On a remark of Makowski about perfect numbers

Transcription

1 Elem. Math. 65 (2010) /10/ DOI /EM/149 c Swss Mathematcal Socety, 2010 Elemente der Mathematk On a remark of Makowsk about perfect numbers Lus H. Gallardo Lus H. Gallardo receved hs Ph.D. from the Unversty of Pars 7 n He s an assocate professor at the Unversty of Brest, Brttany, France. He works on arthmetc problems related to polynomals. 1 Introducton A postve number n > 0 s called perfect f t s equal to the sum σ(n) n of ts proper dvsors. Some classcal papers on perfect numbers were wrtten by Perce [3], Servas [5], and Sylvester [7] n the 19th century. More recently, Wessten [9] lsts many other propertes of the perfect numbers ncludng Makowsk s result [2] that 28 s the only even perfect number that exceeds a cube by one. It s natural to wonder whether the result holds for all perfect numbers. More generally, we can ask whether 28 s the only perfect number n that s a sum of two non-negatve cubes, say, n = x 3 + a 3 wth a 0andx > 0. Ths seems to be a very dffcult queston. The object of ths paper s (a) to prove that 28 s the only even perfect number that s a sum of two postve ntegral cubes; (b) to prove (case a = 0) that there are no perfect numbers that are cubes; (c) to descrbe a one parameter famly of sums s of two cubes such that s s odd and σ(s) 2 (mod 4).. Ene postve natürlchezahl n hesst bekanntlch vollkommen, falls n glech der Summe sener echten postven Teler st; de Zahlen 6 und 28 snd bespelswese vollkommen. Sätze von Eukld und Euler besagen, dass ene gerade Zahl n genau dann vollkommen st, wenn n = 2 p 1 (2 p 1) glt, wobe 2 p 1 und somt auch p Prmzahlen snd. Ausgehend von der Glechung 28 = bewes A. Makowsk m Jahr 1961, dass de Zahl 28 de enzge gerade vollkommene Zahl st, de ene Kubkzahl um ens übertrfft. In der vorlegenden Arbet verallgemenert der Autor deses Ergebns dahngehend, dass er nachwest, dass 28 de enzge gerade vollkommene Zahl st, de Summe zweer Kubkzahlen st.

2 122 L.H. Gallardo Now we recall some basc results about perfect numbers. It s well known that any even perfect number n has only two prme factors, so that n = t(2t 1) (1) where t = 2 p 1 for some prme number p such that 2t 1salsoprme. Observethat 2t 1 > t and that gcd(t, 2t 1) = 1. On the other hand, the most basc result about the form of a possble odd perfect number comes from Euler. Euler [1] proved that odd perfect numbers n have the form n = y 4k+1 z 2 (2) wth k 0 a non-negatve nteger, z > 0 a postve nteger and y a prme number such that gcd(y, z) = 1andy 1 (mod 4). Ths s ndeed an easy consequence of the fact that σ(n) 2 (mod 4);whereσ(n) denotes the sum of all postve dvsors of n. Let n be an odd perfect number. Touchard [8] proved that ether n 1 (mod 12) or n 9 (mod 36). (3) 2 The only even perfect number that s also a sum of two cubes s 28 Assume that the even perfect number n s a sum of two cubes: n = x 3 + a 3 = (x + a)(x 2 ax + a 2 ) for some ntegers x > 0anda 0. In partcular a and x have the same party. Let δ be the dscrmnant of x 2 (a + 1)x + a 2 a. One has δ = (3a 2 6a 1). Observe that x 2 ax + a 2 > x + a for all a > 2snceδ<0 f and only f a > 2. Let us assume now that a > 2. The case a {0, 1, 2} wll be consdered later. Thus, from (1) we see that x + a = 2 p 1 and x 2 ax + a 2 = 2 p 1sothat x 2 ax + a 2 = 2(x + a) 1. (4) Snce (4) has ntegral roots x and a + 2 x the dscrmnant = 3a(a 4) of the quadratc x 2 (a + 2)x + (a 1) 2 must be a perfect square. It s easy to see that s non-negatve exactly when a {0, 1, 2, 3, 4}. So t remans to consder the cases a {3, 4}. If a = 3thenx {4, 1}. But n s even so x = 1. Thus, x = 1anda = 3. In other words, we get the perfect number n = 28. If a = 4thenx = 3. Ths s not possble snce a and x have the same party. Now we dscuss the case a {1, 2}. Observe that f x + a x 2 ax + a 2 then we must have x + a > x 2 ax + a 2 snce the perfect number n = (x + a)(x 2 ax + a 2 ) s never a square. Take a = 2. Thus x + 2 > x 2 2x + 4. But ths s not possble for an nteger x. It remans only the case a = 1. In ths case we have as before x + 1 > x 2 2x + 1. Ths s true only for x = 1. So a = 1andx = 1. But n = x 3 + a 3 = 2 s not perfect. Fnally, observe that a = 0 s not possble snce 3 does not dvde the exponent 1 of the prme 2t 1nn = t(2t 1). Thus, n s not a cube. Ths proves the result.

