Characterizing the properties of specific binomial coefficients in congruence relations

Transcription

1 Eastern Mchgan Unversty Master's Theses and Doctoral Dssertatons Master's Theses, and Doctoral Dssertatons, and Graduate Capstone Projects Characterzng the propertes of specfc bnomal coeffcents n congruence relatons Tyler Robert Russ Follow ths and addtonal works at: Part of the Mathematcs Commons Recommended Ctaton Russ, Tyler Robert, "Characterzng the propertes of specfc bnomal coeffcents n congruence relatons" (2015. Master's Theses and Doctoral Dssertatons Ths Open Access Thess s brought to you for free and open access by the Master's Theses, and Doctoral Dssertatons, and Graduate Capstone Projects at DgtalCommons@EMU. It has been accepted for ncluson n Master's Theses and Doctoral Dssertatons by an authorzed admnstrator of DgtalCommons@EMU. For more nformaton, please contact lb-r@emch.edu.

2 Characterzng the Propertes of Specfc Bnomal Coeffcents n Congruence Relatons by Tyler Russ Thess Submtted to the Department of Mathematcs Eastern Mchgan Unversty n partal fulfllment of the requrements for the degree of MASTER OF ARTS n Mathematcs Thess Commttee: Andrew Wlfong, Ph.D., Char Davd Folk, Ph.D. Jayakumar Ramanathan, Ph.D. July 15, 2015 Ypslant, Mchgan

3 Abstract The number theoretc conjecture we examne n ths paper orgnates when tryng to construct a characterzable generatng set for the complex cobordsm polynomal rng. To date there s no effcent, unversal method for characterzng such a generatng set. Wlfong conjectures that smooth projectve torc varetes can act as these generators [7]. Torc varetes are related to polytopes by a bjectve correspondence. Studyng the combnatoral structure of these polytopes s much more manageable than studyng propertes of torc varetes drectly. Ths gves rse to the number theoretc conjecture consdered here. A proof of ths number theoretc conjecture would n turn prove the conjecture that smooth projectve torc varetes provde a generatng set for the complex cobordsm polynomal rng. Here, we do not provde a complete proof of the number theoretc conjecture, rather we gve more evdence to the conjecture, buldng on pror work of Wlfong and Parry.

4 Contents Abstract Introducton: Wlfong s Conjecture n Context Propostons from Wlfong Cases n whch the Conjecture Holds Generalzaton of Many Propostons and Lemmas of Secton Generalzaton: Two Prme Powers Prmes Far Apart: Partal Generalzaton of Theorem Prmes Close Together: Partal Generalzatons Concluson References

5 Characterzng the Propertes of Specfc Bnomal Coeffcents n Congruence Relatons 1. Introducton: Wlfong s Conjecture n Context The number theoretc conjecture we examne n ths paper orgnates n the topologcal work of Andrew Wlfong [7]. In Wlfong s work on torc varetes, the queston arses of constructng a characterzable generatng set for the complex cobordsm polynomal rng. To date there s no effcent, unversal method for characterzng such a generatng set. It s a prmary conjecture n the work of Andrew Wlfong that a smooth projectve torc varety for each generator s possble [7]. An algebrac varety s the soluton set of a system of polynomal equatons. A torc varety s an algebrac varety that has an added structure comng from a torus acton. For more detals refer to [1, 3]. A nce feature of torc varetes s that many of ther topologcal propertes stand n bjectve correspondence to the combnatoral structure of the polytopes, whch s more easly studed than the topologcal propertes of torc varetes. One topologcal property of torc varetes that can be computed n terms of the combnatorcs of polytopes ncludes determnng whether or not torc varetes are polynomal generators of the complex cobordsm rng, whch leads to the number theory conjecture examned n ths thess. To better understand the complex cobordsm polynomal rng, we formally ntroduce cobordsm. Defnton 1. (Addtonally, refer to [5]. Two smooth compact n-dmensonal manfolds M 1 and M 2 are cobordant f ther dsjont unon M 1 M 2 forms the boundary of an n + 1-dmensonal smooth manfold. A cobordsm of real manfolds conssts of two n-dmensonal manfolds connected by a smooth manfold of dmenson n + 1. We can vsualze a cobordsm usng the followng example. Consder the three crcles defnng the wast and the ankles of a par of pants. The crcles are dsjont. Taken together, these crcles defne the boundary of the surface of the pants. The materal connectng these one-dmensonal crcles s the manfold defned by the cobordsm, the two-dmensonal surface of the pants.

6 The cobordsm relaton has many useful propertes. One of ts prmary propertes s that cobordsm s an equvalence relaton. A second key property s that the set of equvalence classes of cobordant manfolds forms a rng under dsjont unon and cross product. The complex cobordsm rng, denoted Ω U, has a defnton smlar to cobordsm of real manfolds. We modfy the defnton for complex cobordsm snce complex cobordsm of two real n-dmensonal manfolds would smlarly requre that a complex manfold of real dmenson n + 1 could be constructed wth the manfolds of dmenson n defnng the boundares of that manfold. However, complex spaces are always of even dmenson, so the constructon of a cobordsm meetng ths defnton s not possble. Instead, we defne a complex cobordsm to be a cobordsm of stably complex manfolds. Ths s a weakenng of the condton for a manfold to be complex. There are odd-dmensonal stably complex manfolds, so ths constructon provdes a framework n whch to defne complex cobordsm [6, 7]. Wth the defnton of cobordsm n mnd, we ntroduce the followng theorem, whch shows the algebrac structure of the complex cobordsm rng. Theorem 2 (Mlnor, Novkov. Ω U = Z [α 1, α 2, ] s a polynomal rng wth one generator α n n each postve even dmenson 2n. Smooth projectve torc varety polynomal generators for Ω U have been constructed n all dmensons except for those n whch n s even and not one less than a prme power. Wlfong has also verfed the conjecture usng computatonal software for the case when n s even and not one less than a prme power for all n through The case when n s even and not one less than a prme power s the subject of ths thess. If the number theory conjecture s true, t wll prove that there are polynomal generators of Ω U n these remanng dmensons. Refer to [7] for detals. We now turn to the number theory conjecture studed n ths paper. Here we ntroduce the expresson R n (ε = n ε + ( 1 ε ( n 1 ε, whch arses n Wlfong s work on the combnatorcs of polytopes. The number n n ths expresson relates to the complex dmenson of the manfolds connected by the complex cobordsm. Our goal s to 2

7 dentfy choces for ε, for whch the quantty R n (ε and the number n + 1 are relatvely prme. We state ths formally n the followng conjecture. Conjecture 3. (See [7] for background. For a gven even n, n not one less than a power of a prme, the equaton gcd(r n (ε, n + 1 = 1 (1.1 has a soluton ε, where R n (ε s defned as R n (ε = n ε + ( 1 ε ( n 1 ε, and ε s n the range {2,, n 1}. We wll refer to ths conjecture as conjecture 1.1 throughout the rest of ths paper. Conjecture 1.1 s the prmary focus of ths paper. We buld on prevous results from Wlfong and Parry to provde more evdence that the conjecture s true. As the conjecture states, we need a choce for ε makng R n (ε and the number n + 1 relatvely prme. Snce n + 1 s not a prme power, t has at least two prmes n ts prme factorzaton. To verfy that R n (ε and n + 1 are relatvely prme for a gven choce for ε, we can verfy that each prme dvdng n + 1 does not dvde the quantty R n (ε. To do ths, we verfy that the range {2,, n 1} contans at least one choce for ε satsfyng the condton R n (ε 0 mod p for each prme p dvdng n + 1. We note that n 1 mod p for any prme p dvdng n + 1. We can rewrte the expresson R n (ε n a congruence wth a prme number modulus p, where the prme p dvdes n + 1, as ( n 1 R n (ε = n ε + ( 1 ε ε 1 ε + ( 1 ε ( n 1 ε = (ε ( 1 ε ( n 1 ε mod p mod p. Usng the prevous congruence, we see that our task of provng conjecture 1.1 s dentcal to showng that we can guarantee at least one choce for ε n the range {2,, n 1} satsfyng 3

