ODD PERFECT NUMBERS, DIOPHANTINE EQUATIONS, AND UPPER BOUNDS

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1 ODD PERFECT NUMBERS, DIOPHANTINE EQUATIONS, AND UPPER BOUNDS PACE P. NIELSEN Abstract. We obtan a new upper bound for odd multperfect numbers. If N s an odd perfect number wth k dstnct prme dvsors and P s ts largest prme dvsor, we fnd as a corollary that 0 2 P 2 N < 2 4k. Usng ths new bound, and extensve computatons, we derve the nequalty k 0. Introducton One of the oldest unsolved problems n mathematcs s whether there exsts an odd perfect number N. There are many roadblocks to the exstence of such a number. For nstance, from [4] we now know that N > and N has at least 0 prme factors countng multplcty. If k s the number of dstnct prme factors, then as proved n [2, 3] we have k 9 and N < 2 4k. A lst of other restrctons can be found n [3]. Whle work wth odd perfect numbers has been mostly computatonal, the bound N < 2 4k s a purely theoretcal result. Due to ts doubly exponental growth t has not been used serously n calculatons. In ths paper we fnd a way to make ths upper bound an effectve estmaton tool. As an applcaton, we are able to prove that an odd perfect number must have at least 0 dstnct prme factors. The followng s a bref outlne of the paper. In the focus s on generalzng the upper bound N < 2 4k, and mprovng t n a small, but mportant, way. Ths generalzaton works for odd multperfect numbers, for the spoof odd perfect numbers, and n more abstract settngs. In 2, by applyng ths new upper bound we show that f p and q are dstnct prmes and p a q b N then there are reasonably szed bounds on a and b n terms of k, the number of dstnct prme dvsors. Due to a massve computaton, takng a few months, t follows that gcdσp a, σq b has only moderately szed prme dvsors. In 3 we descrbe mprovements to the methods n [3], takng advantage of the new nformaton we have. Ths s followed by another computer calculaton, provng that any odd perfect number has at least 0 dstnct prme factors. Fnally, n 4 we descrbe one of many roadblocks whch prevents us from provng k > 0. Startng n 2, readers should be famlar wth basc facts about odd perfect numbers, ncludng knowledge of congruence restrctons related to the specal prme. As ths paper s an extenson of the methods used n [3], startng n 3 the reader should be famlar wth the deas n that paper.. A better upper bound Let N be a postve nteger. Followng the lterature, N s sad n ncreasng order of generalty to be perfect when σn/n = 2, multperfect when σn/n Z, and n/d-perfect when σn/n = n/d. For smplcty, we wll always assume n, d Z >0. Note that n/d does not need to be n lowest terms. Wrtng N = k pe where p <... < p k are the prme dvsors of N, the equaton σn/n = n/d 200 Mathematcs Subject Classfcaton. Prmary N25, Secondary Y50. Key words and phrases. Dophantne equaton, perfect number.

2 2 PACE P. NIELSEN can be rewrtten as d k e j=0 j=0 p j x j = n Ths motvates us to look at the Dophantne equaton k e k d = n n k varables x,..., x k. It turns out that f we fx k and look for nteger solutons wth the x s greater than and odd, then there are fntely many solutons. In fact, there s an explct upper bound on k xe n terms of n, d, and k, but ndependent of the e. Many of the deas for ths result flow from proofs n [4, 7, 2]. As we wll generalze and mprove these results, and as some of the proofs are scattered n the lterature, we nclude all the needed peces here. k p e. Lemma.. Let x, x 2, w R >0, and assume w <. We have 2 x x2 wx w x 2 f and only f w x 2 /x. Furthermore, equalty holds n 2 f and only f w = x 2 /x. In partcular, f x x 2 then strct nequalty holds n 2. Proof. Ths follows from basc algebrac manpulaton. Note that we can characterze equalty and strct nequalty n 2 under the much weaker assumpton that x, x 2, and w are nonzero. However, there s no need for ths generalty. Lemma.2. Let < x x 2... x n and < y y 2... y n be non-decreasng sequences of real numbers satsfyng m m 3 x for every m n the range m n. Then we have n 4 x n where equalty holds f and only f x = y for every. y x e y, Proof. We follow Cook [4], but mprove a lttle. We wsh to mnmze n 5 y subject to the constrants that the y form a non-decreasng sequence and satsfy 3 for each m. If we set N = n x, then lowerng each y whch s bgger than N down to N wll only decrease 5. Thus, we see that any mnmzng soluton belongs to the compact set nsde the box [x, N] n subject to the constrants gven n 3 and the constrant that the y are non-decreasng. The nequalty y > s an open condton, but the closed condton y x mples t. Thus we may fx the y so that they n fact mnmze 5. Assume, by way of contradcton, that ths mnmzng soluton does not agree wth the x. We let r be the frst ndex where x r y r, and so from 3 we have x r < y r. As the y mnmze 5, we see that r < n. We let t be the largest ndex where y t = y r+.