3 On a remark of Makowsk about perfect numbers A perfect number cannot be a cube We have just seen that an even perfect number cannot be a cube. We assume n the rest of the secton that n = x 3 s an odd perfect number. We wll get a contradcton by consderng the equalty σ(n) = 2n (mod 12). It follows from (2) that n = p 12k+9 r 6 for some non-negatve nteger k 0, for some prme number p 1 (mod 4) and for some postve nteger r > 0suchthatgcd(p, r) = 1. Assume that r can be factored as r = q α 1 1 qα m m wth prme numbers q 1,...,q m.wehavethen 3.1 Case 1: gcd(3, n) = 1 2n = σ(n) = σ(p 12k+9 )σ (q 6α 1 1 ) σ(q 6α m m ). (5) From (3) we get 2n 2 (mod 12). Now we compute the value of the rght hand sde of (5) modulo 12. Frst of all (3) mples p {1, 5} (mod 12) for the Euler prme p. So σ(p 12k+9 ) = 1 + p p 12k+9 12k (mod 12) when p 1 (mod 12) and σ(p 12k+9 ) = ( k+10 ) + 5( k+8 ) (6k + 6) + 5(6k + 5) = 36k (mod 12) when p 5 (mod 12), snce trvally (mod 12). Observe that q > 3 s an odd prme. So q {1, 5, 7, 11} (mod 12). Thus, q 2 1 (mod 12). Inotherwordsq 2k+1 q (mod 12) and q 2k 1 (mod 12) for any nteger k. Thus, one has σ(q 6α ) 6α + 1 (mod 12) when q 1 (mod 12),and σ(q 6α ) = q q 6q + (q 1 + q q 6q 1 ) (3α + 1) + (3α )q 3α (1 + q ) + 1 (mod 12) when q 1 (mod 12). Set a = 3α (1 + q ) + 1. Observe that a 1 (mod 12) when q 7 (mod 12) or when q 11 (mod 12). Wehavealsoa 6α + 1 (mod 12) when q 5 (mod 12). Moreover, observe that for any nteger x, ether 6x (mod 12) or 6x (mod 12). Thus, for all one has σ(q 6α ) {1, 7} (mod 12). (6)

4 124 L.H. Gallardo Observe that z { 2, 7} (mod 12) and y {1, 7} (mod 12) mples zy { 2, 1, 7} (mod 12). Thus, m 2n = σ(n) = σ(p 12k+9 ) σ(q 6α ) { 2, 1, 7} (mod 12). Ths contradcts (3) that gves 2n 2 (mod 12); thereby provng the result. 3.2 Case 2: gcd(3, n) = 0 Observe that Hence, as before, we get =1 σ(3 6α ) = 1 + ( ) ( ) (mod 12). σ(n) { 2, 1, 7} (mod 12). But (3) mples that 2n 6 (mod 12). So we get the contradcton σ(n) = 2n. Ths completes the proof that an odd perfect number cannot be a cube. 4 Our man result about a famly of perfect canddates that are sums of two cubes Frst of all a techncal and useful lemma follows: Lemma 4.1. If for some prme number p, for some non-negatve nteger k 0 and for some postve nteger s 1 one has then p 4k+1 = s 1 (7) k = 0. Proof. We choose r = p, A = 2, B = 3 2s 1,andx = 4k + 1nordertohave A + B = r x. We get PQ = 6sncePQ s the largest squarefree dvsor of AB. Thus, n order to have the condton δ = (AB/P) 1/2 s ntegral fulflled, we are forced to take P = 6andQ = 1. Hence, δ = 3 s 1. It then follows from [4, Theorem 3, p. 219] that x < 1 2 QP1/2 ln(p) = 6 ln(6) < 3. 2 But x s odd. So, x = 1andk = 0. Another proof s to choose x = 2r 1, y = 1, n = p, z = 4k + 1 n [4, Lemma 2, p. 228] so that we have We get as before k = 0. 3 x + 2 y = n z.