8 the condton for each prme p dvdng n + 1. ( n 1 ( 1 ε ε + 1 mod p (1.2 ε If a choce for ε satsfes condton 1.2 for the number n and for each prme p dvdng n + 1, then that choce for ε satsfes conjecture 1.1 for the number n as defned. Ths follows, snce f the quantty R n (ε s not dvsble by any prme p dvdng n + 1, then the quantty R n (ε and the number n + 1 are relatvely prme. For that reason, we can elmnate partcular choces for ε by showng that the gven choce does not satsfy condton 1.2 for at least one prme p dvdng n + 1. If the expresson R n (ε s dvsble by any of the prmes dvdng n + 1, t follows mmedately that ths choce for ε also fals to satsfy conjecture 1.1 for the number n. Ths s the prmary method that we use to establsh that a gven choce for ε satsfes conjecture 1.1 for the number n. Though the general prme factorzaton of such a number can be gven by n + 1 = p m 1 1 p m 2 2 p mt t, there are many hurdles to tacklng ths form of the problem drectly. We begn nstead wth specal forms of the problem and buld n complexty. In Sectons 2 and 3, we restrct ourselves to the case that the prmes dvdng n + 1 are squarefree. The propostons n Secton 2, ncorporatng Lucas Theorem, help us to dentfy choces for ε satsfyng condton 1.2 for ndvdual prmes dvdng the number n + 1. The propostons show ntervals and specfc postons where no sutable choces for ε exst and ntervals n whch choces for ε necessarly satsfy condton 1.2 for a gven prme p dvdng the number n + 1. The long-range goal s to characterze the choces for ε completely and to show that there s always a choce for ε satsfyng conjecture 1.1 for any even n that s not one less than a prme power. Secton 3 analyzes some smplfed versons of the conjecture. Theorem 18 of Secton 3 analyzes numbers n + 1 that are products of two squarefree prmes. The proof that choces for ε exst satsfyng conjecture 1.1 for an even number n, where n + 1 s of ths form, was shown by Wlfong n unpublshed notes. We also analyze numbers n + 1 wth multple squarefree prme factors. These numbers look lke n + 1 = p 1 p 2 p t where each p s prme. 4

9 Our goal n Sectons 4, 5, 6, and 7 s to gradually extend our analyss to even n where n + 1 has the generc prme factorzaton n + 1 = p m 1 1 p m 2 2 p m t t. In Secton 5 we extend Theorem 18 to the general case allowng numbers n + 1 that are products of two nonsquarefree prmes. These numbers look lke n + 1 = p m 1 1 p m 2 2 where p 1 and p 2 are both prme. Ths theorem was proved orgnally by Walter Parry [4], but we offer an alternatve proof here usng methods developed n ths paper. The new propostons proved n ths paper provde extensons to prevous results and verfy the conjecture n new cases, especally where factors n the prme factorzaton of n + 1 are not squarefree. We restrct ourselves to some partal cases and do not offer a general proof of conjecture 1.1. Walter Parry has examned conjecture 1.1 and acheved some sophstcated generalzatons. In hs paper [4], Parry reformulates conjecture 1.1 to gan more leverage on the problem. New, equvalent condtons arse on a choce for ε to satsfy conjecture 1.1 for a number n. Hs reformulaton reveals more of the underlyng structure relatng the number n and the correspondng choces for ε. One of hs major results s ntutvely labeled The Bg Exponents Theorem. In that theorem, the use of contnued fractons leads to a separaton of the orgnal conjecture nto two cases. He s able to show that for any fnte set S of odd prmes, S = {p 1, p 2,, p t }, there exsts a postve nteger r such that f each prme s rased to at least the power of r, then there s a guaranteed choce for ε satsfyng conjecture 1.1 for the number n. That s, f for each number n {r 1, r 2,, r t } we have r > r, ths theorem guarantees a choce for ε satsfyng conjecture 1.1 for the number n where n + 1 = p r 1 1 p r 2 2 p rt t. Currently, ths requres that the number n be very large, snce each prme power must be very large. What makes hs work exctng s that t addresses the general case drectly and can potentally be refned to mprove the bound on the prme powers dvdng n + 1. If the exponents r can be reduced, the approach could eventually provde a general proof to the conjecture. It s possble that a general proof wll requre more advanced methods, lke those used n the work of Parry. Evaluatng the bnomal coeffcent ( n 1 ε n condton 1.2 s the man challenge to dentfyng choces for ε satsfyng conjecture 1.1 for the number n. Lucas Theorem provdes 5

10 a way of rewrtng a bnomal coeffcent ( m n n a congruence wth a prme number modulus as a product of bnomal coeffcents taken from the p -adc expansons of the ntegers m and n. Ths rewrtng strategy s a key component of the approach taken n ths paper to prove conjecture 1.1. We state and prove the theorem below. Theorem 4 (Lucas Theorem. For nonnegatve ntegers m and n and a prme p, the followng congruence relaton holds: ( m n k =0 ( m n mod p where m and n are expressed n ther base p expansons as m = m k p k + m k 1 p k m 1 p + m 0 and n = n k p k + n k 1 p k n 1 p + n 0. Here, we use the conventon that f m < n, then ( m n = 0. Proof. (Ths proof s attrbuted to Nathan Fne. See [2]. If p s prme and n s an nteger satsfyng 1 n p 1, then the numerator of the bnomal coeffcent ( p = n s dvsble by p, but the denomnator s not. p (p 1 (p n + 1 n (n 1 1 Clearly, p must dvde ( p n. We recall the dentty (usng ordnary generatng functons (1 + X p 1 + X p mod p. In fact, for every nonnegatve nteger we have (by nducton (1 + X p 1 + X p mod p. Recall that m s a nonnegatve nteger and p s a prme. As before, we wrte m n ts base p expanson, yeldng m = k =0 m p for some nonnegatve ntegers k and m, n whch the bound 0 m p 1 holds for each m by the defnton of p-adc expanson. 6