3 OPNS, DIOPHANTINE EQUATIONS, AND UPPER BOUNDS 3 Defne the new sequence y f r, t z = wy r f = r w y t f = t for some real number w wth 0 < w < to be further specfed shortly. Consder what happens f we replace the y s wth the z s. Frst, we choose w so that t satsfes w > y r /y r where y r = n case r =, and also w > y t /y t+ where ths condton s vacuously satsfed f t = n. Wth these assumptons on w, the new sequence {z } satsfes < z z 2... z n. Second, the quantty 5 decreases by Lemma.. Thrd, 3 stll holds when m < r or m t, snce n those cases m y = m z. We can make 3 hold for an m n the nterval r m < t, f we have a strct nequalty m x < m y by assumng w > m x /y for each such m. Notce that ths strct nequalty does hold when m = r. However, f the strct nequalty held for all m n the nterval r < m < t, then ths would contradct our assumpton that the y s were a mnmzng choce. Thus, we must have an equalty s s 6 x = for some s satsfyng r < s < t. By the defnton of r we also have the equalty r x = r y, hence s s 7 x = y. =r Recall that x r < y r and that y r+ = y r+2 = = y s = = y t. Thus 7 turns nto s =r+ x > s =r+ y = ys s r. As the x are a non-decreasng sequence, we have s x s+ x s =r+ =r y x /s r > y s = y s+ where the last equalty holds snce r + s < t usng the defnton of t. But then 6 mples s+ x > s+ y, contradctng 3 when m = s +. As we reached a contradcton n every case, ths proves that the only mnmzng soluton s when x = y for every. Before puttng the prevous lemma to good use, we need one more straghtforward result. Lemma.3. Let r Z >0 and defne F r : R R 0 by the rule F r x = x 2r x 2r. The functon F r s monotoncally ncreasng. Proof. Indeed, the dervatve F rx = 2 r x 2r 2 r x 2r = 2 r x 2r 2x 2r s postve when x. The followng s just a slght strengthenng of [2, Lemma ], allowng equalty among the x s. Lemma.4. Let r, a, b Z >0 and let x,..., x r be ntegers wth < x... x r. If r 8 x ab r < x then a r x a + 2r a + 2r.

4 4 PACE P. NIELSEN Proof. Work by nducton on r. Notce that a < b n any case. When r = we have x b/b a whch s maxmzed when b = a +. Thus ax aa + = a + 2 a Now assume that r 2 and also assume that the lemma holds for all ntegers smaller than r and for any choces for a and b. Treatng a as a fxed constant, we can assume that b has been chosen, along wth ntegers < x x 2... x r, so that r x s maxmal and 8 holds. Next, set n = a for < r, and set n r = a + 2r. We have < n < n 2 <... < n r and r n Thus, from our maxmalty assumpton, 9 = a r a + < r n r x. If ax < an = a + 2, then after multplyng 8 by r x =2 r ax bx < =2 n. x x we have x. The nducton hypothess mples ax r =2 x ax + 2r ax + 2r 2 < a+ 2r a+ 2r. Thus we may as well assume n x. x x 2 x x 2 If ax x 2 < an n 2 = a + 4 then multplyng 8 by and performng a smlar computaton yelds the upper bound we seek. Thus we may assume n n 2 x x 2. Repeatng ths argument, we have m n m x for m < r. But ths also holds when m = r by 9. Lemma.2 now apples, so we have r r x n = a a +. But as 8 holds for some b we must have b = a +. Agan appealng to Lemma.2, we have x = n for all. In ths case, we compute r a n = a + 2r a + 2r as desred. Remark. The bound gven n the lemma s the best possble n case b = a +. To smplfy notaton, throughout the paper we let F r be defned as n Lemma.3. Thus, 8 says r a x F r a +. We also defne F 0 x := x. Notaton. Let X be a fnte set of ntegers. We wrte ΠX for x X x, wth the empty product equalng. We wll also wrte Π X for x X x. The followng lemma s an mprovement on the author s work n [2], and also mproves on the recent paper [3]. Whle seemngly nnocuous, the mproved result has far-reachng effects. The man dfference s the denomnator of nequalty.

5 OPNS, DIOPHANTINE EQUATIONS, AND UPPER BOUNDS 5 Lemma.5 cf. [2, Theorem ],[7, Lemma 2]. Let k, n, d Z >0. Suppose holds for some choce of postve nteger exponents {e,..., e k }, and odd ntegers X = {x,..., x k } each greater than. Let S be a possbly empty subset of X. There exst sets S, S X satsfyng S S =, S S S, and such that f we let w = S, v = S, T = S S \ S, δ = d e x S j=0 xj, and ν = n x S xe then: δ x X\S e j=0 xj = ν x X\S xe. δπt < ΠS Π S F v+wdπs +. Proof. We are assumng that the elements n X are odd, and so the fracton n/d, when wrtten n lowest terms, has odd denomnator. In partcular 0 x d n. x S Thus, we have two cases to consder. Case : x S x > d n. In ths case, set d = dπs and n = nπ S. From x S x > d n we see d /n <. Further, we calculate x < x x x X x = x e e = d X x e x X j=0 xj n. Therefore x < d n <. x / S Ths mples that there s a subset S X \S such that f we wrte S = {y,..., y w } wth y... y w then w y d w n < y and so by Lemma.4 we have d ΠS F w d +. Usng the defnton of d we rewrte ths as 2 dπsπs F w dπs +. Notce that we also have 3 x S S x S S x S S x < d n. Ths completes the constructon of S n Case. Case 2: x S x < d n. We set S = so w = 0. We note that nequaltes 2 and 3 stll hold n ths case. The constructon of the set S s the same n both Case and Case 2. We only need to know that nequaltes 2 and 3 hold n both cases, so we contnue wth the general constructon. Put d = dπsπs and n = nπ SΠ S. Inequalty 3 s equvalent to n /d <, whch we wll use shortly. We calculate x e + = e j=0 xj e j=0 xj = n x x e x e d x X

