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## Chapter 2 Introduction to Number Theory and Its applications

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**Chapter 2Introduction to NumberTheory and Its applications**Cheng-Chia Chen October 2002**outline**• Division • Prime • Gcd and Lcm • Modular Arithmetic • Chinese Remainder Theorem • Fermat’s little theorem • The RSA algorithm**Division**Def: a,b Z with a ≠ 0. • We say a divides b (written a | b) if 9k Z s.t. b = ka • a | b => • a is a factor (or divisor) of b and • b is a multiple of a. • Ex: • 3 | 12 ( * 12 = 4 x 3 ) • -4 | 8, • 13 | 0 (0 = 0 x 13) • 3 - 7**Properties of |**• a | b /\ a |c ) a | b + c • a | b ) a | bc for all c Z • | is reflexive ( a | a for all a Z ) • | is transitive ( a | b /\ b | c ) a | c ) • pf: a | b /\ b | c ) • b = k1 a and c = k2 b for some k1, k2Z • ) c = k2 (k1 a) = (k1 k2) a • | is antisymmetric ( a | b /\ b | a ) a = b) • Any relation satisfying 3,4,5 is called a partial order**Primes**• An integer p > 1 is said to be prime if • n N+ ( p | n ! ( n = 1 \/ n = p ). • I.e., the only positive factors of p are 1 and p. • p > 1 is not prime => P is composite. • Examples: • 7 is prime • primes < 20 include : 2,3,5,7,11,13,17,19.**The fundamental theorem of arithmetic (FTA)**• n N+ > 1, there exists a unique increasing sequence of primes p1 ≤ p2 ≤ … ≤ pk ( k ≥ 0) s.t. n = p1 x p2 … x pk. • Ex: • 100 = 2 x 2 x 5 x 5 • 99 = 3 x 3 x 3 x 37.**Proof of FTA**• ( Existence) by Math Ind. • Basis: n = 1, 2 ok. • Ind. n = i + 1. • if n is prime, then n = p1, where p1 = n and k = 1 ). • if n is not prime then n = n1 x n2 with n1,n2 < n. • => by ind. hyp. n1 = q1 x q2 … x qt • n2 = r1 x r2 … rs • => n = n1 x n2 = q1 x … x qt x r1 x … x rs. • => n = p1 x … x ps+t. where p1,…,ps+t is an increasing reordering of q1,…,qt and r1,…,rt. • Uniqueness: • let n = p1 x … x pk x q1 x … x qs • = p1 x … x pk x r1 x … x rt where q1 ≠ r1 • => n – n = p1 x … x pk x (q1 x … x qt – r1 x … rt) • ≠ 0 ( a contradiction !!).**Theorem 3**• If n is composite => 9 a ≤ s.t. a | n. pf: n is composite => n = p x q with p, q > 1. if p > /\ q > => p q > = n. a contradiction Hence n must have a factor ≤ Example: 101 is a prime. pf: x y = 10. But no prime ≤ 10 is a factor of 101.**The division algorithm**• a Z, d N+ i q,r s.t. a = qd + r where 0 ≤ r < d. Def: if a = dq + r Then • d is called the divisor(除數) • a : dividend(被除數) • q: quotient(商數) • r: remainder(餘數) • Examples: • 101 = 11 ∙ 9 + 2 • -11 = -4 ∙ 3 + 1 • Note: d | a iff r = 0.**Proof of the division algorithm**Consider the sequence : … a-3d, a-2d, a-d, a, a-(-d), a-(-2d), a-(-3d), … • Let r = a – qd be the smallest nonnegative number in the sequence. 1. since the sequence is strictly increasing toward infinity such q (and r) must exist and unique. 