Chapter 5. Factorization of Integers

Transcription

1 Chapter 5 Factorization of Integers 51 Definition: For a, b Z we say that a ivies b (or that a is a factor of b, or that b is a multiple of a, an we write a b, when b = ak for some k Z 52 Theorem: (Basic Properties of Divisors Let a, b, c Z Then (1 a 0 for all a Z an 0 a a = 0, (2 a 1 a = ±1 an 1 a for all a Z (3 If a b an b c then a c (4 If a b an b a then b = ±a (5 If a b an b 0 then a b (6 If a b an a c then a (bx + cy for all x, y Z Proof: We leave the proof as an exercise, but we make some remarks Some of these properties hol in all rings while other properties are specific to Z Properties (1, (3 an (6 hol in all rings Properties (2 an (4 hol in Z but not in all rings, for example they o not hol in Q an R Property (2 was proven in Example 337 an the proof use the Discreteness Property of Z Property (4 follows easily from Property (2 Property (5 oes not hol in Q or R, an it oes not even make sense in a ring with no orer relation To prove that Property (5 hols in Z you shoul first use the Discreetness Property show that for all k Z, if k 0 then k 1 53 Theorem: (The Division Algorithm For all a, b Z with b 0, there exist unique integers q an r such that a = qb + r an 0 r < b The integers q an r are calle the quotient an remainer when a is ivie by b Proof: Let a, b Z with b 0 Since b 0, either b > 0 or b < 0 We consier the case that b > 0 Since b > 0 we have b = b For q, r Z we have ( a = qb + r an 0 r < b ( r = a qb an 0 r < b Also, since b > 0 we have ( r = a qb an 0 a qb < b 0 a qb < b 0 a + qg > b a qb > a b a b < qb a a b 1 < q a b q = a b Thus we have a = bq + r an 0 r < b if an only if q = a b an r = a bq We leave it as an exercise to show that, in the case that b < 0 so that b = b, we have a = qb + r an 0 r < b if an only if q = a b an r = a bq 54 Note: For a, b Z, when we write a = qb + r with q, r Z an 0 r < b, we have b a if an only if r = 0 Inee if r = 0 then a = qb so that b a an, conversely, if b a with say a = pb = pb + 0, then we must have q = p an r = 0 by the uniqueness of the quotient an remainer 1

2 55 Definition: Let a, b Z A common ivisor of a an b is an integer such that a an b When a an b are not both 0, we enote the greatest common ivisor of a an b by gc(a, b For convenience, we also efine gc(0, 0 = 0 56 Theorem: (Basic Properties of the Greatest Common Divisor Let a, b, q, r Z (1 gc(a, b = gc(b, a (2 gc(a, b = gc( a, b (3 If a b then gc(a, b = a In particular, gc(a, 0 = a (4 If b = qa + r then gc(a, b = gc(a, r Proof: The proof is left as an exercise 57 Theorem: (The Eucliean Algorithm With Back-Substitution Let a an b be integers an let = gc(a, b Then there exist integers s an t such that as + bt = The proof provies explicit proceures for fining an for fining s an t Proof: We can fin using the following proceure, calle the Eucliean Algorithm If b a then we have = b Otherwise, let r 1 = a an r 0 = b an use the ivision algorithm repeately to obtain integers q i an r i such that r 1 = a = q 1 b + r 1 0 < r 1 < a r 0 = b = q 2 r 1 + r 2 0 < r 2 < r 1 r 1 = q 3 r 2 + r 3 0 < r 3 < r 2 r k 2 = q k r k 1 + r k 0 < r k < r k 1 r n 2 = q n r n 1 + r n 0 < r n < r n 1 r n 1 = q n+1 r n + r n+1 r n+1 = 0 Since r n 1 = q n+1 r n we have r n rn 1 so gc(r n 1, r n = r n Since r k 2 = q k r k 1 + r k we have gc(r k 2, r k 1 = gc(r k 1, r k an so = gc(a, b = gc(b, r 1 = gc(r 1, r 2 = = gc(r n 2, r n 1 = gc(r n 1, r n = r n Having foun using the Eucliean algorithm, as above, we can fin s an t using the following proceure, which is known as Back-Substitution If b a so that = b, then we can take s = 0 an t = ±1 to get as + bt = Otherwise, we let s 0 = 1, s 1 = q n, an s l+1 = s l 1 q n l s l for 1 l n 1 an then we can take s = s n 1 an t = s n to get as + bt =, because, writing k = n l, = r n = r n 2 q n r n 1 = s 1 r n 1 + s 0 r n 2 = = s l r n l + s l 1 r n l 1 = s n k r k + s n k 1 r k 1 = s n k ( rk 2 q k r k 1 + sn k 1 r k 1 = ( s n k 1 q k s n k rk 1 + s n k r k 2 = (s l 1 q n l s l r n l 1 + s l r n l 2 = s l+1 r n l 1 + s l r n l 2 = = s n r 0 + s n 1 r 1 = s n b + s n 1 a 2