6 6 PACE P. NIELSEN and hence 4 x S S z ez+ x e+ n d <. We pck a subset S = {z,..., z v } S S such that z ez+... zv ezv+ and v n v d <. z ez+ By ez we mean the exponent correspondng to z. By Lemma.4, and the nequalty n < d, we have 5 n x e+ F v n + F v d = F v dπsπs. x S Ths completes the constructon of S. We now only need to verfy propertes and. Property s obvous, comng from equaton. For property, we compute δπt = d x e + ΠSΠS x ΠS by the defnton of δ and T = x S ΠS Π S d ΠS Π S n x e+ by the defnton of d x S x S x e+ by nequalty 4 ΠS Π S F vdπsπs by nequalty 5 ΠS Π S F vf w dπs + by nequalty 2 < ΠS Π S F v+wdπs +. Thus, we have establshed the needed nequalty. We are now ready to put all of these lemmas together to prove an mproved upper bound on odd mult-perfect numbers. The mprovement from prevous results s n the denomnator term. Theorem.6. Let k, n, d Z >0, and suppose equaton holds for some choce of postve nteger exponents {e,..., e k } and odd ntegers X = {x,..., x k } each greater than. In ths case k x e < F 2kd + nπxπ X 2k d + 2 < nπxπ X. Proof. Let X 0 = X, n 0 = n, d 0 = d, and S 0 =. Usng Lemma.5, we can construct S 0, S 0, w 0, v 0, ν 0, δ 0, and T 0 usng the same notaton, just wth the extra subscrpt satsfyng propertes and. Thus 6 δ 0 x X 0\S 0 e j=0 x j = ν 0 x X 0\S 0 Puttng X = X 0 \ S 0, S = T 0, n = ν 0, and d = δ 0, we see from equaton 6 that we can agan use Lemma.5. Hence, we can construct S, S, T, and so forth. We contnue ths process of repeatedly usng Lemma.5, ncreasng the ndces at every step. Snce S, we see that X 0 x e

7 OPNS, DIOPHANTINE EQUATIONS, AND UPPER BOUNDS 7 X..., and so ths process must termnate n at least k steps, say X r+ =. Further, we see that r =0 w = r =0 v = k snce for each element x X there are unque ndces j such that x s added nto S, and then put nto S j. Usng property, repeatedly, we have d r+ ΠS r+ < ΠS r Π S r F w r+v r d r ΠS r + ΠS r Π S r F w r+v r < ΠS r Π S r F w r+v r ΠS r Π S r F w r +v r d r ΠS r + + w r +v r + ΠS r Π S r d r ΠS r + 2 Note that w r, and v for each. Also observe that d r ΠS r + 2w r Thus F wr+v r ΠS r Π S r d r ΠS r + 2 w r +v r + F wr+v r ΠS r d r ΠS r + 2 w r +v r ΠS r F 2 w r+v r d r ΠS r + 2w r +v r = ΠS r F 2 w r+v r+w r +v r d r ΠS r +. ΠS r by 2. Contnung our nequalty from before, and repeatng the deas used n the computatons above, we have d r+ ΠS r+ < ΠS r Π S r ΠS r F 2 w r +w r+v r +v r d r ΠS r + < ΠS r S r Π S r S r F w r +w r+v r +v r d r ΠS r + <... < Π r =0 S Π r r =0 S F =0 w+v d 0 ΠS 0 + = ΠXΠ X F 2kd 0 ΠS Now, d 0 = d, S 0 = S r+ =, and d r+ = d x X the nequalty above, we obtan the theorem. x e + x = n x X xe. Pluggng these values nto Remark. It s beleved that there are an nfnte number of even perfect numbers. Such numbers necessarly have exactly two dstnct prme factors. Thus, the prevous theorem should at least conjecturally prove false f we do not stpulate that the x are odd, even f we force the x to be prme. In any case, we do have the nfnte famly of solutons to when k = 2, x = 2, x 2 = 2 m, for n = 2, d =, e = m, and e 2 =, when we remove the hypothess that the x are odd. There also exst more exotc nfnte famles lke x = 3, x 2 = 3, x 3 = 3 m, for n = 2, d =, e =, e 2 = m, and e 3 =. Remark. The hypotheses n the prevous theorem are weak enough to capture the so called spoof odd perfect number constructed by Descartes; N = , where 2202 = s treated as a prme. Accordng to the work of Dttmer [5], there are no other spoofs of ths sort wth k 7. On

8 8 PACE P. NIELSEN the other hand, our condtons are not weak enough to capture spoofs nvolvng negatve ntegers, such as N = whch s attrbuted to Greg Martn. Usng a clever dea from [3] whch was ndependently suggested to me by Mtsuo Kobayash, we can fnd a cleaner upper bound. Corollary.7. Usng the assumptons and notatons of Theorem.6, and settng N = k followng chan of nequaltes holds: 7 N < d F 2kd + < dd + k 2 2. F k+ d + In partcular, when N s an odd multperfect number we acheve 8 N < 2 2k 2. Proof. By hypothess, we have the equalty k x e+ x e x = n d whch we can rewrte n the form k x e+ = nπ X dπx. Applyng Lemma.4 wth a = nπ X and b = dπx, we see that k nπ X x e+ = nπ XΠXN F k nπ X +. xe, the After dvdng both sdes by nπ XΠX, and usng the fact that dπx nπ X +, we arrve at the nequalty 9 N d nπ X + 2k nπ X + 2k nπ X + 2 nπ. X + If nπ X + < d + 2k, then as the quantty on the rght-hand sde of 9 s a strctly ncreasng functon n terms of nπ X+, we obtan the frst nequalty n 7. On the other hand, f nπ X+ d + 2k, then usng the bound found n Theorem.6, we agan acheve the frst nequalty n 7. Now, we prove the second nequalty n 7. As y ab /y a = y ab a + y ab 2a + + y a + < y ab a+ for y 2, we calculate d + 22k d + 22k d = dd + 2 2k 2 k d + 22k d + 2k+ d + 2k d + 2k For the last statement, take d =. < dd + 22k 2 k +2 2k 2 k + = dd + 2k 2. Those famlar wth the bounds n [3, 2] wll notce that these new bounds are not sgnfcantly smaller. For the purposes of ths paper, we want to concentrate on the stuaton where N s dvsble by a large prme. Hence, n what follows we do not want to ncorporate the ΠX and Π X terms nto an upper bound for N. Ths naturally leads to an nequalty that seems much weaker, but wll often suffce for our purposes. Corollary.8. Let N be an odd perfect number wth k dstnct prme factors. If P s the largest prme factor of N, then 0 2 P 2 N < 2 4k.

