Download Odd Perfect Numbers Not Divisible By 3. II

MATHEMATICS OF COMPUTATION
VOLUME 40, NUMBER 161
JANUARY 1983. PAGES 405-411
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 TVis 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])provedthat <o(/V)» 8.
Hagis and McDaniel [3] proved that the largest prime factor of TV> 100129, 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)
> 11.
2. In the remainder of this paper we assume that TVis 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| TVand 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), rAenp ¥=tr.
Lemma 2. Suppose q = 5 or 17 andp" IITV.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, 1982.
1980 MathematicsSubject Classification.Primary 10A20.
*Current address: Department of Mathematics, East Carolina University, Greenville, North Carolina
27834.
©1983 American Mathematical Society
0025-5718/82/0O0O-0742/$02.50
405
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406
MASAOKISHORE
Lemma 3. Suppose pa || TV,q is a prime and qb\\ a + 1. Then TVis 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=100129.Ifp1 2* 103, thenps *£ 113.
Lemma 5. Suppose p, q are odd primes, a, A are positive integers, pb\q+
and 2A > a. 77ien »7j(¡ipa).
I, p > 5
Proof. Since /? and »7are odd primes, »7> 2pb — 1. Suppose a(/?a) = mq for some
integer m. Then a > A and
ff(/>*"')
+ m=a(/7°)
+ w = m(i+
1) =o(pb).
Hence
m >/
- a(/-')
= (y+1
- 2// + 1)/ (p - I),
and
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 > 100129 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* 100129t?xc»?pr
(a) ///? = 5, a = 2,4,6,8,
(b)ifp=
11,a = 2,4.
Proof. We can easily show that 514|fTV and 526jl' TV because a(514) =
11 - 13- 71 - 181- 1741 and a(526) = 19-31•109-271•829-4159-31051.
Then
Corollary 6 follows from Lemmas 1 and 6. Q.E.D.
Lemma 7. If IT || N and a> 8, then p9 > 100129,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= 103and 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 1731a, + 1, and by Lemma 3 TVwould 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~21oipf>), 172\ a¡ + I,
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407
ODD PERFECT NUMBERSNOT DIVISIBLE BY 3. II
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 > 100129.
The same arguments hold if p-, = 103 because 17] o"(/?88). Q-E.D.
Lemma8. If 5" IITVanda> 14, then a > I6,p9> 100129andp]0 > 579281.
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(/),
oip4)\aipb).
then by Lemma 2
Since 131 • 21491 | a(614), 211-2221 | a(714) and
31-391-
13811cr(1014),it is easy to showthat if 5| a(/?*),/? =£61,71, or 101.
Suppose5 | o(41*).Then 5792811a(414).Sincethe order of 5 mod 579281is 72410
and a is even, 579281 \ a(5"), and a(5") has at least one prime factor s* 100129 by
Corollary 6. Hence we may assume that 5\ ai4lb).
Since 3001-3221-241511o(l l24) and 101-4951• 17351| a(3124), 52,a(ll*) and
52Ta(3l*).
Suppose 5|a(ll6). Then 3221|a(ll4), and if 3221C||TV,52fa(3221c) because
151-601• 1301■16011a(322124).Similarly,if 5 | a(31*), then 173511a(314), and if
17351e|| TV,52t a(17351c)because 101-2351 | a(1735124).
Supposepb \\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 || TVwhere q = 3221 or 17351 and 52\ aiqc).
Now we will show that 5"~8177+ 1. 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*100129, 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°"6 | 77 + 1. Suppose
5a_1 IIa(/?^/?ïo0)- Then/?, = 3221 or 17351 and 52\oip^), 5a"2| o(patf),pl0 = 77,
and 5a-41 77+ 1.
Similar arguments show that if two/?, = 1 (5) for 1 < / < 7, then 5°~81 77+ 1 and
that if/?, = 1 (5) for 1 < i < 7, then 5""51 77+ 1.
