A conjecture of De Koninck regarding particular square values of the sum of divisors function

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1 Journal of Number Theory ) Contents lsts avalable at ScenceDrect Journal of Number Theory A conecture of De Konnck regardng partcular square values of the sum of dvsors functon Kevn Broughan, Danel Delbourgo, Qzh Zhou Department of Mathematcs, Unversty of Wakato, Prvate Bag 3105, Hamlton, New Zealand artcle nfo abstract Artcle hstory: Receved 8 March 013 Receved n revsed form 5 August 013 Accepted 0 October 013 Avalable onlne 4 December 013 Communcated by Davd Goss MSC: 11A5 11A41 Keywords: Sum of dvsors Squarefree core De Konnck s conecture Compactfcaton We study ntegers n>1 satsfyng the relaton σn) =γn), where σn)andγn) are the sum of dvsors and the product of dstnct prmes dvdng n, respectvely. If the prme dvdng asolutonn s congruent to 3 modulo 8 then t must be greater than 41, and every soluton s dvsble by at least the fourth power of an odd prme. Moreover at least /5 of the exponents a of the prmes dvdng any soluton have the property that a + 1 s a prme power. Lastly we prove that the number of solutons up to x>1satmostx 1/6+ɛ, for any ɛ>0andall x>x ɛ. 013 Elsever Inc. All rghts reserved. 1. Introducton A decade ago, Jean-Mare De Konnck asked for all nteger solutons n to the equaton σn) =γn) 1) * Correspondng author. E-mal addresses: kab@wakato.ac.nz K. Broughan), delbourg@wakato.ac.nz D. Delbourgo), qz49@wakato.ac.nz Q. Zhou) X/$ see front matter 013 Elsever Inc. All rghts reserved.

2 K. Broughan et al. / Journal of Number Theory ) where σn) denotes the sum of all postve dvsors of n, andγn) s the product of the dstnct prme dvsors of n. The only known solutons wth 1 n are n =1and n = 178, and so De Konnck sensbly conectured that there exst no other solutons. It s ncluded n Rchard Guy s compendum [4, Secton B11] as an unsolved problem. In [] a number of restrctons on the form of Eq. 1) were developed: the two solutons n =1andn = 178 are the only ones havng ωn) 4; furthermore, f an nteger n>1 s fourth power free.e. p 4 n for all prmes p), then t was shown that n cannot satsfy De Konnck s equaton. The am of ths work s to present further tems of evdence n support of De Konnck s conecture, and to ndcate the necessary structure of a hypothetcal counter-example. In fact, upon combnng together the results of [] and ths artcle, then any non-trval soluton other than 178 must be even, have one prme dvsor to power 1 and possbly another prme dvsor to a power congruent to 1 modulo 4, whle all other odd prme dvsors should occur only to even powers. Here we shall establsh that f the prme to power 1 s congruent to 3 modulo 8, then t must be no less than 43 Proposton 1). Moreover, we prove that at least one odd prme dvsor must appear wth an exponent no smaller than 4 Theorem 1). Applyng an dea from [3], weshowncorollary that more than /5 of the exponents a appearng n the prme factorzaton of any soluton of Eq. 1) are such that a +1s a prme or a prme power. We then count the number of potental solutons n up to x, n the followng manner: usng results of Pollack and Pomerance [8], and by extendng a method of [,Thm.1], we shall prove n Theorem that the number of solutons n x to Eq. 1) can be at most x 1/6+ɛ, for any ɛ>0 and every x>x ɛ. Fnally, by explotng the propertes of the product compactfcaton of N, weshow there are only fntely many solutons to 1) supported on any gven fnte set of prmes P. Indeed we wll prove a more general result for the equaton σn) α φn) β = θ n μ γn) τ ) where α, β, μ, τ Z wth θ>0 some fxed ratonal, and α + β>μsee Theorem 3). The argument tself has a rather dfferent flavor from that n [5]. Notatons. If p s prme then v p n) s the hghest power of p whch dvdes n, ωn) wll denote the number of dstnct prme dvsors of n, andk s the set of all solutons to σn) =γn). Lastly, the symbols p, q, p,q are reserved exclusvely for odd prmes.. Prelmnary lemmas We begn by recallng some basc structure theory concernng solutons to Eq. 1). The followng two background results were proved n []. Lemma 1. If n>1 belongs to K, then one has a decomposton s n = e p 1 = p a

