Discrete Logarithm(s) (DLs)

1. Discrete Logarithm(s) (DLs) • Fix a prime p. Let a, b be nonzero integers (mod p). The problem of finding x such that ax ≡ b (mod p) is called the discrete logarithm problem. Suppose that n is the smallest integer such that an ≡1 (mod p), i.e., n=ordp(a). By assuming 0≤x<n, we denote x=La(b), and call it the discrete log of b w.r.t. a (mod p) • Ex: p=11, a=2, b=9, then x=L2(9)=6

2. Discrete Logarithms • In the RSA algorithms, the difficulty of factoring a large integer yields good cryptosystems • In the ElGamal method, the difficulty of solving the discrete logarithm problem yields good cryptosystems • Given p, a, b, solve ax ≡ b (mod p) • a is suggested to be a primitive root mod p

3. One-Way Function • A function f(x) is called a one-way function if f(x) is easy to compute, but, given y, it is computationally infeasible to find x with y=f(x). • La(b) is a one-way function if p is large

4. Primitive Roots mod 13 • a is a primitive root mod p if {ak | 1≦k≦p-1} = {1,2, …,p-1} ♪ 2, 6,7,11 are primitive roots mod 13 • 33 ≡ 1 (mod 13), 46 ≡ 1 (mod 13), • 54 ≡ 1 (mod 13), 84 ≡ 1 (mod 13), • 93 ≡ 1 (mod 13), 106 ≡ 1 (mod 13), • 122 ≡ 1 (mod 13)

5. Solve ax ≡ b (mod p) • An exhaustive search for all 0 ≤ x < p • Check only for even x or odd x according to b(p-1)/2 ≡ (ax)(p-1)/2 ≡(a(p-1)/2)x ≡(-1)x≡ 1 or -1 (mod p), where a is a primitive root (Ex) p=11, a=2, b=9, since b(p-1)/2 ≡95≡1, then check for even numbers {0,2,4,6,8,10} only to find x=6 such that 26 ≡ 9 (mod 11)

6. Solve ax ≡ b (mod p) by Pohlig-Hellman Let p-1 = Πqr for all q|(p-1), write b0 =b,and x=x0 + x1q+x2q2 + … + xr-1qr-1 for 0 ≤ xi ≤ q-1 1. Find 0≤ k ≤q-1 such that (a(p-1)/q)k≡b(p-1)/q , then x0 ≡k, next let b1≡b0a-x0 2. Find 0≤ k ≤q-1 such that (a(p-1)/q)k≡[b1](p-1)/q^2 , then x1 ≡k, next let b2≡b1a-x1 3. Repeat steps 1, 2 until xr-1 is found for a q 4. Repeat steps 1~3 for all q’s, then apply Chinese Remainder Theorem to get the final solution

7. 7x ≡12 (mod 41); p=41, a=7, b=12, • p-1=41-1=40 =23 5 • b0 =12 • For q=2: b0 =12, b1 =31, b2=31, and x = x0 +2x1+4x2 ≡1+2·0+4·1≡ 5 (mod 8) • For q=5: b0 =12, b1 =18, and x = x0 ≡ 3 (mod 5) Solving x ≡ 5 (mod 8) andx≡ 3 (mod 5), We have x≡13 (mod 40)

8. Solve ax ≡ b (mod p) by Index Calculus Let B be a bound and let p1,p2,…, pm be the primes less than B and cover all of the prime Factors of p-1. Then appropriately choose k(j)’s such that ak(j)≡(p1)r1(p2)r2… (pm)rm,i.e., r1*La(p1)+r2*La(p2)+… + rm*La(pm) ≡k(j) for several j’s, solve the linear system to get La(p1), La(p2), … , La(pm), then select R apply baR≡(p1)b1 (p2)b2… (pm)bm , then the solution is La(b)≡-R+ΠbiLa(pi)

9. Solve 2x ≡37 (mod 131) p=131, a=2, b=37, let B=10, then p1=2, p2=3, p3=5, p4=7, since 28≡53 , 212≡5·7, 214≡32 , 234≡3·52 (mod p), we have 3L2(5)≡ 8 (mod 130) L2(5)+ L2(7)≡12 (mod 130) 2L2(3)≡14 (mod 130) L2(3)+2L2(5)≡34 (mod 130)

10. L2([3, 5, 7])=[72, 46, 96] Choose R=43, then 37·243 ≡3·5·7 (mod 131), so we have L2(37) ≡-43+ L2(3)+ L2(5)+ L2(7) ≡ 41 (mod 130) ♪ L2(11) ≡ 56 (mod 130) [R=4] ♪ L2(23) ≡ 23 (mod 130) [R=5]

11. A Lemma on p≡3 (mod 4) Let p≡3 (mod 4), r≥2. Suppose a and g are nonzero integers such that g≡ay(2^r) (mod p). Then g(p+1)/4 ≡ ay[2^(r-1)] (mod p) [Proof] g(p+1)/4 ≡ a(p+1)y[2^(r-2)] ≡ay(2^(r-1))[a(p-1)]y(2^(r-2)) ≡ ay(2^(r-1)) (mod p)

12. A La(b) (mod 4) Machine • Let a be a primitive root (mod p), where p≡3 (mod 4) is large, then Computing La(b) (mod 4) is as difficult as finding the solution of ax ≡ b (mod p) [P.172]

13. The ElGamal Public Key Cryptosystem Alice wants to send a message m to Bob. Bob chooses a large prime p and a primitive root a. Assume m is an integer 0≤m<p, and Bob selects a secret integer x to compute b≡ax (mod p). The information (p,a,b) is made public and is Bob’s public key. Alice does the following procedures.

14. Encryption and Decryption • Downloads (p,a,b) • Chooses a secret random k and computes r≡ak (mod p) • Computes t≡bkm (mod p) • Sends the pair (t,r) to Bob Bob decrypts by computing tr-x (≡m (mod p))

15. Exercises on Pages 175 and 176