Elementary Number Theory An Algebraic Apporach

Transcription

1 Elementary Number Theory An Algebraic Apporach Base on Burton s Elementary Number Theory 7/e 張世杰 bfhaha@gmail.com

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3 Contents 1 Preliminaries Mathematical Induction The Binomial Theorem Divisibility Theory in the Integers 13.1 Early Number Theory The Division Algorithm The Greatest Common Divisor The Euclidean Algorithm The Diophantine Equation ax + by = c Primes and Their Distribution The Fundamental Theorem of Arithmetic The Sieve of Eratosthenes The Goldbach Conjecture The Theory of Congruences Carl Friedrich Gauss Basic Properties of Congruence Binary and Decimal Representations of Integers Linear Congruences and the Chinese Remainder Theorem Fermat s Theorem Pierre de Fermat Fermat s Little Theorem and Pseudoprimes Wilson s Theorem The Fermat-Kraitchik Factorization Method Number-Theoretic Functions The Sum and Number of Divisors The Möbius Inversion Formula The Greatest Integer Function An Application to the Calendar Euler s Generalization of Fermat s Theorem Leonhard Euler Euler s Phi-function Euler s Theorem

4 4 CONTENTS 7.4 Some Properties of the Phi-function Primitive Roots and Indices The Order of an Integer Modulo n Primitive Roots for Primes Composite Numbers Having Primitive Roots The Theory of Indices The Quadratic Reciprocity Law Euler s Criterion The Legendre Symbol and Its Properties Quadratic Reciprocity Quadratic Congruences with Composite Moduli Introduction to Cryptography From Caesar Cipher to Public Key Cryptography The Knapsack Cryptosystem An Application of Primitive Roots to Cryptography Numbers of Special Form Marin Mersenne Perfect Numbers Mersenne Primes and Amicable Numbers Fermat Numbers Certain Nonlinear Diophantine Equations The Equation x + y = z Fermat s Last Theorem Representation of Integers as Sums of Squares Joseph Louis Lagrange Sums of Two Squares Sums of More Than Two Squares Fibonacci Numbers Fibonacci The Fibonacci Sequence Certain Identities Involving Fibonacci Numbers Continued Fractions Srinivasa Ramanujan Finite Continued Fractions Infinite Continued Fractions Farey Fractions Pell s Equation

5 CONTENTS 5 16 Some Modern Developments Hardy, Dickson, and Erdös Primality Testing and Factorization An Application to Factoring: Remote Coin Flipping The Prime Number Theorem and Zeta Function

6 6 CONTENTS

7 Preface (a) Notation and some theorem in introduction to abstract algebra you ought to know ˆ miscellaneous Denote the cardinality of a set S by S. If p is a prime, then p ( p ) i for 1 i p 1. (Theorem 8.9 used it.) ˆ Let G be a multiplicative group, we denote its identity by 1. ˆ Let G be a multiplicative group and g G. Then the multiplicative order ord(g) of g is the smallest positive integer s such that g s = 1. If no such positive integer exists, we define ord(g) =. If r s = 1, then ord(r) s. (This is Theorem 8.1. Theorem 5.5, Corollary 8.4, Corollary 8.7, Theorem 8.9, Theorem 8.1 s Supplement, Theorem 9.1 used it.) ord(r t ) = gcd (t,ord(r)). (This is Theorem 8.3. The application of this theorem is ubiquitous.) ˆ Let G be a multiplicative group and g G. Then the cyclic subgroup generated by g is denoted by g = {g i i Z}. ˆ ring ord(g) = g. Lagrange s Theorem: Let G be a finite group and H G is a subgroup of G. Then H divides G. In particular, if g G, since ord(g) = g and g G, we have ord(g) dividies G. (Theorem 5.1, Theorem 7.5, Theorem 7.4, Corollary 8.4, Corollary 8.7, Theorem 8.9, Theorem 9.1 used it.) Let R be a commutative ring with multiplicative unity and S = {s 1, s,..., s n } R. Then the ideal generated by S is denoted by S = {r 1 s 1 + r s + + r n s n r i R} R. If T S R, then T S. (Lemma.1 used it.) ˆ some algebraic structure Z is a ring. Z n = {0, 1,,..., n 1} is a ring. (Z n, +) = {0, 1,,..., n 1} is an additive cyclic group. ˆ The set of all unit in the ring Z n form a group under multiplication. Which is denoted by Z n. a is a unit in Z n gcd (a, n) = 1, where n. 7

8 8 CONTENTS Z n = {1 a n gcd (a, n) = 1} Z n = φ(n) def. = #{1 a n gcd (a, n) = 1}. 注意到, 我們有時候也會在 Z n 當中做加法運算, ( 例如 Theorem 5.5 及 Corollary 8.4 及 Theorem 8.7,) 不是因為我們違背了運算規則, 而是因為我們可以把 Z n 視為 ring Z n 當中的 unit 構成的子集合 ˆ If Z n is cyclic, then a primitive root modulo n is a generator of Z n. r is a primitive root modulo n r is a generator of Z n Z n = r ord(r) = r = Z n = φ(n) ˆ Let F be a finite field. Then the multiplicative group F {0} must be cyclic. In particular, consider the finite field Z p. Z p {0} = Z p is cyclic. That is, a primitive root modulo p must exists. (Theorem 5.5 s Supplement, Corollary 8.4, Theorem 8.6, Corollary 8.7, Theorem 8.9, Theorem 8.1 s Supplement, Theorem 9.1 used it.) (b) 因為下列的幾個原因, 筆者認為 elementary number theory 是一門很適合數學系及資工系大一學生學習的科目 ˆ 其中大量地使用了演繹法 數學歸納法及反證法 ˆ 定理的論證雖然不像高等微積分那麼地具有幾何直觀, 但是很清楚 很嚴謹 ˆ 程式語言的入門練習題常出現 elementary number theory 中的觀念, 例如 prime, gcd 等等 (c) elementary number theory 不只作為學習 introduction to abstract algebra 的預備科目, 當學生學完 introduction to abstract algebra 後再回頭用代數的方法證明 elementary number theory 中的定理時, 可以 ˆ 熟練 introduction to abstract algebra 中所學的內容, ˆ 複習 elementary number theory 的知識, ˆ 感受到在數學中, 用不同觀點看待同一個觀念的重要性與價值及其美妙之處 所以數學系大三很適合開一門課程 elementary number theory-algebraic approach 給修過 elementary number theory 及 introduction to abstract algebra 的學生來修 (d) 筆者是採用 [4] 來教授基礎數論, 所以筆者班上的學生學完 [4] 及 introduction to abstract algebra 後, 可以直接回頭看 [4], 然後練習用代數的觀點重新證明 elementary number theory 中的定理 而這本書就是列出了 [4] 中的幾個重要定理, 並且用代數的方法證明 (e) 成功大學夏杼教授在 finite field 的課堂上說過, elementary number theory 就像是創造了一把一把小鐵槌, 去解決一堆零散的小問題 ; 而 abstract algebra 就像是一把大鐵鎚, 可以用來一次擊破好幾個小問題 最好的例子就是 Lagrange s Theorem in group theory 與 Fermat s Little Theorem 及 Euler s Theorem 之間的關係 (f) ch.6, 10, 11, 1, 14, 15, 16 不討論, 沒有用 algebraic approach