3 5 K. Broughan et al. / Journal of Number Theory ) where e 1, anda s even for all =3,...,s. Furthermore, ether a s even n whch case p 1 3mod8),ornsteada 1mod4)and p 1 p 1mod4). Lemma. If n>1 s an element of K and does not equal to 178 = , thenn has at least 5 dstnct prme factors, and there exsts a prme ether even or odd) dvdng n to at least a fourth power. The proof of the next result s due Pollack, and can be found n [6]. Lemma 3. If σn)/n = N/D wth gcdn,d) =1, then gven x 1 and d 1: #{n x such that D = d} = x o1) as x. Lastly we wll requre Apéry s soluton to the generalzed Ramanuan Nagel equatons. Lemma 4. See Apéry [1].) The Dophantne equaton x +D = n+, wth gven non-zero nteger D 7, has at most two solutons. In addton: ) f D =3then x, n) {3, 5), 45, 11)}, ) f D has the form m 1 wth m 4, thenx, n) {1,m), m 1, m 1)}. Hence, n both these cases, there are exactly two solutons. 3. Restrctons on prmes dvdng members of K In ths secton, we shall make a prelmnary study of restrctons on the possble values of p 1 and p assocated to elements of K, addtonal to those descrbed n Lemma 1 above. Clearly p cannot be dvsble by any cube, otherwse Eq. 1) s volated. Hence for prme numbers congruent to 3 modulo 8, ths excludes frst 107 and secondly n ncreasng order) 499 from occurrng. We wll henceforth refer to these as bad De Konnck prmes; ndeed there are an nfnte number of prmes p 3 mod 8) such that p + 1 s dvsble by a proper cube. In Proposton 1 below, we shall prove that 3, 11 and 19 are also bad. In the case p 1 p 1mod4)anda = 1, ths same constrant appled to p 1 +1)p +1) excludes for those prmes less than 100, the pars {5, 17}, {5, 53}, {5, 89}, {13, 53}, {13, 97}, {17, 9}, {17, 41}, {17, 53}, {17, 89}, {9, 53}, {9, 89}, {37, 53}, {41, 53}, {41, 89}, {41, 97} called here bad De Konnck pars. LaternCorollary 1, weshow{5, 13} s also bad.

4 K. Broughan et al. / Journal of Number Theory ) Proposton 1. Under the same notatons as Lemma 1, fasolutonn Ksatsfes both ωn) > 4 and p 1 3mod8), then the prme p Proof. Frst suppose that p 1 =3,nwhchcase e+1 1 ) 4 = σp a ) p =4 3. As a drect consequence e+1 1 < 9soe {1, }, andby[, Theorem 3] we can assume a 4. Therefore 3 13 = ) < e+1 1 ) σpa ) p < 3, whch s obvously false, and we conclude that p 1 3. Next f one supposes that p 1 = 11, then e+1 1 ) 1 = σ p a ) =4 11 p, thus 3 e+1 1) < 11 whch mples that 1 e 4. If all of the a were strctly less than 4, then by [, Theorem 3] agan we would have e =4,nwhchcase The latter mples e+1 1 ) p 1 +1) 31 3 = = = σp a ) p =4 11. σp a ) p =11, hence there exsts an wth11 p + p +1;thssmpossblesnce11 1mod3. It follows there s at least one wtha 4, and wthout loss of generalty suppose that t s a say. One therefore obtans an nequalty and consequently, e+1 1 ) p whch s clearly false. Therefore p σpa ) p < 11 3 σ34 ) 3 =11 < 11

