Basic Number Theory

Basic Number Theory • Divisibility Let a,b be integerswith a≠0. if there exists an integer k such that b=ka, we say a divides b which is denoted by a|b 11|143, 1993|3980021 ◇ if a≠0, then a|0 and a|a; 1|b for each b a|b and b|c → a|c a|b and a|c → a|sb+tc for all s, t

Prime Numbers • An integer p>1 that is divisible only by 1 and itself is called a prime number, otherwise it iscalled composite (P.64) • primegen.c generates prime numbers • Let π(x) be the number of primes less than x, then π(x) ≈x/ln(x) as x→∞ • Exercise Plot π(x) vs. x for x=216 to 232

A Plot of π(x)≈x/ln(x) vs. x

Greatest Common Divisor gcd • gcd(343, 63)=7, gcd(12345,11111)=1 gcd(1993,3980021)=1993 • Euclidean Algorithm to compute gcd(a,b) does not require the factorization of the numbers and is fast. • gcd(482,1180)=2

Solving ax+by=1 when gcd(a,b)=1 • Let a,b be integers with a2 +b2 ≠0, and gcd(a,b)=1, then ax+by=1 has an integer solution (x,y) ♪ Euclidean Algorithm • Example 7(-2) + 5(3) =1 • Solving ax+by=d with gcd(a,b)=d can be reduced as solving • a0x + b0y = 1 where a=a0d, b=b0d

Congruences • Let a,b,n be integers with n≠0. We say that a≡b (mod n) {read as a is congruent to b mod n} if n|(a-b) a=b+nk for an integer k is another description • Example 32≡7 (mod 5)

Simple Properties • Let a,b,c,n be integers with n≠0 (1) a≡0 (mod n) iff n|a (2) a≡a (mod n) (3) a≡b (mod n) iff b≡a (mod n) (4) a≡b and b≡c (mod n) → a≡c (mod n) (5) a≡b and c≡d (mod n) → a+c≡b+d, a−c≡b−d, ac≡bd (mod n) (6) ab≡ac (mod n) with n≠0, and gcd(a,n)=1, then b≡c (mod n)

Computational Properties • Finding a-1 (mod n) • Solving ax≡c (mod n) when gcd(a,n)=1 • What if gcd(a,n)>1 ☺Solve 11111x≡4 (mod 12345) ☻Solve 12x≡21 (mod 39) ♫ How to solve x2 ≡a (mod n)? □ Working with fractions (inverse ?)

The Chinese Remainder Theorem • Let m1,m2, …, mk be integers with gcd(mi,mj) = 1, there exists only one solution x (mod m1 m2…mk) to the simultaneous congruences [P.76-78] x≡a1 (mod m1) x≡a2 (mod m2) : : x≡ak (mod mk)

Fermat's Little Theorem • How to fast evaluate 21234 (mod 789)? • How to fast evaluate Xa (mod n)? • If p is a prime and gcd(p,a)=1, then ap-1 ≡ 1 (mod p)

Euler’s φ-Function and Theorem • φ(n)= #{a | 1 ≤ a ≤ n, gcd(a,n)=1}, that is, the number of positive integers which are relatively prime to n Examples: φ(15)=8, φ(16)=8, φ(17)=16 φ(pq)=(p-1)(q-1) if p and q are primes φ(p)=p-1 if p is a prime number φ(pr)=pr-pr-1=pr(1- 1/p) • If gcd(a,n)=1, then aφ(n)≡ 1 (mod n)

Examples and Basic Principle • [Page 82] • What are the last three digits 7803 ? • Compute 243210 (mod 101) • Let a,n,x,y be integers with n≥1 and gcd(a,n)=1. If x≡y (mod φ(n)), then ax ≡ ay (mod n) (Hint) x=y+kφ(n); by Euclidean Theorem

Primitive Roots If p is a prime, a primitive root mod p is a number g whose power yield every nonzero class mod p. {gk|0<k<p}={1,2,…,p-1} Proposition: Let g be a primitive root mod p • gn≡1 (mod p) iff (p-1)|n or n≡0 (mod p-1) • gj≡gk (mod p) iff j≡k (mod p-1) ♪ 3 is a primitive root mod 7 but not for mod 13

Inverting Matrices (mod n) • A matrix M is invertible under (mod n) if gcd(det(M), n)=1 • The inverse of A=[1 2;3 4] (mod 11) is A-1 =[9 1 ; 7 5] and det(A)= -2≡9 (mod 11) • The inverse of M=[1 1 1; 1 2 3; 1 4 9] under (mod 11) is [3 3 6; 8 4 10; 1 4 6], where det(M)= ½ ≡ 6 (mod 11)

Square Roots mod n (1/9) • X2 ≡71 (mod 77) has solutions ±15, ±29 • How to (efficiently) solve X2 ≡b (mod pq), where p,q are (very close) primes? • Every prime p (except 2) must satisfy p≡1 (mod 4) or p≡3 (mod 4) • The square roots of 5 mod 11 are ±4

Square Roots mod n (2/9) • Let p≡3 (mod 4) be prime and y is an integer such that x≡y(p+1)/4 (mod p). ♪ If y has a square root mod p, then the square roots of y mod p are x and –x ♪ If y has no square roots mod p, then –y has a square root mod p, and the square roots of –y are x and –x.

