basic number theory divisibility let a,b be integers with a≠0. if there exists an integer k such...
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Basic Number Theory
• Divisibility
Let a,b be integers with 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 is called 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
Prime Factorization Theorem
• Every positive integer is a product of primes. This factorization into primes is unique, up to reordering the factors
• 49500=22 32 5311
• If a prime p|ab, then either p|a or p|b Moreover, p|x1 x2 … xn →p|xj for some j
• 7|14•30,
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, the
n 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 nExamples: φ(15)=8, φ(16)=8, φ(17)=16φ(pq)=(p-1)(q-1) if p and q are primesφ(p)=p-1 if p is a prime numberφ(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(1) gn≡1 (mod p) iff (p-1)|n or n≡0 (mod p-1)(2) gj≡gk (mod p) iff j≡k (mod p-1) ♪ 3 is a primitive root mod 7 but not for mo
d 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)
2. y ≡ x2 (mod p), the square roots of y are ± x.
3. -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)
(1) x2≡ 1 (mod 7) → x ≡±1 (mod 7)
(2) 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
(1) For any elements a,b in G, a b is in G⊕(2) (a b) c=a (b c) for any a,b,c in G⊕ ⊕ ⊕ ⊕(3) There exists a unit element e such that e
a=a e for any a in G⊕ ⊕(4) 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 bina•ry 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≧