5 54 K. Broughan et al. / Journal of Number Theory ) Fnally suppose p 1 = 19. Usng Lemma 1 we can wrte n = e p 1 m = pa, whence e+1 1 ) p 1 +1) = σp a ) p =4p 1. Thus e+1 1) 5F =19 where F s a postve ratonal value strctly greater than 1. As a consequence e+1 1) < 19 /5, mplyng that 1 e 5. Case 1). If e =5then 9 7 5F =19 and t follows that F<19 /315 < If some exponent a 3thenF σp 3 )/p > 3 whch cannot occur, and therefore one may assume that a =forevery {,...m}. Now by studyng the left hand sde, there must exst a prme p whch we wll call p ) that equals 3. Then σp )= = 13 yelds a new prme, denoted p 3,wth σp 3)= =3 61. One thereby obtans a left hand sde wth at least three 3 s n the numerator but at most two 3 s n the denomnator, whle the rght hand sde has none. Ths contradcton shows e<5. Case ). If e =4then 31 5 = σp a ) p =19. Argung as n the prevous case, wthout loss of generalty assume a =for. Examnng the left hand sde, one of the prmes p must equal 31; let us call t p. Then we have σp )= = 3 331, thence the new prme p 3 = 331 gves σp 3) = = , and ultmately p 4 =7wth7 +7+1=3 19. Hence there are at least three 3 s n the numerator and exactly two n the denomnator, wth none occurrng on the rght hand sde. Ths shows e<4. Case 3). If e =3then15 5F =19 mples )< 19, whch s false. Case 4). If e = then we would get < 19, whch agan s false. Case 5). Henceforth we consder the stuaton where e = 1. It follows that 3 5 = σ p a ) =19 p =

6 K. Broughan et al. / Journal of Number Theory ) mplyng 3 n. One can then take p =3,andbyLemma 1) assume that a s even. Suppose frst that a 6. Then σ3 a ) σ3 6 ) = 1093, n whch case =3 σ p a ) 5 σ 3 a ) m =3 whch s false, whence a {, 4}. However f a =,then ) =3 σ p a ) =19 3 σp a ) p =19 3 and there must exst an odd prme dvdng n whch s greater than 3, and whch dvdes n to a power not less than 4. Ths eventualty n turn mples =3 p ) < 19 3 whch agan s mpossble. Hence the only remanng possblty s that a = 4. Because σ19) = 5and σ3 4 )=11, one may then assume p 3 =5andp 4 = 11, whch gves us the equalty σ 1) σ 19 1) σ 3 4) σ 5 a 3 ) σ 11 a 4 ) Cancelng lke terms yelds =5 σp a ) p = a 3 11 a 4 <σ 5 a 3 ) σ 11 a 4 ) whch s false f ether a 3 4ora 4 4; snce both are even, clearly a 3 = a 4 =. Now / Kso there exsts a prme p 5 7 such that p a 5 5 n wth a 5 even. If a 5 4 then one would have 7 +7 ) 5 11 < 5 11 σpa 5 5 ) p < whch s certanly false; thus all prmes other than, 3, 19 whch dvde n must do so exactly to the power. As a consequence m 5, and we can wrte n = p. =5

