In number theory we will generally be working with integers, though occasionally fractions and irrationals will come into play.

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1 Number Theory Math 5840 otes. Sectio 1: Axioms. I umber theory we will geerally be workig with itegers, though occasioally fractios ad irratioals will come ito play. Notatio: Z deotes the set of all itegers Z = {..., 2, 1, 0, 1, 2, 3,...} ad Z + deotes the set of positive itegers {1, 2, 3,...}. Axioms about additio ad multiplicatio: There exists two operatios o the itegers: additio deoted by + ad multiplicatio deoted by. Strictly speakig + is a map from the cross product of the itegers with itself ito the itegers with certai properties as defied by the axioms, ad similarly for multiplicatio. Note that a b is usually writte as ab. Axioms about additio. A1. If a Z ad b Z the a + b Z. [Closure.] A2. If a Z ad b Z the a + b = b + a. [Commutativity.] A3. If a Z, b Z ad c Z the a + (b + c) = (a + b) + c. [Associativity.] A4. There exists a elemet 0 Z so that if a Z the a + 0 = a. [Idetity elemet.] A5. If a Z the there exits a elemet i Z deoted by a so that a + a = 0. [Additive iverse. Note that the egative umbers are, by defiitio, the additive iverses of the positive umbers.] Defiitio: a b meas a + ( b). Assumig Axioms A1-A5 is equivalet to sayig that that Z is a commutative group uder the operatio of additio. Axioms about multiplicatio. B1. If a Z ad b Z the a b Z. [Closure.] B2. If a Z ad b Z the a b = b a. [Commutivity.] B3. If a Z, b Z ad c Z the a (b c) = (a b) c. [Associativity.] B4. There exists a elemet 1 Z so that if a Z the a 1 = a. [Idetity elemet.] B5. If a Z, b Z, c Z ad ac = bc the a = b. [Cacellatio rule.] 1

2 Axioms o the relatioship betwee additio ad multiplicatio. C1. If a Z, b Z ad c Z the a (b+c) = a b+a c. [Distributive law.] Note the assumptio that the order of operatio is to perform first the +. C (Note that this axiom is really implicit i defiitio of Z ad that axioms A4 ad B4 selected differet specific elemets of Z.) Defiitio: If a Z ad b Z the a < b meas that b a Z +. [Defiitio of order.] Axioms o order. D1. If a Z + ad b Z + the a + b Z + ad a b Z +. [Closure of additio ad multiplicatio for positive itegers.] D2. If a Z the either a < 0, a = 0, or 0 < a. [Trichotomy Law.] Iductio axiom. E1. Every oempty subset of Z + has a least elemet. [Well ordered property.] Axiom E1 is equivalet to the followig: If S is a subset of Z + cotaiig 1 such that if S the + 1 S; the S = Z +. Defiitio: a 0 = 1, a 1 = a, a +1 = a a. Theorem 1.1. Suppose that each of a, b ad c is a iteger. a. (a + b) c = a c + b c. b. a + (b + c) = (c + a) + b. c. (a + b) 2 = a 2 + 2a b + b 2. d. ( 1) a = a. e. ( a) = a. f. ( a) b = (ab) = a ( b). g. ( a) ( b) = a b. h. 0 = 0. i. The additive ad multiplicative idetities are both uique. (i.e. o umber other tha 0 is the additive idetity ad similarly for the multiplicative idetity.) j. 0 a = 0. 2

3 Theorem 1.2. Suppose that each of a, b ad c is a iteger. a. If a < b the a + c < b + c. b. If a < b ad 0 < c the ac < bc. c. If a < b ad 0 > c the ac > bc. d. If a 0, the a 2 > 0. e. If ab = 0 the either a = 0 or b = 0. f. If a, b, c Z + ad ac < bc, the a < b. Theorem 1.3. There is o iteger betwee 0 ad 1. Sectio 2: Iductio. Defiitio. Suppose that a ad r are real umbers. The the sequece: a, ar, ar 2, ar 3,... is called a geometric sequece with commo ratio r. Theorem 2.1. If a ad r are real umbers the. i=0 ar i = ar+1 a. r 1 Defiitio. A sum i the form x i+1 x i is called a telescopig series. Exercise 2.2. Verify that x i+1 x i = x +1 x 1. Exercise 2.3. Express 1 i(i+1) as a partial fractio (remember your itegra- is a telescopig series ad calculate tio techiques?), show that its sum. 1 i(i+1) Exercise 2.4. What happes to the geometric series ad the series of exercise 2.3 as? Use mathematical iductio to prove the ext few theorems. Theorem 2.5. i = ( + 1). 2 [The umbers T = i are called triagular umbers. (Why?)] 3