Square Roots mod n (3/9) Proof: x4 ≡yp+1≡y2 . yp-1 ≡y2 (mod p) → (x2 + y) (x2 - y) ≡ 0 (mod p) Suppose both y and –y are squares mod p This is impossible.

Square Roots mod n (4/9) • Lemma: Let p ≡ 3 (mod 4) be prime, then X2 ≡ -1 (mod p) has no solutions. Proof: Let p = 4q+3 X2 ≡ -1→ Xp-1 ≡ -1(p-1)/2≡ -12q+1 ≡-1 But Xp-1 ≡ 1 (Fermat’s theorem)

Square Roots mod n (5/9) • Suppose both y and –y are squares mod p, say y ≡ a2 and -y ≡ b2. Then (a/b)2 ≡ -1 (mod p) But according to the previous lemma, (a/b)2 ≡ -1 (mod p) is impossible

Square Roots mod n (6/9) • y ≡ x2 (mod p), the square roots of y are ± x. • -y ≡ x2 (mod p), the square roots of -y are ± x.

Examples for Square Roots (7/9) • x2 ≡ 5 (mod 11) • (p+1)/4 = 3 • x≡ 53 ≡ 4(mod 11) • Since 43 ≡ 5 (mod 11), the square root of 5 mod 11 are ±4

Examples for Square Roots (8/9) ◎ To solve x2≡ 71 (mod 77) • x2≡ 1 (mod 7) → x ≡±1 (mod 7) • x2≡ 5 (mod 11) → x ≡±4 (mod 11) By Chinese remainder theorem x ≡±15 , x ≡±29 (mod 77)

Square Roots mod n (9/9) • Suppose n=pq is the product of two primes congruent to 3 mod 4 (type 4k+3), and let y with gcd(y,n)=1 has a square root mod n. Then finding the four solutions x=±a, ±b to x2≡ y (mod n) is computationally equivalent to factoring n which is regarded as extremely difficult when n is large, say n has a length of 256 bits or higher

Group Theory • Let G be a nonempty set and let ⊕ be a binary operation defined on GxG. G is said to be a group if • For any elements a,b in G, a⊕b is in G • (a⊕b)⊕c=a⊕(b⊕c) for any a,b,c in G • There exists a unit element e such that e⊕a=a⊕e for any a in G • For each a in G, there exists an inverse a-1 such that a-1⊕a=a⊕a-1 = e

Field (Informal Definition) • (F, +,‧) is a nonempty set F with two binary operations +, ‧such that (1) (F,+) is a commutative group with unit element 0 (2) (F’, ‧) is a commutative group with unit element 1, where F’=F\{0} (3) a‧(b+c)=(a‧b) + (a‧c) for any a,b,c

Examples Groups • (Z,+) is a group, Z is the set of all integers • Zp ={0, 1, 2, …, p-1} with + under (mod p) • Zp-1={1,2,…,p-1} with x under (mod p) Fields • (R,+,*) • (Zp,+,x) under (mod p)

Finite Fields with Applications • A field with finite elements • Suppose we need to work in a field whose range is 0 to 28-1 • Z256={0,1, ‥‥, 255} is not a field since 256 is not a prime GF(4)={0,1, ω, ω2} • Zp (p is prime) • GF(pn) (p is prime)

Galois Field GF(pn) • Z2[X] be the set of polynomials whose coefficients are integers mod 2. e.g., X+1, X6+X3+1 are in this set • GF(pn) has pn elements, where p is prime • Zp[X] mod an irreducible polynomial whose degree is pn. • GF (28) = Z2[X] (mod X8+X4+X3+X+1)

Galois Field • For every power pn of a prime p, there is exactly one finite field with pn elements • It can be proved that two fields with pn elements constructed by two different polynomials of degree n are isomorphic

Multiplication of GF(2n) • (X7+ X6 + X3 + X + 1)(X)=? (mod X8+ X4 + X3 + X + 1) • 11001011 b7=1 • Left shift one bit, we have b6 b5 b4 b3b2 b1 b00 = 10010110 • ?=110010110 + 100011011 = 10001101 =X7+X3+X2+1

Linear Feedback Shift Register • Xn+4 ≡ Xn + Xn+1 (mod 2) A recurrence Eq. • If the initial values are X0 X1 X2 X3 = 1101, • The sequence is 1101011110001001101... • Associated with the recurrence Eq. is • X4 +X+1 which is irreducible (mod 2) • The k-th bit can be obtained by • Xk(1+X+X3) (mod X4 +X+1) for k≧4

Basic Number Theory