The Lucas and Babbage congruences

Transcription

1 The Lucas ad Baage cogrueces Dar Grerg Feruary 26, 2018 Cotets 01 Itroducto 1 1 The cogrueces 2 11 Bomal coeffcets 2 12 Negatve 3 13 The two cogrueces 4 2 Proofs 5 21 Basc propertes of omal coeffcets modulo prmes 5 22 Restatg Vadermode covoluto 6 23 The cogruece lemma 9 24 Proof of the Lucas theorem Two lemmas for Baage s theorem Proof of Baage s theorem 17 3 The sums of the frst p powers The cogruece Powers ad power sums va Strlg umers of the secod d Fshg the proof Itroducto I ths expostory ote, we prove the Lucas ad Baage cogrueces for omal coeffcets The proof s elemetary y ducto ad fuctos for artrary teger parameters as opposed to merely for oegatve tegers Afterwards, we also prove the cogruece s ot a postve multple of p p 1 l0 l 0 mod p for ay prme p ad ay N that 1

2 The Lucas ad Baage cogrueces page 2 1 The cogrueces 11 Bomal coeffcets Let us frst recall the stadard defto of omal coeffcets: 1 Defto 11 Let N ad m Q The, the omal coeffcet ratoal umer defed y m m m 1 m 1! m s a Ths defto s equvalet to the defto gve [Gre17, Defto 21] deed, ths s show [Gre17, Proposto 23] The followg propertes of omal coeffcets are well-ow ad appear [Gre17]: Proposto 12 We have for every m Q m Proposto 12 s [Gre17, Proposto 24 ] wth Z replaced y Q Proposto 13 We have for every m N ad N satsfyg m < Proposto 13 s [Gre17, Proposto 27] m 0 2 Proposto 14 We have for every m N m 1 3 m Proposto 14 s [Gre17, Proposto 210] Proposto 15 We have m m 1 1 m 1 4 for ay m Z ad {1, 2, 3, 1 We use the otato N for the set {0, 1, 2,

3 The Lucas ad Baage cogrueces page 3 Proposto 15 s [Gre17, Proposto 212 ] Proposto 16 We have m Z for ay m Z ad N Proposto 16 s [Gre17, Proposto 220] Proposto 17 For every x Z ad y Z ad N, we have x y 0 x y Proposto 17 s the so-called Vadermode covoluto detty, ad appears [Gre17, Proposto 227 a] 12 Negatve m We have so far defed the omal coeffcet oly for N For the sae m of coveece, let us exted the defto of to artrary tegers To do m so, we eed to defe whe s a egatve teger We do so the smplest possle way: Defto 18 Let e a egatve teger Let m Z The, the omal m m coeffcet s a ratoal umer defed y 0 Ths coveto s the oe used y Graham, Kuth ad Patash [GrKPa] Other authors use other covetos m Hece, the omal coeffcet s defed for all m Z ad Z Namely, t s defed Defto 18 whe s egatve, ad t s defed Defto 11 whe s oegatve The followg fact s easy: m Proposto 19 We have Z for ay m Z ad Z m Proof of Proposto 19 Whe s egatve, Proposto 19 follows from 0 Thus, we WLOG assume that s oegatve Hece, N Thus, Proposto 19 follows from Proposto 16

4 The Lucas ad Baage cogrueces page 4 We ca also exted Proposto 15 to artrary teger values of : Proposto 110 We have m for ay m Z ad Z m 1 1 m 1 Proof of Proposto 110 Let m Z ad Z We are oe of the followg three cases: Case 1: We have < 0 Case 2: We have 0 Case 3: We have > 0 Let us frst cosder Case 1 I ths case, we have < 0 Thus, oth ad m m 1 1 are egatve tegers Hece, all three omal coeffcets, 1 m 1 ad equal 0 y Defto 18 Therefore, the clam of Proposto 110 rewrtes as 0 0 0, whch s clearly true Thus, Proposto 110 s prove Case 1 Let us ext cosder Case 2 I ths case, we have 0 Hece, 1 1 s a m 1 egatve teger Therefore, Defto 18 yelds 0 Also, from 0, 1 m m m 1 m 1 we ota 1 ad 1 Hece, 0 0 m 1 m {{{{ 0 m m m 1 m 1 Comparg ths wth 1, we ota Thus, 1 Proposto 110 s prove Case 2 Let us fally cosder Case 3 I ths case, we have > 0 Thus, {1, 2, 3, m m 1 m 1 Hece, Proposto 15 yelds Thus, Proposto s prove Case 3 We have ow prove Proposto 110 each of the three Cases 1, 2 ad 3 Ths completes the proof 13 The two cogrueces m Proposto 19 shows that s a teger wheever m Z ad Z We shall use ths fact tactly It allows us to state cogrueces volvg omal coeffcets 1