7 56 K. Broughan et al. / Journal of Number Theory ) Substtutng ths form nto the equaton σn) =γn) and then cancelng, one deduces ) p + p +1 = =5 p Therefore the set of p wth 5 m ncludes {31, 131} and none out of {3, 5, 19}. However σ31 )=3 331, σ131 ) = 1793 and σ1793 )= , hence 3 = 9 dvdes the numerator of the product on the left and does not cancel wth any denomnator. Ths crcumstance s mpossble, as 9 does not dvde the rght hand sde. The above contradcton completes the proof that p Proposton. If p 1 1mod4)and a 5, thenp Proof. Applyng Lemma 1 one knows p 5, and we can wrte e+1 1 ) σpa ) p =3 σp a ) p = 4p 1 p However σ5 5 )/5 = 906/5 σp a )/p n whch case e+1 1) < 4p 1 p 1 +1 < 4p 1; the latter nequalty s only satsfed by prmes p Proposton 3. If n Ks a soluton wth p 1 3mod8)such that n s not dvsble by the fourth power of any odd prme, then p 1 cannot dvde e+1 1. Proof. Usng [, Theorem 3], one can express n = e p 1 p = and moreover e+1 1 4p 1/p 1 +1) < 4p 1. Thus under the assumpton that p 1 e+1 1, ether p 1 = e+1 1or3p 1 = e+1 1. Frst suppose that p 1 = e+1 1. From the proof of [, Theorem 3], one has 1 m 4 = p + p +1 p 0.73 consequently p 1 1) 0.73 p 1. The latter nequalty mples p 1 < 3, whch s false. Alternatvely f 3p 1 = e+1 1, because 9 e+1 1 for any value of e, clearly 3 p 1, so we can nstead set p = 3. Smlarly 13 = p 1,and =3 61 wth 61 p 1. However gvng at least three powers of 3 dvdng the left hand sde of σn) =γn), whch agan yelds a contradcton. The followng three techncal lemmas are key ngredents n the proof of Theorem 1.

8 K. Broughan et al. / Journal of Number Theory ) Lemma 5. If n Ks dvsble by 3, there exsts an odd prme p such that p 4 n. Proof. Assume hypothetcally) n s not dvsble by the fourth power of an odd prme. If p 1 3 mod 8) then usng Lemma 1, wecanwrte e+1 1 ) p 1 +1) p + p +1 ) p m + p m +1 ) =4p 1p p m. By Lemma 1 once more, we know p 1 3 so nstead put p = 3. Consder the system: = 13; 13 5mod8, 13 p 1, 13 = p =3 61; 61 5mod8, 61 p 1, p 4 = = ; 97 1mod8, 97 p 1, p 5 =97. We observe that the left hand sde of the prevous equaton must be dvsble by 3 3 =7 whlst the rght hand sde s only dvsble by 3 = 9, yeldng a contradcton. If p 1 1 mod 4), one has the decomposton e+1 1 ) p 1 +1) p +1) p 3 + p 3 +1 ) p m + p m +1 ) =4p 1p p m. Nether p 1 nor p canbe3,thuswemaytakep 3 =3. If p 1 =13thenp 1 +1= 7, and we set p 4 = 7; therefore =3 19 and =3 17, agan gvng too many 3 s. If nether p 1 nor p s 13, we can choose p 4 = 13 and thereby obtan = If 61 = p 1 or p let s say p 1 = 61), we can wrte n = e 61 p p 3 p m and so p 1 +1 = 31 wth 31 p. Consequently we can choose p 4 = 31, leadng to the equaton σ31 )= =3 331 and agan too many 3 s. Fnally f 61 p 1,p then we stll pck up an addtonal 3, snce Lemma 6. If a soluton n Ks not dvsble by the fourth power of an odd prme and p 1 3mod8),then3 n. Proof. Suppose n Kbut 3 n. In general, f a prme q p + p + 1 then ether q =3, or we must have q 1mod3)sothat3 q + q + 1. Now from the expresson e+1 1 ) p 1 +1) p + p +1 ) p m + p m +1 ) =4p 1p p m we can defne Q := m = p + p + 1). Because 3 n, each prme number p wth 1 m whch appears as a factor of Q does not appear n the form p + p +1; ths means we must have Q p 1. However by Lemma, thentegerq has at least three quadratc factors, gvng rse to a contradcton. Lemma 7. If n Ksatsfes p 1 1mod4)and 3 n, thenn s dvsble by the fourth power of an odd prme.