9 OPNS, DIOPHANTINE EQUATIONS, AND UPPER BOUNDS 9 Proof. Snce N s perfect we take n = 2 and d =. Clearly, 2P > P. It s known, due to work of Iannucc [8, 9] that the second largest prme factor of N s bgger than 0 4, and the thrd largest prme factor s bgger than 0 2. The corollary now follows by specalzng the man result of Theorem.6 to ths case. In the past, these types of theoretcal upper bounds on odd perfect numbers were of lttle use n calculatons due to ther doubly exponental growth. However, n the next secton we fnd a way to explot the exstence of very large prme dvsors of N when they occur, whch then makes the upper bound a feasble computatonal tool. 2. Usng the GCD algorthm In [3] t was proved that f N s an odd perfect number wth k dstnct prme factors, p s a Fermat prme, and p a N wth a large, then the specal prme factor of N s large. In partcular, we can use ths fact n conjuncton wth Corollary.8 to obtan an upper bound on the sze of the specal prme, and hence on the sze of a. To do so, we frst need to recall a few well-known results. The next three lemmas state some lmtatons on the exponent a. In partcular, t cannot have too many prme dvsors or be too large. Lemma 2.. Let p be a prme and let N be an odd perfect number wth k dstnct prme dvsors. If p a N wth a Z >0 then σ 0 a + k. Proof. Weaken [3, Lemma 4]. Lemma 2.2. Let p be a Fermat prme, let N be an odd perfect number, and let π be the specal prme factor. Let a, m Z >0. If p m π + and p a N wth a < 3m then π σp a. Proof. Ths s a specalzaton of [, Lemma ]. Lemma 2.3. Let q be a Fermat prme, let N be an odd perfect number wth k dstnct prme factors, let π be the specal prme factor, and suppose q a N for some a Z >0. If a > k k 2 then 2q a k 2k 3 k 2/2 π +. Proof. Ths s a sgnfcant weakenng of [3, Proposton 7]. In the notaton of that proposton, we take b = 0 and assume q s the only known prme. Thus k 2 = k and k = l = 0. Compare wth [6, Proposton 3.]. The followng proposton takes these lemmas and n conjuncton wth the man theorem of the prevous secton fnds strong upper bounds on the exponent a. Proposton 2.4. Let p and q be odd, dstnct prmes less than 80, and let T = {3, 5, 7}. Let N be an odd perfect number wth k 9 dstnct prme dvsors. Suppose p a N, q b N, and 0 00 < p a < q b for some a, b Z >0. If p, q T, then a < 4 9 log logq 2 log logp If p T and q / T, then a < 4 9 log2 + 2 logq 2 log0 6 logp If p / T and q T, then a < 4 9 log logq 2 log0 6 logp

10 0 PACE P. NIELSEN v If p, q / T, then a < log2 + 2 logq 2 log0 logp + 2. Proof. A quck computer search, restrctng to odd prmes p and q less than 80, demonstrates that f q p mod p n then n 3 except for the par p, q = 3, 63. For those pars p, q 3, 63, by Lemma 2. and [3, Lemma 2] we have v p σq b k + 3. Ths nequalty also holds for the exceptonal par p, q = 3, 63 snce the multplcatve order of 63 modulo 3 s and 63 cannot be the specal prme. Smlarly, v q σp a k+3. We also note that there s only one par wth q p mod p 3 for whch smlar reasonng does not allow us to use the slghtly better bound k +2, namely, p, q = 3, 53. But such an mprovement s not sgnfcant, and we wll not pursue t here. By Corollary.8 we fnd 2 49 > 0 2 2N 0 2 2p a q b σp a q b 2p k+3 q k+3 2 p4a 2 > 0 q 2. Note that the term nsde the large parentheses conssts of dvsors of N relatvely prme to p and q possbly wth extra factors n the denomnator, whch s how we obtan the second nequalty. Solvng for a yelds the bound n part v. Now suppose for a moment that p s a Fermat prme n T. Snce k 9, we have k k But 7 80 < 0 00 so a > 80. Lemma 2.3 then yelds 2p a 45 2p a k 2k 3 k 2/2 π +. From a > 80 we obtan 3a 45 > a, so by Lemma 2.2 we conclude that π σp a. If q s a Fermat prme n T, then by the same analyss we obtan π σq b and 2q b 45 π +. In ether case, we see that nether p nor q can be the specal prme. In case, when q s a Fermat prme n T and p s not, we compute 2 49 > 0 2 P 2 N 0 2 π 2 p a q b σpa q b p k+3 q k+3 > 02 q 4b 02 p 2a 2 > 0 2 p 6a 2 q 02. When p s a Fermat prme n T but q s not we smlarly fnd 2 49 > 0 2 P 2 N 0 2 π 2 p a q b σpa q b p k+3 q k+3 > 02 p 6a 02 q 2. Fnally, n case we know 2p a 45 q b 45 π +. So we have 2 49 > 0 2 P 2 N 0 2 π 3 p a q b σpa q b p k+3 q k+3 > 02 p 0a 47 q 47. The extra π comes from N, snce π σp a q b n ths case. Solvng for a n the above nequaltes yelds the stated bounds. The numbers n ths proposton are not chosen to be the strongest possble, but rather to be convenent for the case k = 9. If some of the hypotheses are strengthened then the proposton wll work wth modfed bounds for larger k and larger prmes. The real strength of the proposton s n the fracton out front. We can use, wth lttle loss n computatonal speed, the more unform bound ɛ 4 k log2 logp + Ck, q where Ck, q s some constant dependng only on k and q, and where ɛ s 4, 6, or 0 dependng on the number of Fermat prmes among {p, q}. Smlar statements apply to the followng proposton.