5 The Lucas ad Baage cogrueces page 5 Next, let us state two classcal results o the ehavor of omal coeffcets modulo prmes: Theorem 111 Let p e a prme Let a ad e two tegers Let c ad d e two elemets of {0, 1,, p 1 The, ap c a c mod p p d d Theorem 111 s ow uder the ame of Lucas s theorem, ad s prove may places eg, [Mestro14, 21] or [Hause83, Proof of 4] or [ABeRo05, proof of Lucas s theorem] or [GrKPa, Exercse 561] the case whe a ad are oegatve tegers The stadard proof of Theorem 111 ths case uses geeratg fuctos; t s ot hard to twea ths proof so that t apples mutats mutads the geeral case as well But we are gog to gve a dfferet, more elemetary proof of Theorem 111 Aother classcal result aout omal coeffcets ad prmes s the followg fact: Theorem 112 Let p e a prme Let a ad e two tegers The, ap p a mod p 2 I the case whe a ad are oegatve tegers, Theorem 112 s a ow result, due to Charles Baage see, eg, [Sta11, Exercse 114 c] or [GrKPa, Exercse 562] Notce that f p 5, the the modulus p 2 ca e replaced y p 3 or depedg o a, ad p y eve hgher powers of p; see [Mestro14, 22 ad 23] for the detals We shall prove Theorem 112 later 2 Proofs 21 Basc propertes of omal coeffcets modulo prmes Let us frst state a smple fact: Proposto 21 Let p e a prme Let {1, 2,, p 1 The, p p Proposto 21 s [Gre16, Corollary 56] ad [BeQu03, Theorem 13]

6 The Lucas ad Baage cogrueces page 6 22 Restatg Vadermode covoluto Let us also derve oe more smple corollary of Proposto 17: Corollary 22 Let x Z ad y N ad Z The, y x y x 0 y 5 Proof of Corollary 22 We are oe of the followg two cases: Case 1: We have < 0 Case 2: We have 0 I Case 1, Corollary 22 s easy to chec, ecause oth sdes of 5 are 0 ths case 2 Let us ow cosder Case 2 I ths case, we have 0 Hece, N sce Z Defe a g N y g max {y, Thus, g max {y, y ad g max {y, We have g y 0 sce y N, so that 0 y g Hece, we ca splt the sum g 0 x y at y We thus ota g y g x y x y x 0 0 y1 y x y 0 y x 0 y x 0 {{ 0 g y1 y {{ 0 y Proposto 13 sce y< sce y1>y 6 O the other had, g 0, so that 0 g Hece, we ca splt the sum 2 Ideed, the left-had sde of 5 s 0 sce < 0, ad the rght-had sde of 5 s 0 sce each x {0, 1,, y satsfes < 0 ad thus 0

7 The Lucas ad Baage cogrueces page 7 g x y at We thus ota 0 g x y x y 0 0 x y 0 Comparg ths wth 6, we fd Proposto 17 yelds x y 0 0 x 0 y g 1 y x y x 0 0 x y y 7 Ths proves Corollary 22 x {{ 0 sce <0 sce 1> g y 0 1 {{ 0 x 0 y y 7 y {{ y here, we have susttuted for the sum x y x y y 0 Let us state a few cosequeces of Corollary 22: Corollary 23 Let x Z ad Z Let y e a postve teger The, x y x y 1 x 1 y x y Proof of Corollary 23 We ow that y s a postve teger Thus, 0 ad y are two

8 The Lucas ad Baage cogrueces page 8 dstct elemets of {0, 1,, y Corollary 22 yelds y x y x y 0 y 1 x y x y x 0 0 y {{ {{ 1 x 1 x Ths proves Corollary 23 here, we have splt off the addeds for 0 ad for y from the sum sce 0 ad y are two dstct elemets of {0, 1,, y y y 1 1 x x y Corollary 24 Let x Z ad Z Let p e a prme The, x p x x mod p p y y {{ 1 y Proposto 14 appled to my Proof of Corollary 24 We have p 0 mod p for each {1, 2,, p The umer p s a prme, ad thus a postve teger Hece, Corollary 23 appled to y p yelds p 1 x p x x p x p 1 x Ths proves Corollary 24 p 1 1 x 0 {{ 0 {{ 0 mod p y 8 x p x x mod p p 3 Proof of 8: Let {1, 2,, p 1 Proposto 21 appled to yelds p p words, 0 mod p Ths proves 8 p I other