3 58 Example: Let a = 5151 an b = 1632 Fin = gc(a, b an then fin integers s an t so that as + bt = Solution: The Eucliean Algorithm gives 5151 = = = = so = 51 Using the quotients q 1 = 3, q 2 = 6 an q 3 = 2, Back-Substitution gives s 0 = 1 s 1 = q 3 = 2 s 2 = s 0 q 2 s 1 = 1 6( 2 = 13 s 3 = s 1 q 1 s 2 = 2 3(13 = 41, so we take s = s 2 = 13 an t = s 3 = 41 (It is a goo iea to check that inee we have (1632( 41 + (5151(13 = Example: Let a = 754 an b = 3973 Fin = gc(a, b then fin integers s an t such that as + bt = Solution: The Eucliean Algorithm gives 3973 = , 754 = , 203 = , 145 = , 58 = so that = 29 Then Back-Substitution gives rise to the sequence 1, 2, 3, 11, 58 so we have (754(58 + (3973( 11 = 29, that is (754(58 + ( 3973(11 = 29 Thus we can take s = 58 an t = Theorem: (More Properties of the Greatest Common Divisor Let a, b, c, Z (1 If c a an c b then c gc(a, b (3 We have gc(a, b = 1 if an only if there exist x, y Z such that ax + by = 1 (4 If = gc(a, b 0 then gc ( a, b = 1 (5 If a bc an gc(a, b = 1 then a c Proof: We prove Part (5 Suppose that a bc an gc(a, b = 1 Since a bc we can choose k Z so that bc = ak Since gc(a, b = 1, by the Eucliean Algorithm with Back- Substitution, we can choose s, t Z with as + bt = 1 Then we have an so a c, as require c = c 1 = c(as + bt = acs + bct = acs + akt = a(cs + kt, 3