9 CONTENTS 9 (g) 同樣的書名已經有另一本書 [3] 了, 為何筆者還要脫褲子放屁? 原因如下 ˆ [3] 沒有講一些很能夠展現 algebraic approach 的威力的定理, 例如 Theorem 7.4 Theorem 9.4 ˆ [3] 的某幾個重要定理依然是用 elemementary number theory 的方法來證明, 例如 quadratic reciprocity law (Theorem 9.9) Fermat s the sum of two square theorem (Theorem 13.) ˆ [3] 的 primitive root theorem (Theorem 8.10), 雖然是用代數的方法證明, 但還是 [7] 的解說比較清楚 ˆ [3] 並沒有證明 Dirichlet s theorem on arithmetic progressions (Theorem 3.7), 本書雖然也沒有證, 不過給出了 reference [5] 或 [10] 另外, 下面幾個定理都是該定理的 corollary, 還有一些出現在 [4] 的習題裡 Theorem 3.4 Theorem 3.6 Theorem 9.3 Theorem 9.8 (h) 沒有用到 algebraic approach 的證明, 但給出了基本且簡潔的證明 ˆ Corollary.4 and Theorem.5 ˆ Theorem.8 ˆ Theorem.9 ˆ Theorem 4.8 ˆ Theorem 7.1 ˆ Theorem 7.6 ˆ Theorem 8.1 and Theorem 8. ˆ Theorem 8.3 (i) 打灰的定理代表 ˆ 該定理原本是用來作為 elementary 證法的引理, 現在改用 algebraic approach, 所以不需要用到了, 例如 Lemma 3.1 for Theorem 3.6 Lemma 5.1 for the failure of the converse of the Fermat s Little Theorem (Theorem 5.1). Lemma 7.1 for Theorem 7. Lemma 7. for Theorem 7.5 Theorem 8.5, Corollary 8.4, Theorem 8.6 and Corollary 8.6 for Z p is cyclic Lemma 8.1 for Corollary 8.7 Lemma 8. for Theorem 8.9

10 10 CONTENTS Theorem 9.5 for Theorem 9.9 Lemma 9.1 for Theorem 9.9 Lemma 13.1 for Theorem 13. Lemma 13. for Theorem 13. Lemma 13.4 and Corollary 13.4 for Theorem 13.6 ˆ 該定理只是簡化的結果, 雖然在計算上很方便, 但還是鼓勵讀者用最根本的原理去思考, 所以不用記該定理, ( 這個論點行不太通, 有一些定理可能是因為我不會用所以沒有顯其大用,) 例如 Theorem.7 Corollary.5 Theorem 4.3 Corollary 9. Theorem 9.6 Theorem 9.7 Corollary 9.5 Corollary 9.6 Theorem 9.10 ˆ 該定理在代數裡面用不到 (?), 例如 Theorem 3.3 Theorem 3.5 Corollary 3.4 Theorem 3.8 Theorem 4.4 Corollary 4.3 Theorem 4.5 Theorem 4.6 Theorem 5. Theorem 5.3 Theorem 7.7 Theorem 7.8 Theorem 13.4 Corollary 13.3 Theorem 13.5 (j) 不滿意的證明, 沒有用到 algebraic approach, 或是用了 algebraic approach 依然很難 ˆ Corollary 8.7 沒有用到 algebraic approach, 不過很短 ˆ Theorem 8.9 沒有用到 algebraic approach, 而且很長 ˆ Theorem 9.11 沒有用到 algebraic approach, 雖然很短, 但很 tricky ˆ Theorem 13. 雖然用了 algebraic approach, 但太長了

11 Chapter 1 Preliminaries 1.1 Mathematical Induction Theorem 1.1 (Archimedean property). If a and b are any positive integers, then there exists a positive integer n such that na b. Theorem 1. (First Principle of Finite Induction). Let S be a set of positive integers with the following properties: (a) The integer 1 belongs to S. (b) Whenever the integer k is in S, the next integer k + 1 must also be in S. Then S is the set of all positive integers. 1. The Binomial Theorem 11

12 1 CHAPTER 1. PRELIMINARIES

13 Chapter Divisibility Theory in the Integers.1 Early Number Theory. The Division Algorithm Theorem.1 (Division Algorithm). Given integers a and b, with b > 0, there exists unique integers q and r satisfying a = qb + r, 0 r < b. The integers q and r are called, respectively, the quotient and remainder in the division of a by b. Proof. Z is an Euclidean domain. Supplement. There is an application of Division Algorithm, see Theorem 8.1 Corollary.1. If a and b are integers, with b 0, then there exist unique integers q and r such that a = qb + r, 0 r < b..3 The Greatest Common Divisor Definition.1. An integer b is said to be divisible by an integer a 0, in symbols a b, if there exists some integer c such that b = ac. We write a b to indicate that b is not divisible by a. Theorem.. For integers a, b, c, the following hold: (a) a 0, 1 a, a a. (b) a 1 if and only if a = ±1. (c) If a b and c d, then ac bd. (d) If a b and b c, then a c. (e) a b and b a if and only if a = ±b. 13

14 14 CHAPTER. DIVISIBILITY THEORY IN THE INTEGERS (f) If a b and b 0, then a b. (g) If a b and a c, then a (bx + cy) for arbitrary integers x and y. Definition.. Let a and b be given integers, with at least one of them different from zero. The greatest common divisor of a and b, denoted by gcd (a, b), is the positive integer d satisfying the following: (a) d a and d b. (b) If c a and c b, then c d. Theorem.3. Given integers a and b, not both of which are zero, there exist integers x and y such that gcd (a, b) = ax + by. Proof. Consider the ideal {ax + by x, y Z} = a, b in Z which is generated by a and b. Then a, b = gcd (a, b) because Z is a principal ideal domain. See [6], p.74, prop.. Corollary.. If a and b are given integers, not both zero, then the set T = {ax + by x, y are integers} is precisely the set of all multiples of d = gcd (a, b). Proof. It follows immediately from the proof of Theorem.3. Definition.3. Two integers a and b, not both of which are zero, are said to be relatively prime whenever gcd (a, b) = 1. Theorem.4. Let a and b be integers, not both zero. Then a and b are relatively prime if and only if there exist integers x and y such that 1 = ax + by. Proof. ( ) It follows immediately from Theorem.3. ( ) Supplement. Note that gcd (a, b) = d gcd (a, b) = 1 gcd (a, b) a, gcd (a, b) b Theorem. gcd (a, b) (ax + by) = 1 Theorem. gcd (a, b) = 1 Theorem.3 x, y such that ax + by = d. Theorem.4 x, y such that ax + by = 1. But notice that gcd (a, b) = d / x, y such that ax + by = d.

15 .3. THE GREATEST COMMON DIVISOR 15 Corollary.3. If gcd (a, b) = d, then gcd (a/d, b/d) = 1. Proof. Theorem.3 gcd (a, b) = d ax + by = d a d x + b d y = 1 Theorem.4 gcd ( a d, b d ) = 1 Corollary.4. If a c and b c, with gcd (a, b) = 1, then ab c. Proof. gcd (a, b) = 1 x,y Z ax + by = 1 multiplying c a c and b c imply c=as and c=bt for some s and t in Z cax + cby = c (bt)ax + (as)by = c (ab)tx + (ab)sy = c ab c Theorem.5 (Euclid s lemma). If a bc, with gcd (a, b) = 1, then a c. Proof. gcd (a, b) = 1 x,y Z ax + by = 1 multiplying c a bc implies bc=as for some s Z cax + cby = c cax + asy = c a(cx + sy) = c a c Supplement. Note this proof is similar to the proof of Corollary.4. Theorem.6. Let a, b be integers, not both zero. For a positive integer d, d = gcd (a, b) if and only if

16 16 CHAPTER. DIVISIBILITY THEORY IN THE INTEGERS (a) d a and d b. (b) Whenever c a and c b, then c d. Supplement. Note that this theorem states that every common divisor divides gcd. Proof. By Theorem.3, suppose d = ax + by. If c a and c b, then by Theorem..(g), c (ax + by) = d..4 The Euclidean Algorithm Lemma.1. If a = qb + r, then gcd (a, b) = gcd (b, r). Proof. By the minimality of the ideal generated by some subset, that is, if X Y, then X Y, we have gcd (a, b) = a, b = qb + r, b r=1 (qb+r) q b qb+r,b = r, b = gcd (b, r) Z. Thus, gcd (a, b) gcd (b, r) and gcd (b, r) gcd (a, b). Then by Theorem..(e). Theorem.7. If k > 0, then gcd (ka, kb) = k gcd (a, b). Corollary.5. For any integer k 0, gcd (ka, kb) = k gcd (a, b). Definition.4. The least common multiple of two nonzero integers a and b, denoted by lcm (a, b), is the positive integer m satisfying the following: (a) a m and b m. (b) If a c and b c, with c > 0, then m c. Theorem.8. For positive integers a and b, gcd (a, b) lcm (a, b) = ab. Supplement. The most intuitive proof follows from the theorem: If m = p k 1 1 pk pkr r and n = p j 1 1 p j p jr r, then gcd (m, n) = p min (k 1,j 1 ) 1 p min (k,j ) p and lcm (m, n) = p max (k 1,j 1 ) 1 p max (k,j ) p min (kr,jr) r, max (kr,jr) r. Corollary.6. For any choice of positive integers a and b, lcm (a, b) = ab if and only if gcd (a, b) = 1. Proof. It follows immediately from Theorem.8.

