Odd Perfect Numbers Not Divisible By 3. II

Transcription

1 MATHEMATICS OF COMPUTATION VOLUME 40, NUMBER 161 JANUARY PAGES Odd Perfect Numbers Not Divisible By 3. II By Masao Kishore* Abstract. We prove that odd perfect numbers not divisible by 3 have at least eleven distinct prime factors. 1. N is called a perfect number if a(7v) = 27V, where a(tv) is the sum of positive divisors of TV. Twenty-seven even perfect numbers are known; however, no odd perfect (OP) numbers have been found. Suppose TV is OP and <o(tv) is the number of distinct prime factors of N. Gradstein (1925), Kühnel (1949), Weber (1951), and the author (1978,[5]) proved that co(tv) > 6. Pomerance (1972, [7]) and Robbins (1972) proved that «(TV) > 1. Hagis (1975,[2]) and Chein (1978,[1]) proved that <o(/v)» 8. Hagis and McDaniel [3] proved that the largest prime factor of TV > , and Pomerance [8] proved that the second largest prime factor of TV > 139. If 31 TV, then Kanold (1949) proved that co(tv) s= 9, and the author (1977,[4]) proved that to(tv) > 10. In this paper we prove Theorem. If N is OP and3\n, then <o(tv) > In the remainder of this paper we assume that TV is OP and 10 TV= ]\pf, i=i where p/s are primes, 5 </>,< <pxo and a/s are positive integers, and we will get a contradiction. We write pf11 TV and a = V (N). The following lemmas were proved in [4] and [7]: Lemma 1. Supposepa \\ TV. Then (a) All a's are even except for one a in which case a =p = 1(4). We write tr for p. (b)//; l(3),os2(3), (c) Ifp = 1 (4) andp = 2 (3), raen p = tr. Lemma 2. Suppose q = 5 or 17 andp" II TV. Then Vqia+l) Vq(oiP )) = \vq(p+l) ifp = l(q), + Vq(a + 1) ifp = -\(q) andp = it, 0 otherwise. Received November 2, 1981; revised June 21, Mathematics Subject Classification. Primary 10A20. * Current address: Department of Mathematics, East Carolina University, Greenville, North Carolina American Mathematical Society /82/0O0O-0742/$02.50

2 406 MASAO KISHORE Lemma 3. Suppose pa TV, q is a prime and qb\\ a + 1. Then TV is divisible by at least c distinct primes = liq) other than p, where c = b if qb = a + 1, and c = 2b if qb^a+ I. The proof of the next lemma is similar to that of Lemma 6 in [4]. Lemma 4. px = 5, p2 = 7, p3 = 11, p4 < 17, p5 < 23, p6 < 37, p-, < 107, p9 > 139, pw 3= Ifp1 2* 103, thenps * 113. Lemma 5. Suppose p, q are odd primes, a, A are positive integers, pb\q+ I, p > 5 and 2A > a. 77ien»7j ( ipa). Proof. Since /? and»7 are odd primes,»7 > 2pb 1. Suppose a(/?a) = mq for some integer m. Then a > A and Hence ff(/>*"') + m=a(/7 ) + w = m(i+ 1) =o(pb). and m >/ - a(/-') = (y+1-2// + 1)/ (p - I), a(/y) = mq> (2pb - l)(pb+x - 2pb +l)/(p-l) = (2p2b+x - 4p2b-pb+x + 4p» - l)/(p - 1) >(P2b+x-l)/iP-l)=o(p2b)>o(p ), because p > 5 and 2A > a, a contradiction. Q.E.D. Remark. Lemma 5 also holds if p = 3 and 2A > a. The next lemma is due to Hagis. Lemma 6. Suppose p = 5 or 17 andp" is a component of an OP number. Then oipa) has at least one prime factor > except (a) ifp = 5, a =1,2,4,5,6,8,9,13,14,17,26,29. (b)ifp= 17, a =1,2,4,5,9. Corollary 6. Suppose p = 5 or 17 andp" \\ TV. TAen a(/>") Aas ai /easf one prime factor s* t?xc»?pr (a) ///? = 5, a = 2,4,6,8, (b)ifp= 11, a = 2,4. Proof. We can easily show that 514 ftv and 526 jl' TV because a(514) = and a(526) = Then Corollary 6 follows from Lemmas 1 and 6. Q.E.D. Lemma 7. If IT N and a> 8, then p9 > , pxo>2- IT~3-1 > 2 106, a«i/17a-3 77-r- 1. Proof. If p is a prime andp < 113, then/? z ± 1 (17) except forp = 103. Hence by Lemmas 1, 2 and 4 if 17 oipf) for 1 «i < 7, then i = 7,/>7 = 103 and 171 <*(/>88)- Suppose /?7 = 103. Then 17 oip^p^p"^). Since a > 8, 173 a(/>,a') for some 8 < ; < 10. If p = 1 (17), then 1731 a, + 1, and by Lemma 3 TV would have at least three more primes = 1 (17), a contradiction. Hence p. = -I (17) and />, = 77. Then by the same lemma 172f a(/>/>) for j = i, 8 <j < 10, 17a~21 oipf>), 172 \ a + I,