9 The Lucas ad Baage cogrueces page 9 23 The cogruece lemma We ow show a geeral lemma that helps us attac cogrueces volvg omal coeffcets: 4 Lemma 25 Let A : Z Z Z e ay map Let N e a teger Let u Z Assume that the followg four codtos hold: Every a Z ad Z satsfy A a, A a 1, A a 1, 1 mod N 9 We have A 0, 0 u mod N 10 Every a Z ad every egatve Z satsfy A a, 0 mod N 11 Every postve teger satsfes A 0, 0 mod N 12 The, every a Z ad Z satsfy a A a, u mod N Proof of Lemma 25 Let us frst show the followg fact: 0 Oservato 1: Let Z We have A 0, u mod N [Proof of Oservato 1: We are oe of the followg three cases: Case 1: We have < 0 Case 2: We have 0 Case 3: We have > 0 Let us frst cosder Case 1 I ths case, we have < 0 Thus, s a egatve teger sce Z Hece, Defto 18 yelds Thus, u {{ 0 0, 4 We shall oly use Lemma 25 the case whe N s a postve teger For the sae of geeralty, we are evertheless statg t for artrary tegers N Mae sure to correctly terpret the otato u v mod N whe N s 0: If u ad v are two tegers, the u v mod 0 holds f ad oly f u v

10 The Lucas ad Baage cogrueces page so that 0 u But 11 appled to a 0 yelds A 0, 0 u mod N Thus, Oservato 1 s prove Case Let us ow cosder Case 2 I ths case, we have 0 Thus, u u 0 {{ u, so that u u But 10 yelds A 0, 0 u u mod N From 0, we ota A 0, {{ 0 A 0, 0 u mod N Thus, Oservato 1 s prove 0 Case 2 Fally, let us cosder Case 3 I ths case, we have > 0 Thus, N ad 0 0 < Hece, Proposto 13 appled to m 0 ad yelds 0 Hece, u 0, so that 0 u But 12 yelds A 0, 0 u mod N Thus, {{ 0 Oservato 1 s prove Case 3 We have ow prove Oservato 1 each of the three Cases 1, 2 ad 3 Sce these three Cases cover all possltes, we thus coclude that Oservato 1 always holds] Next, we clam the followg fact: a Oservato 2: We have A a, u mod N for each a N ad Z [Proof of Oservato 2: We shall prove Oservato 2 y ducto over a 0 Iducto ase: We have A 0, u mod N for each Z accordg to Oservato 1 I other words, Oservato 2 holds for a 0 Ths completes the ducto ase Iducto step: Let c e a postve teger Assume that Oservato 2 holds for a c 1 We must prove that Oservato 2 holds for a c We have assumed that Oservato 2 holds for a c 1 I other words, we have c 1 A c 1, u mod N for each Z 13

11 The Lucas ad Baage cogrueces page 11 For each Z, we have c c 1 c 1 1 y Proposto 110 appled to m c ad c 1 c Now, for each Z, we have A c, A c 1, {{ c 1 u y 13 mod N A c 1, 1 {{ c 1 u 1 mod N y 13 appled to 1 stead of y 9 appled to a c c 1 c 1 c 1 c 1 u u u 1 1 {{ c c u mod N y 14 I other words, Oservato 2 holds for a c Ths completes the ducto step Thus, Oservato 2 s prove] Our ext step shall e to prove the followg fact: a Oservato 3: Let h N We have A a, u mod N for each a Z ad Z satsfyg a < h [Proof of Oservato 3: We shall prove Oservato 3 y ducto over h: a Iducto ase: We have A a, u mod N for each a Z ad Z satsfyg a < 0 5 I other words, Oservato 3 holds for h 0 Ths completes the ducto ase a 5 Proof Let a Z ad Z e such that a < 0 We must prove that A a, u mod N If a N, the ths follows mmedately from Oservato 2 Thus, for the rest of ths proof, we WLOG assume that we do t have a N Hece, a / N Comg a Z wth a / N, we ota a Z \ N { 1, 2, 3, Hece, a < 0 But from a < 0, we ota < a < 0 a Thus, s a egatve teger sce Z Therefore, Defto 18 yelds 0 Hece,