4 511 Definition: A iophantine equation is a polynomial equation in which the variables represent integers Some iophantine equations are fairly easy to solve while others can be extremely ifficult 512 Theorem: (Linear Diophantine Equations Let a, b, c Z with (a, b (0, 0 Let = gc(a, b an note that 0 Consier the Diophantine equation ax + by = c (1 The equation has a solution (x, y Z 2 if an only if c, an (2 if (u, v Z 2 is one solution to the equation then the general solution is given by (x, y = (u, v + k ( b, a for some k Z Proof: Suppose that the equation ax+by = c has a solution (x, y Z 2 Choose (s, t Z 2 so that as + bt = c Since a an b, it follows that (ax + by for all x, y Z, so in particular (as + bt, that is c Conversely, suppose that c, say c = l with l Z Use the Eucliean Algorithm with Back-Substitution to fin s, t Z such that as + bt = Multiply by l to get a(sl + b(tl = l = c Thus we can take x = el an y = tl to obtain a solution (x, y Z 2 to the equation ax + by = c This proves Part (1 Now suppose that (u, v Z 2 is a solution to the given equation, so we have au+bv = c To prove Part (2, we nee to prove that for all k Z, if we let (x, y = (u, v + k ( b, a then (x, y is a solution to ax + by = c an, conversely, that if (x, y is a solution then there exists k Z such that (x, y = (u, v + k ( b, a Let k Z an let (x, y = (u, v + k ( b, a Then x = u kb ax + by = a ( ( u kb + b v + ka = (au + bv kab + kab ka an y = v + = au + bv = c an so Conversely, let (x, y be a solution to the given equation, so we have ax + by = c Suppose that a 0 (we leave the case a = 0 as an exercise Since ax + by = c an au + bu = c we have ax + by = au + bv an so a(x u = b(y v Divie both sies by to get a (x u = b (y v Since a b (y v an gc ( a, b = 1, it follows that a (y v Choose k Z so that y v = ka kab Since a 0 an a(x u = b(y v =, we have x u = kb an so (x, y = (u, v + k( b, a, as require 513 Example: Let a = 426, b = 132 an c = 42 Fin all x, y Z such that ax + by = c Solution: The Eucliean Algorithm gives 426 = , 132 = , 30 = , 12 = so that = gc(a, b = 6 Note that c, inee c = l with l = 7, so a solution oes exist Back-Substitution gives the sequence 1, 2, 9, 29 so we have a(9 + b( 29 = Multiply by l = 7 to get a(63 + b( 203 = c, so one solution is given by (x, y = (63, 203 Since a = = 71 an b = = 22, The general solution is (x, y = (63, k( 22, Exercise: Let a = 4123, b = an c = 798 Fin all x, y Z with 0 y 100 such that ax + by = c 4

5 515 Definition: Let n be a positive integer We say that n is a prime number when n 2 an n has no factor a Z with 1 < a < n We say that n is composite when n 2 an n is not prime, that is when n oes have a factor a Z with 1 < a < n 516 Theorem: (Basic Properties of Primes Let p be a prime number (1 For all a Z we have gc(a, p {1, p} with gc(a, p = p if an only if p a (2 For all a, b Z, if p ab then either p a or p b Proof: The proof is left as an exercise Part (2 follows from Part (5 of Theorem Theorem: Every integer n 2 has a prime factor Every composite integer n 2 has a prime factor p with p n Proof: Let n 2 Suppose, inuctively, that every integer k with 2 k < n has a prime factor If n is prime, then n is a prime factor of itself, so n has a prime factor Suppose that n is composite Let a be a factor of n with 1 < a < n By the inuction hypothesis, a has a prime factor Let p be a prime factor of a Since p a an a n we have p n, an so p is a prime factor of n It follows, by inuction, that every integer n 2 has a prime factor Now suppose that n is composite Write n = ab where a, b Z with 1 < a b < n Note that a n because if we ha a > n then we woul also have b a > n so that n = ab > n n = n which is impossible Let p be a prime factor of a Since p a an a n we have p n so that p is a prime factor of n Since p a an a n we have p a n 518 Note: Given an integer n 2, we can list all primes p with p n using the following proceure, which is calle the Sieve of Eratosthenes We begin by listing all the integers from 1 to n, an we cross off the number 1 (1 is a unit; it is not a prime We circle the smallest remaining number p 1 (namely p 1 = 2, which is prime then we cross off all other multiples of p 1 (which are composite We circle the smallest remaining number p 2 (namely p 2 = 3, which is prime then we cross off all other multiples of p 2 (which are all composite At the k th step of the proceure, when we circle the smallest remaining number p k, it must be prime because if p k was composite then it woul have a prime factor p i with p i < p k, but we have alreay foun all primes p i < p k an we have alreay crosse off all their multiples We continue the proceure until we have circle a prime p l with p l n an crosse off its multiples At this stage we circle all of the remaining numbers in the list because they are all prime Inee, if a remaining number m was composite then it woul have a prime factor p with p m n, but we have alreay foun all primes p with n an crosse off all their multiples 519 Exercise: Use the Sieve of Eratosthenes to list all primes p with p Theorem: (Eucli There exist infinitely many prime numbers Proof: Suppose, for a contraiction, that there exist finitely many prime numbers Let p 1, p 2,, p l be all of the prime numbers Consier the number n = p 1 p By Theorem 517, the number n has a prime factor an so p k n for some inex k But p k is not a factor of n because when we write n = qp k + r as in the Division Algorithm, we fin that the remainer is r = 1 0 (an the quotient is q = p i 521 Example: Note that there exist arbitrarily large gaps between consecutive prime numbers because, given a positive integer m 2, we have 2 (m!+2, 3 (m!+3, 4 (m!+4 an so on, so the consecutive numbers m!+2, m!+3, m!+4,, m!+m are all composite i k 5