9 58 K. Broughan et al. / Journal of Number Theory ) Proof. Suppose p 1 p 1 mod 4). Then n the notaton of Lemma 6, t follows that there are two quadratc factors for Q = m = p +p +1) and followng cancelaton) three possble forms for the equaton σn) =γn). We shall treat each of these separately. Case 1): p 3 + p 3 +1=p 1 p 4 + p 4 +1=p e+1 1 ) ) ) p 1 +1 p +1 = p 1 p p 3p 4. Note that p +1 has at least one prme dvsor, and at most three prme dvsors. 1.1) If p +1 has only one prme dvsor then p +1 = p 1 ; under ths scenaro, there are seven possbltes for p ) If p 1+1 = p then e+1 1=p 3p 4, whch s mpossble. 1.1.) If p 1+1 = p 3 then p 3 p 1 + 1; however p 3 p 1 1sop 3 gcdp 1 +1,p 1 1) =, whch s mpossble ) If p 1+1 = p 4 then p 3 + p 3 +1= p +1 = p 4 + p 4 + mplyng both p 3 p 4 +1andp 4 p 3 + 1, whch s clearly false ) If p 1+1 = p 3 p 4 then p 3 p 1 + 1; however p 3 p 1 1 hence p 3 p 1 +1,p 1 1) =, whch s agan false ) If p 1+1 = p p 3 then e+1 1=p 4, whch s mpossble ) If p 1+1 = p p 4 then e+1 1=p 3, whch s mpossble ) If p 1+1 = p p 3 p 4 then p 3 p 1 +1; now p 3 p 1 1thusp 3 gcdp 1 +1,p 1 1) =, whch s false. 1.) If p +1 has two prme dvsors, ether p +1 = p 3;orp 4;orp 3 p ) If p +1 = p 3, then ether p 1+1 = p or p 1+1 = p 4: ) If p 1+1 = p then one has p 4 p 4 +1) = p 3 p 3 + 1), whch mples p 3 p 4 +1andp 4 p 3 + 1; the last two condtons are ncompatble ) If p 1+1 = p 4 then p 4 p 4 +1)=p 3 +1)p 3 1), whch mples that p 4 <p 3 ; further p 3 p 3 +1)=p 4 +1)p 4 1) whch mples p 3 <p 4, mpossble! 1..) If p +1 = p 4, then ether p 1+1 = p or p 1+1 = p 3: 1...1) If p 1+1 = p then p 4 =, whch s false. 1...) If p 1+1 = p 3 then p 3 =, whch s false.

10 K. Broughan et al. / Journal of Number Theory ) ) If p +1 = p 3 p 4, then ether p 1+1 = p or p 1+1 = p 3 p 4 : ) If p 1+1 = p then p 3 p 4 +1andp 4 p 3 + 1, whch s mpossble. p 1..3.) If 1 +1 = p 3 p 4 then p 1 = p, whch s false as they are dstnct prmes. 1.3) If p +1 has three prme dvsors, ether p +1 = p 1 p 3;orp 1 p 3 p 4 ;orp 1 p ) If p +1 = p 1 p 3 then one deduces e+1 1=p 4, whch s false. 1.3.) If p +1 = p 1 p 3 p 4 then p 1+1 = p, whch mples that p 4 p 3 +1andp 3 p 4 +1; the latter condtons are ncompatble ) If p +1 = p 1 p 4 then we fnd e+1 1=p 3, whch s false. Combnng 1.1), 1.), and 1.3) together, clearly Case 1) s mpossble n ts entrety. Case ): p 3 + p 3 +1=p 1 p 4 + p 4 +1=p 1 p e+1 1 ) ) ) p 1 +1 p +1 = p 3p 4. Here p 3 p 4 mod3), p 1+1 p +1 1 mod 3), and there are at least two prme factors n e+1 1 whch beng congruent to 3 modulo 4 cannot nclude p,andbeng congruent to 1 modulo 3 cannot nclude p 3 or p 4 ). It follows that there s at least one prme factor n p 1+1 and p +1 respectvely, whch leaves us only p 1+1 = p 3 or p 1+1 = p 4, and these are both mpossble. Case 3): p 3 + p 3 +1=p 1 p 4 + p 4 +1=p 1 p e+1 1 ) ) ) p 1 +1 p +1 = p p 3p 4. Note that t cannot happen that one of p,p 3,p 4 s the only prme dvsor of p +1. Furthermore e+1 1 must have at least two prme dvsors, and t cannot be a square; n addton e+1 1 p p 1+1 p +1 1mod3)andp 3 p 4 mod3).one therefore deduces p 1 +1 = p e+1 1=p 3 p 4 p +1 = p 3 p 4.