4 Theorem 2.6. i 2 = ( + 1)(2 + 1). 6 Theorem 2.7. i 3 ( + 1) = [ ] 2. 2 Exercise 2.8. Note that (i + 1)3 i 3 is a telescopig series ad use exercise 2.2 to fid the sum. The expad each term of the sum by distributig the summatio symbol to obtai a equatio ad solve for i 2 i terms of i. The use theorem 2.5 to obtai the formula of theorem 2.6. Exercise 2.9. Repeat exercise 2.8 oly replace 2 with 3, the with 4 ad 5 ad obtai the summatio formulae for i 3 ad i 4. Verify the secod formula by iductio if you are so iclied. Defiitio: We defie 1! = 1; for > 1, we defie ( + 1)! = ( + 1)!; ad we defie 0! = 1. Exercise Verify for each positive iteger : i i! = ( + 1)! 1. Exercise a. Verify for each positive iteger : i 1 +.

5 b. Verify for each positive iteger, (2)! < 2 2 (!) 2. Theorem If is a positive iteger ad x ad y are real umbers the x y is a factor of x y. Corollary: x y x y = x i 1 y i = x 1 + x 2 y y 1. i=0 Defiitio. The Fiboacci umbers {F } are defied as follows: F 1 = 1. F 2 = 1. For Z, > 2, F +1 = F + F 1. Exercise Calculate: a. F i, (aswer: F 2 1;) b. F 2i 1. (aswer: F 2.) Theorem For each positive iteger : a. F i 2 = F F +1 ; b. If > 1, the F +1 F 1 F 2 = ( 1) ; c. If > 2, the F +1 F F 1 F 2 = F 2 1 ; d. 2 1 F i F i+1 = [F 2 ] 2. Exercise Let be a positive iteger. a. F 2 1, b. F 7 4 1, c. If β = , the F β 1. [Note: β 2 = β + 1.] Theorem Let α ad β be the roots of x 2 = x + 1 with α < β, the: Exercise Let M = F = β α The M 1 0 F+1 F =. F F 1 Exercise Verify: 5

6 2i 1 = 2. Sectio 3: Biomial coefficiets ad combiatios. Defiitio. If ad r are oegative iteger so that r, the! = r r!( r)!. Theorem( 3.1. ) Let ad r be positive itegers so that r. The: a. = = 1. ( 0 ) ( ) b. = ( r ) ( r ) + 1 c. + =. r 1 r r Theorem 3.2. [The biomial theorem.] If is a positive iteger ad x ad y are umbers, the: (x + y) = x i y i. i i=0 Fact: [Which may be assumed.] The quatity is the umber of i differet possible i-card hads that ca be dealt from a deck of cards. Assume for the followig theorems that the various ( quatities ) all take o the reasoably expected values. (E.g. if the symbol is used the i assume i ad are o-egative itegers with i.) Theorem 3.3. i=0 = 2. i Theorem

7 ( 1) i = 0. i i=0 Exercise 3.5. Show that: r = r i Exercise 3.6. Show that: r i=0 ( i r + i = r ) i. r i + 1. r + 1 Exercise 3.7. Determie the value of k that makes the followig equatio true ad prove the idetity. k i=0 Sectio 4: Divisibility. i = F i +1. Defiitio. If a ad b are itegers the a is said to divide b if ad oly if there is a iteger q so that b=aq. The stadard otatio is: a b. Theorem 4.1. If each of a, b, ad c is a iteger so that a b ad b c, the a c. Theorem 4.2. If each of a, b ad c is a iteger so that a b ad a c the for arbitrary itegers x ad y, a (xb + yc). Theorem 4.3. [The divisio algorithm.] If each of a ad b is a iteger with 0 < b the there exist uique itegers q ad r with 0 r < b so that: a = bq + r. I the followig exercises ad theorems, as usual assume that the quatities are appropriately defied. 7