Since a > 16 and 50-81 tt + 1, 77to(5°) by Lemma 5, 77>2-5a_81 >7-105
> 579281, and Lemma 8 follows from Corollary 6. Q.E.D.
Corollary
8. Suppose 5" \\N, a> 14, and 579281\ TV// 41 | TV.Then a > 16,
p9 > 100129,pw > 2 ■5"-* - 1 > 7 • 105, and 5a's | 77+ 1.
The next lemma is due to McDaniel [6].
Lemma 9. Suppose a>2,a+
ia)Ifp2\ai5a),thenp>229.
1 is a prime, andp is a prime.
(b) Ifp21 0(17"), then p > 227orp = 48947.
Lemma 10. Suppose pa IITV,q\ a(/?fc), A + 11a + 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.
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Q.E.D.
408
MASAOKISHORE
Lemma 11. Suppose IT\\N,
a ^ 8, and 307\ TV.If ps < 1000, rAe/7a > 10,
p9 > 25646167, a«d/?10 > 8 ■108.
Proo/. Since 307 | a(178), a s* 10, and by Lemma 7, 17a"31 77+ 1 and pxo >
2 ■IT'3 - 1 > 8 • 108. Suppose /?8 < 1000 and 100129 < p9 < 25646167 < 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 3071a(172), 88741| a(174) and 256461671a(176). Hence
A> 10. If 1 < /' < 7 and /?,| a(176), then by Lemmas 4, 10 1 = 7 and p1 = 47, 59, or
83. Since 77= pxo\ n(17a) by Lemma 5, we have a(17*) | /?7/?8/?9by Lemma 9. Then
a(17'°) < 83 • 1000■p9, or p9 > 25646167, a contradiction. Hence p9 > 25646167.
Q.E.D.
Corollary
11. If p1 < 29 and ps < 6203, then a > 10, p9 > 25646167, an¿
/?10>8-108.
Proof. As in Lemma 11 a(176) | psp9. Then a(1710) < 6203 •/?<,,orp9 > 25646167,
a contradiction. Q.E.D.
Lemma12. Suppose5" \\N,a> 14, 579281{TVif 411TV,ana /?}TV;//? = 31, 71,
191,409, or 19531.If p%< 41, then a > 22,p9> 12207031,andpw > 2- 5a~8 - 1 >
10'°.
Proof, a > 22 because a =¿=14 as before, 409 | a(516), 191 | a(518), and
19531 I tr(520). The rest of the proof is similar to that of Lemma 11 using a(510) =
12207031and a(512)= 307175781. Q.E.D.
Corollary
pw>2-5a-*
12. If p7< 29 and pg < 6203, then a > 22, p9 > 12207031, and
- 1 > 10'°.
Lemma 13. Suppose 5" IITV,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,ana1
5792811TVZ/411TV.7/Aen/?9> 3 • 106,andpx0 > 12207031.
Proof. Suppose 3011 < p9 < 3 • 106.Then/?10 = a(5a) = 12207031or 305175781,
and 5\oipaxtf) because 1311a(122070314) and 3011| a(3051757814). Suppose
5 I aipf-), 1 < j < 8. Since 3221| a(ll4), 17351| a(314), 579281| a(414) and
104670301I a(1514),/?¡ £ 1 (5). Since /?, ==*
77if /?, = 19, 29, 59, 79, 89, and 149 by
Lemma 1, we have/>8 = 109 = 77by Lemmas 2, 4. If 54| a(/?88), then 531 a8 + 1 by
Lemma 2, and TVwould 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 TVwould 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 > 1010where/?," satisfies the restrictions
implied by Lemma 1},
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409
ODD PERFECT NUMBERSNOT DIVISIBLE BY 3. II
_ ¡a,
'
ifa,<a(/?,),
\a(pi)
if a^aipi).
Lemma 14. SupposeM = ll)ixpb'. Then 0 < log2 - logS(M) < 10■10'10.