11 60 K. Broughan et al. / Journal of Number Theory ) From these three equatons, we obtan e p 3 = and by the result of Apéry n Lemma 4, ths s clearly an mpossble occurrence. We are now ready to gve the man result of ths secton. Theorem 1. If n Kthen n s dvsble by the fourth power of an odd prme. Proof. Frstly applyng Lemma 5, fn Kand 3 n then p 4 n for some odd prme p. Wthout loss of generalty, we may therefore assume n Kand 3 n. If p 1 1 mod 4) then the result s covered by Lemma 7. Lkewse f p 1 3mod8) then the result s covered by Lemma 6. Fnally the remanng case p 1 7mod8)s already excluded courtesy of Lemma 1. Corollary 1. If {p 1,p } = {5, 13} then a 5. Proof. Assume that a =1.Snceσn) =γn), settng p 3 =3andp 4 = 7 mples e+1 1 ) 3) 7) =3 σ p a ) = p. Usng the dvsblty of n by the fourth power of an odd prme whch mnmally s 3): e+1 1 ) < 1 =5 so e+1 1) < 1, whch s false for e N. Therefore a > 1, n whch case a The exponents for members of K We now study the exponents a occurrng n the decomposton of a De Konnck number. The frst step s to adapt an dea of Chen and Chen [3], n order to relate ωn) wth m =0 da +1), where dx) s defned to be the number of dvsors of an nteger x 1. The second step s to apply the AM/GM nequalty, then further analyse the exponents. Lemma 8. Let a soluton n Kbe represented as the product n = e p 1 m = pa.if we set p 0 =, a 0 = e and a 1 =1, then there are nequaltes ωn) m da +1) 3ωn). =0

12 K. Broughan et al. / Journal of Number Theory ) Proof. One need only derve the upper bound, snce the lower bound follows from 1). Frst consder the case where anda s even, so p s odd. Put w = da +1) 1 and wrte n,1,...,n,w to denote all the postve nteger dvsors of a + 1 other than 1. Let q, be a prmtve prme dvsor of p n, 1)/p 1) for 0 m and 1 w. In partcular, there are dvsbltes p n, 1 q, p 1 1 p 1 pa +1 and f Ωx) counts the number of prme factors of x wth multplcty, then w ω σ p a )) Ω σ p a )). Alternatvely, f = 0 then prmtve dvsors exst except for e +1=6, and n that case w 0 = de +1) 1=3=Ω 6 1 ). Lastly f =1ora =1,thenwehave1=da +1) 1 < Ωp +1)=Ωσp a )). Therefore n all cases da +1) 1 Ωσp a )), hence there s an nequalty m m da +1) ωn) = da +1) 1 ) =0 =0 m =0 Ω σ p a )) ) = Ω σn) = Ω γn) ) =ωn) thereby completng the dervaton of the upper bound. Corollary. If n Kthen n the notaton of Lemma 8, a proporton of more than /5 of the numbers a +1 must be ether prme or prme powers. Proof. Because ωa +1) da +1), usng the arthmetc geometrc mean and Lemma 8: m =0 ωa +1) ) m 1 ωn) =0 ωa +1) 3. ωm) Moreover, takng the logarthm of both sdes, one deduces m ωa +1) =0 ) log 3 ωn). log For an nteger 1, let n := #{: ωa +1)=}. Then the above nequalty becomes ) log 3 n 1 +n +3n 3 + n 1 + n + ) log

