > 1 thatare positive but less thanN arethe aliquotparts of N. It is usualto denotetheir sum by s(N). ClassicalGreek mathematicians singled out the aliquotparts of N fromamong the integersless thanN, by notingthat N canbe "built"additively from multiple copies of any one of the aliquotparts. For Euclid [4], a primenumber is "thatwhich is measuredby a unit alone"(Book VII, definition11), thatis, a numberwhich has 1 as its only aliquotpart, so N is primeif s (N) = 1. Again, a perfect numberis "thatwhich is equal to its own parts"(Book VII, definition22), thatis, a numberthat can be "built"additively from a single copy of each of its aliquot parts,so N is perfectif s(N) = N. Others,such as Theon,added that N is deficientif s(N) < N, and abundant if s(N) > N. The numbers 6, 10, and 12 are examples from the threeclasses. VOL. 77, NO. 3, JUNE2004 191 Euclidknew thatthere are infinitelymany primes (Book IX, Theorem20) andthat an even number of the form 2k-1(2k - 1) is perfect when 2k - 1 is prime (Book IX, Theorem36). In moremodem times it has been provedthat every even perfectnumber musthave this form, and currently40 such perfectnumbers have been found,corre- spondingto the knownMersenne primes [12], but it is not yet knownwhether there are infinitelymany perfect numbers. Indeed, it is not knownwhether any odd perfect numberexists, thoughmany constraints on the possible form of such a numberhave been proved.By contrast,infinitely many positive integers are deficientand infinitely manyare abundant;there can be no doubtthat the Greeksknew easy proofs of these facts. Interestin such mattersunderlies sophisticated modem computationalstudies of aliquot sequences, the sequences ao, al, a2, ... beginning at a chosen positive integer a0 = N, witheach subsequentterm found by computingthe sumof the aliquotparts of the current term: ak+l = s(ak) for k > 0. (See [2, 7] as entry points to current knowl- edge aboutaliquot sequences.) The aliquotsequence of a given N behavesin one of threepossible ways: (1) aftera finite numberof termsit arrivesat 1; (2) aftera finite numberof termsit entersa finitecycle, whichrepeats forever; (3) it continuesforever withoutrepetition. Sequences that arriveat a perfectnumber are of type (2), as are those that arriveat eithermember of a pair of amicablenumbers, namely solutions to s(a) = b, s(b) = a. Pythagoras knew that a = 220 and b = 284 are the smallest amicablepair. Members of largercycles are called sociable numbers,and severalex- ampleshave been found in modem times. Intriguingly,it is not yet known whether there are any sequencesof type (3); currentlythere arejust five possible candidates with N < 1000, the first being N = 276. It can be checked that the ratio s(N)/N is 2 when N = 120, and is 3 when N = 30240. Indeed,it turs out thats(N)/N has no absoluteupper bound, and vari- ous simpleproofs are known. When we haveproved the key inequalitywe need in this article,we shall show thatit also providesan elementaryproof of this fact. (A recent Note in the MAGAZINEby Ryan [10] concernsthe densenessof the set of numbersof the forms(N)/N, andof the complementaryset, in the positivereals.) Likethe aliquotparts of a positiveinteger N, the principaldivisors are a rathernatu- ral subsetof the divisorsof N. Indeed,if N is not a prime-power,its principaldivisors area propersubset of its aliquotparts. Thus s*(N), the sum of principaldivisors of N, satisfiess*(N) < s(N) wheneverN is not a prime-power.In contrastto s(N), it turns out in fact thats*(N) neverexceeds N. We shall provethis as our firsttheorem. Fol- lowing commonpractice, we writed N when d is a positivedivisor of N, and pa IIN when pa is a principaldivisor of N. The FundamentalTheorem of Arithmeticimplies thatany positiveinteger N can be built multipicativelyfrom a single copy of each of its principaldivisors: N= H p, p IIN where the notationalconvention is that the productranges over all principaldivisors of N. If H is replacedby E, we havethe sumof all principaldivisors of N. It is simple andinstructive to prove THEOREM 1. Every positive integer N satisfies > N = H pa E pt = s*(N), (1) palIN p"IIN and (1) holds with equality just when N is a prime-power. 