6 522 Remark: Here are a few facts about prime numbers which are ifficult to prove (1 Bertran s Postulate: for every integer n 1 there exists a prime p with n p 2n (2 Dirichlet s Theorem: for all positive integers a, b with gc(a, b = 1, there exist infinitely many primes of the form p = a + kb for some k N (3 The Prime Number Theorem: for x R, let π(x be the number of primes p with π(x p x Then lim x x/ ln x = Remark: Here are a few statements about prime numbers which are conjecture to be true, but for which no proof has, as yet, been foun (1 Legenre s Conjecture: for every n Z + there exists a prime p with n 2 p (n +1 2 (2 Golbach s Conjecture: every even integer n 4 is the sum of two prime numbers (3 Twin Primes Conjecture: there exist infinitely many p for which p an p + 2 are prime (4 The n Conjecture: there exist infinitely many primes p = n with n Z + (5 Mersenne Primes Conjecture: there exist infinitely many primes p = 2 k 1 with k Z + (6 Fermat Primes Conjecture: there exist finitely many primes p = 2 k + 1 with k N 524 Theorem: (The Unique Factorization Theorem Every integer n 2 can be written uniquely in the form n = p 1 p 2 p l l k=1 p k = p 1 p 2 where l Z + an the p k are primes with Proof: First we prove the existence of such a factorization Let n be an integer with n 2 an suppose, inuctively, that every integer k with 2 k < n can be written in the require form If n is prime then we can write n = l k=1 p k = p 1 with l = 1 an p 1 = n Suppose that n is composite Write n = ab where a, b Z with 1 < a < n an 1 < b < n By the inuction hypothesis, we can write a = q 1 q 2 q l an b = r 1 r 2 r m where l, m Z + an the p i an q i are primes with p 1 p 2 p l an q 1 q 2 q m Then n = q 1 q 2 q l r 1 r 2 r m = p 1 p 2 +m where the orere (l+m-tuple (p 1, p 2,, p l+m is obtaine from the orere (l + m-tuple (q 1, q 2,, q l, r 1, r 2,, r m by rearranging the terms into non-ecreasing orer Let us prove uniqueness Suppose that n = p 1 p 2 = q 1 q 2 q m where l, m Z + an the p i an q j are primes with p 1 p 2 p l an q 1 q 2 q m We nee to prove that l = m an that p i = q i for every inex i Since n = p 1 p 2 we see that p 1 n an so p q1 q 2 q m By applying Part (2 of Theorem 512 repeately, it follows that p 1 q i for some inex i Since p 1 q i an q i is prime, we must have p 1 {±1, ±q i } Since p 1 is prime, we have p 1 > 1 Since p 1 > 1 an p 1 {±1, ±q i } it follows that p 1 = q i A similar argument shows that q 1 = p j for some inex j Since p 1 = q i q 1 = p j p 1, it follows that p 1 = q 1 Since p 1 p 2 = q 1 q 2 q m an p 1 = q 1, we can ivie both sies by p 1 to get p 2 p 3 = q 2 q 3 q m By repeating the above argument, we can show that p 2 = q 2, then we can ivie both sies by p 2 = q 2 to get p 3 = q 3 q m an so on If we ha l m, say l < m, repeating the above proceure woul eventually yiel p l = q l q l+1 q m with p l = q l an then 1 = q l+1 q m which is not possible since each q i > 1 Thus we must have l = m an repeating the above proceure gives p i = q i for all inices i, as require 6