11 We can then wrte m n=0 ( m X n = (1 + X m = n k ((1 + X p m =0 k ((1 + X p m k = =0 k =0 ( p 1 n =0 ( m n =0 X n p = ( m n =0 ( m n ( m k n=0 =0 X n p ( m n X n mod p, where n the fnal product the n represent the coeffcents of the p-adc expanson of n, whch s unque. Snce the expanson of n s unque, the theorem s proved. Before movng to the propostons, we ntroduce a small lemma that wll prove useful for smplfyng expressons resultng from the applcaton of Lucas Theorem. In order to apply Lucas Theorem, we often need to characterze the value of ε accordng to ts p -adc expanson for the prme p. We use the conventon ε = b 0 + b 1 p + where the nonnegatve ntegers {b 0, b 1, } are less than p. Ths conventon of usng b 0 for the frst coeffcent n the p -adc expanson for ε motvates the choce of the nteger b 0 n the followng lemma. Lemma 5. For nteger b 0 bounded by 0 b 0 < p, the bnomal coeffcent ( p 2 b 0 s equvalent to ( 1 b 0 (b modulo the prme p. Proof. We note that each of p 2 and b 0 s less than the prme p tself. Notng ths, we can rewrte the bnomal coeffcent ( p 2 b 0 modulo the prme p as ( p 2 b 0 = (p 2! b 0! (p 2 b 0! = (p 2 (p 3 (p b 0 (p (b (p (b 0 + 2! 2 3 b 0 (p (b 0 + 2! ( 2 ( 3 ( b 0 ( (b b 0 mod p ( 1b 0 (2 (3 (b 0 (b b 0 mod p. 7

12 Snce each factor 2, 3,, b 0 s less than p, we can assocate factors n the numerator and denomnator and smplfy as follows: ( p 2 b 0 ( 1 b 0 ( 2 2 ( 3 3 ( 1 b 0 (b mod p. ( b0 b 0 (b mod p From the precedng, we have proved the dentty ( p 2 ( 1 b 0 (b mod p. b 0 We can smplfy the bnomal coeffcent ( p 2 b 0 modulo the prme p n ths way whenever the nonnegatve nteger b 0 s also less than p. 8

13 2. Propostons from Wlfong We gve credt to Wlfong for the followng propostons and lemmas, proved n unpublshed notes. The results of ths secton are used to prove the theorems of Secton 3. Some propostons dentfy partcular choces for ε, as well as ranges contanng choces for ε, that can never satsfy conjecture 1.1 for an even number n not one less than a prme power. See especally Propostons 6 and 9, Corollary 10, Lemma 12, and Corollary 13. Other results establsh ranges contanng choces for ε that satsfy condton 1.2 for any prme p dvdng n + 1. See especally Proposton 11, Corollares 14 and 15, Lemma 16, and Corollary 17. Here, we ntally restrct ourselves to the case n + 1 = p 1 p 2 p t where n + 1 s the product of t dstnct, squarefree prmes. To standardze our method, we unfalngly adhere to the conventon p 1 < p 2 < < p t. As stated earler, n order to utlze the notatonal convenence of Lucas Theorem to evaluate choces for ε more effcently, we often need to rewrte the number n 1 n ts p -adc expanson for a gven prme p dvdng n + 1. We wrte n + 1 n ts expanded p -adc form as n + 1 = a 1 p + a 2 p 2 +, where t s guaranteed that a 1 0, snce p n + 1, but p 2 n + 1. Subtractng two, we can wrte n 1 n expanded p -adc form as n 1 = (p 2 + (a 1 1 p + a 2 p 2 +. From ths expresson, we can read off the p -adc coeffcents of n 1 drectly. The followng proposton shows that choces for ε that are too small can never satsfy conjecture 1.1 for an even number n not one less than a prme power. Recall that n + 1 s of the form p 1 p 2 p t wth the conventon p 1 < < p t, and ( a b 0 mod p, whenever a < b for ntegers a and b. 9

14 Proposton 6. Let n + 1 = p 1 p 2 p t wth the conventon p 1 < < p t. If ε 1 mod p for a gven prme p, that choce for ε does not satsfy condton 1.2 for the prme p and the number n. Proof. For the prme p dvdng n + 1, condton 1.2 requres that the statement ( 1 ε ( n 1 ε ε + 1 mod p hold for a sutable choce for ε. We consder any choce for ε n any sutable range. We can wrte ε n ts expanded p -adc form as ε = (p 1 + b 1 p + b 2 p 2 + where the ntegral coeffcents {b 1, b 2, } are n the range 0 b 1, b 2, < p. Applyng Lucas Theorem to the bnomal coeffcent ( n 1 ε, we can wrte ( ( ( ( n 1 p 2 a1 1 a2 mod p ε p 1 0 mod p. b 1 b 2 Ths holds snce p 2 < p 1 and the value of the bnomal coeffcent ( p 2 p 1 s equal to zero by conventon. It follows that ( n 1 ( 1 ε ε + 1 mod p. ε For any prme p dvdng n + 1 and any choce for ε, where ε 1 mod p, that choce for ε never satsfes condton 1.2 for the prme p and the number n. The followng corollary emphaszes the fact that any choce ε 1 mod p not only fals to satsfy condton 1.2 for a prme p dvdng n + 1, but also fals to satsfy conjecture 1.1 for the number n. Corollary 7. Let n + 1 = p 1 p 2 p t wth the conventon p 1 < < p t. For any prme p dvdng n + 1, a choce ε 1 mod p cannot satsfy conjecture 1.1 for the even number n. The followng proposton s especally useful to gan nsght nto the work of Walter Parry. Ths result enables the constructon of choces for ε satsfyng condton 1.2 for partcular prmes dvdng the number n + 1. Ths result s a crucal component of Parry s Bg Exponents Theorem. The proposton also guarantees that a choce for ε that fals to satsfy 10

15 condton 1.2 for a partcular prme p dvdng n + 1 cannot have p -adc coeffcent b 1 equal to p 1 when b 0 p 1. An example of ths s seen n Lemma 16. Proposton 8. Let n + 1 = p 1 p 2 p t wth the conventon p 1 < < p t. For a prme p dvdng n + 1, any choce for ε wth b 1 = p 1, b 0 p 1 and p -adc expanson gven by ε = b 0 + b 1 p + b 2 p 2 + necessarly satsfes condton 1.2 for the prme p dvdng n + 1. Proof. In ths case, by Lucas Theorem and the conventon on bnomal coeffcents stated there, we have ( n 1 ε ( ( p 2 a1 1 b 0 p 1 ( p 2 (0 b 0 0 mod p. ( a2 ( a2 b 2 b 2 mod p mod p Ths s true snce a 1 < p mmedately mples a 1 1 < p 1. It follows that ( n 1 ( 1 ε 0 mod p. ε Proposton 6 requres that ε 1 mod p, and snce ε b 0, we know that b 0 1 mod p. It follows that Taken together we see that ε + 1 b mod p. ( n 1 ( 1 ε 0 ε + 1 mod p. ε Therefore, any choce for ε wth b 1 = p 1 always satsfes condton 1.2 for the prme p. The precedng proposton plays an ntegral role n Parry s Bg Exponents Theorem, enablng the constructon of choces for ε satsfyng condton 1.2 for partcular prmes dvdng n