17 .5. THE DIOPHANTINE EQUATION AX + BY = C 17.5 The Diophantine Equation ax + by = c Theorem.9. The linear Diophantine equation ax + by = c has a solution if and only if d c, where d = gcd (a, b). If x 0, y 0 is any particular solution of this equation, then all other solutions are given by where t is an arbitrary integer. Proof. x = x 0 + ( b d ) t, y = y 0 ( a d ) t, ax + by = c c a, b = gcd (a, b) Z gcd (a, b) c. All the integer points on the line ax + by = c will be the integer points on the line ax + by = 0 by translation (x 0, y 0 ), and vice verse. So, all the integer solutions of ax + by = c will be all the integer solutions of ax + by = 0 by add (x 0, y 0 ), and vice versa. All the integer solutions of ax + by = 0 are t ( b d, a ). d So, all the integer solutions of ax + by = c are (x 0, y 0 ) + t ( b d, a). d ax + by = c ax + by = 0 (x 0, y 0 ) ( b d, a b ) (b, a) Supplement. 這個過程很類似於我們在 linear algebra 中也是先解 homogeneous Ax = 0, 再解 Ax = b; 也很類似於我們在 differential equation 中先解 homogeneous a n (x)y (n) + a n 1 (x)y (n 1) + + a 1 (x)y + a 0 (x)y = 0, 再解 a n (x)y (n) + a n 1 (x)y (n 1) + + a 1 (x)y + a 0 (x)y = c(x) Corollary.7. If gcd (a, b) = 1 and if x 0, y 0 is a particular solution of the linear Diophantine equation ax + by = c, then all solutions are given by for integral values of t. x = x 0 + bt, y = y 0 at

18 18 CHAPTER. DIVISIBILITY THEORY IN THE INTEGERS

19 Chapter 3 Primes and Their Distribution 3.1 The Fundamental Theorem of Arithmetic Definition 3.1. An integer p > 1 is called a prime number, or simply a prime, if its only positive divisors are 1 and p. An integer greater than 1 that is not a prime is termed composite. Theorem 3.1. If p is a prime and p ab, then p a or p b. Supplement. p is a prime ideal in Z. 這是很典型的證明 P (Q or R) 的題目, 這類題目的證明手法通常就是 (P and Q) R, 也就是先假設 P 且 Q, 然後證得 R Corollary 3.1. If p is a prime and p a 1 a a n, then p a k for some k, where 1 k n. Corollary 3.. If p, q 1, q,..., q n are all primes and p q 1 q q n, then p = q k for some k, where 1 k n. Theorem 3. (Fundamental Theorem of Arithmetic). Every positive integer n > 1 is either a prime or a product of primes; this representation is unique, apart from the order in which the factors occur. Proof. Z is a U.F.D.. Corollary 3.3. Any positive integer n > 1 can be written uniquely in a canonical form n = p k 1 1 pk pkr r, where, for i = 1,,..., r, each k i is a positive integer and each p i is a prime, with p 1 < p < < p r. Theorem 3.3 (Pythagoras). The number is irrational. 19

20 0 CHAPTER 3. PRIMES AND THEIR DISTRIBUTION 3. The Sieve of Eratosthenes Theorem 3.4 (Euclid). There is an infinite number of primes. Theorem 3.5. If p n is the nth prime number, then p n n 1. Corollary 3.4. For n 1, there are at least n + 1 primes less than n. 3.3 The Goldbach Conjecture Lemma 3.1. The product of two or more integers of the form 4n + 1 is of the same form. Theorem 3.6. There are an infinite number of primes of the form 4n + 3. Theorem 3.7 (Dirichlet). If a and b are relatively prime positive integers, then the arithmetic progression a, a + b, a + b, a + 3b,... contains infinitely many primes. Proof. 參考 [5], ch. 或是 [10], ch.10, 頗難, 用到一點 character, 算是 representation theory 的東西 Theorem 3.8. If all the n > terms of the arithmetic progression p, p + d, p + d,..., p + (n 1)d are prime numbers, then the common difference d is divisible by every prime q < n.

21 Chapter 4 The Theory of Congruences 4.1 Carl Friedrich Gauss 4. Basic Properties of Congruence Definition 4.1. Let n be a fixed positive integer. Two integers a and b are said to be congruent modulo n, symbolized by a b (mod n) if n divides the difference a b; that is, provided that a b = kn for some integer k. Supplement. a = b Z n or a = b Z n if gcd (a, n) = gcd (b, n) = 1. Note that is an equivalence relation. Theorem 4.1. For arbitrary integers a and b, a b (mod n) if and only if a and b leave the same nonnegative remainder when divided by n. Theorem 4.. Let n > 1 be fixed and a, b, c, d be arbitrary integers. Then the following properties hold: (a) a a (mod n). (b) If a b (mod n), then b a (mod n). (c) If a b (mod n) and b c (mod n), then a c (mod n). (d) If a b (mod n) and c d (mod n), then a + c b + d (mod n) and ac bd (mod n). (e) If a b (mod n), then a + c b + c (mod n) and ac bc (mod n). (f) If a b (mod n), then a k b k (mod n) for any positive integer k. Theorem 4.3. If ca cb (mod n), then a b (mod n/d), where d = gcd (c, n). Corollary 4.1. If ca cb (mod n) and gcd (c, n) = 1, then a b (mod n). 1

22 CHAPTER 4. THE THEORY OF CONGRUENCES Proof. It follows immediately from ca cb (mod n) n (ca cb) and the Euclid s Lemma (Theorem.5). Corollary 4.. If ca cb (mod p) and p c, where p is a prime number, then a b (mod p). Proof. It follows immediately from p c gcd (p, c) = 1 and the Euclid s Lemma (Theorem.5). 4.3 Binary and Decimal Representations of Integers Theorem 4.4. Let P (x) = m k=0 c k x k be a polynomial function of x with integral coefficients c k. If a b (mod n), then P (a) P (b) (mod n). Corollary 4.3. If a is a solution of P (x) 0 (mod n) and a b (mod n), then b also is a solution. Theorem 4.5. Let N = a m 10 m + a m 1 10 m a a 0 be the decimal expansion of the positive integer N, 0 a k < 10, and let S = a 0 + a a m. Then 9 N if and only if 9 S. Theorem 4.6. Let N = a m 10 m + a m 1 10 m a a 0 be the decimal expansion of the positive integer N, 0 a k < 10, and let T = a 0 a 1 + a + ( 1) m a m. Then 11 N if and only if 11 T. 4.4 Linear Congruences and the Chinese Remainder Theorem Theorem 4.7. The linear congruence ax b (mod n) has a solution if and only if d b, where d = gcd (a, n). If d b, then it has d mutually incongruent solutions modulo n. Supplement. 這其實只是 Z n ( 或 modulo n) 觀點下的 Theorem.9 Corollary 4.4. If gcd (a, n) = 1, then the linear congruence ax b (mod n) has a unique solution modulo n. Proof. If gcd (a, n) = 1, then a is a unit in Z n and ax = b Z n has a solution x = a 1 b Z n.