192 MATHEMATICSMAGAZINE Proof When N = 1, the set of all principal divisors of N is empty, so by standard conventions for empty sums and products, (1) holds with strict inequality in this case. Clearly (1) holds with equality when N has exactly one principal divisor (so N is a prime-power). Next suppose N has exactly two principal divisors, say N = p"qP. The inequality paqf > p" + qf is equivalent to (pa - 1)(qt - 1) > 1, and the latter is satisfied because 2 < pa < qt holds without loss of generality. Now suppose for some k > 2 that (1) holds with strict inequality for every positive integer with k principal divisors. Let N be any integer with exactly k + 1 principal divisors, let q' be one of them, and let N* := N/q5. Then N* has exactly k principal divisors, so N = N*q =pq H pa>>, q pa = pOqfi p" IIN* pa IIN* pa IIN* E > p = p. > y (pa+q q)=kqI + pa qf + E pa IIN* pa IIN* pa IIN* pa IIN Hence (1) again holds with strict inequality. The theorem now follows by induction onk. U From the proof of Theorem 1, we see that the inequality (1) will usually be very weak when N has several principal divisors, especially if any of them is relatively large. So could it be that N is usually at least twice as large as the sum of its principal divisors? It certainly can! This is Alspach's observation, which we mentioned at the outset: THEOREM2. Let N be an odd positive integer with at least two distinct prime factors. If N > 15, then N-1N > pa = s*(N). (2) pa IN We shall now briefly recall a classic inequality for real numbers, and then use it to prove Theorem 2. The Bernoulli-Weierstrassinequality Let R+ := {x E R : x > 0} and, for any n > 1, let a := (al, a2,..., a) E (R+)n be a sequence of nonnegative real numbers. Weierstrass [11] reasoned: (1 + al)(l +a2) = 1 +al +a2 +ala2 > 1 +al +a2, (1 + al)(l + a2)(1 + a3) > (1 + al + a2)(1 + a3) > 1 + + a2+2 a3, and so on. Modulo attention to when equality may hold, this is essentially an inductive proof of the following theorem. THEOREM3. (WEIERSTRASS)If a E (R+)n and n > 1, then n n f(l+ ai) > l+ ai (3) i=1 i=1 and (3) holds with equality if and only if at most one of the numbers ai is nonzero. This classical elementary inequality (3) is the key tool underlying our arguments in this article. Some authors, such as Durell and Robson [3], call it Weierstrass's inequal- ity but it is not clear whether Weierstrasswas the first to establish it. Hardy,Littlewood, VOL. 77, NO. 3, JUNE2004 193 andP6lya [8] notedit as Theorem58 withoutattribution, though they creditedJacques Bernoulliwith the special case in which a is a constantsequence with termsgreater than -1. We shall refer to (3) as the Bernoulli-Weierstrass inequality. Earlierwhen discussingaliquot parts we remarkedthat the ratios(N)/N is known to have no absoluteupper bound. It is of interesthere to see how this can be derived fromthe Beroulli-Weierstrassinequality. THEOREM4. For any integer N > 2, the sum of aliquot parts s(N) satisfies s(N) CE I (4) N plN P and (4) holds with equality if and only if N is prime. Proof. The sum of all positivedivisors of N is N+ s(N) = n pEa1p > (pa + pa JINIIpot pfipII pa N PIINpoIN pa IIN P ) pN ( p/ The secondstep holds with equalityif andonly if all principaldivisors of N areprime, so if and only if N is squarefree.After dividingby N, the Beroulli-Weierstrassin- equality(3) now gives s(N) 1 + >l-I> ( 111+ 1+ E- NI7!N (?ppIN / pIN with equalityat the second step just when N has only one primedivisor.
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