1

Prelude to Public-Key Cryptography

Rocky K. C. Chang, February 2014

2

The next 2 sets of slides address

Secret keyfunctions

Hashfunctions

Secrecyservice

Authenticationservice

Messageintegrity service

Nonrepudiationservice

Public keyfunctions

3

Outline

Motivations for public-key cryptography Affine Cipher Generalizing Affine Cipher to multiplicative

groups. Computing the multiplicative inverses using

Euclidean algorithms The Chinese Remainder Theorem Other useful Group Theory results

Multiplication modulo prime Primitive elements

4

Public-key cryptography Drawbacks of the symmetric key

cryptosystems: Require a secret key established before sending

ciphertext. Cannot be used for digital signatures.

Main ideas behind the public-key cryptosystems: It is computationally infeasible to determine DK()

given EK(). Therefore, EK() can be public and DK() must be

private.

5

Public-key cryptography Key people behind the public-key

cryptography: Diffie and Hellman Rivest, Shamir, and Adleman

The RSA algorithm is based on the difficulty of factoring large integers.

ElGamal, Elliptic Curve, and Diffie-Hellman are based on the difficulty of solving the discrete logarithm problem.

6

The Affine Cipher

7

Recall that the Affine Cipher is: Let M = C = Z26 = {0, 1, 2, …, 25} K = (a, b), where a, b {0, 1, 2, …, 25}. Encryption and decryption functions:

EK(m) = am + b mod 26 DK(c) = a-1(c b) mod 26

EK(m) is not an one-to-one function for all a. When a = 1, Affine Cipher is the same as a Shift

Cipher. Affine Cipher is still a special case of the

Substitution Cipher.

8

EK(m) is not an one-to-one function for all a.

Not all (a, b) can be used as keys. E.g., a = 2 and b = 1: E(m) = 2m + 1 mod 26. But E(0) = E(13) = 1.

For any c Z26, the decryption is possible iff the congruence am c (mod 26) has a unique solution for m. Decryption is possible iff there is a unique solution m in

am + b c (mod 26) or am c b (mod 26). Note that b just shifts c to the left hand side by b, which

gives the same set of values for c. Thus, decryption is possible iff there is a unique solution

m in am c (mod 26).

9

The values of a: gcd(a,26) = 1. The congruence am c (mod 26) has a unique

solution for any c Z26 iff gcd(a,26) = 1 (i.e., a and 26 are relative prime). Assume that gcd(a,26) = d > 1.

Without loss of generality, take c = 0. Then am 0 (mod 26) has two solutions: m = 0 and m =

26/d. The congruence does not have a unique solution.

Assume that gcd(a,26) = 1. Consider some m1 and m2 for which am1 am2 (mod 26) or

a(m1m2) 0 (mod 26).

That is, 26 | a(m1m2) (i.e., 26 divides a(m1m2)).

Since gcd(a,26) = 1, we have 26 | (m1m2).

By definition, m1 m2 (mod 26).

Therefore, a unique solution m Z26.

10

What is the size of the key space? How many a Z26 for which gcd(a,26) = 1?

All odd numbers except for 13 (i.e., 12 of them). Thus, the size of the key space = 1226 = 312.

Define a-1 to be the multiplicative inverse of a for which aa-1 a-1a 1 (mod 26).

Inverses of a Z26

a a-1

1 1 3 9 5 21 7 15 9 3 11 19

a a-1

15 7 17 23 19 11 21 5 23 17 25 25

Multiplicative inverses for the set of a for which gcd(a,26) = 1:

11

Multiplicative inverses do not exist for the set of a for which gcd(a,26) ≠ 1.

12

Decryption function c am + b (mod 26) am c b (mod 26) Assuming that the a-1 exists, we have a-1(am)

a-1(c b) (mod 26) The left side is a-1(am) (a-1a)m 1m m

(mod 26). Therefore, m = a-1(c b) mod 26.