23 4.4. LINEAR CONGRUENCES AND THE CHINESE REMAINDER THEOREM 3 Supplement. How to solve the equation ax b (mod n) if gcd (a, n) = 1? You can use the Euclid s Algorithm to find the integers s, t Z such that as + nt = 1. Then as 1 = ( t)n and as 1 (mod n). Hence, sax 1 x x sb (mod n) is the solution. For example, since gcd (31, 69) = 1, we use Euclid Algorithm to find integers s and t such that 1 = 69 s + 31 t. 69 = , (4.1) 31 = , (4.) 7 = (4.3) by (4.3), 1 = 7 3 (4.) = 7 (31 7 4) = = (4.1) = (69 31 ) 9 31 = = s = 9, t = 0. Theorem 4.8 (Chinese Remainder Theorem). Let n 1, n,..., n r be positive integers such that gcd (n i, n j ) = 1 for i j. Then the system of linear congruences x a 1 (mod n 1 ) x a (mod n ) x a r (mod n r ) has a simultaneous solution, which is unique modulo the integer n 1 n n r. Supplement. 我們先來學習怎麼解兩個方程式, 三個或更多方程式的解法就是反覆施行這個方法 Suppose that gcd (m, n) = 1, there exists s, t Z such that ms+nt = 1. Verify directly that x ant + bms (mod mn) are the solution of the following simultaneous congruence equations x a (mod m) { x b (mod n) 這你在小學的時候就學過了, 有一堆糖果, 如果一次拿 m 個, 最後會剩下 a 個 ; 如果一次拿 n 個, 最後會剩下 b 個, m, n 是互質的, 請問這堆糖果可能有幾個? 這類問題也被稱為剩餘 (Remainder) 問題或是韓信點兵, 或是 Chinese Remainder Theorem 分享一個我在大學時記下這個公式的方法, 因為 ms + nt = 1, 將 s 跟 t 分別註記在 m 跟 n 的旁邊, 然後利用交叉相乘的方法, 記下 x ant + bms (mod mn) x a (mod m)s x b (mod n)t

24 4 CHAPTER 4. THE THEORY OF CONGRUENCES Theorem 4.9. The system of linear congruences ax + by r (mod n) cx + dy s (mod n) has a unique solution modulo n whenever gcd (ad bc, n) = 1. Proof. { d to first b to second { ax + by r (mod n) cx + dy s (mod n) adx + bdy dr (mod n) bcx + bdy bs (mod n) subtract (ad bc)x dr bs (mod n) (ad bc)x = dr bs Z n Then by gcd (ad bc, n) = 1 and Corollary 4.4.

25 Chapter 5 Fermat s Theorem 5.1 Pierre de Fermat 5. Fermat s Little Theorem and Pseudoprimes Theorem 5.1 (Fermat s Theorem). Let p be a prime and suppose that p a. Then a p 1 1 (mod p). Proof. Applies Lagrange s Theorem on Z p. Corollary 5.1. If p is a prime, then a p a (mod p) for any integer a. Lemma 5.1. If p and q are distinct primes with a p a (mod p) and a q a (mod p), then a pq a (mod pq). Theorem 5.. If n is an odd pseudoprime, then M n = n 1 is a larger one. Theorem 5.3. Let n be a composite square-free integer, say, n = p 1 p p r, where the p i are distinct primes. If p i 1 n 1 for i = 1,,..., r, then n is an absolute pseudoprime. 5.3 Wilson s Theorem Theorem 5.4 (Wilson). If p is a prime, then (p 1)! 1 (mod p). Proof. By Fermat s Little Theorem (Theorem 5.1), all nonzero elements in Z p is a root of the polynomial x p 1 1 Z p [x]. Hence, by Factor Theorem, x p 1 1 = (x 1)(x ) (x (p 1)). Compare the constant term of these two polynomials. We have 1 = ( 1) p 1 (p 1)! = (p 1)! Z p. That is, (p 1)! 1 (mod p). 5

26 6 CHAPTER 5. FERMAT S THEOREM Theorem 5.5. The quadratic congruence x (mod p), where p is an odd prime, has a solution if and only if p 1 (mod 4). Proof. It follows immediately from Euler s Criterion (Theorem 9.1). Supplement. 另解 Recall that Z p is a cyclic group. Suppose that Z p = r and ord(r) = r = Z p = φ(p) = p 1. ( ) If p = 4k + 1, suppose that ( 1) = r s, then r s = 1 and 4k = (p 1) = ord(r) s and k s. Thus, ( 1) = r s = r t = (r t ). ( ) If x = 1 Z p, then x 4 = 1 and ord(x) = 4. Suppose that x = r u. Then 4 = ord(x) = ord(r u ) = (k+1) gcd (u,4k+) p 1 gcd (u,p 1) = 4. If p = 4k + 3, then we have p 1 gcd (u,p 1) = 4k+ gcd (u,4k+) = = 4. Which is impossible because the left hand side of the equation doesn t have the divisor The Fermat-Kraitchik Factorization Method

27 Chapter 6 Number-Theoretic Functions 6.1 The Sum and Number of Divisors Definition 6.1. Given a positive integer n, let τ(n) denote the number of positive divisors of n and σ(n) denote the sum of these divisors. Theorem 6.1. If n = p k 1 1 pk pkr r is the prime factorization of n > 1, then the positive divisors of n are precisely those integers d of the form where 0 a i k i (i = 1,,..., r). Theorem 6.. If n = p k 1 1 pk pkr r (a) τ(n) = (k 1 + 1)(k + 1) (k r + 1), and (b) σ(n) = pk p k +1 1 p 1 1 p 1 pkr+1 r 1 p. r 1 d = p a 1 1 pa par r, is the prime factorization of n > 1, then Definition 6.. A number-theoretic function f is said to be multiplicative if whenever gcd (m, n) = 1. f(mn) = f(m)f(n) Theorem 6.3. The function τ() and σ() are both multiplicative functions. Lemma 6.1. If gcd (m, n) = 1, then the set of positive divisors of mn consists of all products d 1 d, where d 1 m, d n and gcd (d 1, d ) = 1; furthermore, these products are all distinct. Theorem 6.4. If f is a multiplicative function and F is defined by then F is also multiplicative. F (n) = f(d), d n Corollary 6.1. The functions τ() and σ() are multiplicative functions. 7

28 8 CHAPTER 6. NUMBER-THEORETIC FUNCTIONS 6. The Möbius Inversion Formula Definition 6.3. For a positive integer n, define µ by the rules µ (n) = 1 if n = 1 0 if p n for some prime p ( 1) r if n = p 1 p p r, where p i are distinct primes Theorem 6.5. The function µ () is a multiplicative function. Theorem 6.6. For each positive integer n 1, µ (d) = { 1 if n = 1 0 if n > 1 d n Theorem 6.7 (Möbius inversion formula). Let F and f be two number-theoretic functions related by the formula F (n) = f(d). d n Then f(n) = µ (d) F ( n d n d ) = µ ( n d n d ) F (d). Supplement. 這個在 advanced finite field 的時候才會學到應用, 參考 [9], sec.3. Theorem 6.8. If F is a multiplicative function and then f is also multiplicative. F (n) = f(d), d n 6.3 The Greatest Integer Function Definition 6.4. For an arbitrary real number x, we denote by [x] the largest integer less than or equal to x; that is, [x] is the unique integer satisfying x 1 < [x] x. Theorem 6.9. If n is a positive integer and p a prime, then the exponent of the highest power of p that divides n! is [ n p ], k k=1 where the series is finite, because [n/p k ] = 0 for p k > n. Theorem If n and r are positive integers with 1 r < n, then the binomial coefficient ( n r ) = n! r!(n r)! is also an integer. Corollary 6.. For a positive integer r, the product of any r consecutive positive integers is divisible by r!.

29 6.4. AN APPLICATION TO THE CALENDAR 9 Theorem Let f and F be number-theoretic functions such that Then, for any positive integer N, N n=1 F (n) = f(d). d n F (n) = N k=1 Corollary 6.3. If N is a positive integer, then N n=1 τ(n) = f(k) [ N k ]. N n=1 Corollary 6.4. If N is a positive integer, then N n=1 σ(n) = N n=1 [ N n ]. n [ N n ]. 6.4 An Application to the Calendar Theorem 6.1. The date with month m, day d, year Y = 100c + y where c 16 and 0 y < 100, has weekday number w d + [(.6)m 0.] c + y + [ c 4 ] + [y ] (mod 7) 4 provided that March is taken as the first month of the year and January and February are assumed to be the eleventh and twelfth months of the previous year.