3 ODD PERFECT NUMBERS NOT DIVISIBLE BY 3. II 407 and 17a"3 \ p,+ 1 by Lemma 2. By Lemma 5 p \ a{lt), and by Corollary 6 a(17") has at least one prime factor > The same arguments hold if p-, = 103 because 17] o"(/?88). Q-E.D. Lemma 8. If 5" II TV anda> 14, then a > I6,p9> andp]0 > Proof. We showed that a = 14 in the proof of Corollary 6. Suppose pb \\ TV, p = 1 (5)and/?< 107. Then p= 11, 31, 41, 61, 71, or 101. If 5 a(/), then by Lemma 2 oip4)\aipb). Since a(614), a(714) and cr(1014), it is easy to show that if 5 a(/?*),/? = 61, 71, or 101. Suppose 5 o(41*). Then a(414). Since the order of 5 mod is and a is even, \ a(5"), and a(5") has at least one prime factor s* by Corollary 6. Hence we may assume that 5\ ai4lb). Since o(l l24) and a(3124), 52,a(ll*) and 52Ta(3l*). Suppose 5 a(ll6). Then 3221 a(ll4), and if 3221C TV, 52fa(3221c) because a(322124). Similarly, if 5 a(31*), then a(314), and if 17351e TV, 52t a(17351c) because a( ). Suppose pb \\N,p= -1 (5), and p < 107. Then p = 19, 29, 59, 79, or 89, and by Lemma 1 p = it. Hence by Lemma 2 5 { a( /?*). In summary if pb TV, p < 107, and if 5 oipb), then/? = 11 or 31, in which case qc TV where q = 3221 or and 52\ aiqc). Now we will show that 5"~ Suppose three/?, = 1 (5) for 1 < i < 7. Then /?3 = 11, p6 = 31 and p1 = 41, and it is easy to show that 41 </?8 < 61. Hence 5tf(/?88). Smce T^io5* , the above summary shows that 52t a(n,8=1/??') Suppose 5" a(/?99/? ). By a similar argument used in the proof of Lemma 7 we have for i = 9 or 10 5"~4 aipf), /?, = 77, and 5 " Suppose 5a_1 II a(/?^/?ïo0)- Then/?, = 3221 or and 52\oip^), 5a"2 o(patf),pl0 = 77, and 5a Similar arguments show that if two/?, = 1 (5) for 1 < / < 7, then 5 ~ and that if/?, = 1 (5) for 1 < i < 7, then 5"" Since a > 16 and tt + 1, 77to(5 ) by Lemma 5, 77>2-5a_8-1 >7-105 > , and Lemma 8 follows from Corollary 6. Q.E.D. Corollary 8. Suppose 5" \\N, a> 14, and \ TV // 41 TV. Then a > 16, p9 > , pw > 2 5"-* - 1 > 7 105, and 5a's The next lemma is due to McDaniel [6]. Lemma 9. Suppose a>2,a+ 1 is a prime, andp is a prime. ia)ifp2\ai5a),thenp>229. (b) Ifp21 0(17"), then p > 227 orp = Lemma 10. Suppose pa II TV, q\ a(/?fc), A + 11 a + 1, q < 107, and q, b + 1 are primes. Then (a) Ifp = 5, q = 11, 31, 59, or 71. (b) If p = 17, q = 47, 59, or 83. Proof. Suppose p = 5 or 17, and d is the order of /? mod q. Then pd = 1 iq), and d\ A + 1. Since p z 1 iq), d > 1, and d = b + I because A + 1 is an odd prime. The order d is not an odd prime except for those q stated above. Q.E.D.