8 Exercise 4.4. Fid itegers a, b, ad c so that a bc but a b ad a c. Theorem 4.5. If c 0 the a b iff ac bc. Theorem 4.6. If a b the a b. Exercise 4.7. If a b the a b for each positive iteger. Theorem 4.8. If a b ad b a the a = b. Exercise 4.9. Show that if a is a iteger the, 3 (a 3 a). Exercise Show that 6 a(a + 1)(a + 2) for ay iteger a. Exercise Show that 5 (a 5 a) for every positive iteger a. Exercise The sum of the cubes of three successive itegers is divisible by 9. Exercise Show that if m > 1 the F +m = F m F +1 + F m 1 F. Exercise Show that if m the F F m. [Hit: use exercise 4.13.] Sectio 5: Prime Numbers. Defiitio. Suppose that is a positive iteger. If is oly divisible by oe positive iteger the is called a uit; if there are oly two positive itegers that divide the is called a prime umber; if is divisible by three positive itegers the is called a composite umber. Observatio. The positive iteger p is a prime umber iff it is ot equal to 1 ad it divisible oly by itself ad 1. Theorem 5.1. If is a positive iteger the there exists a prime umber p so that p. Theorem 5.2. The set of prime umbers i ifiite. 8

9 Theorem 5.3. If is a composite umber the there exists a prime umber p so that p. Defiitio. If x is a umber the π(x) deotes the umber of primes less tha or equal to x. The proof of the followig importat theorem uses techiques that will ot be covered i this course. The Prime Number Theorem: lim π() l() = 1. Exercise 5.4. Use the prime umber theorem to estimate the umber of primes less tha 1, 000, 000. How good a estimate is it for the umber of primes less tha 100? Theorem 5.5. If is a positive iteger the at some poit i the umber system, there exists cosecutive composite itegers. Ope problem: Goldbach s Cojecture: If > 2 is a eve positive iteger the is the sum of two primes. Exercise 5.6. Determie the itegers for which the quatity is prime. Exercise 5.7. Show that if p > 3 is prime, the oe of p + 2 or p + 4 is ot prime. Exercise 5.8. Show that x 2 x + 41 is prime for all positive itegers x < 41. [Do t do this by had, use a spreadsheet or somethig.] Show that for x = 41 this polyomial does ot produce a prime. Theorem 5.9. [Geeralizatio of exercise 5.8.] Show that if P (x) is a polyomial with iteger coefficiets the there is a positive iteger so that P () is a composite umber. 9

10 Exercise Let > 1; show that if a 1 is prime the a = 2 ad is prime. Exercise Suppose that N is a iteger ad A is a set of primes less tha N ad B is the set of primes less tha N that are ot i A. The a A a + b B b is ot divisible by a prime less tha N. Exercise Suppose that p is the smallest prime factor of the iteger ad that 3 < p the either or is prime. p Exercise Show that every iteger greater tha 11 is the sum of two composite umbers. ( Exercise ) Suppose that is a positive iteger. Does it follow that is divisible by? What if is prime? r Theorem The set S = {4k + 3 k Z + } cotais ifiitely may primes. [Hit: ote that is a ad b are both i the form the so is ab.] Sectio 6: Prime Factorizatio. Defiitio. If each of a ad b is a iteger the the greatest commo divisor of a ad b is the largest positive iteger that divides both a ad b. The commo otatio for the greatest commo divisor d of a ad b is: d = (a, b) or d = gcd(a, b). We will use the latter, though you should be aware that the former is more commo. Defiitio. If each of a ad b is a iteger the a ad b are said to be relatively prime iff gcd(a, b) = 1. Defiitio. The greatest commo divisor for a set of umbers {a i } is similarly defied: gcd({a i } ) is the largest positive iteger that divides all the elemets of the set {a i }. 10

11 Example: Fid three itegers so that gcd(a, b, c) = 1 but that are pairwise ot relatively prime. Theorem 6.1. Suppose that each of a, b ad c is a positive iteger ad d = gcd(a, b). The: a. gcd( a d, b d ) = 1; b. gcd(a + cb, b) = gcd(a, b). Theorem 6.2. Suppose that each of a ad b is a iteger ad at least oe of them is ot 0. Let S = {a + mb Z, m Z, 0 < a + mb}. The gcd(a, b) is the least elemet of the set S. Exercise 6.3. Fid the gcd of the followig pairs: a. gcd(a, a 2 ) =?, b. gcd(a, a + 1) =?, c. gcd(a, a + 2) =?, d. gcd(ca, cb) =?. Exercise 6.4. Show that if gcd(a, b) = 1, the gcd(a + b, a b) is 1 or 2. Exercise 6.5. If a ad b are eve the gcd(a, b) = 2gcd( a, b ). If a is eve 2 2 ad b is odd the gcd(a, b) = gcd( a, b). 2 Theorem If gcd(e, f) = 1, e a ad f a, the ef a. Exercise 6.7. a. If gcd(a, b) = 1 ad c (a + b) the gcd(a, c) = 1. b. If a bc the a gcd(a, c)gcd(a, b). c. If gcd(a, b) = 1 the gcd(a, b ) = 1. d. If gcd(a, b) = gcd(a, c) = 1 the gcd(a, bc) = 1. e. gcd(3k + 2, 5k + 3) = 1. Theorem 6.8. [The Euclidea Algorithm to calculate the Greatest Commo Divisor.] Suppose that a ad b are two positive itegers. Let r 0 = a ad r 1 = b, the recursively defie r +1 ad q usig the divisio algorithm by: r 1 = r q + r