11 OPNS, DIOPHANTINE EQUATIONS, AND UPPER BOUNDS Proposton 2.5. Let p and q be odd, dstnct prmes less than 80, and let T = {3, 5, 7}. Let N be an odd perfect number wth k 9 dstnct prme dvsors. Suppose p a N, q b N, and 0 00 < p a < q b for some a, b Z >0. If p, q T, then b < log2 5a 47 logp 2 log0 logq If p T and q / T, then b < 4 9 log2 4a 02 logp 2 log0 2 logq If p / T and q T, then b < 4 9 log2 a 2 logp 2 log0 5 logq v If p, q / T, then b < log2 2a 2 logp 2 log0 logq Proof. One does an analyss as n the prevous proposton. The only dffculty s decdng to use the lower bound 0 2 P 2 N 0 2 π 2 p a q b σp a q b /p k+3 q k+3 when p T and q s not, and to use 0 2 P 2 N 0 2 π 3 p a q b σq b /p k+3 when q T and p s not. In case we fnd 2 49 > 0 2 P 2 N > 0 2 π 3 p a q b σpa q b p k+3 q k+3 > 02 p 5a 47 q 5b 47. Solvng for b gves the needed bound. The other cases are smlar and are left to the reader. Remark. One also has the nequalty b > a logp logq n all cases, as pa < q b. When searchng through canddate odd perfect numbers N, one often can reduce to the case when N s dvsble by a prme power p a N wth a large. Usng congruence condtons, when a s large enough one can show that there exsts a very bg prme factor Q of N. Hstorcally, t was consderatons such as ths whch led to the proofs n [5] and then [2, 6] that odd perfect numbers must have seven, and then eght, dstnct prme factors. In [3], the nsght whch mproved the number of dstnct prme factors to nne was that not only do congruence condtons yeld a very large prme factor Q, but there must also be another large prme Q > 0 whch, n practce, s not qute as large as Q that dvdes σp a. If one can show the exstence of a thrd large prme, further mprovements can be made. The propostons above are the key tool to fndng a possble thrd large prme dvsor. Frst, reduce to the case where we have two dfferent prme powers p a N and q b N wth a, b large. The szes of a and b are bounded above. We know that there should be a large prme dvsor Q of σp a and a large prme dvsor Q 2 for σq b. Our am s to show that Q and Q 2 are not equal. Thus, we compute gcdσp a, σq b for a, b lmted to the ranges gven n the propostons above, and fnd that there are no common large prmes. Ths computaton was run n Mathematca, on a sngle core, over the course of a few months. We summarze the results of ths computaton as follows. Theorem 2.6. Let p and q be odd, dstnct prmes, less than 05. Let N be an odd perfect number wth k 9 dstnct prme dvsors. Suppose p a N, q b N, and 0 00 < p a < q b for some a, b Z >0. Then the largest prme whch dvdes both σp a and σq b s smaller than 0.