7 525 Note: Here are two alternate ways of expressing the above theorem (1 Every integer n 2 can be written uniquely in the form n = l m p i m1 m i = p 1 l where l Z + an the p i are istinct primes with p 1 < p 2 < < p l an each m i Z + (2 Given istinct primes p 1, p 2,, p l, every n Z + whose prime factors are inclue in {p 1,, p l } can be written uniquely in the form n = l m p i m1 m i = p 1 l with m i N 526 Theorem: (Unique Factorization an Divisors Let n = p 1 m 1 p 2 m2 m l where l Z +, the p i are istinct primes, an each m i N Then the positive ivisors of n are the numbers of the form a = p 1 j 1 p 2 j2 j l where each j i Z with 0 j i m i m Proof: Suppose that n = p 1 m2 m 1 p 2 l j an a = p 1 j2 j 1 p 2 l where p 1, p 2,, p l are istinct primes an 0 j i m i for all inices i k Let b = p 1 k2 k 1 p 2 l where k i = m i j i (note that k i 0 since j i m i Then ab = (p 1 j1 jl (p 1 k1 k l = p 1 j 1 +k1 j l +k l = p 1 m1 m l = n an so a n m Conversely, suppose that n = p 1 m2 m 1 p 2 l, as above, an let a be a positive ivisor of n Let p be any prime factor of a Since p a an a n we have p n Since p n m an n = p 1 m2 m 1 p 2 l we have p p i for some inex i Since p an p i are both prime an p p i, we have p = p i This proves that every prime factor of a is among the primes j p 1, p 2,, p l It follows that a can be written in the form a = p 1 j2 j 1 p 2 l with each j i N It remains to show that j i m i Since a n we can choose b Z so that n = ab Since n an a are positive, so is b Since b is a positive factor of n, the above argument shows that every prime factor of b is k among the primes p 1, p 2,, p l an so we can write b = p 1 k2 k 1 p 2 l for some k i N Since n = ab we have p 1 m 1 p 2 m2 m l = n = ab = (p 1 j1 j l (p 1 k1 k l = p 1 j 1 +k1 j l +k l By the uniqueness of prime factorization, it follows that m i = j i +k i for all inices i Since k i 0 it follows that j i = m i k i m i, as require 527 Definition: For a, b Z, a common multiple of a an b is an integer m such that a m an b m When a an b are nonzero, we efine lcm(a, b to be the smallest positive common multiple of a an b For convenience, we also efine lcm(a, 0 = lcm(0, a = 0 for a Z 528 Theorem: Let a = l j p i i an b = l an j i, k i N Then (1 gc(a, b = l min{j p i,k i } i, (2 lcm(a, b = l max{j p i,k i } i, an (3 gc(a, b lcm(a, b = ab Proof: The proof is left as an exercise k p i i where l Z +, the p i are istinct primes, 7

8 529 Definition: For a prime p an a positive integer n, the exponent of p in (the prime factorization of n, enote by e(p, n, is efine as follows We write n in the form m n = p 1 m2 m 1 p 2 l where the p i are istinct primes an each m i N, then we efine e(p, n = m i if p = p i an we efine e(p, n = 0 if p p i for any inex i 530 Exercise: Let n Z + Show that if e(p, n is o for some prime p then n is irrational / 531 Exercise: Show that e(p, n! = n p + n p + n 2 p + an that n = n p 3 p k+1 p k 532 Exercise: Fin the number of zeros which occur at the en of the number 100!, in its ecimal representation 533 Definition: For a positive integer n, we write τ(n to enote the number of positive ivisors of n, we write σ(n to enote the sum of the positive ivisors of n, an we write ρ(n to enote the prouct of the positive ivisors of n 534 Exercise: Let n = l k p i i where p 1, p 2,, p l are istinct primes an each k i N Show that τ(n = l (k i + 1, σ(n = l p i k i +1 1 p i 1 an ρ(n = n τ(n/2 535 Exercise: Determine which positive integers n have the property that τ(n o, an which have the property that σ(n is o 536 Exercise: Let m Z + For 0 n m, efine a n = (a n,1, a n,2,, a n,m {0, 1} m recursively as follows Let a 0 = (0, 0,, 0 an for n 0 let { 1 an,k if n k, a n+1,k = a n,k otherwise Fin m a m,k k=1 8