12 The there exists a iteger k so that r k = 0 ad if k is the first such iteger the gcd(a, b) = r k 1. Defiitio Let a ad b be two positive itegers the the least commo multiple of a ad b is the smallest positive iteger m so that a m ad b m. The commo otatio for the least commo multiple m of a ad b is: m = [a, b] or d = lcm(a, b). We will use the latter, though you should be aware that the former is more commo. Theorem 6.9. Let a ad b be two positive itegers. The: lcm(a, b) = ab gcd(a, b). Theorem If gcd(a, b) = 1 ad a bc the a c. Theorem Suppose that p is a prime, that for each i, a i is a positive umber ad p the there exists a k so that p a k. a i, Theorem [The Fudametal Theorem of Arithmetic.] Every positive iteger greater that 1 has a uique represetatio as a product of primes: such that p i p i+1. = p 1 p 2 p 3...p k, Exercise [Applicatios of the Fudametal Theorem of Arithmetic.] Assume as usual that each of r, s, are positive iteger, a, b,... etc take o reasoable values; assume that p deotes a prime. a. If a 3 b 2 the a b. b. If p r a ad p s b the p r+s ab. c. If p r a the p r a. d. If a 2 b 2 the a b. 12

13 e. If m is a commo multiple of a ad b the lcm(a, b) m. f. If a r = b r the a = b. Exercise [More applicatios.] a. Show that log 2 (3) is irratioal. b. Show that 2 is irratioal. c. Show that if p is a prime umber the p is irratioal. d. Show that 6 is irratioal. e. Geeralize b - d as much as you ca. Theorem Suppose that each of a ad b is a positive iteger ad that, by the fudametal theorem of arithmetic, a = pr i i ad b = ps i i ; where each p i is a prime ad each of r i ad s i is a positive iteger or zero. The: Exercise gcd(a, b) = lcm(a, b) = p mi(r i,s i ) i, p max(r i,s i ) i. gcd(a, b) lcm(a, b). Exercise lcm(a, b) c if ad oly if a c ad b c. Exercise Let p be a prime, a. If p a 2 the p a. b. If p a the p a. Exercise gcd(a, b) = gcd(a + b, lcm(a, b)). Exercise

14 a. Show that there are ifiitely may primes i the set {4k + 3 k Z + }. [Hit: prove the followig lemma first: the set {4k + 1 k Z + } is closed uder multiplicatio. b. Show that there are ifiitely may primes i the set {6k + 5 k Z + }. c. Geeralize a ad b as much as possible. [The geeralizatio is called Dirichlet s Theorem o Primes.] Exercise If a 2 b 2 the a b. Exercise If a, b, ad c are positive itegers so that gcd(a, b) = 1 ad ab c, the there exists itegers u ad v so that a = u ad b = v. Exercise Show that: a. log 2 (3) is irratioal. b. log p (q) is irratioal where p ad q are primes. c. Ca b be geeralized to the case where gcd(p, q) = 1? Exercise Show that: a. 2 is irratioal. b. p is irratioal, where p is a prime. c. 6 is irratioal. d. Geeralize b as much as possible. e. 3 5 is irratioal. f. Geeralize as much as you ca. Defiitio. Suppose that each of a, b ad c is a iteger. The the followig equatio i which iteger solutios i x ad y are sought is called a liear Diophatie equatio i two variables: ax + by = c. 14