2. if r ≥ d r’ =r-d =a – (q+1) d ≥ 0 is another nonnegative number in the sequence smaller than r. That’s a contradiction. Hence r must < d. QED**gcd and lcm**• a,b Z, ab ≠ 0. if d | a and d | b d is a common divisor of a and b. • gcd(a,b) =def the greatest common divisor of a and b. Note: The set cd = {x > 0 |, x | a and x | b} is a finite subset of N+ (∵ {1} cd {1,… min(a,b)} gcd(a,b) must exist. • Example: • gcd(24,36) = ? • factors of 24 : 1,2,3,4,6,12,24 • factors of 36: 1,2,3,4,6,9,12,18,36 • cd(24,36) = {1,2,3,4,6,12} • gcd(24,36) = 12.**Relatively prime**• If gcd(a,b) = 1 we say a and b are relatively prime(r.p.). • Ex: gcd(17,22) = 1. • a1,a2,…an are pairwise r.p. if gcd(ai,aj) = 1 for all 1 ≤ i < j ≤ n. • Ex: • 10,17,21 are p.r.p. • 10,19,24 are not p.r.p since gcd(10,24) = 2. • Proposition 1: If a = p1x1 p2x2 … pnxn , b = p1y1 p2y2 … pnyn, where p1 < p2 …< pn are primes and all xi, yj ≥ 0, then gcd(a,b) = s =def p1z1 p2z2 … pnzn where zi = min(xi,yi) for all 0 ≤ i ≤ n.**The proof**1. s cd(a,b). • what are the quotients of a and b when divided by s ? 2. t t cd(a,b) t = p1d1 p2d2 … pndn for some d1,…dn with di ≤ xi , di ≤ yi ,and di ≤ zi. • Ex: • 120 = 23 ∙31 ∙51 • 500 = 22 ∙53 • gcd(120,500) = 22 ∙30 ∙51 = 20**lcm ( least common multiplier)**• a,b Z cN+ if a|c and b|c d is a common multiplier of a and b. • lcm(a,b) =def the least common multiplier of a and b. Note: The set cm = {x > 0 |, a|x and b|x} ≠ ∅ (∵ { a∙b} cm lcm(a,b) must exist. Proposition 2: If a = p1x1 p2x2 … pnxn , b = p1y1 p2y2 … pnyn, where p1 < p2 …< pn are primes and all xi, yj ≥ 0, then lcd(a,b) = t =def p1z1 p2z2 … pnzn where zi = max(xi,yi) for all 0 ≤ i ≤ n. Theorem 5: gcd(a,b) ∙ lcm(a,b) = a b.**Modular Arithmetic**Def 8: m N+, a Z. a mod m =def the remainder of a when divided by m. • Ex: • 17 mod 5 = 2 • -133 mod 9 = 2. Def 9: a,b Z, m N+. a ≡ b (mod m) means m | (a-b). • i.e., a and b have the same remainder when divided by m. • i.e., a mod m = b mod m • we say a is congruent to b (module m). • Ex: • 17 ≡ 5 (mod 6) ? • 24 ≡ 14 (mod 6) ?**Properties of congruence**Theorem 6:a ≡ b (mod m) iff a = km + b for some k Z. pf: a ≡ b (mod m) (a-b) = km a = km + b. Theorem 7:If m > 0, a ≡ b (mod m) and c ≡ d (mod m), then (1) a + c ≡ b + d (mod m) (2) ac ≡ bd (mod m). pf: By the premise, a = km + b and c = sm + d for some k,s. a + c = (b + d) + (k + s) m and ac = bd + (kd + sb + skm) m (1) and (2) hold. Ex: 7 ≡ 2 (mod 5), 11 ≡ 1 (mod 5) 18 ≡ 3 and 77 ≡ 2.**The Euclidean Algorithm**Lemma 1: a = bq + r gcd(a,b) = gcd(b,r). pf: it suffices to show that cd(a,b) = cd(b,r). But • d|a /\ d | b d | (a-bq) = r, and • d | b /\ d | r d | bq + r = a. Hence cd(a,b) = cd(b,r). Note: if a = bq + 0 gcd(a,b) = gcd(b,0) = b. • A simple algorithm: gcd(a,b) // a ≥ b ≥ 0. if (b == 0) return a; else return gcd(b, a mod b); Note: this algorithm is very efficient.**gcd(662, 414) = ?**∴ gcd(662,414) = gcd(414,248) = … = gcd(2,0) = 2.