30 30 CHAPTER 6. NUMBER-THEORETIC FUNCTIONS

31 Chapter 7 Euler s Generalization of Fermat s Theorem 7.1 Leonhard Euler 7. Euler s Phi-function Definition 7.1. For n 1, let φ(n) denote the number of positive integers not exceeding n that are relatively prime to n. Theorem 7.1. If p is a prime and k > 0, then φ(p k ) = p k p k 1 = p k (1 1 p ). Proof. φ(p k ) = #{1 a p k gcd (a, p k ) = 1} = p k #{1 a p k gcd (a, p k ) 1} = p k #{1 a p k p a} = p k #{1p, p, 3p,..., p k 1 p} = p k p k 1. Lemma 7.1. Given integers a, b, c, gcd (a, bc) = 1 if and only if gcd (a, b) = 1 and gcd (a, c) = 1. Theorem 7.. The function φ() is a multiplicative function. Proof. We show that θ Z mn Z m Z n given by θ(a) = (a, a) is a bijection, where gcd (m, n) = 1. Then φ(mn) = Z mn = Z m Z n = Z m Z n = φ(m)φ(n). 31

32 3 CHAPTER 7. EULER S GENERALIZATION OF FERMAT S THEOREM Well-defined and One-to-One are easy to verify. For onto, suppose (s, t) Z m Z n. Then by Chinese Remainder Theorem (Theorem 4.8), there exists a solution of the simultaneous linear congruences { x s (mod m) x t (mod n) It follows that θ(x) = (x, x) = (s, t). Theorem 7.3. If the integer n > 1 has the prime factorization n = p k 1 1 pk pkr r, then φ(n) = (p k 1 1 pk ) (p k pk 1 ) (p kr r p kr 1 r ) = n (1 1 p 1 ) (1 1 p ) (1 1 p r ) Proof. It follows immediately from Theorem 7.1 and Theorem 7.. Theorem 7.4. For n >, φ(n) is an even integer. Proof. Consider the element ( 1) Z n. Since ord( 1) = and Z n = φ(n), by Lagrange s Theorem, = ord(( 1)) divides Z n = φ(n). Supplement. An application of this theorem see Theorem Euler s Theorem Lemma 7.. Let n > 1 and gcd (a, n) = 1. If a 1, a,..., a φ(n) are the positive integers less than n and relatively prime to n, then aa 1, aa,..., aa φ(n) are congruent modulo n to a 1, a,..., a φ(n) in some order. Theorem 7.5 (Euler). If n 1 and gcd (a, n) =, then a φ(n) 1 (mod n). Proof. Applies Lagrange s Theorem on Z n. Corollary 7.1 (Fermat). If p is a prime and p a, then a p 1 1 (mod p). 7.4 Some Properties of the Phi-function Theorem 7.6 (Gauss). For each positive integer n 1, n = φ(d), d n the sum being extended over all positive divisorsof n.

33 7.4. SOME PROPERTIES OF THE PHI-FUNCTION 33 Proof. n = # { 1 n, n, 3 n,..., n n } r 11 = #, r 1,..., r 1φ(d 1 ), r 1, r,..., r φ(d ),..., r τ(n)1, r τ(n),..., r τ(n)φ(d τ(n)) d 1 d 1 d 1 d d d d τ(n) d τ(n) d τ(n) φ(d 1 ) terms φ(d ) terms φ(d τ(n) ) terms reduce to lowest terms, where d i n = φ(d), d n where τ(n) is the number of divisors of n. Theorem 7.7. For n > 1, the sum of the positive integers less than n and relatively prime to n is 1 nφ(n). Theorem 7.8. For any positive integer n, µ (d) φ(n) = n d n d.

34 34 CHAPTER 7. EULER S GENERALIZATION OF FERMAT S THEOREM

35 Chapter 8 Primitive Roots and Indices 8.1 The Order of an Integer Modulo n Definition 8.1. Let n > 1 and gcd (a, n) = 1. The order of a modulo n (in older terminogy: the exponent to which a belongs modulo n) is the smallest positive integer k such that a k 1 (mod n). Supplement. The order of a, ord(a), in Z n. Theorem 8.1. Let the integer a have order k modulo n. Then a h 1 (mod n) if and only if k h; in particular, k φ(n). Supplement. Consider a h = 1 Z n. By Division Algorithm (Theorem.1), write h = ord(a)q + r, where r = 0 or 0 < r < ord(a). Then 1 = a h = a ord(a)q+r = (a ord(a) ) q a r = a r. If r 0, then contrary to the minimality of the ord(a). Therefore, r = 0 and ord(a) h. Theorem 8.. If the integer a has order k modulo n, then a i a j (mod n) if and only if i j (mod k). Supplement. Note that a i = a j Z n if and only if a i j = 1 Z n. Then by Theorem 8.1. Corollary 8.1. If a has order k modulo n, then the integers a, a,..., a k are incongruent modulo n. Supplement. If ord(a) = k, then a = {a, a,..., a k = 1} Z n. Theorem 8.3. If the integer a has order k modulo n and h > 0, then a h has order k/ gcd (h, k) modulo n. Supplement. The most intuitively proof is given as following. Note that if G = a and lcm (h,k) h Theorem.8 k = gcd (h,k). As the following G = ord(a) = k, then the order of a h in G is figure indicates. 這其實就是國小的用最小公倍數求在圓形操場上賽跑, 同時抵達終點的問題 35

36 36 CHAPTER 8. PRIMITIVE ROOTS AND INDICES h h 3h lcm (h,k) h h k k lcm (h,k) k k Corollary 8.. Let a have order k modulo n. Then a h also has order k if and only if gcd (h, k) = 1. Definition 8.. If gcd (a, n) = 1 and a is of order φ(n) modulo n, then a is a primitive root of the integer n. Supplement. If Z n is cyclic, then a primitive root modulo n is a generator r of Z n. That is, Z n = r and ord(r) = r = Z n = φ(n). Theorem 8.4. Let gcd (a, n) = 1 and let a 1, a,..., a φ(n) be th positive integers less than n and relatively prime to n. If a is a primitive root of n, then a, a,..., a φ(n) are congruent modulo n to a 1, a,..., a φ(n), in some order. Proof. Z n = a. Corollary 8.3. If n has a primitive root, then it has exactly φ(φ(n)) of them. Proof. Let r be a primitive root modulo n. That is, Z n = r and ord(r) = r = Z n = φ(n). #{g Z g is a primitive root modulo n} = #{g Z n g is a generator of Z n} = #{r s Z n r s is a generator of Z n} = #{1 s Z n ord(r s ) = Z n } = # {1 s φ(n) ord(r s ) = ord(r) gcd (s, ord(r)) = φ(n)} φ(n) = # {1 s φ(n) gcd (s, φ(n)) = φ(n)} = #{1 s φ(n) gcd (s, φ(n)) = 1} = φ(φ(n)) 8. Primitive Roots for Primes Theorem 8.5 (Lagrange). If p is a prime and f(x) = a n x n + a n 1 x n a 1 x + a 0, a n / 0 (mod p), is a polynomial of degree n 1 with integral coefficients, then the congruence f(x) 0 (mod p) has at most n incongruent solutions modulo p.