16 Proposton 9. Let n + 1 = p 1 p 2 p t wth the conventon p 1 < < p t. For a prme p dvdng n + 1, any choce for ε n the range {2,, p 2} never satsfes condton 1.2 for the prme p and the number n. Proof. For any prme p dvdng n + 1, the possble choces for ε, where ε < p, are restrcted to the range {2,, p 2} by the statement of conjecture 1.1 and Proposton 6. The p -adc expanson of any choce for ε n ths range conssts solely of a constant term snce the prme p s larger than ε. Specfcally, we have ε = b 0 where b 0 s a nonnegatve nteger less than p 1. We wsh to determne whether any choce for ε n ths range satsfes condton 1.2 for the prme p. We apply Lucas Theorem and express the bnomal coeffcent ( n 1 ε as ( ( ( ( ( ( n 1 p 2 a1 1 a2 a3 p 2 mod p. ε b 0 By Lemma 5 we can rewrte the bnomal coeffcent ( p 2 b 0 as ( 1 b 0 (b mod p. We multply by the factor ( 1 ε on the left and rght hand sdes of the congruence above to match the full expresson n condton 1.2. We recall that b 0 s dentcal wth ε, and we wrte ( n 1 ( 1 ε ε ( p 2 ( 1 ε b 0 mod p ( 1 ε ( 1 b 0 (b ( 1 ε ( 1 ε (ε + 1 mod p ε + 1 mod p. b 0 It follows from ths that any choce for ε, where ε < p, never satsfes condton 1.2 for the prme p and the number n. The followng corollary follows drectly by replacng the generc prme p n the prevous proposton wth the prme p t, the largest prme factor of n + 1. Corollary 10. Let n + 1 = p 1 p t wth the conventon p 1 < < p t. No choce for ε less than p t satsfes conjecture 1.1 for the even number n. That s, there s no choce for ε n the range {2,, p t 1} satsfyng conjecture 1.1 for the number n. 12

17 In the followng propostons and lemmas, we characterze ranges contanng choces for ε that always satsfy condton 1.2 for a prme p dvdng n + 1. In order for a choce for ε to satsfy conjecture 1.1 for the number n, t s necessary that that choce for ε satsfes condton 1.2 for each prme p dvdng n + 1. It s therefore helpful to dentfy ranges contanng choces for ε satsfyng condton 1.2 for ndvdual prmes dvdng n + 1. Once these ranges are establshed for each ndvdual prme, we know that choces for ε n the ntersecton of these ranges where these ntervals overlap satsfy conjecture 1.1 for the number n. For a prme p dvdng n + 1, Proposton 11 shows that all choces for ε n the range {p,, 2p 2} satsfy condton 1.2 for the prme p and the number n. For a prme p dvdng n + 1, Corollary 15 shows that f a multple of p, call t kp where k s an nteger, satsfes condton 1.2 for the prme p and the number n, then all choces for ε n the range {kp,, (k + 1 p 2} also satsfy the condton for the prme p and the number n. For ndvdual prmes dvdng n + 1, these results guarantee entre ranges contanng choces for ε that satsfy condton 1.2 for the prme p and the number n. Ths s a crucal step n establshng the exstence of a choce for ε satsfyng conjecture 1.1 for the number n. For notatonal consstency and ease of readng, we always characterze the p -adc expanson of ε as ε = b 0 + b 1 p + b 2 p 2 + where the nonnegatve nteger coeffcents {b 0, b 1, } are less than p. We nclude the restrcton b 0 p 1 snce these choces for ε fal to satsfy conjecture 1.1 by Corollary 7. We contnue wth the assumpton that n + 1 s a product of squarefree prmes wth factorzaton n + 1 = p 1 p 2 p t, followng the conventon p 1 < p 2 < < p t. For each prme dvdng n + 1, the followng proposton establshes a range contanng choces for ε, all of whch satsfy condton 1.2 for that prme and the number n. Proposton 11. Let n + 1 = p 1 p 2 p t wth the conventon p 1 < < p t. For a prme p dvdng n + 1, all choces for ε n the range p ε < 2p 1 satsfy condton 1.2 for the prme p and the number n. Proof. Any choce for ε n the range {p,, 2p 2} contans two nonzero terms n ts p -adc expanson. Namely, we have ε = b 0 + p. Snce ε s strctly less than 2p 1, the 13

18 constant term b 0 n the p -adc expanson of ε s strctly less than p 1. So, we have b 0 n the bound 0 b 0 < p 1. We apply Lucas Theorem and Lemma 5 to rewrte the bnomal coeffcent ( n 1 ε as ( n 1 ε ( ( p 2 a1 1 b 0 1 ( a2 0 mod p ( 1 b 0 (b (a 1 1 mod p. We multply both sdes of the congruence by the factor ( 1 ε to match the expresson n condton 1.2. We have ( n 1 ( 1 ε ε ( 1 ε ( 1 b 0 (b (a 1 1 mod p (b (a 1 1 mod p (b (1 a 1 mod p. For a prme p dvdng n + 1, we wsh to verfy that any choce for ε n the range {p,, 2p 2} satsfes the condton ( 1 ε ( n 1 ε ε + 1 mod p for the number n. Snce we have ( 1 ε ( n 1 ε (b0 + 1 (1 a 1 mod p, we need to check whether ( 1 ε ( n 1 ε ε + 1 (b0 + 1 (1 a 1 mod p. We can wrte ε as b 0 + p. Ths leads drectly to the congruence ε + 1 b mod p. Therefore, ( 1 ε ( n 1 ε ε + 1 mod p f and only f a 1 = 0. Snce a 1 0 by assumpton, ths means that all choces for ε n the range {p,, 2p 2} satsfy condton 1.2 for any prme p dvdng n + 1. Lemma 12. Consder n + 1, where n s even, and n + 1 s the product of squarefree prme factors. For a prme p dvdng n + 1, any choce for ε wth ts p -adc coeffcent b 0 fxed n the range {0,, p 2} fals to satsfy condton 1.2 for the prme p dvdng n + 1 f and only f ε b 0 also fals to satsfy condton 1.2 for the prme p and the number n. Proof. For a prme p dvdng n + 1, we consder an arbtrary choce for ε n any sutable range, where ε 1 mod p, and ε = b 0 + b 1 p +. To determne whether the choce for ε satsfes condton 1.2 for the prme p, we use Lucas Theorem and Lemma 5 to rewrte the 14

19 bnomal coeffcent ( n 1 ε modulo the prme p as ( ( ( n 1 p 2 a1 1 ε b 0 b 1 ( a2 ( ( 1 b 0 a1 1 (b b 1 mod p b 2 ( a2 b 2 mod p. Multplyng both sdes of the congruence above by a factor of ( 1 ε and rewrtng ε n terms of ts p -adc expanson, we have ( n 1 ( 1 ε ( 1 (2 b 0+b 1 p +b 2 p 2+ ( a1 1 (b ε It follows that ( 1 ε ( n 1 ε ε + 1 b0 + 1 mod p f and only f ε + 1 ( 1 (b 1p +b 2 p 2+ ( a1 1 (b b 1 b 1 ( a2 b 2 ( a2 b 2 mod p. mod p. Snce b 0 s strctly less than p 1, we know that b s strctly less than p and we can dvde both sdes of the congruence by b Smplfyng, we see that ( 1 ε ( n 1 ε ε + 1 mod p f and only f 1 ( 1 (b 1p +b 2 p 2 + ( a1 1 b 1 ( a2 b 2 mod p. The above congruence shows that ths property s completely ndependent of the value of b 0. For any prme p dvdng n + 1, any choce for ε wth ts p -adc coeffcent b 0 fxed n the range {0,, p 2} fals to satsfy condton 1.2 f and only f the choce ε b 0 also fals to satsfy the condton for the prme p and the number n. The followng corollary follows mmedately from the precedng lemma. Corollary 13. If ε = kp, where k s an nteger, fals to satsfy condton 1.2 for the prme p dvdng n + 1, then all choces for ε n the range {kp,, (k + 1 p 2} also fal to satsfy condton 1.2 for the prme p and the number n. The followng corollary s dentcal to Lemma 12 but stated n a postve sense, descrbng ranges contanng choces for ε that satsfy condton 1.2 for the prme p dvdng 15