4 408 MASAO KISHORE Lemma 11. Suppose IT\\N, a ^ 8, and 307 \ TV. If ps < 1000, rae/7 a > 10, p9 > , a«d/?10 > Proo/. Since 307 a(178), a s* 10, and by Lemma 7, 17a" and pxo > 2 IT'3-1 > Suppose /?8 < 1000 and < p9 < < 227. Choose A such that A + 1 is a prime and A + 11 a + 1. Then 0(17*) o(17"), and A = =2, 4, or 6 because 3071 a(172), a(174) and a(176). Hence A > 10. If 1 < /' < 7 and /?, a(176), then by Lemmas 4, 101 = 7 and p1 = 47, 59, or 83. Since 77 = pxo\ n(17a) by Lemma 5, we have a(17*) /?7/?8/?9 by Lemma 9. Then a(17' ) < p9, or p9 > , a contradiction. Hence p9 > Q.E.D. Corollary /?10> If p1 < 29 and ps < 6203, then a > 10, p9 > , an Proof. As in Lemma 11 a(176) psp9. Then a(1710) < 6203 /?<,, orp9 > , a contradiction. Q.E.D. Lemma 12. Suppose 5" \\N,a> 14, { TV if 411 TV, ana /?} TV ;//? = 31, 71, 191, 409, or If p%< 41, then a > 22, p9> , andpw > 2-5a~8-1 > 10'. Proof, a > 22 because a = = 14 as before, 409 a(516), 191 a(518), and I tr(520). The rest of the proof is similar to that of Lemma 11 using a(510) = and a(512) = Q.E.D. Corollary pw>2-5a-* 12. If p7< 29 and pg < 6203, then a > 22, p9 > , and - 1 > 10'. Lemma 13. Suppose 5" II TV, a = 10 or 12, p% < 151, p9 > 3011, at most two p = 1 (5) for 1 < i < 8, /?, = 11, 31, 41, or 151 <//?, = 1 (5) and 1 < /' < 8, /?,. = 19, 29, 59, 79, 89, 109, or 149 ///?, = -1 (5) and 1 < i < 8,/?f TV «//? = 131, 3221, or 17351, ana TV Z/411 TV. 7/Aen /?9 > 3 106, andpx0 > Proof. Suppose 3011 < p9 < Then/?10 = a(5a) = or , and 5\oipaxtf) because 1311 a( ) and 3011 a( ). Suppose 5 I aipf-), 1 < j < 8. Since 3221 a(ll4), a(314), a(414) and I a(1514), /? 1 (5). Since /?, ==* 77 if /?, = 19, 29, 59, 79, 89, and 149 by Lemma 1, we have/>8 = 109 = 77 by Lemmas 2, 4. If 54 a(/?88), then 531 a8 + 1 by Lemma 2, and TV would have at least six prime factors = 1 (5), a contradiction. Hence 54\ o(/? 8)» and s» 5 "31 a(/?^«). P9 ^ w, /?9 = 1 (5) and TV would have at least 7 more prime factors = 1 (5), a contradiction. Hence 51 tr(pf') for 1 < /' < 8. Then 5a o(p$>), p9 = 77, 52 a9+l, 5a_11 p9 + 1, and /?9 > 2-5a"' - 1 > 3 106, a contradiction. Q.E.D. Definition. Suppose M = Wrj=x pf'. Then 5(M) = a(m)/m, ai Pi) = min a, /?,a+1 > 1010 where/?," satisfies the restrictions implied by Lemma 1},