12 The Lucas ad Baage cogrueces page 12 Iducto step: Let g N Assume that Oservato 3 holds for h g We must prove that Oservato 3 holds for h g 1 We have assumed that Oservato 3 holds for h g I other words, we have a A a, u mod N for each a Z ad Z satsfyg a < g 15 Now, let a Z ad Z e such that a < g 1 The, 9 appled to a 1 stead of a shows that Hece, A a 1, A a 1 1, A a 1 1, 1 {{{{ a a A a, A a, 1 mod N A a, A a 1, A a, 1 mod N 16 Also, Proposto 110 appled to m a 1 ad yelds a 1 a 1 1 a 1 1 a 1 1 sce a 1 1 a a a 1 Hece, a a 1 a a 17 1 But a 1 {{ a 1 < g 1 1 g Hece, we ca apply 15 to a 1 <g1 stead of a We thus ota a 1 A a 1, u mod N Also, 1 a {{ a 1 < g 1 1 g Hece, we ca apply 15 to 1 <g1 stead of We thus ota a A a, 1 u mod N 1 a a a u 0, so that 0 u But 11 yelds A a, 0 u mod N Hece, we have {{ 0 a prove that A a, u mod N Qed

13 The Lucas ad Baage cogrueces page 13 Thus, 16 ecomes A a, A a 1, {{ a 1 u a 1 u mod N A a, 1 {{ a u mod N 1 a 1 a a u u mod N 1 {{ a a u 1 y 17 Now, forget that we fxed a ad We thus have show that we have A a, a u mod N for each a Z ad Z satsfyg a < g 1 I other words, Oservato 3 holds for h g 1 Ths completes the ducto step Thus, Oservato 3 s prove] a Now, let a Z ad Z e artrary We must prove that A a, u mod N Defe h Z y h max {0, a 1 Thus, h max {0, a 1 0, so that h N sce h Z Also, h max {0, a 1 a 1 > a, so that a a < h Hece, Oservato 3 shows that we have A a, u mod N Ths completes the proof of Lemma Proof of the Lucas theorem We are ow ready to prove Theorem 111: Proof of Theorem 111 Let us forget that we fxed a ad c Defe a teger u Z y u Defe a map A : Z Z Z y d ap c A a, for each a, Z Z p d Let us ow prove some propertes of the map A: Oservato 1: Every a Z ad Z satsfy A a, A a 1, A a 1, 1 mod p [Proof of Oservato 1: Let a Z ad Z The, the defto of A yelds a 1 p c A a 1, 18 p d

14 The Lucas ad Baage cogrueces page 14 Also, the defto of A yelds A a 1, 1 a 1 p c 1 p d a 1 p c p d p sce 1 p d p d p Corollary 24 appled to x a 1 p c ad p d yelds a 1 p c p a 1 p c a 1 p c p d p d p d p {{{{ Aa 1, Aa 1, 1 y 18 y 19 Now, the defto of A yelds ap c a 1 p c p A a, p d p d A a 1, A a 1, 1 mod p y 20 Ths proves Oservato 1] Oservato 2: We have A 0, 0 u mod p 19 A a 1, A a 1, 1 mod p 20 sce ap c a 1 p c p [Proof of Oservato 2: The defto of A yelds 0p c c A 0, 0 sce 0p c c ad 0p d d 0p d d c u sce u d u mod p Ths proves Oservato 2] Oservato 3: Every a Z ad every egatve Z satsfy A a, 0 mod p [Proof of Oservato 3: Let a Z Let Z e egatve The, p d s a ap c egatve teger 6 Hece, Defto 18 yelds 0 Now, the defto p d of A yelds ap c A a, 0 0 mod p p d Ths proves Oservato 3] 6 Proof We have d {0, 1,, p 1, so that d p 1 < p But s a egatve teger sce Z s egatve; hece, { 1, 2, 3,, so that 1 Hece, p 1 p sce p s postve Thus, p {{{{ d < 1 p p 0 Therefore, p d s a egatve teger sce <p 1p p d Z Qed

15 The Lucas ad Baage cogrueces page 15 Oservato 4: Every postve teger satsfes A 0, 0 mod p [Proof of Oservato 4: Let e a postve teger Thus, > 0 Also, d {0, 1,, p 1, so that d 0 Hece, {{ p {{{{ d > 0 Thus, p d s a >0 >0 0 postve teger sce p d Z, so that p d N Also, c {0, 1,, p 1 N Moreover, > 0, so that 1 sce s a teger Usg p > 0, we thus fd {{ p p But c {0, 1,, p 1, so that c p 1 < p Hece, p > c 1 Now, p {{ d p p > c, so that c < p d Hece, Proposto 13 appled 0 c to m c ad p d yelds 0 p d Now, the defto of A yelds 0p c c A 0, sce 0p c c p d p d 0 0 mod p Ths proves Oservato 4] We have ow prove the four Oservatos 1, 2, 3 ad 4 I other words, the four codtos Lemma 25 hold f we set N p Thus, Lemma 25 appled to N p yelds that every a Z ad Z satsfy a A a, u mod p 21 Now, let a ad e two tegers Thus, a Z ad Z Hece, 21 yelds a c a a c A a, {{ u mod p d d c I vew of ths rewrtes as A a, Ths proves Theorem 111 d ap c p d ap c p d a y the defto of A, c mod p d