20 n + 1. From the earler proof, the congruence 1 ( 1 (b 1p +b 2 p 2 + ( a1 1 b 1 ( a2 b 2 mod p does not contan the p -adc coeffcent b 0. Ths shows that the value of b 0, assumng b 0 p 1, does not affect whether a choce for ε satsfes condton 1.2 for a prme p dvdng n + 1. So, f any choce for ε wth b 0 fxed n the range {0,, p 2} satsfes condton 1.2 for the prme p, then all choces for ε wth b 0 fxed n that range also satsfy the condton for the prme p and the number n. Corollary 14. For any prme p dvdng n + 1, any choce for ε wth ts p -adc coeffcent b 0 fxed n the range {0,, p 2} satsfes condton 1.2 for the prme p and the number n f and only f ε b 0 also satsfes condton 1.2 for the prme p and the number n. The followng corollary s dentcal to Corollary 13 but stated n a postve sense. Corollary 15. Let n + 1 = p 1 p 2 p t wth the conventon p 1 < < p t. If ε = kp, where the prme p dvdes n + 1 and k s an nteger, satsfes condton 1.2 for the prme p, then all choces for ε n the range {kp,, (k + 1 p 2} also satsfy condton 1.2 for the prme p and the number n. Lemma 12 and the related corollares allow us to characterze the sutablty of entre ranges contanng choces for ε. In the followng lemma, we use Lemma 12 and Corollares to smplfy the p -adc expanson of our choce for ε, lettng ε 0 mod p so that the p -adc coeffcent b 0 s dentcally equal to zero. Lemma 16. If a choce ε 1 mod p fals to satsfy condton 1.2 for the prme p dvdng n + 1, t then follows that ε = ε + p necessarly satsfes the condton for the prme p. In other words, f ( n 1 ( 1 ε ε + 1 mod p ε, 16

21 then ε fals to satsfy condton 1.2 for the prme p by defnton, and t then necessarly follows that ( ( 1 (ε +p n 1 (ε + p ε + 1 mod p, + p so that ε = ε + p satsfes condton 1.2 for the prme p and the number n. Proof. Smplfyng by the property of Lemma 12, we can let ε 0 mod p. We characterze ε wth the p -adc expanson ε = b 1 p + b 2 p 2 +. Applyng Lucas Theorem, we can rewrte the bnomal coeffcent ( n 1 ε as ( ( ( ( ( ( n 1 p 2 a1 1 a2 a1 1 a2 mod p ε. 0 By assumpton ε fals to satsfy condton 1.2 for the prme p, so b 1 b 2 ( n 1 ( 1 ε ε mod p ε. Usng the form for ( n 1 ε derved above, we can wrte ( 1 ε ( n 1 ε ( 1 ε ( a1 1 b 1 ( a2 b 2 b 1 b 2 1 mod p. The choce ε fals to satsfy condton 1.2, and we wsh to establsh that ε = ε + p satsfes condton 1.2 for the prme p. We note that ε has the p -adc expanson gven by ε = ε + p = (b p + b 2 p 2 +. Proposton 8 guarantees that b s less than p snce b 1 = p 1 would mply that the choce for ε necessarly satsfes condton 1.2 for the prme p. We have ( ( n 1 ε = n 1 ε +p, and we can wrte ( ( ( n 1 p 2 a1 1 ε + p 0 b ( a2 b 2 ( a1 1 b ( a2 b 2 mod p. 17

22 In order to evaluate the rght hand sde completely, we wll need to rewrte some of the factors. In partcular, notce that we can expand and rewrte the bnomal coeffcent ( a 1 1 b 1 +1 as ( a1 1 b = (a 1 1! (a 1 1 (b 1 + 1! (b 1 + 1! (a 1 1 b 1 (a 1 1! = (a 1 1 b 1 (a 1 1 (b 1 + 1! (b 1! (b = a ( 1 (b a1 1. (b Contnung to smplfy the bnomal coeffcent ( ( n 1 ε = n 1 ε +p, we have ( n 1 ε + p ( a1 1 b ( a2 b 2 a 1 (b (b a 1 (b (b b 1 mod p ( a1 1 b 1 ( n 1 ε ( a2 b 2 mod p. mod p So we see that ( ( n 1 n 1 = ε ε + p a 1 (b (b ( n 1 ε mod p. Note that ε + 1 = (ε + p mod p. We ncorporate the factor ( 1 ε to match the expresson n condton 1.2. We have ( n 1 ( 1 ε ε ( = ( 1 (ε +p n 1 ε + p ( 1 a ( ε 1 (b n 1 mod p b ε a ( ( 1 (b n 1 ( 1 ε mod p b ε a 1 (b b mod p. From ths, we can conclude that ( 1 ε ( n 1 ε s congruent to 1 modulo the prme p f and only f the condton a 1 (b 1 +1 b mod p holds for the expresson derved above. Rewrtng, we see that ths happens f and only f a 1 (b (b mod p, whch

23 holds f and only f a 1 0 mod p. Snce we have assumed that a 1 s not dentcally zero, ths condton cannot hold. Ths tells us that ε = ε + p does ndeed satsfy condton 1.2 for the prme p dvdng n + 1. So we see that f a partcular choce ε 1 mod p fals to satsfy condton 1.2 for the prme p dvdng n + 1, then the choce ε = ε + p satsfes condton 1.2 for the prme p along wth choces for ε wth b 0 n the range {0,, p 2} by Corollary 15. The precedng lemma holds for any choce for ε where ε s a multple of the prme p. For convenence, call t ε = kp where k s an nteger. We make ths explct n the corollary drectly below, whch follows mmedately from the lemma. Corollary 17. Let n + 1 = p 1 p 2 p t wth the conventon p 1 < < p t. If ε = kp, where the prme p dvdes n + 1 and k s an nteger, fals to satsfy condton 1.2 for the prme p and the number n, then the choce ε = ε + p = (k + 1 p satsfes condton 1.2 along wth all values n the range {ε,, ε + p 2} = {(k + 1 p,, (k + 2 p 2}. 19

24 3. Cases n whch the Conjecture Holds The followng theorems rely on the propostons and lemmas of Secton 2 for ther proofs. These theorems were proved ndependently by Walter Parry [4] and n unpublshed notes by Andrew Wlfong. To buld up to the complexty of the general case, where odd n + 1 has an arbtrary prme factorzaton, we begn wth smplfed versons of the conjecture as presented n ths secton. The followng theorem consders the specfc case where n + 1 s the product of two prmes. We begn wth ths most reduced form of the conjecture and gradually buld to more general forms of n + 1. Theorem 18. If n + 1 = p q, where p < q and p and q are both prme, there s a choce for ε n the range {q,, n 2} that satsfes the equaton gcd (R n (ε, n + 1 = 1, whch s conjecture 1.1. Proof. (Ths proof s attrbuted to Andrew Wlfong. Case 1. p = 3 and q = 5. In the specal case that p = 3 and q = 5, we note that ε = 7 s a partcular choce for ε satsfyng conjecture 1.1 for the number n = 14. For the remanng cases, assume (p, q (3, 5. Case 2. q < 2p 1. For the number n + 1 and the prme p, we know from Proposton 11 that all choces for ε n the range {p,, 2p 2} satsfy the condton ( 1 ε ( n 1 ε ε + 1 mod p. We also know that the correspondng condton s satsfed for the prme q by all choces for ε n the range {q,, 2q 2}. Snce p < q < 2p 1, and p and q are odd, the prme number q s n the range {p + 2,, 2p 2}. By Proposton 11, q s a partcular choce for ε satsfyng the condton on R n (ε for both the prme p and the prme q. Ths guarantees a choce for ε ensurng that the expressons R n (ε and n + 1 are relatvely prme, whch s the clam of conjecture 1.1. Case 3. q = 2p 1. Gven ths specfc relaton for the prmes p and q, we can express the value of 2q as 4p 2. By Proposton 11, we know that all choces for ε n the range 20

