cs555spring 2012/topic 61 cryptography cs 555 topic 6: number theory basics

CS555 Spring 2012/Topic 6 1

CryptographyCS 555

Topic 6: Number Theory Basics


Outline and Readings

• Outline– Divisibility, Prime and composite

numbers, The Fundamental theorem of arithmetic, Greatest Common Divisor, Modular operation, Congruence relation

– The Extended Euclidian Algorithm – Solving Linear Congruence

• Readings:• Katz and Lindell: 7.1.1, 7.1.2


Divisibility (cont.)

Theorem (Division algorithm)Given integers a, b such that a>0, a<b then there exist two unique integers q and r, 0 r < a s.t. b = aq + r.

Proof:Uniqueness of q and r: assume q’ and r’ s.t b = aq’ + r’, 0 r’< a, q’ integerthen aq + r=aq’ + r’ a(q-q’)=r’-r q-q’ = (r’-r)/a as 0 r,r’ <a -a < (r’-r) < a -1 < (r’-r)/a < 1So -1 < q-q’ < 1, but q-q’ is integer, therefore q = q’ and r = r’


Prime and Composite Numbers

DefinitionAn integer n > 1 is called a prime number if its positive divisors are 1 and n.

DefinitionAny integer number n > 1 that is not prime, is called a composite number.

ExamplePrime numbers: 2, 3, 5, 7, 11, 13, 17 …Composite numbers: 4, 6, 25, 900, 17778, …


Decomposition in Product of Primes

Theorem (Fundamental Theorem of Arithmetic)Any integer number n > 1 can be written as a product of prime numbers (>1), and the product is unique if the numbers are written in increasing order.

Example: 84 = 2237

ekk

ee pppn ...221

1


Classroom Discussion Question (Not a Quiz)

• Are the total number of prime numbers finite or infinite?


Greatest Common Divisor (GCD)

Definition Given integers a > 0 and b > 0, we define gcd(a, b) =

c, the greatest common divisor (GCD), as the greatest number that divides both a and b.

Example gcd(256, 100)=4

Definition Two integers a > 0 and b > 0 are relatively prime if gcd(a, b) = 1.

Example 25 and 128 are relatively prime.


GCD as a Linear Combination TheoremGiven integers a, b > 0 and a > b, then d = gcd(a,b) is the least positive integer that can be represented as ax + by, x, y integer numbers.

Proof: Let t be the smallest positive integer s.t. t = ax + by.We have d | a and d | b d | ax + by, so d | t, so d t.We now show t ≤ d. First t | a; otherwise, a = tu + r, 0 < r < t; r = a - ut = a - u(ax+by) = a(1-ux) + b(-uy), so we foundanother linear combination and r < t. Contradiction.Similarly t | b, so t is a common divisor of a and b, thus t ≤

gcd (a, b) = d. So t = d.Example

gcd(100, 36) = 4 = 4 100 – 11 36 = 400 - 396


GCD and Multiplication

TheoremGiven integers a, b, m >1. Ifgcd(a, m) = gcd(b, m) = 1, then gcd(ab, m) = 1

Proof idea:ax + ym = 1 = bz + tmFind u and v such that (ab)u + mv = 1


Finding GCD

Using the Theorem: Given integers a>0, b, q, r, such that b = aq + r, then gcd(b, a) = gcd(a, r).

Euclidian Algorithm

Find gcd (b, a)

while a 0 do

r b mod a

b a

a r

return b

Q ui ckTi me™ and a TI FF (Uncompressed) decompressor are needed to see thi s pi cture.


Euclidian Algorithm Example

Find gcd(143, 110)

gcd (143, 110) = 11

143 = 1 110 + 33110 = 3 33 + 1133 = 3 11 + 0


Modulo Operation

rnqaqrna s.t. , mod

where 0 rn 1

Definition:

Example:7 mod 3 = 1-7 mod 3 = 2


Congruence Relation

Definition: Let a, b, n be integers with n>0, we say that a b (mod n),

if a – b is a multiple of n.Properties: a b (mod n) if

and only if n | (a – b) if and only if n | (b – a) if and only if a = b+k·n for some integer k if and only if b = a+k·n for some integer k

E.g., 327 (mod 5), -1237 (mod 7), 1717 (mod 13)


Properties of the Congruence Relation

Proposition: Let a, b, c, n be integers with n>01. a 0 (mod n) if and only if n | a2. a a (mod n)3. a b (mod n) if and only if b a (mod n)4. if a b and b c (mod n), then a c (mod n)Corollary: Congruence modulo n is an equivalence

relation.Every integer is congruent to exactly one number in

{0, 1, 2, …, n–1} modulo n


Equivalence Relation

DefinitionA binary relation R over a set Y is a subset of Y Y. We denote a relation (a,b) R as aRb.•example of relations over integers?