13 6 K. Broughan et al. / Journal of Number Theory ) whch mples log 3 ) n + n 3 + ) log 3 ) n + 3 log 3 ) ) log 3 n 3 + log log log log 1 n 1. Rearrangng the n s yelds n 1 + n + log 3 log 1 ) log 3 +1 n 1 log and as the bracketed term equals.41 to two decmal places, we conclude that 5 < 1.41 n 1 n 1 n 1 + n + + n m as requred. 5. Countng the elements n K [1,x] For every real x>0, we wll from now on use the notaton Kx) :=K [1,x]. In [, Theorem 4], t was shown that the sze of the solutons Kx) s asymptotcally bounded by x 1/4+o1) as x tends to nfnty and ths result was tself an mprovement on the work of Pomerance and Pollack [8], whch nstead gave an upper bound of x 1/3+o1) ). In ths secton we wll sharpen the bound stll further, as descrbed drectly below. Theorem. The estmate holds as x. #Kx) x 1/6+o1) Proof. Let n>1benkx), so we may express t as n = A B where gcda, B) =1, wth A squarefree and B squarefull. Explotng Lemma 1, thena {p 1, p 1,p 1 p, p 1 p } and B s dvsble by at least one prme to the fourth power or greater. Under the notaton of Lemma 3, one can wrte N D = σn) n = γn) = γa) n A γb) B = A B/γB) > 1 wth gcda, B/γB) ) = 1. It follows that B/γB) <A, whence B γb) <AB= n x = B γb) x.

14 K. Broughan et al. / Journal of Number Theory ) Now by Lemma 1, we can always decompose B = δ C D where δ {1, 3 }, C s a squarefree nteger, D s a 4-full nteger, and such that δ, C and D are parwse coprme. As a consequence, B γb) = δ γδ) C D γd) = D γd) x. In addton B γb) = δ γδ) D γd) so that B γb) = D γd) or B γb) = D γd). Moreover one knows that D/γD) x above, whch means D/γD) x. Now f two 4-full numbers D 1 and D satsfy D 1 /γd 1 )=D /γd ), then we must also have D 1 /γd 1 ) = D /γd ). Hence the number of choces for D/γD) x wth D/γD) x and D 4-full, s less than or equal to the number of choces for D/γD) x whch s of type x 1 6 +o1). Therefore the number of choces for B/γB) s also x 1 6 +o1), and the proof s completed upon applyng Lemma Applcatons of the product compactfcaton For each prme p, letn p denote the one pont compactfcaton of N; npartcular, each fnte pont n N s tself an open set, and a bass for the neghborhoods of the pont at nfnty, p say, s gven by the open sets U p ɛ) = {p e N: e 1/ɛ} {p } wth ɛ>0. If P ndcates the set of prme numbers, let us wrte ˆN := p P N p for the product of these ndexed spaces, endowed wth the standard product topology. Then ˆN s a compact metrzable space so t s sequentally compact, hence every sequence n ˆN has a convergent subsequence. Remark. We shall call ˆN equpped wth ts topology the product compactfcaton of N. A nce account detalng propertes of the so-called supernatural topology n attackng the odd perfect number problem, s gven by Pollack n [7]. Consder now the more general equaton σn) α φn) β = θ n μ γn) τ 3)

15 64 K. Broughan et al. / Journal of Number Theory ) where α, β, μ, τ Z and θ>0 s a ratonal number. Wrte K = K α,β,μ,τ for the set of solutons K α,β,μ,τ = { n N: σn) α φn) β = θ n μ γn) τ } whch clearly depends on the ntal choce of quntuple α, β, μ, τ, θ). Theorem 3. Let P P denote a fxed fnte set of prmes, and assume that α + β>μ. Then there exst only fntely many n Kwth support n P. Before we gve the demonstraton, we pont out that choosng α =1,β =0,μ =0, τ =andθ = 1 mples there exst only fntely many solutons to De Konnck s equaton 1), supported on any prescrbed fnte set of prmes P. Proof. Gven A, B, M, T Z, defne a multplcatve functon h = h A,B,M,T : N Q >0 by the formula hn) := σn)a φn) B n M γn) T. For every r 1 and at each prme p, one calculates that h p r) = p A B T p 1) B A 1 p r 1) A p r ) A+B M whle h1) = 1. Ths naturally leads us to the defnton ĥ p ) f A + B>M := 0 f A + B<M p A B T p 1) B A f A + B = M, and provdes a unque extenson ĥ : ˆN R { }of the orgnal arthmetc functon h. In fact f A + B = M and T = 0, one can then show ĥ s contnuous on the monod ˆN. Fx a fnte set of prmes L = {l 1,...,l k }, and put N L := { n N: n = l e 1 1 le k k,e 1 }. Key clam. If A + B M then h NL s monotonc ncreasng wth respect to dvsblty. To establsh ths clam suppose that n = n l e wth n N L\{l },andsetm = n l e +1.Thenhm) =hn ) hl e +1 )and h l e ) σl e ) A φl e +1 = l e +1)M γl e +1 ) B ) T