25 {q,, 2q 2} = {2p 1,, 4p 4} satsfy condton 1.2 for the prme q. By ts structure, ths range contans the numbers 2p and 3p, so long as (p, q (3, 5. (For ths crcumstance, see Case 1. By Lemma 16, ether ε = 2p or ε = 3p must satsfy condton 1.2 for the prme p. Both choces for ε satsfy condton 1.2 for the prme q. Ths guarantees a choce for ε satsfyng conjecture 1.1 for the number n. Case 4. q 2p + 1. In ths case the nterval {q,, 2q 2} contans at least 2p elements, snce the nterval {q,, 2q 2} s dentcal to {2p + 1,, 4p} at ts smallest when q = 2p + 1. In general, ths guarantees at least two multples of p n the nterval. For convenence, call them kp and (k + 1 p where k s an nteger. By Lemma 16, ether the choce ε = kp or ε = (k + 1 p must satsfy condton 1.2 for the prme p. Both choces satsfy the condton for the prme q. Ths guarantees a choce for ε satsfyng conjecture 1.1 for the number n. These cases taken together show that there s a guaranteed choce for ε satsfyng conjecture 1.1 for n when n + 1 s the product of two squarefree prmes. We nformally characterze the followng theorem by the property that the prme factors of the number n + 1 are far apart. Theorem 19. Let n + 1 = p 1 p 2 p t 1 p t wth the conventon p 1 < < p t. If the dfference p k p k 1 s large for all prme factors p k of n + 1 specfcally, f the bound p k 3p k s satsfed for all consecutve prme factors p k and p k 1 of n + 1 then there s a choce for ε satsfyng conjecture 1.1 for the number n. Proof. By Corollary 10, the number p t s the smallest possble choce for ε satsfyng conjecture 1.1 for the number n. By Proposton 11, we know that all choces for ε n the range {p t,, 2p t 2} satsfy condton 1.2 for the prme p t. The range {p t,, 2p t 2} contans p t 1 elements. By assumpton, consecutve prmes p t and p t 1 satsfy the bound p t 3p t Ths means that the nterval {p t,, 2p t 2} contans at least 3p t 1 elements, whch guarantees at least three multples of p t 1 n the nterval. Equvalently, the 21

26 nterval must contan at least two complete, consecutve ntervals of the form {lp t 1,, (l + 1 p t 1 2} where l s an nteger. Usng ths labelng, the next consecutve guaranteed nterval s {(l + 1 p t 1,, (l + 2 p t 1 2}. Corollares 15 and 17 guarantee that n at least one of the gven ranges, {lp t 1,, (l + 1 p t 1 2} or {(l + 1 p t 1,, (l + 2 p t 1 2}, all choces for ε satsfy condton 1.2 for the prme p t 1. The choces for ε n both of these ranges satsfy the condton for the prme p t, so ths guarantees a range contanng choces for ε that smultaneously satsfy condton 1.2 for the prmes p t and p t 1. Consderng analogously the prme p t 2, we can smlarly establsh a range contanng choces for ε that smultaneously satsfy condton 1.2 for the prmes p t, p t 1, and p t 2. Contnung nductvely, we can establsh a nonempty nterval contanng choces for ε that satsfy condton 1.2 for all prme factors p of n + 1. Ths guarantees at least one choce for ε satsfyng conjecture 1.1 for the number n. The followng theorem s nformally characterzed by the property that the prme factors of n + 1 are close together. Theorem 20. Let n + 1 = p 1 p 2 p t 1 p t wth the conventon p 1 < < p t. If the prme factors p of n + 1 are suffcently close together specfcally, f the bound p t < 2p 1 1 s satsfed then we can guarantee the exstence of at least one choce for ε satsfyng conjecture 1.1 for the number n. Proof. By Proposton 11, all choces for ε n the range {p,, 2p 2} satsfy condton 1.2 for the prme p dvdng n + 1. The strng of nequaltes p 1 p p t < 2p 1 1 2p 2 holds for all prmes p dvdng n + 1. The partcular choce ε = p t s n the nterval {p,, 2p 2} for each prme p dvdng n + 1, smultaneously satsfyng condton 1.2 for each prme factor of n + 1. Ths guarantees the exstence of at least one choce for ε satsfyng conjecture 1.1 for the number n when the squarefree prme factors of n + 1 are suffcently close together. 22

27 4. Generalzaton of Many Propostons and Lemmas of Secton 2 In ths secton, we prove generalzatons of many of the propostons and lemmas of Secton 2. We ncrease the complexty of the prme factorzaton of n + 1, consderng the general prme factorzaton n + 1 = p m 1 1 p m 2 2 p m t t. We assume that the exponents m are nteger-valued. We adhere to the conventon p m 1 1 < < p mt t, so that p mt t s the largest prme power dvdng n + 1. We do not place any restrctons on the relatve szes of the prmes p themselves. For example, we do not requre that the prme p 1 be the smallest, nor that the prme p t be the largest, of the prmes p dvdng n + 1. These generalzatons provde tools to prove conjecture 1.1 n cases where n + 1 s not squarefree. We provde a generalzaton of Corollary 10, showng that no choce for ε less than the smallest prme power dvdng n + 1 satsfes conjecture 1.1 for the number n. We generalze Proposton 11, provng that for any prme power p m dvdng n + 1, all choces for ε n the nterval {p m,, 2p m 2}, except where ε 1 mod p, satsfy condton 1.2 for the prme p and the number n. We prove generalzatons of Corollares 14 and 15, showng that f a partcular multple of the prme power p m k p m dvdng n + 1, a number of form where k s an nteger, satsfes condton 1.2 for the prme p, then so do all choces for ε n the range {kp m,, (k + 1 p m 2}, except those choces where ε 1 mod p. We also generalze Lemma 16 and Corollary 17, showng that f a partcular choce ε 1 mod p fals to satsfy condton 1.2 for the prme p and the number n, then the choce ε = ε + p m must satsfy the condton for the prme p dvdng n + 1. Ths result entals the property that f the choces for ε contaned n a range of the form {kp m,, (k + 1 p m 2} fal to satsfy the condton on R n (ε for the prme p, then the choces for ε n the next consecutve range {(k + 1 p m,, (k + 2 p m 2} do satsfy the condton for the prme p, excludng those choces for whch ε 1 mod p. In the followng propostons and lemmas, we wll rewrte the number n 1 n ts p -adc expanson for a prme p dvdng n + 1 many tmes, so we choose to nclude here a summary of the steps taken to rewrte n 1 n ths way. 23