13

Multiplicative group

14

Abelian Group or Commutative Group A group G is a set of numbers together with

an operation that satisfies the following requirements: (Closure) For all a, b G, a b G. (Associative) For all a, b, c G, a (b c) = (a b)

c. (Identity) Exists some unique e G such that for all

a G, a e = e a = a. (e is the identity element) (Inverse) For all a G, there exists an a-1 G, such

that a a-1 = a-1 a = e. (a-1 is the inverse of a). (Commutative) For all a, b G, a b = b a.

15

For example, The set of real numbers under addition is a

(additive) group. e = 0 and a-1 = -a.

The set of non-zero real numbers under multiplication is a (multiplicative) group. e = 1 and a-1 = 1/a.

The set of integers under addition is a group, but the set of integers under multiplication is not a group.

Zn = {0, 1, 2, …, n–1} under addition modulo n is a group.

16

Multiplicative group Let Z*

26 = {1, 3, 5, 7, 9, 11, 15, 17, 19, 21, 23, 25} under multiplication modulo 26 forms a group. Z*

26 is the set of residues modulo 26 that are relatively prime to 26.

We can generalize the modulo 26 to any modulo p. am c (mod p) has a unique solution m Zp for

every c Zp iff gcd(a,p) = 1. The number of integers in Zp that are relatively

prime to p is denoted by (p). (26) = ? There is a formula to compute (p).

17

Multiplicative group Suppose a Zp, a-1 exists iff gcd(a,p) = 1.

If a-1 exists, it is unique. It is not difficult to prove that Z*

p forms a group under multiplication modulo p.

As a special case, if p is prime, then every nonzero element of Zp has a multiplicative inverse. Therefore, (p) = p – 1. Z*

p = Zp \ {0}.

How to compute the multiplicative inverse?

Use the Euclidean algorithm to compute gcd(a,b). E.g., gcd(108,42) = gcd(42,24) = gcd(24,18) = gcd(18,6)

= 6. E.g., gcd(75,28) = gcd(28,19) = gcd(19,9) = gcd(9,1) = 1. Can determine whether a positive integer a < p has a

multiplicative inverse modulo p.public static int gcd(int a, int b) { int c; while (a % b != 0) { c = b; // temporarily store b b = a % b; // update b, the second argument a = c; // update a, the first argument } return b; // note that b is the gcd value. } // end gcd

18

19

The Extended Euclidean algorithm Use the Extended Euclidean algorithm to

compute r, s, t, such that sa + tb = r = gcd(a,b).

For example, a = 108, b = 42 (i.e., gcd(a,b) > 1), 108 = 242+24 (24 = a–2b) 42 = 124+18 (b=1(a–2b)+18 or -a+3b=18) 24 = 118+6 (a–2b=1(-a+3b)+6 or 2a–5b=6) 18 = 36+0 Therefore, 2a–5b=6 (s = 2, t = -5, and r = 6).

20

The Extended Euclidean algorithm For example, a = 75, b = 28 (i.e., gcd(a,b) =

1), 75 = 228+19 (19 = a–2b) 28 = 119+9 (b=1(a–2b)+9 or -a+3b=9) 19 = 29+1 (a–2b=2(-a+3b)+1 or 3a–8b=1) 9 = 91+0 Therefore, 3a–8b=1 (s = 3, t = -8, and r = 1).

21

Compute the multiplicative inverse Consider a Zp and gcd(p,a) = 1. From the Extended Euclid. Algorithm, we have

sp + ta = 1. Reducing the above modulo p, we have ta 1

(mod p). In other words, t is the multiplicative inverse of a.

Note that it is also unique. E.g., for a =28 and Z75, a-1 = -8 mod 75 = 67.

Check aa-1 mod 75 = 1876 mod 75 = 1!

22

The Chinese Remainder Theorem

23

The Chinese Remainder Theorem The CRT is a method of solving the followings

for x, where gcd(pi, pj) = 1 for i j. x a1 (mod p1) x a2 (mod p2) … x ar (mod pr),

The CRT asserts that there is a unique solution in {0, 1, …, p1 … pr – 1}.

To see why, consider mapping x to x mod pi (called X).