5 ODD PERFECT NUMBERS NOT DIVISIBLE BY 3. II 409 _ a, ifa,<a(/?,), ' \a(pi) if a^aipi). Lemma 14. Suppose M = ll)ixpb'. Then 0 < log2 - logs(m) < 10 10'10. Proof. Suppose/?0 TV and a > a(/?). Then Hence 0 < log Sip")- log S( pa(p)) < log P-r - log ^ /?-l *pa{p)tp_x} pa'p)+\ I, log JL,, = log 1 -,,a(/>)-h _ ] ^ _a(/>)+l _ <-!-<io-10. pac>+1-1 0<logS(TV)-logS(M) 10 < 2 I log S(/?,"') - log S(/>,?') < '0. =1 Q.E.D. The proof of the next two lemmas is easy. Lemma 15. // q is a prime, q\ a(/?," ) for some Ki^7 q < p-j, then q = 2 or q /?,/or some 1 < t < 7. with a, < a(/?,), and if Lemma 16. Suppose M = W1i=xpb' and L = M \\rj=xq'j' where q is a prime, 1j > Pn Qj I aipf') for some 1 < i < 7 vví'r A, < a(/? ),»/,< <»7,., and Cj is the minimum allowable power of q determined by Lemma 1. If there is no such q, then r = 0 and the product is defined tobe 1. 77it?«(a)r^3 analog 5(L) < log 2. (b) Ifr = 3, thenp% = qx,p9 = q2,pl0 = a3 and log2 < log S(M) log V (?, - O- 3 7=1 (c) Ifr = 2 and q2 < , thenp% = qx,p9 = q2 and 2 log2<logs(m) ' + 2 log?, /(«7; - 1) +log / Lemma 17. /?8 < Proof. Suppose /?8 > We used a computer (PDP 11/70 at the University of Toledo) to find M = U]=xpf' satisfying Lemmas 1, 4, 15, 16, log 5(A/) < log2, and log2 < logs(m) log3011/ log 3019/ log /

6 410 MASAO KISHORE The results were: where a6 = 6 or 8. Since «296, ^296, "<>296, 5, '296, ll a<296, < ' 3011 </?8< By Corollaries 11 and 12 p9 > min{ , } and pxo > min{8 108,1010}. Then TV is not OP because The proof of the next lemma is also easy. Lemma 18. Suppose M = II?=,/?*', q = max{/? /? is a prime and log S(M) + log Sip") > log2 where a is the minimum allowable power of p) and r = min{p \ p is a prime and log SiM) "9 + log /?/(/? 1) < log2}. Then q </?I0 < r; in particular, if there are no primes between q and r, TV is not OP. Lemma 19./?9< < QED- Proof. Suppose/?8 < 3011 < p9. We used a computer to find M = \\%i=x /?* satisfying Lemmas 1, 4, 7, 8, 15, 16, log 5(M) < log2, and log2 < logs(m) "10 + log3011/ log / There were seventy-two such M's. However, none of them satisfied Lemmas 11, 12 and 13 except It is easy to show that 7753 </?9 < 8389, a2 > 22 (o(712) is a prime), a3 ^ 16, a4 s* 16, a5 = 10 or > 16 (if a(1910) is a prime, a5 > 16), a6 > 12, a7 > 12, and a8 > 12. Then for each/?9 with 7753 </?9 < 8389 Lemma 18 is not satisfied. Hence TV is not OP. Q.E.D. Proof of Theorem. By Lemma 19 /?9<3011. We used a computer to find M = n9=1 /?*' satisfying Lemmas 1, 4, 7, 8, 15, 16, log SiM) < log2, and log2 < logs(m) "10 + log / There were thirty-nine such M's; however, none of them satisfied Lemma 18. Hence TV is not OP. Q.E.D. Acknowledgement. The author would like to thank Professor P. Hagis, Jr. for supplying the table in Lemma 6. Department of Mathematics The University of Toledo Toledo, Ohio 43606

7 ODD PERFECT NUMBERS NOT DIVISIBLE BY 3. II J. E. Z. Chein, An Odd Perfect Number Has at Least 8 Prime Factors, Ph.D. Thesis, Pennsylvania State University, P. Hagis, Jr., " Outline of a proof that every odd perfect number has at least eight prime factors," Math. Comp., v. 35, 1980, pp P. Hagis, Jr. & W. L. McDaniel, "On the largest prime divisor of an odd perfect number. II," Math. Comp., v. 29, 1975, pp M. Kishore, "Odd perfect numbers not divisible by 3 are divisible by at least ten distinct primes," Math. Comp., v. 31, 1977, pp M. Kishore, "Odd integers N with five distinct prime factors for which 2-10~12 < o(n)/n < '2," Math. Comp., v. 32, 1978, pp W. L. McDaniel, "On multiple prime divisors of cyclotomic polynomials," Math. Comp., v. 28, 1974, pp C. Pomerance, "Odd perfect numbers are divisible by at least seven distinct primes," Acta Arith., v. 25, 1973/74, pp C. Pomerance, "The second largest prime factor of an odd perfect number," Math. Comp., v. 29, 1975, pp