37 8.. PRIMITIVE ROOTS FOR PRIMES 37 Proof. This is a special case of the Fundamental Theorem of Algebra on the finite field Z p. Corollary 8.4. If p is a prime number and d p 1, then the congruence has exactly d solutions. x d 1 0 (mod p) Proof. Recall that Z p is a cyclic group. Since d (p 1), by the Fundamental Theorem of Finite Cyclic Group, there exists a cyclic subgroup H Z p with order d. We claim that H = {x Z n x d = 1}. For every t d, by the Fundamental Theorem of Finite Cyclic Group again, there exists a subgroup K t H with order t. Note that K t is also a subgroup of Z p. ( ) If x d = 1 Z p, then by Theorem 8.1, ord(x) d and x is a subgroup of Z p with order ord(x) d. By the uniqueness of the subgroup, x = K ord(x) H. It follows that x H. ( ) On the other hand, for every x H, by Lagrange s Theorem, ord(x) divides H = d. So x d = 1. Theorem 8.6. If p is a prime number and d p 1, then there are exactly φ(d) incongruent integers having order d modulo p. Proof. Recall that Z p is a cyclic group. Since d (p 1), by the Fundamental Theorem of Finite Cyclic Group, there exists a unique cyclic subgroup H = h Z p with order d. If x Z p with ord(x) = d, by the uniqueness of H, x = H and x H. That is, every element of order d in Z p must be in H. #{x H ord(x) = d} = # {h s H ord(h s ) = d = # {1 s d gcd (s, d) = d} = #{1 s d gcd (s, d) = 1} = φ(d) ord(h) gcd (s, ord(h)) = d} Corollary 8.5. If p is a prime, then there are exactly φ(p 1) incongruent primitive roots of p. Proof. It follows immediately from Corollary 8.3 or Theorem 8.6.

38 38 CHAPTER 8. PRIMITIVE ROOTS AND INDICES 8.3 Composite Numbers Having Primitive Roots The proof is the same as [7]. The following chart is the sketch of the proof of Theorem Z n is cyclic, if n = l, l {0, 1, }, verify directly; is not cyclic, if n = l, l 3, Theorem 8.7; is not cyclic, if n = p k l, p is an odd prime, k 1, l, Corollary 8.6; is not cyclic, if n = p k q l, p and q are distinct odd primes, k 1, l 1, Corollary 8.6; is cyclic, if n = p k, p is an odd prime, k 1, Theorem 8.9; is cyclic, if n = p k, p is an odd prime, k 1, Corollary 8.8; Theorem 8.7. For k 3, the integer k has no primitive roots. Proof. If Z is cyclic, then by the Fundamental Theorem of Finite Cyclic Group, there k is exactly one cyclic subgroup of order. We show that there are two distinct elements in Z with order. Then there are two distinct cyclic subgroups of order in Z. So k k is not cyclic. Z k Observe the cases k = 4 and k = 5. 1 = 1 Z, 4 3 = 9, 5 = 9, 7 = 1, 9 = 1, 11 = 9, 13 = 9, 15 = 1.

39 8.3. COMPOSITE NUMBERS HAVING PRIMITIVE ROOTS 39 1 = 1 Z, 5 3 = 9, 5 = 5, 7 = 17, 9 = 17, 11 = 5, 13 = 9, 15 = 1, 17 = 1, 19 = 9, 1 = 5, 3 = 17, 5 = 17, 7 = 5, 9 = 9, 31 = 1. We note that ( k 1) = ( k + 1) = 1 Z k. Indeed, ( k k ± 1) = ( k 1 ± 1) = k ± k + 1 = k k ± k + 1 = 1 Z. k Note that when k =, k 1 = 1 Z necessarily. and ord( k 1) = 1. So the condition k 3 is Theorem 8.8. If gcd (m, n) = 1, where m > and n >, then the integer mn has no primitive roots. Proof. If gcd (m, n) = 1, it is easy to show that the mapping θ Z mn Z m Z n given by θ(a) = (a, a) is an one-to-one group homomorphism. Then by Chinese Remainder Theorem (Theorem 4.8), θ is onto. (See the proof of Theorem 7..) Thus, Z mn Z m Z n. By the Fundamental Theorem of Finite Abelian Group, Z mn is cyclic if and only if gcd ( Z m, Z n ) = 1 and Z m, Z n both are cyclic. Since m >, n >, by Theorem 7.4, both Z m and Z n are even. Hence, gcd ( Z m, Z n ) 1 and Z mn is not cyclic. Corollary 8.6. The integer n fails to have a primitive root if either (a) n is divisible by two odd primes, or (b) n is of the form n = m p k, where p is an odd prime and m. Proof. It follows immediately from Theorem 8.8. Lemma 8.1. If p is an odd prime, then there exists a primitive root r of p such that r p 1 1 (mod p ).

40 40 CHAPTER 8. PRIMITIVE ROOTS AND INDICES Corollary 8.7. If p is an odd prime, then p has a primitive root; in fact, for a primitive root r of p, either r or r + p (or both) is a primitive root of p. Proof. 注意在這題中, 一個元素的 order 會因為其所在的 group 而不同, 要特別注意 Recall that Z p is cyclic. Suppose that Z p = r. Let h t be the order of r + tp in Z p, where t {0, 1}. Then (r + tp) ht 1 (mod p ) and (r + tp) ht 1 (mod p). It follows that 1 = (r + tp) ht = r ht + ( h t 1 )rht 1 tp + ( h t )rht (tp) + + (tp) ht = r ht Z p. Thus, the order of r in Z p divides h t, that is, (p 1) h t. On the other hand, by Lagrange s Theorem, h t Z = p(p 1). Hence, h p t {p 1, p(p 1)}. We show that there is at least one of t = 0 or t = 1 such that h t = p(p 1) by contradiction. If h 0 = h 1 = p 1, then and r p 1 = (r + p) p 1 = 1 Z p 1 = (r + p) p 1 = r p 1 + ( p 1 1 )rp p + ( p 1 Z p )rp 3 p + + p p 1 = 1 + (p 1)r p p. Which implies that 0 = (p 1)r p p Z and p p (p 1)r p p, a contradiction. Lemma 8.. Let p be an odd prime and let r be a primitive root of p with the property that r p 1 / 1 (mod p ). Then for each positive integer k, r pk (p 1) / 1 (mod p k ). Theorem 8.9. If p is an odd prime number and k 1, then there exists a primitive root for p k. Proof. Recall that Z p is a cyclic group. That is, there is a primitive root modulo p. This prove the case k = 1. Now, we know that it must has a primitive root modulo p. By Corollary 8.7, there is a primitive root modulo p. We show that a primitive root modulo p is also a primitive root modulo p k. We use induction on k. Suppose that Z = r and r = Z for every p p i i k. In particular, r = Z. Note that the order of r Z is φ(p k ) = p k 1 (p 1) p k p k and the order of r Z is φ(p k 1 ) = p k (p 1). 1 Let h be the order of r in Z. p k 1 p k+1 We show that h = p k (p 1) = φ(p k+1 ). r h = 1 Z p k+1 r h 1 (mod p k+1 ) r h 1 (mod p k ) r h = 1 Z p k the order of r Z p divides h k p k 1 (p 1) h 1 Clever reader may notice that when k =, does r = Z p implies that r = Z p? It is true. If r = Z p, then gcd (r, p ) = 1 and gcd (r, p) = 1. That is, r Z p. For every a Z p, gcd (a, p) = 1 implies that gcd (a, p ) = 1. Hence, a Z p. Since a = r s Z p for some s Z, we have a = r s Z p.

41 8.4. THE THEORY OF INDICES 41 On the other hand, by Lagrange s Theorem, h divides Z p k+1 = p k (p 1). Thus, h {p k (p 1), p k 1 (p 1)}. We are going to show that r pk 1 (p 1) 1 Z p k+1. Then h p k 1 (p 1) and it must be h = p k (p 1). By the induction hypothesis, r = Z p k = Z p k 1, so r pk (p 1) 1 Z p k and r pk (p 1) = 1 Z p k 1. It follows that r pk (p 1) = 1 + bp k 1 for some b which is not divisible by p and r pk 1 (p 1) = (r pk (p 1) ) p = (1 + bp k 1 ) p = 1 + ( p 1 )bpk 1 + ( p )b p k + + ( p p 1 )bp 1 p (p 1)(k 1) + b p p pk p pk p k+1 and in Z p k+1 = 1 + ( p 1 )bpk 1 + ( p )b p k + + ( p p 1 )bp 1 p (p 1)(k 1) k k, p divides ( p 1 ),(p ),...,( p p 1 ) = 1 + ( p 1 )bpk 1 = 1 + bp k Z p k+1 Since p b, we have r pk 1 (p 1) 1 Z p k+1, as required. Corollary 8.8. There are primitive roots for p k, where p is an odd prime and k 1. Proof. By the proof of Theorem 8.8 and the result of Theorem 8.9. Theorem An integer n > 1 has a primitive root if and only if where p is an odd prime. n =, 4, p k, or p k, Proof. See the beginning of the section. 8.4 The Theory of Indices Definition 8.3. Let r be a primitive root of n. If gcd (a, n) = 1, then the smallest positive integer k such that a r k (mod n) is called the index of a relative to r. Supplement. Note that the index notion is similar to the logarithm. And by Theorem 8., we have a b (mod n) ind r a ind r b (mod φ(n)). Theorem If n has a primitive root r and ind r a denotes the index of a relative to r, then the following properties hold: (a) ind r (ab) ind r a + ind r b (mod φ(n)). (b) ind r a k k ind r a (mod φ(n)) for k > 0. (c) ind r 1 0 (mod φ(n)), ind r r 1 (mod φ(n)).