16 K. Broughan et al. / Journal of Number Theory ) e σl +1 ) ) A = σl e ) l B M e l + = l B M 1 l e +1 1 σle )A φl e )B l e M ) A h l e ). γl e )T However e l + l B M 1 l e +1 1 ) A e l + = l B M l l e +1 = l B M snce A + B M. It follows that hl e +1 hm) =h n ) h l e l 1 l e +1 1 l + l 1 l e +1 1 ) >hl e ), n whch case ) >h n ) h l e ) = hn). ) A ) A >l A+B M 1 The proof of the clam then follows by nducton on the number of prmes wth multplcty) whch dvde the quotent of a general par n and m, wthn m. Now let us take A = α, andchooseb,m Z such that μ β<m B α. Suppose there exsts a sequence of elements n K supported on P whch are all dstnct. Under the supernatural topology, there exsts a subsequence N ) 1 and a lmt N o ˆN such that N N o. The element N o s supported on P, otherwse at least one of the N would also not be supported on P. We may therefore wrte N o = A B where suppa) P, suppb) P, and gcda, B) =1wthB squarefree. Furthermore suppa) suppb) =L = {l 1,...,l k }, say. Then there exsts a subsequence N ) 1 of the sequence N ) 1 satsfyng for all 1: ) suppn )=L, ) N properly dvdes N +1,and ) A N. Each N Kand h s monotonc on the monod N, ), hence for all one has 0 <hn 1 ) <hn ) by 3) = θ φn ) B β γn ) T τ N M μ

17 66 K. Broughan et al. / Journal of Number Theory ) = φn ) B β N M μ N B β) M μ) θ k s=1 l s) T τ θ k s=1 l s) T τ whch tends to zero as snce M B>μ β. Ths mmedately yelds a contradcton, and completes the proof of the theorem. 7. Fnal comments In Theorem, we beleve t should be possble to reduce the upper bound to x o1). Moreover extendng the lst of bad De Konnck prmes, for example by fndng addtonal nfnte sets, seems readly achevable. In the fundamental Lemma 1, showng that the exponent e of the power of equals 1 or at least s odd) looks lke a reasonable goal, but we have been unable to prove ths. Lastly, extendng the method of Theorem 3 to nclude subsets of K wth prme support of bounded sze, seems altogether more challengng. References [1] R. Apéry, Sur une équaton dophantenne, C. R. Acad. Sc. Pars Sér. A ) , [] K.A. Broughan, J.M. De Konnck, I. Káta, F. Luca, On ntegers for whch the sum of dvsors s the square of the squarefree core, J. Integer Seq ) 1 1. [3] F.J. Chen, Y.G. Chen, On odd perfect numbers, Bull. Aust. Math. Soc ) [4] R.K. Guy, Unsolved Problems n Number Theory, thrd edton, Sprnger, 004. [5] F. Luca, On numbers n for whch the prme factors of σn) are among the prme factors of n, Results Math ) [6] P. Pollack, The greatest common dvsor of a number and ts sum of dvsors, Mchgan Math. J ) [7] P. Pollack, Fnteness theorems for perfect numbers and ther kn, Amer. Math. Monthly ) [8] P. Pollack, C. Pomerance, Prme-perfect numbers, Integers 1 6) 009)