16 The Lucas ad Baage cogrueces page Two lemmas for Baage s theorem Before we start provg Theorem 112, let us show two more auxlary results The frst oe s a cosequece of Theorem 111: Lemma 26 Let p e a prme Let r Z ad s Z Let {1, 2,, p 1 rp The, p sp Proof of Lemma 26 From {1, 2,, p 1, we coclude that s a postve teger Thus, > 0 Hece, 0 < Also, {1, 2,, p 1 N Thus, Proposto 0 13 appled to m 0 ad yelds 0 We have {1, 2,, p 1 {0, 1,, p 1 Also, 0 {0, 1,, p 1 sce p 1 N sce p s a postve teger Thus, Theorem 111 appled to a r, s, c 0 ad d yelds rp 0 r 0 0 mod p sp s {{ 0 rp 0 rp I other words, p I vew of rp 0 rp, ths rewrtes as p sp sp Hece, Lemma 26 s prove The ext auxlary result s smlar to Corollary 24, ad also follows from Corollary 23: Corollary 27 Let r Z ad Z Let p e a prme The, r 1 p rp rp mod p 2 p p 1 p Proof of Corollary 27 We have rp p 0 mod p 2 p for each {1, 2,, p p 7 Proof of 22: Let {1, 2,, p 1 Proposto 21 appled to yelds p I other p words, there exsts a x Z such that px Cosder ths x O the other had, from {1, 2,, p 1, we ota p {1, 2,, p 1 Hece,

17 The Lucas ad Baage cogrueces page 17 The umer p s a prme, ad thus a postve teger Hece, Corollary 23 appled to x rp, y p ad p yelds rp p p rp p p 1 rp p p 1 rp 1 1 p p {{ 0 mod p 2 y 22 0 {{ 0 rp 1 p rp p p {{ rp 1 p sce p p 1p rp p rp 1 p I vew of rp p r 1 p, ths rewrtes as r 1 p rp rp mod p 2 p p 1 p Ths proves Corollary 27 mod p 2 26 Proof of Baage s theorem We are ow ready to prove Theorem 112: Proof of Theorem 112 Let us forget that we fxed a ad Defe a teger u Z y u 1 Defe a map A : Z Z Z y A a, ap p Let us ow prove some propertes of the map A: Oservato 1: Every a Z ad Z satsfy for each a, Z Z A a, A a 1, A a 1, 1 mod p 2 rp rp Lemma 26 appled to s 1 ad p yelds p 1 p p p rp sce 1 p p p I other words, there exsts a y Z such that py p Cosder ths y We have rp p pypx p 2 xy 0 mod p 2 p {{{{ py px Ths proves 22

18 The Lucas ad Baage cogrueces page 18 [Proof of Oservato 1: Let a Z ad Z The, the defto of A yelds a 1 p A a 1, 23 p Also, the defto of A yelds A a 1, 1 Corollary 27 appled to r a 1 yelds a 1 1 p a 1 p a 1 p p p 1 p {{{{ Aa 1, Aa 1, 1 y 23 y 24 a 1 p 24 1 p A a 1, A a 1, 1 mod p 2 25 Now, the defto of A yelds ap a 1 1 p A a, sce a a 1 1 p p A a 1, A a 1, 1 mod p 2 y 25 Ths proves Oservato 1] Oservato 2: We have A 0, 0 u mod p 2 [Proof of Oservato 2: The defto of A yelds 0p 0 A 0, 0 1 u u mod p 2 0p 0 Ths proves Oservato 2] Oservato 3: Every a Z ad every egatve Z satsfy A a, 0 mod p 2 [Proof of Oservato 3: Let a Z Let Z e egatve The, p s a egatve ap teger sce s egatve ut p s postve Hece, Defto 18 yelds 0 p Now, the defto of A yelds ap A a, 0 0 mod p 2 p Ths proves Oservato 3]