**Theorem 1**• a ≥ b ≥ 0 gcd(a,b) = sa + tb for some s,t Z. • i.e., gcd(a,b) is a linear combination of a and b. Pf: By induction on b. Basis: b = 0. gcd(a,b) = a = 1 ∙ a + 0 ∙ b. Inductive case: b > 0. case1: b | a gcd(a,b) = b = 0 a + 1 b. case2: b ∤ a gcd(a,b) = gcd(b,r) where 0 ≤ r = mod(a,b) < b. By I.H. gcd(b,r) = sb + t r. But r = a - bq ∴ gcd(a,b) = gcd(b,r) = sb + tr = sb + t(a – bq) = t a + (s – qt) b. QED**Example**• gcd(252, 198) = 18 = ___∙ 252 + ___ ∙ 198. Sol: Exercise: Let L(a,b) = {sa + tb | s,t Z } is the set of all linear combinations of a and b. Show that gcd(a,b) = the smallest positive number of L(a,b). Hint: 1. By Induction 2. L(a,b) = L(b,r) if a = bq + r.**Lemma 1 and Lemma 2**Lemma 1:gcd(a,b) = 1 /\ a | bc a | c. pf: gcd(a,b) = 1 1 = sa + tb for some s,t Z c = sac + tbc = sac + tka ∵ a | bc = (sc + tk) ∙ a ∴ a | c. Lemma 2’: p : prime /\ p ∤ a gcd(p,a) = 1. Pf: cd(p,a) factors of p = {1,p}. but p is not a factor of a. Hence gcd(p,a) = 1. Lemma 2: p : prime /\ p | a1 a 2 … an p | ai for some i. Pf: By ind. on n. Basis: n = 1. trivial. Ind. case: n = k + 1. p | a1 a 2 … ak a k+1. If p | a1 we are done. O/W p ∤ a1 and gcd(p, a1) = 1 by lem2’. By Lem 1 : p | ( a 2 … ak+1 ) p | ai for some 2 ≤ i ≤ k+1 by IH.**Uniqueness of FTA**Pf: Suppose two distinct sequences p1 , … , ps and q1 , … , qt with n = p1 x … x ps = q1 x … x qt Removing all common primes on both sides : m =def pi1 x … piu = qj1x … x qjv where pi ≠ qj for all pi and qj. pi1 | m = qj1x … x qjv pi1 | qj for some j ( a contradiction!!).**Theorem 2**m > 0 /\ ac ≡ bc (mod m) /\ gcd(m,c) = 1 a ≡ b (mod m). Pf: ac ≡ bc (mod m) m | (ac – bc) = (a – b) c. ∵ gcd(m,c) = 1 ∴ m | (a – b) ∴ a ≡ b (mod m).**Linear Congruence**Ex: Find all x such that 7 x ≡ 2 (mod 5). Def: Equations of the form ax ≡ b (mod m) are called linear congruence equations. Def: Given (a,m), any integer a’ satisfying the condition: a a’ ≡ 1 (mod m) is called the inverse of a(mod m). Proposition:a a’ ≡ 1 (mod m) x = a’ b + km is the general solution of the congruence equation ax ≡ b (mod m) Pf: 1. a’b + km is a solution for any k Z. 2. y is a solution ay ≡ b (mod m) => y ≡ a’b (mod m) => m | (y – a’b) y = a’b + k’ m for some k.**Theorem 3**• m > 0, gcd(a,m) = 1. Then bZ s.t. • 1. ab ≡ 1 (mod m) • 2. if ab ≡ ac [≡ 1] b ≡ c (mod m). Pf: 1. gcd(a,m) = 1. Then b,t with ba + tm =1. since m | ba –1 and hence ab ≡ 1 (mod m). 2. Direct from Theorem 2. Note: Theorem 3 means That the inverse of a mod m uniquely exists (and hence is well defined) if a and m are relatively prime.**Examples**Ex: Find a s.t. 3a ≡ 1 (mod 7). Sol: since gcd(3,7) = 1. the inverse of 3 (mod 7) exists and can be computed by the Euclidean algorithm: 7 = 3 X 2 + 1 1 = 7 + 3 (-2). 3 (-2 ) ≡ 1 (mod 7) a = -2 + 7k for all k Z. EX: Find all solutions of 3x ≡ 4 (mod 7). Sol: -2 is an inverse of 3 (mod 7). Hence x = 4 (-2) + 7k where k Z are all solutions of x.**The Chinese Remainder Theorem**• EX: Find all integer x satisfying the equations simultaneously: • x ≡ 2 (mod 3) • x ≡ 3 (mod 5) • x ≡ 2 (mod 7) • Theorem 4: m1,m2,…,mn : pairwise relatively prime. The system of congruence equations: • x ≡ a1 (mod m1) • x ≡ a2 (mod m2) • … • x ≡ an (mod mn) • has a unique solution modulo m = m1 m2 … mn.