24

For example, Consider p1 = 5 p2 = 3, P = p1p2 = 15, and x

{0, 1, 2, …, 14}. X(0) = (0,0), X(1) = (1,1), X(2) = (2,2), X(3) = (3,0), X(4) = (4,1), X(5) = (0,2), X(6) = (1,0), X(7) = (2,1), X(8) = (3,2), X(9) = (4,0), X(10) = (0,1), X(11) = (1,2), X(12) = (2,0), X(13) = (3,1), X(14) = (4,2)

The mapping X(x) is bijective => a unique solution to x a1 (mod p1) x a2 (mod p2).

25

The Chinese Remainder Theorem Suppose p1, …, pr are pairwise relatively prime, and

a1, …, ar are integers. Then the system of r congruences x ai (mod pi) has a unique solution modulo P = p1… pr, which is given by x = a1P1y1 mod P + … + arPryr mod P,

where Pi = P/pi and yi = Pi-1 mod pi, i=1, …, r.

For example, (p1,p2,p3) = (7,11,13) and (a1,a2,a3)=(5,3,10). P = 1001. From the Extended Euclid. Algorithm, y1 = 5, y2 = 4, and y3

= 12. From the CRT, x = ( 5(1113)5 + 3(713)4 +

10(711)12 ) mod 1001 = 894.

26

Multiplicative group modulo prime

27

Lagrange’s theorem For a finite multiplicative group G under modulo p,

define The order of G is (p) (i.e., the number of elements in G) The order of an element g G to be the smallest +ve

integer n such that gn mod p = 1. E.g., for Z*

26 = {1, 3, 5, 7, 9, 11, 15, 17, 19, 21, 23, 25}, recall that (p) = 12. The order of 1 is 1. The order of 3 is 3, because 33 mod 26 = 1. The order of 5 is 4, because 54 mod 26 = 1. …

(Lagrange) Suppose G is a multiplicative group of order n, and g G. Then the order of g divides n.

28

Multiplicative group modulo prime From the Lagrange’s theorem, we immediately

have If b Z*

p, then b(p) 1 (mod p). If p is a prime and b Z*

p, then bp b (mod p).

If p is prime, then Z*p is a cyclic group.

There exists at least an element g Z*p having

order equal to (p) = p – 1. Such element is called the primitive element

modulo p. E.g., for Z*

7, 3 is a primitive, because 3i mod 7 1, i=1,…,5, and 37-1 mod 7 = 1.

29

Properties of the primitive elements An element g is a primitive element modulo p iff

gi, i = 0, 1, …, p–2, generate Z*p. E.g., for p = 7

30 mod 7 = 1, 31 mod 7 = 3, 32 mod 7 = 2, 33 mod 7 = 6, 34 mod 7 = 4, 35 mod 7 = 5.

The order of an element a = gi is given by (p–1)/gcd(p–1,i). Thus, a = gi is a primitive element iff gcd(p–1,i) = 1. In other words, the number of primitive elements is (p–

1).

30

For example, For p = 7, p–1 = 6 = 23. Therefore, (6) = (21–

21-1)(31–31-1) = 2. Test for primitive elements:

gcd(6,0) = 6 gcd(6,1) = 1 31 is a primitive element. gcd(6,2) = 2 gcd(6,3) = 3 gcd(6,4) = 2 gcd(6,5) = 1 35 mod 7 = 5 is another primitive

element.

31

A quicker method for testing for primitive elements

Suppose that p is prime and a Z*p. Then a is a

primitive element modulo p iff a(p–1)/q 1 (mod p) for all primes q such that q | (p–1).

Back to p = 7, all primes, for which q | (p–1), are 2 and 3. 1 is clearly not a primitive element. 26/2 1 (mod 7). 36/2 6 (mod 7) and 36/3 2 (mod 7) 3 is a primitive

element. 46/2 1 (mod 7). 56/2 6 (mod 7) and 56/3 4 (mod 7) 5 is a primitive

element.

32

Conclusions We have laid down some foundations for

understanding the public-key cryptography. Affine Cipher Multiplicative groups (Diffie-Hellman) The Chinese Remainder Theorem (RSA) Multiplicative groups modulo prime (Diffie-

Hellman)

33

Acknowledgments The notes are prepared mostly based on

D. Stinson, Cryptography: Theory and Practice, Chapman & Hall/CRC, Second Edition, 2002.