19 The Lucas ad Baage cogrueces page 19 Oservato 4: Every postve teger satsfes A 0, 0 mod p 2 [Proof of Oservato 4: Let e a postve teger Thus, > 0 Hece, p > 0 sce p s postve Thus, p s a postve teger sce p Z, so that p N Moreover, 0 < p sce p > 0 Hece, Proposto 13 appled to m 0 ad 0 p yelds 0 p Now, the defto of A yelds A 0, 0p p 0 p 0 0 mod p 2 sce 0p 0 Ths proves Oservato 4] We have ow prove the four Oservatos 1, 2, 3 ad 4 I other words, the four codtos Lemma 25 hold f we set N p 2 Thus, Lemma 25 appled to N p 2 yelds that every a Z ad Z satsfy a A a, u mod p 2 26 Now, let a ad e two tegers Thus, a Z ad Z Hece, 26 yelds a a A a, {{ u mod p 2 1 I vew of ths rewrtes as A a, Ths proves Theorem 112 ap p ap p y the defto of A, a mod p 2 3 The sums of the frst p powers 31 The cogruece Next, we shall prove a well-ow cogruece cocerg the sum 1 p 1 for a prme p: p 1 l 0 l0

20 The Lucas ad Baage cogrueces page 20 Theorem 31 Let p e a prme Let N Assume that s ot a postve multple of p 1 The, p 1 l 0 mod p l0 Theorem 31 has othg to do wth omal coeffcets Nevertheless, we shall prove t usg omal coeffcets 32 Powers ad power sums va Strlg umers of the secod d We shall frst troduce aother famly of tegers: the so-called Strlg umers of the secod d They have varous equvalet deftos; we defe them y recurso: { m Defto 32 For each m N ad Z, we defe a teger as follows: We proceed y recurso o m: We set { { 0 1, f 0; 0, f 0 { m Ths defes for m 0 for all Z 27 For each postve teger m ad each Z, we set { { { m m 1 m { { m m Thus, a famly of tegers s defed These tegers m, N Z are called the Strlg umers of the secod d { m These Strlg umers have a well-ow comatoral terpretato: { m Namely, f m N ad N, the s the umer of set parttos of the set {1, 2,, m to oempty susets Ths s actually ot hard to prove y ducto o m for example, the proof s setched [Sta11, 19] ad [GrKPa, 61] 8 ; ut we do t eed ths Istead, let us prove the followg algerac facts: { m 8 To e more precse, oth [Sta11, 19] ad [GrKPa, 61] defe as the umer of set

21 The Lucas ad Baage cogrueces page 21 Proposto { 33 For each m N ad Z satsfyg / {0, 1,, m, we have m 0 Proof of Proposto 33 We shall prove Proposto 33 y ducto o m: Iducto ase: Proposto 33 holds whe m 0 as follows easly from 27 Ths completes the ducto ase Iducto step: Let e a postve teger Assume that Proposto 33 holds whe m 1 We must ow prove that Proposto 33 holds whe m We have assumed that Proposto 33 holds whe m 1 I other words, for each Z satsfyg / {0, 1,, 1, we have { Now, let Z e such that / {0, 1,, { Thus, 1 / {0, 1,, 1 1 Hece, 29 appled to 1 stead of yelds 0 Also, / {0, 1,, 1 1 { 1 ths aga follows from / {0, 1,, Hece, 29 yelds 0 Now, 28 appled to m yelds { { { {{{{ 0 0 Now, forget that we fxed { We thus have show that for each Z satsfyg / {0, 1,,, we have 0 I other words, Proposto 33 holds whe m Ths completes the ducto step Thus, the proof of Proposto 33 s complete Lemma 34 Let N ad x Q The, x x! x 1! 1 Proof of Lemma 34 We have N Thus, the defto of x x x 1 x 1! x yelds parttos of the set {1, 2,, m to oempty susets, ad the prove that 27 ad 28 hold wth ths defto Ths s exactly the opposte of what we are dog; ut of course, t s equvalet

22 The Lucas ad Baage cogrueces page 22 Hece,! x x! x {{ x x 1 x 1! x x 1 x 1! x x x 1 x 1 x x 1 x 1 x x x 1 x 30 O the other had, 1 N sce N Hece, the defto of yelds x 1 x x 1 x 1 1 1! Hece, x 1! x x 1 x x x 1 x sce x 1 1 x x! x y 30 Ths proves Lemma 34 Proposto 35 Let m N ad x Q The, x m m! 0 { m x x 1 Proof of Proposto 35 We shall prove Proposto 35 y ducto o m: Iducto ase: It s straghtforward to see that Proposto 35 holds whe m 0 Ths completes the ducto ase Iducto step: Let e a postve teger Assume that Proposto 35 holds whe m 1 We must ow prove that Proposto 35 holds whe m We have assumed that Proposto 35 holds whe m 1 I other words, we have x 1 1! 0 { 1 x