5 On a remark of Makowsk about perfect numbers 125 Our man result (whose proof s an mmedate corollary of Lemma 4.1 and a check of the formulae) s then: Theorem 4.2. Let r > 0 be a postve nteger. Defne w(r) = 3 4r 2. Let k 0 be a non-negatve nteger such that the dophantne equaton p(r) 4k+1 = r 1, has an ntegral soluton p(r). Letalso,a(r) = 3w(r) 2 1,m(r) = 3w(r) 2 + 3w(r) + 1, n(r) = p(r) 4k+1 3 4r m(r) 2,x(r) = 3 4r 1 p(r) 4k+1 a(r). Then, k = 0 provded p(r) s a prme number. In all cases we have w(r) 1 (mod 4), so that p(r) 1 (mod 4), and a) p(r) 4k+1 = 3w(r) + 2, b) x(r) = 2m(r) 1, x(r) = 6w(r) 2 + 6w(r) + 1, c) x(r) + a(r) = 3w(r)(3w(r) + 2), d) x(r) 2 a(r)x(r) + a(r) 2 = 3(3w(r) 2 + 3w(r) + 1) 2, e) x(r) 3 + a(r) 3 = n(r). In other words there s a one parameter famly {n(r)} of ntegers such that n(r) 1 (mod 4). Moreover f p(r) s prme then n(r) satsfes the necessary condton (2) to be a perfect number. Furthermore, n(r) s a sum of two cubes for each r. It s then of some nterest to obtan the r s for whch p(r) s a prme number. We have n(r) 1 (mod 4) and σ(n(r)) 2 (mod 4). A quck computer check gves a) f r 0 (mod 3) then p(r) 0 (mod 7); b) f r 2 (mod 5) then p(r) 0 (mod 11);etc. Usng a) we constructed a lst L of r s for whch p(r) = r 1 s a prme number. For the moment L ={1, 4, 16, 31, 35, 59, 61, 79, 91, 98, 283, 376, 1801, 10948, 11384, 26536} (8) contans sxteen elements. We have examned all possble r s up to Ths took some tme, e.g., about 2 hours CPU to test each possble canddate r when r s close to, say, Ths was done on a 8 processor lnux machne runnng command lne cmaple 11. We do not know f a r = p(r) s a prme number for an nfnty of r s. Compare wth sequence A n Sloane s database [6]. Observe that n(r) s perfect f and only f 2n(r) = 2 p(r) 3 4r m(r) 2 = (p(r) + 1) σ(3 4r ) σ(m(r) 2 ) = σ(n(r)). (9)

6 126 L.H. Gallardo We deduce from (9) that f n(r) s perfect then the prme p(r) dvdes the product σ(3 4r ) σ(m(r) 2 ). Indeed, we checked (n about only 15 seconds) that for all r L the correspondng n(r) s not perfect snce the prme p(r) does not dvde σ(m(r) 2 ). In order to show that ths suffces we clam that f n(r) s perfect then p(r) does not dvde ρ = σ(3 4r ). To prove the clam assume, to the contrary, that p(r) does dvde ρ. Then from the defnton of w(r) n Theorem 4.2 we obtan that But from Theorem 4.2 part a) we get 9w(r) = 3 4r. 27w(r) 18 (mod p(r)). So ρ = 3(34r ) 1 27w(r) 1 19 (mod p(r)) Thus, p(r) = 19. But 19, contrary to p(r), s not congruent to 1 (mod 4). Thsproves the clam. References [1] Euler, L.: De numers amcablbus. Reprnted n: Opera posthuma, Euler archve [E798]. euler 1 (1862), [2] Makowsk, A.: Remark on perfect numbers. Elem. Math. 17 (1962), 109. [3] Perce, B.: On perfect numbers. New York Math. Dary 2, XIII (1832), [4] Scott, R.; Styer, R.: On p x q y = c and related three term exponental Dophantne equatons wth prme bases. J. Number Theory 105 (2004), [5] Servas, C.: Sur les nombres parfats. Mathess 8 (1888), 135. [6] Sloane, N.J.A.: The On-Lne Encyclopeda of Integers Sequences. Publshed onlne at: njas/sequences, [7] Sylvester, J.J.: Sur les nombres parfats. Comptes Rendus Pars 106 (1888), [8] Touchard, J.: On prme numbers and perfect numbers. Scrpta Math. 19 (1953), [9] Wessten, E.W.: Perfect Number. Lus H. Gallardo Mathematcs Unversty of Brest 6, Avenue Le Gorgeu, C.S F Brest Cedex 3, France e-mal: Lus.Gallardo@unv-brest.fr