28 Remark 21. The number n + 1 has generc prme factorzaton gven by n + 1 = p m 1 1 p m 2 2 p m t t. In order to focus our attenton on a partcular prme power p m prme factorzaton of n + 1, we choose to rewrte n + 1 n ts p -adc expanson. Ths s necessary to make use of the notatonal convenence provded by Lucas Theorem. In the followng, we rewrte p m dvsors of n + 1 by a sngle nteger. We wrte as p m and represent the product of the other t 1 prme power n the n + 1 = p m 1 1 p mt t = ( p m 1 1 p m 2 2 p m 1 1 pm p mt t p m = q p m. In order to smplfy ths expresson further, we wrte the number q n p-adc form as q = c 0 + c 1 p +. Usng these expressons, we rewrte n + 1 n the followng way n + 1 = q p m = p m (c 0 + c 1 p + = c 0 p m + c 1 p m+1 +. In the p-adc expanson of n + 1 drectly above, the terms p 0,, p m 1 are absent, so the coeffcents on these terms are dentcally zero. Subtractng two from the expresson above, we can now wrte the p-adc expanson of n 1 as n 1 = (p 2 + (p 1 p + + (p 1 p m 1 + (c 0 1 p m + c 1 p m+1 +. We use ths form repeatedly n the followng propostons and lemmas. Before proceedng to the generalzatons of the propostons and lemmas of Secton 2, we nclude a small lemma remnscent of Lemma 5 that provdes a way of rewrtng the bnomal coeffcent ( p 2 b 0 n congruences wth a prme number modulus. The followng lemma helps to smplfy bnomal coeffcents of the form ( p 1 b when evaluated n a congruence wth a prme number modulus. 24

29 Lemma 22. For the prme p, the value of the bnomal coeffcent ( p 1 b, where 0 b < p, s congruent to ( 1 b mod p. That s, ( p 1 b ( 1 b mod p. Proof. We note that p 1 and b are both less than p, so we can reduce fractonal expressons n a congruence base p wthout worry. Rewrtng the value of the bnomal coeffcent ( p 1 b modulo the prme p, we have ( p 1 b = = (p 1! (p 1 b! b! (p 1 (p 2 (p b (p b 1! (p b 1! b! ( 1 ( 2 ( b (p b 1! b! (p b 1! ( 1b b! b! mod p ( 1 b mod p. mod p Ths proves the congruence ( p 1 b ( 1 b mod p where 0 b < p. Proposton 23. [Generalzaton of Proposton 9]. Let n + 1 = p m 1 1 p m 2 2 p m t t wth the conventon p m 1 1 < < p m t t. For a prme power p m dvdng n + 1, any choce for ε less than p m does not satsfy condton 1.2 for the prme p and the number n. Proof. We rewrte n 1 accordng to Remark 21, for a prme power p m have dvdng n + 1. We n 1 = (p 2 + (p 1 p + + (p 1 p m 1 Consder a choce for ε that s less than the prme power p m + (c 0 1 p m + c 1 p m wth b 0 p 1. Let ε have generc p -adc expanson ε = b 0 + b 1 p + + b m 1 p m 1, where each b κ s n the range 0 b κ < p, and κ s an nteger. Applyng Lucas Theorem, we can now rewrte the bnomal 25

30 coeffcent ( n 1 ε modulo the prme p as ( ( ( ( ( n 1 p 2 p 1 p 1 c0 1 ε b 0 b 1 b m 1 0 ( ( ( p 2 p 1 p 1 mod p. ( p 1 b κ ( p 2 b 0 b 1 b m 1 ( c1 mod p 0 Applyng Lemmas 5 and 22, we can neatly rewrte the bnomal coeffcents ( p 2 b 0 and n the congruence where κ s an nteger. By Lemma 5, we have b 0 ( 1 b 0 (b mod p, and, by Lemma 22, we have ( p 1 b κ ( 1 b κ mod p for each b κ. Smplfyng the expanson for ( n 1 ε above, we have ( ( ( ( n 1 p 2 p 1 p 1 ε b 0 b 1 b m 1 mod p ( 1 b 0 (b ( 1 b1 ( 1 b m 1 mod p ( 1 b0 ( 1 b1 ( 1 b m 1 (b mod p. We wsh to express the exponents n the expresson above n terms of ε, and we note that takng each term ( 1 bκ and rasng t to the power of p κ preserves the value of the factor. ( Ths s true snce p κ s odd, so ( 1 κ p b κ ( bκ = ( 1 pκ = ( 1 b κ. Usng ths property, we can rewrte the expresson above as ( n 1 ε ( 1 (b 0+b 1 + +b m 1 (b0 + 1 mod p ( 1 ( b 0 +b 1 p + +b m 1 p m 1 (b0 + 1 mod p ( 1 ε (b mod p. Fnally, we ncorporate the factor ( 1 ε to match the expresson n condton 1.2, gvng ( n 1 ( 1 ε ε ( 1 ε ( 1 ε (b mod p b mod p, 26

31 whch s congruent to ε + 1 mod p snce ε + 1 = ( b 0 + b 1 p + b 2 p b m 1 p m b mod p. Ths means that for any prme power p m dvdng n + 1, the congruence ( 1 ε ( n 1 ε ε + 1 mod p holds for any ε < p m. Ths shows that any choce for ε n the range {2,, p m 2} fals to satsfy condton 1.2 for the prme p dvdng n + 1. The corollary below follows mmedately from the precedng proposton. Corollary 24. Let n + 1 = p m 1 1 p m 2 2 p m t t where max {p m 1 1, p m 2 2,, p m t t } = p m t t. Any choce for ε less than the number p mt t fals to satsfy conjecture 1.1 for the number n. As a matter of notaton, the precedng corollary shows that all choces for ε n the range {2,, p mt t 1} fal to satsfy the conjecture for the number n. The followng proposton generalzes Proposton 11, whch states that all choces for ε n the range {p,, 2p 2} satsfy condton 1.2 for any prme p dvdng n + 1 = p 1 p 2 p t. Ths generalzaton consders the number n + 1 wth the generc prme factorzaton n + 1 = p m 1 1 p m 2 2 p m t t. Proposton 25. [Generalzaton of Proposton 11]. Let n + 1 = p m 1 1 p m 2 2 p mt t wth the conventon p m 1 1 < < p m t t. We are guaranteed that all choces for ε n the range {p m,, 2p m 2}, provded that ε 1 mod p, satsfy condton 1.2 for the prme p and the number n. Proof. We rewrte n 1 accordng to Remark 21 for an arbtrary prme power p m n + 1. We have dvdng n 1 = (p 2 + (p 1 p + + (p 1 p m 1 + (c 0 1 p m + c 1 p m

32 We consder choces for ε n the range {p m,, 2p m 2}, whch we characterze wth the generc p -adc expanson ε = b 0 + b 1 p + + b m 1 p m p m. Applyng Lucas Theorem, we can rewrte the bnomal coeffcent ( n 1 ε as ( ( ( ( ( n 1 p 2 p 1 p 1 c0 1 ε ( c1 0 b 0 b 1 b m 1 1 ( ( ( p 2 p 1 p 1 (c 0 1 mod p. b 0 b 1 b m 1 mod p By Lemma 5, we can rewrte the bnomal coeffcent ( p 2 b 0 as ( 1 (b0 + 1 mod p and by Lemma 22, we can rewrte the bnomal coeffcent ( p 1 b κ as ( 1 b κ mod p for each b κ. Ths allows us to rewrte our expanson above as ( n 1 ε ( 1 b0 (b ( 1 b1 ( 1 b m 1 (c 0 1 mod p ( 1 b0 ( 1 b1 ( 1 b m 1 (b (c 0 1 mod p. We can rewrte the above expresson usng smlar methods to those used n Proposton 23. We collect the factors ( 1 b0 ( 1 b1 ( 1 b m 1, reconstructng the p -adc expanson of ε n the exponents of these factors by rasng each factor ( 1 bκ to the power p κ. Snce each power p κ s odd, the value of each factor s unchanged. We rewrte the factors ( 1 b0 ( 1 b1 ( 1 b m 1 n the expresson above as ( 1 b0 ( 1 b 1p ( 1 b m 1p m 1. To fully represent the value for ε we must nclude the term ( 1 pm. Ths term s dentcal to 1, so we nclude an addtonal factor of 1 to preserve the orgnal value of the expresson. 28