23 The Lucas ad Baage cogrueces page 23 Multplyg oth sdes of ths equalty y x, we ota x 1 x 1! { 1 x x 1! 0 { 1 { 1 x! x { {{ { 1 x 1 x! { { 1 x 1 x! { 1 x 1 { 1 x! 0! 1! 0 1! 0 x x x x {{ x x 31 { 1 But Proposto 33 appled to m 1 ad yelds 0, whereas { 1 Proposto 33 appled to m 1 ad 1 yelds 0 1 But 28 appled to m ad yelds { { { We have N sce s a postve teger, so that {0, 1,, Now, { 1! 0 1! 0 1! 0 x { { 1 x 1! {{ 0 x here, we have splt off the added for from the sum { 1 x sce {0, 1,, x 1 { 1!0! {{ 0 0 x, so that 1! 0 { 1 x! 0 { 1 x 33

24 The Lucas ad Baage cogrueces page 24 Also, 0 {0, 1,, sce N Now, 1 0 { 1 x! x {{ { x x x! x {{ x 1! 1 y Lemma ! 0 x! 1 { 1 { {{ sce 1 1 x 1 { { 1 1 here, we have susttuted for 1 the sum Comparg ths wth x! 0! 1 { 1 1 { x 1 1 { x 1 0! { {{ here, we have splt off the added for 0 from the sum! 1 x { 1 1 sce 0 {0, 1,, x { x 1 0! 0 0! 1 {{ 1 0, we fd 1! 0 { 1 x x! 0 x {

25 The Lucas ad Baage cogrueces page 25 Comparg the equalty 31 wth x 1 x x, we fd x 1 { 1 x! 0 { {{ 1 x! y 33 { 1 x! x 0 { {{ x 1 1! 0 y 34 1 { 1 x { x 1!! 1 0 { { 1 x x 1!! 1 {{ { { 1 1 x! 1 { { 1 1 x! 1 { {{ y 32! 0 { x I other words, Proposto 35 holds whe m Ths completes the ducto step Thus, Proposto 35 s prove Next, we shall prove aother asc detty aout omal coeffcets, sometmes ow as the hocey-stc detty ths or aother equvalet form: Proposto 36 Let N ad h N The, h x0 x h 1 1 Proof of Proposto 36 For each x N, we have x x 1 x Proof of 35: Let x N The, Proposto 110 appled to m x 1 ad 1 yelds x 1 1 x x x x 1

26 The Lucas ad Baage cogrueces page 26 h1 x0 Proposto 13 appled to m 0 ad 1 yelds 0 < 1 But x 1 h1 x 0 1 x sce 1 {{ 0 here, we have splt off the added for x 0 from the sum h1 x 1 x1 h x0 sce 0 {0, 1,, h 1 x 1 1 here, we have susttuted x 1 for x the sum Hece, x 1 x h x h h x0 h1 x0 x0 36 here, we have splt off the added for x h 1 from the sum sce h 1 {0, 1,, h 1 sce h 1 N Now, h x0 h x0 x {{ x 1 x 1 1 y 35 x 1 1 h x x0 1 Ths proves Proposto 36 x 1 h 1 1 h x0 h x x0 1 h x 1 x0 1 {{ x h y 36 h 1 1 h x0 x 1 We ca ow ota a reasoaly smple formula for sums of the form sce x 1 1 x ad 1 1 Solvg ths equato for x 1 x Ths proves h x m : x0 x, we ota x

27 The Lucas ad Baage cogrueces page 27 Theorem 37 Let m N ad h N The, h x m m! x0 0 { m h 1 1 Proof of Theorem 37 We have h x m h m { m {{! x0 { x0 m m x 0 {{! 0 y Proposto 35 Ths proves Theorem 37 h 0 x0 m m h 0 x0 m! 0 { m { m! x x h 1 1 m { m! 0 h x x0 {{ h 1 1 y Proposto 36 We are ow ready to prove the followg partcular case of Theorem 31: Lemma 38 Let p e a prme Let N Assume that < p 1 The, p 1 l 0 mod p l0 Proof of Lemma 38 We have < p 1, so that 1 < p Sce 1 ad p are tegers, ths yelds 1 p 1 For each {0, 1,,, we have p 0 mod p Proof of 37: Let {0, 1,, Thus, 0 From 0, we ota 1 1 Comg {{ 0 ths wth 1 1 p 1, we ota 1 1 p 1 Hece, 1 {1, 2,, p 1 {{ p p Thus, Proposto 21 appled to 1 stead of yelds p I other words, mod p Ths proves 37