15 Theorem Suppose that each of a, b ad c is a iteger ad d = gcd(a, b) the the liear diophatie equatio i two variables ax + by = c has a solutio (amog the itegers) if ad oly if d c. Furthermore, if it has a solutio x = x 0 ad y = y 0 the it has ifiitely may solutios ad all of the solutio are give by: for k Z. x = x 0 + b d k, y = y 0 a d k, Sectio 7: Cogruece Relatio. Defiitio: Suppose that Z is a set ad Z Z is a relatio. The is called a equivalece relatios provided it satisfies the followig coditios: Let a Z the a a; [Reflexive property.] Let a Z ad b Z, if a b the b a; [Symmetric property.] Let a Z, b Z ad c Z, if a b ad b c the a c. [Trasitive property.] Defiitio: If a Z, b Z ad m Z + the a is said to equal b modulo m meas that: m b a. Notatio: a equals b modulo m is deoted by: a m b or a = b mod m. Theorem 7.1. The relatio a = b mod m is a equivalece relatio o Z. Theorem 7.2. If a Z, b Z ad m Z + the a = b mod m if ad oly if there is a iteger k Z so that a = b + km. Theorem 7.3. If each of a, b, r ad m > 0 is a iteger ad a = b mod m, the: a. a + r = b + r mod m; b. ar = br mod m. 15

16 Theorem 7.4. If each of a, b, r ad m > 0 is a iteger, 1 = gcd(r, m) ad ar = br mod m, the a = b mod m. Theorem 7.5. If each of a, b, r, s ad m > 0 is a iteger, a = b mod m ad r = s mod m, the: a. a + r = b + s mod m; b. ar = bs mod m. Defiitio. Suppose m Z +. The set S = {r i } is called a complete set of residues modulo m provided that for ay iteger k there is exactly oe elemet r ik so that, k = r ik mod m. Theorem 7.6. Suppose m Z +. The set S = {i} m 1 i=0 is a complete set of residues modulo m. = {0, 1, 2,..., m 1} Theorem 7.7. If {r i } is a complete set of residues modulo m the: a. if a is a iteger, the {r i + a} is a complete set of residues modulo m; b. if k is a iteger so that gcd(k, m) = 1,the {kr i } is a complete set of residues modulo m. Exercise 7.8. a. If a is eve the a 2 = 0 mod 4; b. If a is odd the a 2 = 1 mod 4. c. If a is odd the a 3 = 1 mod 8. Exercise 7.9. If k m ad a = b mod m the a = b mod k. Exercise If a = b mod m the ak = bk mod mk. Exercise If a = b mod m, the gcd(a, m) = gcd(b, m). 16

17 Exercise If p is a prime ad a 2 = b 2 mod p the a = b mod p or a = b mod p. [Hit: use the least residues modulo p of a ad b ad prove the statemet for the residues first.] Exercise [Hit: use iductio.] a = 0 mod 3; b. 4 = 1 mod 3; c. 4 = mod 9; d. 5 = mod 16. Exercise a. Fid a complete set of residues modulo 7 cosistig of odd itegers. b. Fid a complete set of residues modulo 7 cosistig of eve itegers. Exercise If p is a prime ad x 2 = x mod p, the x = 0 mod p or x = 1 mod p. Sectio 8: Cogruece ad the Chiese Remaider Theorem. Theorem 8.1. If a Z, b Z, m Z + ad gcd(a, m) = d, the the equatio ax = b mod m has a solutio if ad oly if d b. Furthermore, if d b the the equatio has exactly d may pairwise o-cogruet modulo m solutios. Defiitio. If a Z ad x is a umber such that ax = 1 mod m the x is called the iverse of a modulo m. Notatio. If a Z the a 1m deotes the iverse of a modulo m. Theorem 8.2. gcd(a, m) = 1. The umber a has a iverse modulo m if ad oly if Theorem 8.3. The set S = {i} 1 = {1, 2,..., 1} is a abelia group uder multiplicatio mod if ad oly if is prime. Exercise 8.4. If a Z ad b Z the a 1m b 1m = (ab) 1m. 17

18 Theorem 8.5. If p is prime, the x 2 = 1 mod p if ad oly if either x = 1 mod p or x = 1 mod p. Theorem 8.6. If If a, b Z, m, Z + ad gcd(m, ) = 1, the the followig system of equatios: x = a mod m, x = b mod, has the solutio x = a 1m + bmm 1. Furthermore the solutio is uique modulo m. Exercise 8.7. Geeralize Theorem 8.6 to solve the followig system of equatios. x = a mod m, x = b mod, x = c mod k. Theorem 8.8 [Chiese Remaider Theorem]. Suppose that for each positive iteger i N, a i Z, m i Z + ad gcd(m i, m j ) = 1 for i j. The the system of equatios: x = a i mod m i, i N, has the uique solutio x = [Exercise 8.8: fill i the blak.] modulo N m i. Aswer: Let M = N m i, the: x = N 1 M M Mmi a i m i m i + km. 18

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