42 4 CHAPTER 8. PRIMITIVE ROOTS AND INDICES Proof. Note that and r indrab = ab = r indra r indra = r indra+indrb r indrak = a k = (r indra ) k = r k indra. Then by Theorem 8.. Furthermore, ind r 1 = ord(r) = φ(n) and ind r r = 1. Theorem 8.1. Let n be an integer possessing a primitive root and let gcd (a, n) = 1. Then the congruence x k a (mod n) has a solution if and only if a φ(n)/d 1 (mod n), where d = gcd (k, φ(n)); if it has a solution, there are exactly d solutions modulo n. Proof. take index a φ(n)/ gcd (k,φ(n)) 1 (mod n) Theorem 8.11 φ(n) gcd (k, φ(n)) ind ra ind r 1 0 (mod φ(n)) φ(n) φ(n) gcd (k, φ(n)) ind ra gcd (k, φ(n)) ind r a Theorem 4.7 k ind r x ind r a (mod φ(n)) has solution take index x k a (mod n) has solution Supplement. 上面的證法是比較好的, 充分地展現了 index 的威力 另解 Let r be a promitive root modulo n. That is, Z n = r and ord(r) = φ(n). Suppose that a = r t for some t Z. ( ) ( ) Let x = r s. φ(n) φ(n) a gcd (k,φ(n)) = 1 φ(n) (r t ) gcd (k,φ(n)) = 1 φ(n) φ(n) = ord(r) t gcd (k, φ(n)) φ(n) t = φ(n) u for some u Z gcd (k, φ(n)) t = u gcd (k, φ(n)) a = r t = r u(kc+φ(n)d) = (r uc ) k (r φ(n) ) d = (r uc ) k. φ(n) φ(n) a gcd (k,φ(n)) = (x k ) gcd (k,φ(n)) = ((r s ) k ) gcd (k,φ(n)) = ((r k ) ord(r k ord(r) )= gcd (k,ord(r)) = φ(n) gcd (k,φ(n)) φ(n) gcd (k,φ(n)) ) s = 1. Corollary 8.9. Let p be a primitive and gcd (a, p) = 1. Then the congruence x k a (mod p) has a solution if and only if a (p 1)/d 1 (mod p), where d = gcd (k, p 1).

43 Chapter 9 The Quadratic Reciprocity Law 9.1 Euler s Criterion Definition 9.1. Let p be an odd prime and gcd (a, p) = 1. If the quadratic congruence x a (mod p) has a solution, then a is said to be a quadratic residue of p. Otherwise, a is called a quadratic nonresidue of p. Theorem 9.1 (Euler s Criterion). Let p be an odd prime and gcd (a, p) = 1. Then a is a quadratic residue of p if and only if a (p 1)/ 1 (mod p). Proof. ( ) By Lagrange s Theorem, every nonzero element c Z p satisfying c p 1 = 1. Suppose x 0 = a. Then a(p 1)/ = (x 0 )(p 1)/ = x p 1 0 = 1. ( ) Let r be a primitive root modulo p. That is, Z p = r. Suppose a (p 1)/ = 1 and a = r s for some s Z (r s ) (p 1)/ = r s(p 1)/ = 1 ord(r) = p 1 s(p 1)/ s(p 1)/ = (p 1)k for some k Z s = k a = (r k ) Corollary 9.1. Let p be an odd prime and gcd (a, p) = 1. Then a is a quadratic residue or nonresidue of p according to whether a (p 1)/ 1 (mod p) or a (p 1)/ 1 (mod p). 9. The Legendre Symbol and Its Properties Definition 9.. Let p be an odd prime and let gcd (a, p) = 1. The Legendre symbol (a/p) is defined by (a/p) = { 1 if a is a quadratic residue of p; 1 if a is a quadratic nonresidue of p. 43

44 44 CHAPTER 9. THE QUADRATIC RECIPROCITY LAW Theorem 9.. Let p be an odd prime and let a and b be integers that are relatively prime to p. Then the Legendre symbol has the following properties: (a) If a b (mod p), then (a/p) = (b/p). (b) (a /p) = 1. (c) (a/p) a (p 1)/ (mod p). (d) (ab/p) = (a/p)(b/p). (e) (1/p) = 1 and ( 1/p) = ( 1) (p 1)/. Proof. (a) Easy. (b) a a (mod p). (c) Euler s Criterion (Theorem 9.1). (d) See the textbook. (e) ( 1) 1 (mod p) or Euler s Criterion (Theorem 9.1). Corollary 9.. If p is a odd prime, then ( 1/p) = { 1 if p 1 (mod 4) 1 if p 3 (mod 4) Theorem 9.3. There are inifinitely many primes of the form 4k + 1. Theorem 9.4. If p is an odd prime, then p 1 a=1 (a/p) = 0. Hence, there are precisely (p 1)/ quadratic residues and (p 1)/ quadratic nonresidues of p. Proof. Define θ Z p (Z, +) ({1, 1}, ) by θ(a) = (a/p). Then θ is a group homomorphism by Theorem 9..(d). By the First Isomorphism Theorem, Z p/ ker θ (Z, +). Then ker θ = Z p / Z = Z p /. Corollary 9.3. The quadratic residues of an odd prime p are congruent modulo p to the even powers of a primitive root r of p; the quadratic nonresidues are congruent to the odd powers of r.

45 9.. THE LEGENDRE SYMBOL AND ITS PROPERTIES 45 Proof. Let r be a primitive root modulo p. That is, Z p = r. (r/p) can not be 1, otherwise, for all a Z p, (a/p) = 1. Therefore, (r/p) = 1 and (r k /p) = 1 and (r k+1 /p) = 1 by Theorem 9..(b) and (d). Theorem 9.5 (Gauss s lemma). Let p be an odd prime and let gcd (a, p) = 1. If n denotes the number of integers in the set S = {a, a, 3a,..., ( p 1 ) a}, whose remainders upon division by p exceed p/, then (a/p) = ( 1) n. Theorem 9.6. If p is an odd prime, then (/p) = { 1 if p 1 (mod 8) or p 7 (mod 8) 1 if p 3 (mod 8) or p 5 (mod 8) Corollary 9.4. If p is an odd prime, then (/p) = ( 1) (p 1)/8. Proof. See [1], A.10 or [8], thm Let K be the splitting field of x 8 1 over Z p. Let E be the set of all the root of x 8 1 in K. Then E is a cyclic group under the multiplication of K. (See [9], sec..4.) A generator of the cyclic group E is called a primitive root 8th root of unity over Z p. Let ω be a primitive 8th root of unity over Z p and set τ = ω + ω 1 in K. Since E is a cyclic group, by the Fundamental Theorem of Finite Cyclic Group, there is exactly one element with order. That is, ( 1). Thus, (ω 4 ) = ω 8 = 1 ω 4 = 1 (ω + ω ) = ω 4 + ω 4 + = ( 1) + ( 1) 1 + = ( 1) + ( 1) + = 0 ω + ω = 0 τ = (ω + ω 1 ) = ω + ω + = τ p = (τ ) (p 1)/ τ = (p 1)/ τ Euler s Criterion = (/p)τ. On the other hand, ( 1)(ω 3 + ω 3 ) = ω 4 (ω 3 + ω 3 ) = ω 7 + ω = ω + ω 1 and τ p = (ω + ω 1 ) p = ω p + ω p = { ω8s±1 + ω (8s±1) = ω ±1 + ω 1 = ω 1 + ω 1 = τ, if p ±1 (mod 8); ω 8s±3 + ω (8s±3) = ω ±3 + ω 3 = ( 1)(ω 1 + ω 1 ) = ( 1)τ, if p ±3 (mod 8); = ( 1) (p 1)/8 τ.