Proof. Suppose/?0 || TVand a > a(/?). Then
0 < log Sip")-
log S( pa(p)) < log —P-r - log ^
/?-l
*pa{p)tp_x}
pa'p)+\
log,,a(/>)-H
JL,
I
_
,] = log
1°^
,
_a(/>)+l
_
|
<-!-<io-10.
pac>+1 -1
Hence
0<logS(TV)-logS(M)
10
< 2 IlogS(/?,"')- logS(/>,?')|< 10-10-'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 QjI aipf') for some 1 < i < 7 vví'rA
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
3
log2 < logS(M) + 1 - 10-10+ 2 logV (?, - O7=1
(c) Ifr = 2 and q2 < 100129,thenp%= qx,p9 = q2 and
2
log2<logS(M)+
7-10-'°+
2 log ?,■/(«7;- 1) +log 100129/100128.
Lemma 17. /?8< 3011.
Proof. Suppose /?8 > 3011. 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) + 7 • 10-10+ log3011/3010
+ log 3019/3018 + log 100129/100128.
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MASAOKISHORE
410
The results were:
514712111013917823°«296,
514712111013917623^296,
514712111013617823"<>296,
5,2712111013917823°'296,
510712ll1013917823a<296,
where a6 = 6 or 8. Since
5 7 11 13 17 23 29 6203 6211 100129
4 6 10 12 16 22 28 6202 6210 100128< '
3011 </?8< 6203. By Corollaries 11 and 12 p9 > min{25646167,12207031}and
pxo > min{8 • 108,1010}.Then TVis not OP because
5 7 11 13 17 23 29 3011 12207031800000000
4 6 10 12 16 22 28 3010 12207030799999999<
QED-
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 • 10"9 + log /?/(/? — 1) < log2}. Then q </?I0 < r; in
particular, if there are no primes between q and r, TVis not OP.
Lemma 19./?9<3011.
Proof. Suppose/?8 < 3011 < p9. We used a computer to find M = \\%i=x
/?*■satisfy-
ing Lemmas 1, 4, 7, 8, 15, 16, log 5(M) < log2, and
log2 < logS(M) + 8 • 10"10+ log3011/3010 + log 100129/100128.
There were seventy-two such M's. However, none of them satisfied Lemmas 11, 12
and 13 except
52712111013101910238316596.
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
TVis 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) + 9 ■10"10+ log 100129/100129.
There were thirty-nine such M's; however, none of them satisfied Lemma 18. Hence
TVis 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
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ODD PERFECT NUMBERS NOT DIVISIBLE BY 3. II
411
1. J. E. Z. Chein, An Odd Perfect Number Has at Least 8 Prime Factors, Ph.D. Thesis, Pennsylvania
State University, 1979.
2. P. Hagis, Jr., " Outline of a proof that every odd perfect number has at least eight prime factors,"
Math. Comp.,v. 35, 1980,pp. 1027-1032.
3. P. Hagis, Jr. & W. L. McDaniel,
"On the largest prime divisor of an odd perfect number. II,"
Math. Comp.,v. 29, 1975,pp. 922-924.
4. M. Kishore, "Odd perfect numbers not divisible by 3 are divisible by at least ten distinct primes,"
Math. Comp.,v. 31, 1977,pp. 274 - 279.
5. M. Kishore, "Odd integers N with five distinct prime factors for which 2 - 10~12< o(N)/N < 2
+ 10-'2," Math. Comp.,v. 32, 1978,pp. 303-309.
6. W. L. McDaniel,
"On multiple prime divisors of cyclotomic polynomials," Math. Comp., v. 28,
1974,pp. 847-850.
7. C. Pomerance, "Odd perfect numbers are divisible by at least seven distinct primes," Acta Arith., v.
25, 1973/74,pp. 265-300.
8. C. Pomerance, "The second largest prime factor of an odd perfect number," Math. Comp., v. 29,
1975,pp. 914-921.
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