12 2 PACE P. NIELSEN The number 0 was chosen to be compatble wth the bounds developed n [3], and t was fortunate that t was suffcently large to preclude the exstence of counter-examples. Whle the exstence of large common prme dvsors s very scarce, there are stll some close calls, such as dvdng gcdσp a, σq b wth p = 59, a = 2874, q = 7, and b = Another close call occurred wth the prme , for nputs p = 03, a = 3598, q = 6, and b = 833. Even so, f we were to reduce 0 to 0 0 then we could stll easly deal wth those examples that arse. The two just mentoned are the only two larger than 0 0. The assumpton n Theorem 2.6 that N s an odd perfect number s used n two ways. Frst, we can use the bounds gven n the prevous propostons. Second, we may lmt the exponents a and b even further accordng to whether or not p or q can be the specal prme. The code for these and other computatons s avalable on the author s webste. 3. Improvements to lemmas n the lterature For the remander of ths paper we wll assume that the reader s famlar wth the paper [3]. We wll freely use the notatons, defntons, and results gven there. Some of the lemmas n that paper are streamlned to work for odd perfect numbers wth k 8 dstnct prme factors. Some further computaton s needed to allow the case k = 9. Below s a full lst of the changes we make to the methods employed n [3]. Frst, the pre-compled table of factorzatons of σp a, when p and a are small, requres extenson. When p < 30 the factorzaton of σp a s entered nto the table as long as p a < If 30 < p < 05 then we lst factorzatons when p a < For the prmes 05 < p < 0000, we use the upper bound p a < Fnally, for prmes p > 0 4, factorzatons are not put n a table, but rather we compute them as needed, and only compute them when p a < The cutoffs 0 200, 0 50, 0 50, and 0 30 are used to decde when a prme power becomes nfnte n the sense of [3]. For the prmes p < 30, ths bound s gettng close to the best upper bound possble usng current factorng algorthms. Second, we mprove Lemma 9 from that paper as follows: Lemma 3.. Let p be an odd prme and let q {3, 5, 7}. If q p mod p 2 then ether q, p = 3,, q, p = 7, 3, or q opq has a prme dvsor greater than 0 4. Proof. If p > 0 4 then p s the needed prme dvsor. If p < 0 4 then there are only twelve pars p, q wth q p mod p 2, due to a computaton reported n Mossnghoff s paper []. Two cases are exceptonal, and they appear n the statement of ths lemma. For each of the other cases we compute all the prme dvsors of q opq less than 0 4, and see that the remanng cofactor s not. Our thrd change s that Proposton 0 can be mproved as follows. Proposton 3.2. Let N be an odd perfect number wth k, k, and k 2 havng ther usual meanngs. Suppose q {3, 5, 7} s a known prme dvsor of N, q n N, q π, and π σq n. Suppose p,..., p k are the other known prme factors of N, besdes q. For each =, 2,..., k defne 0 f o p q = 0 ɛ = maxs + t, f o p q 0, where s = v p σq op q and t Z + s mnmal so that p t > 00. Set V = k pɛ. Suppose π s among the k 2 unknown prme factors are greater than 00. unknown prme factors. Fnally, assume that all If mn σq n /V, σq 00 /V >

13 OPNS, DIOPHANTINE EQUATIONS, AND UPPER BOUNDS 3 then k 2 >. In that case, f σq mn 0 4 n, V k 2, σq 00 V k 2 >, then σq n has a prme dvsor among the unknown prmes at least as bg as ths mnmum. Proof. There are two dfferences between ths proposton and the orgnal result. The easer change s that we replaced both 0 and 0 3 by 0 4, whch follows by ctng Lemma 3. above. The other change s that nstead of merely assumng k 2 >, we clam t s a consequence of the nequalty mn σq n /V, σq 00 /V >. Ths new mplcaton follows mmedately from the deas present n the orgnal proof. The fourth change s that after a short computaton Lemma 2 now holds wth n the range q < 0 4 and p mnq n 2, The exact statement of the mproved result s the followng. Lemma 3.3. Let q < 0 4 be an odd prme. If p q mod q n for some n Z + and some odd prme p, then p mnq n 2, except when p, q = , 7. Note that the specal case of ths lemma causes no problems when k = 9 snce o q p = 6 and σp 5 gves rse to twelve addtonal dstnct prme factors of N. Ffth, we mprove Proposton 4 by replacng 0 50 wth Ths change merely nvolves a quck computaton. For completeness we nclude the full statement here. Proposton 3.4. Let N be an odd perfect number, and let q < 000 be a prme dvsor of N wth q n N. Suppose b, k, k, k 2, l, l 2, k, and l, have the same meanngs as n [3, Proposton 7]. Suppose further that the exceptonal case of the prevous lemma doesn t hold. Let T be the set of known prmes wth unknown component, dfferent from q, and mod q. Let τ = n b p T, o q p 0 v q p oqp + k l + k 2 k + k 2. k + k 2 σ 0 o q p If τ > 0 then one of the unknown prmes s not congruent to mod q. Further, n ths case, one of the unknown prmes s at least as large as mnq τ 2, where k τ = mn n b v q p oqp + + k 2 m m k 2 σ 0 o q p p T, o q p 0 k k l + k 2 mk + k 2 m m + k 2 m /m. 2 Sxth, we can replace Lemma 5 wth the followng verson wth the same proof, mutats mutands. Lemma 3.5. Let q be a prme wth 7 q < 000. Suppose a q mod q n for some n Z >0 and some postve nteger a wth q a. Then a mnq n 4, At ths pont, all of the changes we have made have conssted, n the man, of mnor mprovements n a few constants. These are acheved by performng computatons whch take a day or two on a sngle core. The seventh change s a sgnfcant mprovement to Lemma 6, whch takes a great deal of machne power. I wsh to thank Wllam Lpp for helpng dstrbute the needed factorng over the nternet, and also thank those who helped n that effort. The mproved lemma reads as follows:

14 4 PACE P. NIELSEN Lemma 3.6. Let p and q be prmes wth 0 2 < p < 0 and 7 q < 80. If q p mod p 2 then σq opq s dvsble by two prmes greater than 0. Proof. The paper [0] lsts all 6 pars p, q satsfyng the condtons of the lemma. It should be mentoned that Rchard Fscher has recently mproved the range of these computatons, and the nterested reader s drected to hs webste: For 56 of those pars, we have two explct prme factors of A = σq opq, each greater than 0. For another 3 pars, we have one such prme factor P for whch P 2 A. Further, after computng all prme factors of A/P less than 0 we fnd that the cofactor s not, so there must exst some other prme factor greater than 0. The explct factors are lsted on Lpp s webste earthlnk.net/~oddperfect/fermatquotents.html and are also avalable on the author s personal webste. The remanng two pars are p, q = , 4 and q, p = 22327, 57. In the frst case o p q = Exhaustvely removng all of the prme factors of less than 0, we fnd a non-trval cofactor. Smlarly, also has a prme dvsor larger than 0. Both of these prmes dvde A. Ths deals wth the frst case. The second case s dealt wth smlarly, as o p q = It should be mentoned that ths lemma can be mproved further. Indeed, f p and q are odd prmes n the larger ranges q < 000 and p < 0, stll satsfyng q p mod p 2, then once agan all possble pars p, q are known. In many cases we can agan fnd two explct prme factors greater than 0, or fnd one explct factor and prove the exstence of a second such prme. However, there are 29 cases, such as p, q = 0, 8, for whch one, and only one, prme factor larger than 0 exsts. There are also 66 cases, such as p, q = 3, 9, for whch no large prme factors exst. In those cases usually p s very small, whch s why n the lemma above we restrcted to the range p > 00. Fnally, there are 4 cases, such as p, q = , 28, where t s lkely that there exst two large prme factors, but we ddn t explctly prove t because there was no need. However, one can use a varant of the upper bound P 2 N < 2 4k to easly deal wth these remanng cases, at least when k = 9. The eghth change s that the statement of Proposton 7 can be modfed a great deal, wth the only sgnfcant change to the proof that we now cte Lemma 3.6. Here s the precse mproved result. Proposton 3.7. Let N be an odd perfect number and let 7 q < 80 be a known prme dvsor of N, wth q n N. Let τ, τ be as n Proposton 3.4, suppose all the hypotheses of that proposton are met, and let p be the guaranteed unknown prme. Let p,..., p k be the known prmes dfferent from q. Let ɛ be defned as before, and put V = k pɛ. Fnally, assume that all unknown prme factors are greater than 00. If mn σq τ 4 /V, σq 96 /V >, then k 2 >. In that case, f q mn 0 96, q V k 2, q τ 4 q V k 2 > then σq n has a prme dvsor, dfferent from p, among the unknown prmes, at least as bg as the above mnmum. Proof. The range of the prme q has been mproved from q {7,, 3} to the range 7 q < 80, whch we can do by Lemma 3.6. However, the quantty q τ 2 from the orgnal result has been replaced by the slghtly worse q τ 4 essentally, to deal wth q = 7. The constant 0 50 has been superseded by q 96 notng q 96 < Smlar to Proposton 3.2, the assumpton k 2 > has been replaced wth the statement f mnσq τ 4 /V, σq 96 /V >, then k 2 > whch s proved n an analogous manner.

15 OPNS, DIOPHANTINE EQUATIONS, AND UPPER BOUNDS 5 Note that f mnσq τ 4 /V, σq 96 /V > holds but k 2 =, then we reach a contradcton. So we add ths contradcton the lst found n [3, Secton 7]. A smlar statement holds for Proposton 3.2. The fnal change n our mplementaton of the algorthm n [3] s that when applyng Lemma 20 to fnd upper bounds on the next unknown prme, we use Theorem 2.6 above to contrbute more large prmes. However, Theorem 2.6 does not automatcally prove the exstence of a thrd large prme. In practce, we have one prme Q comng from congruence condtons relatve to p, and another prme Q > 0 dvdng σp a wth Q Q. Smlarly we have one prme Q 2 from congruence condtons relatve to q, and another prme Q 2 > 0 dvdng σq b wth Q 2 Q 2. Theorem 2.6 asserts Q Q 2, but t could stll be the case that Q 2 = Q and Q = Q 2. When computng the bounds comng from Lemma 20 we are thus led to consder two stuatons, whch we descrbe now. Suppose our algorthm has reached a pont where we have a lst {q a, qa2 2,..., qan n } of nfnte prme powers n a suspected odd perfect number. Let Q be the large prme comng from q from the congruence condtons n [3, Proposton 7] or Proposton 3.4 accordng to whether q {3, 5, 7}, or not. Smlarly, let Q be the second large prme comng from q, usng Proposton 3.2 or 3.7. We drop from our lst any q for whch Q < 0, so that Theorem 2.6 wll apply. There are now two man cases. We apply Lemma 20 n both cases, and then use the lesser of the two bounds acheved. One opton s that Q = Q 2 = = Q n. In ths case, as each Q Q = Q and Q Q j by Theorem 2.6, we can apply Theorem 20 wth the bounds {P = Q, P 2 = 0, P 3 = 0,..., P n+ = 0 }. The second opton s that the Q are not all equal. Let Q be the largest element n {Q, Q 2,..., Q n }, and let Q n be the smallest. The worst possble case would be that among the dstnct prmes {Q, Q 2,..., Q n} the two largest prmes are equal to Q and Q n, and the rest are close to 0. Thus, we apply Theorem 20 wth the lst of bounds {P = Q, P 2 = Q n, P 3 = 0,..., P n = 0 }. Wth all of these changes n place, we rerun the algorthm descrbed n [3]. The computaton takes just over one day on a sngle core, coverng a lttle over thrty mllon cases, and we acheve: Theorem 3.8. There are no odd perfect numbers wth less than 0 dstnct prme factors. There are a few problem cases requrng specal treatment, whch we descrbe now. These cases also llustrate some of the benefts and lmtatons n the changes we made above. Intally, the plan had been to make the prmes p = 0, 03 become nfnte when p a > However, wth ths choce the prme power 03 never satsfes the bounds n Proposton 3.4. Ths left cases such as p where p Here we needed another large prme, snce the bounds on the next unknown prme were already bgger than 0. Ths problem was solved by expandng the factorzaton table for p = 0, 03 up to the level p a Ths stll leaves fve problem cases: We wll only dscuss the frst case, as the other four are dealt wth smlarly. In that case, the next unknown prme p s gven nsde an nterval < p < Ths nterval contans more than nne bllon prmes, whch s too many to check one at a tme. The reason for the extremely large nterval s that the upper and lower bound on the nterval s larger than 0, and thus falls outsde the scope of the bounds n Proposton 3.7. When ths project was begun, the number 0 was the bound ntally chosen when provng Lemma 3.6, and thus subsequently used n many other lemmas and propostons.