46 46 CHAPTER 9. THE QUADRATIC RECIPROCITY LAW Theorem 9.7. If p and p + 1 are both odd primes, then the integer ( 1) (p 1)/ is a primitive root of p + 1. Theorem 9.8. There are infinitely many primes of the form 8k 1. Lemma 9.1. If p is an odd prime and a and odd integer, with gcd (a, p) = 1, then (a/p) = ( 1) (p 1)/ k=1 [ka/p]. 9.3 Quadratic Reciprocity Theorem 9.9 (Quadratic Reciprocity Law). If p and q are distinct odd primes, then (p/q)(q/p) = ( 1) p 1 q 1. Proof. See [1]. The idea of the proof is to express the product of all the element of the group (Z p Z q)/{(1, 1), ( 1, 1)} by two ways. ˆ Note that Z p Z q = = (1, 1), (, 1),, (p, 1), (p 1, 1) (1, ), (, ),, (p, ), (p 1, ) (1, q 1 q 1 q 1 q 1 ), (, ),, (p, ), (p 1, ) (1, q ), (, q ),, (p, q ), (p 1, q ) (1, q 1), (, q 1),, (p, q 1), (p 1, q 1) (1, 1), (, 1),, (p, 1), (p 1, 1) (1, ), (, ),, (p, ), (p 1, ) (1, ), q 1 (, ),, q 1 (p, ), q 1 (p 1, ) (1, ), (, ),, (, ), ( 1, ) (1, 1), (, 1),, (, 1), ( 1, 1) Let U = {(1, 1), ( 1, 1)} be a normal subgroup of Z p Z q quotient group (Z p Z q)/u. Then and consider the (Z p Z q)/u = (1, 1)U, (, 1)U,, (p, 1)U, (p 1, 1)U (1, )U, (, )U,, (p, )U, (p 1, )U (1, q 1 )U, (, q 1 )U,, (p, q 1 q 1 )U, (p 1, )U

47 9.3. QUADRATIC RECIPROCITY 47 ˆ Furthermore, note that q 3 q 1 s = 1 q 1 Then s = ( 1) q 1 (q 1)! and 1 = ( 1)(q 1) Z q, = ( 1)(q ) Z q, = ( 1) ( q + 3 ) Z q = ( 1) ( q + 1 ) Z q product up = ( 1) q 1 q + 1 q + 3 (q 1) [( q 1 )!] p 1 = s p 1 = (s ) p 1 = [(q 1)!] p 1 ( 1) q 1 p 1. ˆ Let m be the product of all the elements in (Z p Z q)/u. That is, m = q 1 p 1 i=1 j=1 (i, j)u = ([(p 1)!] q 1, [( q 1 p 1 )!] ) U = ([(p 1)!] q 1, [(q 1)!] p 1 ( 1) q 1 p 1 ) U. ˆ On the other hand, by Chinese Remainder Theorem, there is a group isomorphism θ Z pq Z p Z q. (See the proof of Theorem 8.8.) Let V = θ 1 (U) = {1, 1}. Then Z pq/v γ (Z p Z q)/u. Note that Z pq/v = {kv = {k, k} 1 k pq 1 and gcd (k, pq) = 1}. Therefore, γ 1 (m) = kv = kv Z pq/v 1 k (pq 1)/ gcd (k,pq)=1 kv and m = 1 k (pq 1)/ gcd (k,pq)=1 γ(kv ) = 1 k (pq 1)/ gcd (k,pq)=1 (k, k)u. ˆ We denote the floor function by [r]. Note that [( pq 1 ) /q] = [ pq 1 ] = [ p q 1 q ] = [p p 1 q ] Z, 1 q 1 <1 = p 1.

48 48 CHAPTER 9. THE QUADRATIC RECIPROCITY LAW Similarly, [( pq 1 q 1 ) /p] =. Hence, 1 k (pq 1)/ gcd (k,pq)=1 k = = = (mod p) = Euler s Criterion = p p 3p [( pq 1 p p 3p ( q 1 ( pq 1 )! pq 1 ) /p] p q q 3q [( ) /q] q pq 1! ) p q q 3q ( p 1 ) q ( p 1 i=1 i) ( p 1 i=1 p + i) ( p 1 i=1 ( q 1 1) p + i) ( p 1 i=1 q 1 times q q 3q ( p 1 ) q (p 1)! (p 1)! (p 1)! ( p 1 )! q q 3q ( p 1 ) q [(p 1)!] q 1 ( p 1 )! ( p 1 [(p 1)!] q 1 (q/p) p 1 )!q 1 (q/p) =(q/p) = [(p 1)!] q 1 (q/p) by a similar process = (q 1)! p 1 (p/q). q 1 p + i) Therefore, m = 1 k (pq 1)/ gcd (k,pq)=1 ˆ Compare the two expressions of m. (k, k)u = ((p 1)! q 1 (q/p), (q 1)! p 1 (p/q)) U. ([(p 1)!] q 1 (q/p), [(q 1)!] p 1 (p/q)) U = ([(p 1)!] q 1, [(q 1)!] p 1 ( 1) q 1 p 1 ) U If then If then ([(p 1)!] q 1, [(q 1)!] p 1 ( 1) q 1 p 1 ) 1 ([(p 1)!] q 1 (q/p), [(q 1)!] p 1 (p/q)) U = U ((q/p), [( 1) p 1 q 1 ] 1 (p/q)) U = {(1, 1), ( 1, 1)} (q/p) = 1 and [( 1) p 1 q 1 ] 1 (p/q) = 1, (p/q) = ( 1) p 1 q 1 and (p/q)(q/p) = ( 1) p 1 (q/p) = ( 1) and [( 1) p 1 q 1 ] 1 (p/q) = ( 1), (p/q) = ( 1)( 1) p 1 q 1 q 1. and (p/q)(q/p) = ( 1) p 1 q 1.

49 9.4. QUADRATIC CONGRUENCES WITH COMPOSITE MODULI 49 Corollary 9.5. If p and q are distinct odd primes, then (p/q)(q/p) = { 1 if p 1 (mod 4) or q 1 (mod 4) 1 if p q 3 (mod 4) Corollary 9.6. If p and q are distinct odd primes, then (p/q) = { (q/p) if p 1 (mod 4) or q 1 (mod 4) (q/p) if p q 3 (mod 4) Theorem If p 3 is an odd prime, then (3/p) = { 1 if p ±1 (mod 1) 1 if p ±5 (mod 1) 9.4 Quadratic Congruences with Composite Moduli Theorem If p is an odd prime and gcd (a, p) = 1, then the congruence has a solution if and only if (a/p) = 1. x a (mod p n ), n 1 Proof. ( ) x a (mod p n ) x a (mod p). ( ) Suppose that (a/p) = 1. Then x a (mod p) has a solution. Use induction on n. Suppose that x a (mod p n ) has a solution x n for n 1. Then x n = a + cp n for some c Z. p a x n = a + cp n p x n gcd (x n, p) = 1 p is odd gcd (x n, p) = 1 ( c) the linear Diophantince equation (x n )y c (mod p) has a solution y there exists y such that p (x n y + c) We show that x n + yp n is a solution of x n a (mod p n+1 ). Indeed, (x n + yp n ) = x n + x n yp n + y p n = (a + cp n ) + x n yp n + y p n = a + (c + x n y)p n + y p n a (mod p n+1 ).