DefinitionA relation is an equivalence relation on a set Y, if R is

Reflexive: aRa for all a R Symmetric: for all a, b R, aRb bRa . Transitive: for all a,b,c R, aRb and bRc aRc

Example“=“ is an equivalence relation on the set of integers


More Properties of the Congruence Relation

Proposition: Let a, b, c, n be integers with n>0

If a b (mod n) and c d (mod n), then:

a + c b + d (mod n),

a – c b – d (mod n),

a·c b·d (mod n)

E.g., 5 12 (mod 7) and 3 -4 (mod 7), then, …


Multiplicative Inverse

Definition: Given integers n>0, a, b, we say that b is a multiplicative inverse of a modulo n if ab 1 (mod n).

Proposition: Given integers n>0 and a, then a has a multiplicative inverse modulo n if and if only if a and n are relatively prime.


Towards Extended Euclidian Algorithm

• Theorem: Given integers a, b > 0, then d = gcd(a,b) is the least positive integer that can be represented as ax + by, x, y integer numbers.

• How to find such x and y?


The Extended Euclidian Algorithm

First computes

b = q1a + r1

a = q2r1 + r2

r1 = q3r2 + r3

rk-3 =qk-1rk-2+rk-1

rk-2 = qkrk-1

Then computes

x0 = 0

x1 = 1

x2 = -q1x1+x0

xk = -qk-1xk-1+xk-2

And

y0 = 1

y1 = 0

y2 = -q1y1+y0

yk = -qk-1yk-1+yk-2

We have axk + byk = rk-1 = gcd(a,b)


Extended Euclidian Algorithm

Extended_Euclidian (a,b)

x=1; y=0; d=a; r=0; s=1; t=b;

while (t>0) {

q = d/t;u=x-qr; v=y-qs; w=d-qt;

x=r; y=s; d=t;

r=u; s=v; t=w;

}

return (d, x, y)

end

ax + by = d

ar + bs = t

Invariants:


Another Way

Find gcd(143, 111)

gcd (143, 111) = 1

143 = 1 111 + 32111 = 3 32 + 1532 = 2 15 + 215 = 7 2 + 1

32 = 143 1 111 15 = 111 3 32 = 4111 3 143 2 = 32 2 15 = 7 143 9 111 1 = 15 - 7 2 = 67 111 – 52 143


Linear Equation Modulo n

If gcd(a, n) = 1, the equation

has a unique solution, 0< x < n. This solution is often represented as a-1 mod n

Proof: if ax1 1 (mod n) and ax2 1 (mod n), then a(x1-x2) 0 (mod n), then n | a(x1-x2), then n | (x1-x2), then x1-x2=0

How to compute a-1 mod n?

ax 1 mod n


Examples

Example 1:• Observe that 3·5 1 (mod 7). • Let us try to solve 3·x+4 3 (mod 7). • Subtracts 4 from both side, 3·x -1 (mod 7). • We know that -1 6 (mod 7). • Thus 3·x 6 (mod 7). • Multiply both side by 5, 3·5·x 5·6 (mod 7). • Thus, x 1·x 3·5·x 5·6 30 2 (mod 7).• Thus, any x that satisfies 3·x+4 3 (mod 7) must satisfy

x 2 (mod 7) and vice versa.Question: To solve that 2x 2 (mod 4). Is

the solution x1 (mod 4)?


Linear Equation Modulo (cont.)

To solve the equation

When gcd(a,n)=1, compute x = a-1 b mod n.When gcd(a,n) = d >1, do the following

• If d does not divide b, there is no solution.• Assume d|b. Solve the new congruence, get x0

• The solutions of the original congruence are x0, x0+(n/d), x0+2(n/d), …, x0+(d-1)(n/d) (mod n).

nbax mod

) (mod /)/( dndbxda


Solving Linear Congruences

Theorem: • Let a, n, z, z’ be integers with n>0. If gcd(a,n)=1,

then azaz’ (mod n) if and only if zz’ (mod n).• More generally, if d:=gcd(a,n), then azaz’ (mod

n) if and only if zz’ (mod n/d).

Example:• 5·2 5·-4 (mod 6)• 3·5 3·3 (mod 6)

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

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 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

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


• 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.


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.


• 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)


• 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


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)


• 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 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≧

cs555spring 2012/topic 61 cryptography cs 555 topic 6: number theory basics

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