28 The Lucas ad Baage cogrueces page 28 Now, p 1 N sce p 1 sce p s a prme Thus, Theorem 37 appled to h p 1 ad m yelds p 1 x { p 1 1 { p! x0 1! sce p 1 1 p {{ 0 mod p y 37 Now,! 0 { 0 0 mod p p 1 p 1 l x here, we have reamed the summato dex l as x l0 x0 0 mod p Ths proves Lemma Fshg the proof The last gredet we eed for the proof of Theorem 31 s Fermat s lttle theorem: Proposto 39 Let p e a prme Let a Z The, a p a mod p Proposto 39 s, of course, oe of the fudametal facts of umer theory, ad shall ot e prove here We shall use the followg corollary of Proposto 39: Corollary 310 Let p e a prme Let a Z Let e a postve teger Let r e the remader of upo dvso y p 1 Assume that s ot a multple of p 1 The, a a r mod p Proof of Corollary 310 The defto of r shows that r {0, 1,, p 1 1 ad r mod p 1 ad r sce 0 But 0 mod p 1 sce s ot a multple of p 1 Hece, r 0 mod p 1, so that r 0 From r mod p 1, we ota p 1 r Thus, there exsts a q Z such that r p 1 q Cosder ths q We have p 1 q r 0 sce r ad thus q 0 sce p 1 > 0 I other words, q N Proposto 39 yelds a p a mod p Hece, a p 1m a mod p for each m N Applyg ths to m q, we ota a p 1q1 a mod p sce q N Multplyg oth sdes of ths cogruece wth a r 1, we fd a p 1q1 a r 1 aa r 1 a r mod p Thus, a r a p 1q1 a r 1 a p 1q1r 1 a mod p 11 Proof of 38: We shall prove 38 y ducto o m:

29 The Lucas ad Baage cogrueces page 29 sce p 1 q 1 r 1 r 1 r 1 Ths proves Corollary {{ r 310 Proof of Theorem 31 If < p 1, the the clam of Theorem 31 follows from Lemma 38 Hece, for the rest of ths proof, we WLOG assume that we do t have < p 1 Thus, p 1 > 0 Hece, s postve Thus, s ot a multple of p 1 ecause s ot a postve multple of p 1, ut s postve Let r e the remader of upo dvso y p 1 Thus, r {0, 1,, p 1 1, so that r N ad r p 1 1 < p 1 Each l {0, 1,, p 1 satsfes l l r mod p y Corollary 310, appled to a l Hece, p 1 l0 l {{ l r mod p p 1 l r 0 mod p l0 y Lemma 38, appled to r stead of Ths proves Theorem 31 Refereces [Ager07] Mart Ager, A Course Eumerato, Graduate Texts Mathematcs #238, Sprger 2007 [ABeRo05] Peter G Aderso, Arthur T Beam ad Jeremy A Rouse, Comatoral Proofs of Fermat s, Lucas s, ad Wlso s Theorems, The Amerca Mathematcal Mothly, Vol 112, No 3 Mar, 2005, pp [BeQu03] Arthur T Beam ad Jefer J Qu, Proofs that Really Cout: The Art of Comatoral Proof, The Mathematcal Assocato of Amerca, 2003 [Comtet74] Lous Comtet, Advaced Comatorcs: The Art of Fte ad Ifte Expasos, D Redel Pulshg Compay, 1974 Iducto ase: We have a p 101 a 1 a a mod p I other words, 38 holds for m 0 Ths completes the ducto ase Iducto step: Let N Assume that 38 holds for m We must prove that 38 holds for m 1 We have assumed that 38 holds for m I other words, a p 11 a mod p Now, p p 1 1 p 1 Hece, a p 111 a p 11p 1 a p 11 {{ a p 1 aa p 1 a p a mod p a mod p I other words, 38 holds for m 1 Ths completes the ducto step Hece, 38 s prove

30 The Lucas ad Baage cogrueces page 30 [GrKPa] Roald L Graham, Doald E Kuth, Ore Patash, Cocrete Mathematcs, Secod Edto, Addso-Wesley 1994 [Gre16] Dar Grerg, Flec s omal cogruece usg crculat matrces, 4 Decemer [Gre17] Dar Grerg, Notes o the comatoral fudametals of algera, 25 Decemer [Hause83] Melv Hauser, Applcatos of a Smple Coutg Techque, The Amerca Mathematcal Mothly, Vol 90, No 2 Fe, 1983, pp [Mestro14] Romeo Meštrovć, Lucas theorem: ts geeralzatos, extesos ad applcatos , arxv: v1 [Sta11] Rchard Staley, Eumeratve Comatorcs, volume 1, Secod edto, verso of 15 July 2011 Avalale at [Stato16] Des Stato, Addedum to Usg the Freshma s Dream to Prove Comatoral Cogrueces, freshmadesstatopdf