33 Incorporatng these factors, we have ( 1 b0 ( 1 b 1p ( 1 b m 1p m 1 = ( 1 b0 ( 1 b 1p ( 1 b m 1p m 1 ( 1 pm ( 1 = ( 1 ( b 0 +b 1 p + +b m 1p m 1 +p m ( 1 = ( 1 ε. Collectng together the varous expressons above, we can smplfy our expanson of the bnomal coeffcent ( n 1 ε as ( n 1 ε ( 1 b0 ( 1 b1 ( 1 b m 1 (b (c 0 1 mod p ( 1 ε (b (c 0 1 mod p ( 1 ε (b (1 c 0 mod p. Incorporatng the factor ( 1 ε to match the expresson n condton 1.2, we have ( n 1 ( 1 ε ε ( 1 2ε (b (1 c 0 mod p (b (1 c 0 mod p. Choces for ε n the range {p m,, 2p m 2} fal to satsfy condton 1.2 for the prme p dvdng n + 1 f and only f Ths happens f and only f ( n 1 ( 1 ε ε + 1 b mod p. ε (b (1 c 0 b mod p. Snce we are consderng all choces for ε n the range p m ε 2p m 2 wth b 0 p 1, we know that b 0 p 2, so we necessarly have b p 1. Snce b < p, we can 29

34 dvde by b n the above congruence, and we have 1 c 0 1 mod p c 0 0 mod p. Ths tells us that a choce for ε n ths range fals to satsfy condton 1.2 for the prme p f and only f c 0 0 mod p. We have wrtten n + 1 as the product q p m accordng to Remark 21 where q and p are relatvely prme. If c 0 0 mod p, then p q, whch s not possble. Ths proves that all choces for ε n the range {p m,, 2p m 2} wth b 0 p 1 satsfy condton 1.2 for the prme p dvdng n + 1. Lemma 26. [Generalzaton of Lemma 12]. Let n + 1 = p m 1 1 p m 2 2 p m t t wth the conventon p m 1 1 < < p m t t. For a prme p dvdng n + 1, we gve ε the generc p -adc expanson ε = b 0 + b 1 p + + b m 1p m 1 + b m p m +. Any choce for ε wth ts p -adc coeffcent b 0 fxed n the range {0,, p 2} and p -adc coeffcents {b 1,, b m 1} fxed n the range {0,, p 1} fals to satsfy condton 1.2 for the prme p and the number n f and only f ε ( b 0 + b 1 p + + b m 1 p m 1 also fals to satsfy the condton for the prme p and the number n. Proof. We rewrte n 1 accordng to Remark 21 for an arbtrary prme power p m n + 1, and we have dvdng n 1 = (p 2 + (p 1 p + + (p 1 p m 1 + (c 0 1 p m + c 1 p m We gve ε the generc p -adc expanson ε = b 0 + b 1 p + + b m 1p m 1 + b m p m + b m +1p m +1 +, 30

35 where b 0 p 1, so that ε 1 mod p. Applyng Lucas Theorem, Lemma 5, and Lemma 22, we can rewrte the bnomal coeffcent ( n 1 ε as ( ( ( ( ( ( n 1 p 2 p 1 p 1 c0 1 c1 mod p ε b 0 b 1 b m 1 b m b m +1 ( ( ( 1 b0 (b ( 1 b1 ( 1 b m 1 c0 1 c1 mod p ( ( 1 b0 ( 1 b1 ( 1 b m 1 c0 1 (b b m b m +1 ( c1 Here, as n Propostons 23 and 25, we wsh to unfy the factors b m b m +1 mod p. ( 1 b0 ( 1 b1 ( 1 b m 1 by rewrtng the exponents of these factors n terms of ε. The p -adc expanson for ε gven as b 0 + b 1 p + has an unspecfed fnal term. We can stll nclude the remanng factors, but we do not know whether the factor ( 1 ε s postve or negatve. To be specfc, we reconstruct ε n the exponents of the factors of ( 1 as we dd before, wrtng ( 1 b0 ( 1 b1 ( 1 b m 1 = (±1 ( 1 b0 ( 1 b 1p ( 1 b m 1p m 1 = (±1 ( 1 ( b 0 +b 1 p + +b m 1p m 1 +b m p m + = (±1 ( 1 ε. ( 1 bm pm The factor (±1 s ncluded to preserve the orgnal value of the expresson, snce the value of ( 1 ε may be postve or negatve. Usng the above expresson, we now ncorporate the factor ( 1 ε to match the expresson n condton 1.2, and we have ( n 1 ( 1 ε ε ( ( 1 ε (±1 ( 1 ε c0 1 (b ( c0 1 (±1 (b b m ( c1 b m +1 b m ( c1 b m +1 mod p. mod p 31

36 We recall that ε + 1 b mod p. It follows that ( 1 ε ( n 1 ε ε + 1 b0 + 1 mod p f and only f whch holds f and only f ( c0 1 ε + 1 (±1 (b ( c0 1 ±1 b m b m ( c1 b m +1 ( c1 b m +1 mod p, mod p. When ths congruence holds, we know that the choce for ε fals to satsfy condton 1.2 for the prme p. We note that ths property s completely ndependent of the frst m of ε s p -adc coeffcents {b 0, b 1,, b m 1} as s seen n the congruences above. Ths shows that a choce ε 1 mod p fals to satsfy condton 1.2 for a prme p f and only f ε ( b 0 + b 1 p + + b m 1 p m 1 also fals to satsfy the condton for the prme p and the number n. The followng corollary follows mmedately from the precedng lemma. Corollary 27. Let n + 1 = p m 1 1 p m 2 2 p m t t wth the conventon p m 1 where the prme power p m 1 < < p m t t. If ε = kp m, dvdes n + 1 and k s an nteger, fals to satsfy condton 1.2 for the prme p, then all choces for ε n the range {kp m,, (k + 1 p m 2} also fal to satsfy condton 1.2 for the prme p and the number n. The followng corollary s dentcal to Lemma 26 but stated n a postve sense, descrbng ranges contanng choces for ε that satsfy condton 1.2 for the prme p dvdng n + 1. From the earler proof, the congruence ( c0 1 ±1 b m ( c1 b m +1 mod p does not contan the p -adc coeffcents {b 0,, b m 1}. Ths shows that the value of these coeffcents do not affect whether a choce ε 1 mod p satsfes condton 1.2 for a prme p dvdng n + 1. So, f any choce for ε wth coeffcent b 0 fxed n the range {0,, p 2} and coeffcents {b 1,, b m 1} fxed n the range {0,, p 1} satsfes condton 1.2 for 32