16 6 PACE P. NIELSEN There are at least two ways to deal wth ths case. Frst, one could redo the computaton of Theorem 2.6 to nclude the pars p, q {, 2633, 73, 2633, 2633,, 2633, 73}, and then modfy Lemma 3.6, and all subsequent results, to nclude the new prme A second opton s to agan modfy Lemma 3.6, and all subsequent results, but ths tme just for the prmes and 73, by replacng the bound 0 wth 0 2. The second opton s straghtforward and easy, except for extendng the computatons n [0]. Fortunately, as we already mentoned, such an extenson has already been performed by Rchard Fscher up to the even better bound Ths fnshes the frst specal cases. The other four are dealt wth smlarly. 4. Can we prove k > 0? The smple answer to the queston n the ttle of ths secton s no. We wll gve an llustratve example, demonstratng one of the many dffcultes one would face n tryng to extend the computatons further. Consder the case The next unknown prme les n the nterval < p < It would be qute dffcult f not mpossble wth current computng resources to mprove Lemma 3. for the prme q = 5 to acheve a bound , much less also mprove Lemma 3.6 smlarly to for the prmes q = 6, 63. We would need to do ths for at least two of these three prmes, to use the prevous methods to decrease the sze of the nterval. A related problem s whether these technques can be used to show that when 3 N, the bound k 2 can be mproved. Ths may be possble, but t would requre an mprovement to Theorem 2.6 from k 9 to k 2, whch roughly translates nto a sxty-four fold ncrease n computaton tme. As the orgnal calculaton to prove Theorem 2.6 took several months, ths mprovement would only be feasble usng a massve effort dstrbuted over many computers. Acknowledgements The author thanks Wllam Lpp for many useful comments on manuscrpts, and for spearheadng the factorzaton efforts whch led to Lemma 3.6. The author also thanks Samuel Dttmer and Mtsuo Kobayash for conversatons whch led to some mprovements n Lemma.2, and Yong-Gao Chen for provdng a preprnt verson of the paper [3]. Many of the ntal computatons used n ths paper were performed whle the author was partally supported by the Unversty of Iowa Department of Mathematcs NSF VIGRE grant DMS Ths work was partally supported by a grant from the Smons Foundaton #35828 to Pace Nelsen. Fnally, the author thanks the anonymous referees for comments and suggestons whch mproved the paper mmensely. References. R. P. Brent, G. L. Cohen, and H. J. J. te Rele, Improved technques for lower bounds for odd perfect numbers, Math. Comp , no. 96, MR c: Joseph E. Z. Chen, An odd perfect number has at least 8 prme factors, ProQuest LLC, Ann Arbor, MI, 979, Thess Ph.D. The Pennsylvana State Unversty. MR Yong-Gao Chen and Cu-E Tang, Improved upper bounds for odd multperfect numbers, Bull. Aust. Math. Soc. to appear 204, 8 pp. 4. R. J. Cook, Bounds for odd perfect numbers, Number theory Ottawa, ON, 996, CRM Proc. Lecture Notes, vol. 9, Amer. Math. Soc., Provdence, RI, 999, pp MR d:00 5. Samuel Dttmer, Spoof odd perfect numbers, Math. Comp. to appear 204, 8 pp. 6. Peter Hags, Jr., Outlne of a proof that every odd perfect number has at least eght prme factors, Math. Comp , no. 5, MR k: D. R. Heath-Brown, Odd perfect numbers, Math. Proc. Cambrdge Phlos. Soc , no. 2, MR b:30 8. Douglas E. Iannucc, The second largest prme dvsor of an odd perfect number exceeds ten thousand, Math. Comp , no. 228, MR :200

17 OPNS, DIOPHANTINE EQUATIONS, AND UPPER BOUNDS 7 9., The thrd largest prme dvsor of an odd perfect number exceeds one hundred, Math. Comp , no. 230, MR :20 0. Wlfrd Keller and Jörg Rchsten, Solutons of the congruence a p mod p r, Math. Comp , no. 250, electronc. MR :004. Mchael J. Mossnghoff, Weferch pars and Barker sequences, Des. Codes Cryptogr , no. 3, MR c: Pace P. Nelsen, An upper bound for odd perfect numbers, Integers , A4, 9. MR k:009 3., Odd perfect numbers have at least nne dstnct prme factors, Math. Comp , no. 260, MR g:53 4. Pascal Ochem and Mchaël Rao, Odd perfect numbers are greater than 0 500, Math. Comp , no. 279, MR Carl Pomerance, Odd perfect numbers are dvsble by at least seven dstnct prmes, Acta Arth /74, MR # John Voght, On the nonexstence of odd perfect numbers, MASS selecta, Amer. Math. Soc., Provdence, RI, 2003, pp MR j:006 Department of Mathematcs, Brgham Young Unversty, Provo, UT 84602, Unted States E-mal address: pace@math.byu.edu