**Proof of the Chinese remainder theorem**Pf: Let Mk = m / mk for 1 ≤ k ≤ n. Note: 1. gcd(mk, Mk) = 1 and 2. mi | Mk if i ≠ k. Hence s, yk s.t. s mk + yk Mk = 1. Hence yk is an inverse of Mk mod mk. Now Mk yk ≡ 1 (mod mk) and Mk yk ≡ 0 (mod mj) for all j ≠ k. Let x = a1 M1 y1 + … + an Mn yn then x ≡ a1 M1 y1 + … + an Mn yn ≡ ak Mk yk ≡ ak (mod mk) for all 1 ≤ k ≤ n.**Proof of the uniqueness part**If x and y satisfying the equations, then x-y ≡ 0 (mod mk) for all k = 1..n. => s1,…,sn with x-y = s1 m1 = … = sn mn. since gcd(mi, mk) = 1 for all i ≠ k and mk | s1 m1, we have mk | s1 for all k ≠ 1. Hence s1 is a multiple of m2 m3 … mn and x-y = s1 m1 is a multiple of m = m1 m2 … mk. Hence x ≡ y (mod m). QED**Example**• Find x ≡ (2,3,2) (mod (3,5,7)) respectively. • Sol:**Fermat’s little theorem**• p: prime, a N. Then 1. if (p - a) then a p-1 ≡ 1 (mod p). Moreover, 2. for all a, ap ≡ a (mod p). Ex: 1. p = 17, a = 2 216 = 65536 = 3855 x 17 + 1 216 ≡ 1 (mod 17). 2. p = 3, a = 20 203 – 20 = 8000 –20 = 7980 is a multiple of 3. Hence 203 ≡ 20 (mod 3).**Proof of Fermat’s little theorem**Lemma:1≤i<j≤p-1, ia ≢ ja (mod p) and ia ≢ 0 (mod p). Pf: ia ≡ ja (mod p) p | (j-i) a. Since p - a, p |(j-i). But 0 < j-i < p, p - (j-i), a contradiction. 1. Note the above lemma means ia and ja have different remainders when divided by p. Hence a x 2a x … (p-1) a ≡ 1 x 2 … x (p-1) = (p-1)! (mod p) (p-1)! ap-1 ≡ (p-1) ! (mod p). Then p | (p-1)! (a p-1 –1). ∵ p - (p-1)!, p | ap-1 –1, and hence a p-1 ≡ 1 (mod p). 2. if p | a p | a (ap-1–1) = ap – a ap ≡ a (mod p). if p - a ap-1 ≡ 1 (mod p) ap ≡ a (mod p).**public key**Encryption (加密) Decryption (解密) M C cipher text M’ (plain text) Public key encryption and RSA private key • Public key can be known to the public • Private key is kept secret.**The RSA algorithm**• p.q: two large primes ( > 200 digits, 1024 digits recommended now), • n = pq • e = any number with gcd(e, (p-1)(q-1)) = 1. • d = inverse of e (mod (p-1)(q-1)). (i.e., de ≡ 1 (mod (p-1)(q-1))) public key = (n,e) private key = (n,d) note : public and private keys are symmetric. C = Me (mod n) and M’ = Cd (mod n). Theorem : M’ ≡ M (mod n).**Proof of the correctness of the RSA algorithm**• M’ = Cd ≡ (Me)d ≡ Mde ≡ M1 +k(p-1)(q-1) (mod n) [∵ de ≡ 1 (mod (p-1)(q-1)) ] Assume gcd(M,p) = gcd(M,q) = 1. (i.e., p - M and q - M its probability is (p-1)(q-1)/pq ≈ 1. or we can let M < min(p,q)). Then Cd = M ∙ (M(p-1))k(q-1) ≡ M ∙ 1 k(q-1) (mod p) = M ∙ (M(q-1))k(p-1) ≡ M ∙ 1 k(p-1) (mod q) ( by Fermat’s little theorem) M’ = Cd ≡ M (mod n). ∵ Cd-M is a multiple of p and q**Example**p = 43, q = 59 n = pq = 43 ∙ 59 = 2537. choose e = 13 with gcd(13, (43-1)(59-1)=2436)=1. d = 937 is an inverse of 13 mod 2436. 1. To transmit ‘STOP’=1819 1415 : 2 blocks of length 4. 181913 mod 2537 = 2081, 141513 mod 2537 = 2182 C = 2081 2182. 2. Receive 0981 0461 M’1 = 0981937 (mod 2537) =0704 M’2 = 0461937 (mod 2537) = 1115 M’ = 0704 1115 = ‘HELP’.