Assignment #3 Solutions

1. Assignment #3Solutions January 24, 2006

2. Problem #1 Use Fermat’s Little Theorem and induction on k to prove that xk(p–1)+1 mod p = x mod p for all primes p and k ≥ 0. Practical Aspects of Modern Cryptography

3. Answer #1 By induction on k … Base case k = 0: xk(p–1)+1 mod p = x0+1 mod p = x mod p Base case k = 1: xk(p–1)+1 mod p = x(p-1)+1 mod p = xp mod p = x mod p (by Fermat’s Little Theorem) Practical Aspects of Modern Cryptography

4. Answer #1 (cont.) Inductive step: Assume that xk(p–1)+1 mod p = x mod p. Prove that x(k+1)(p–1)+1 mod p = x mod p. Practical Aspects of Modern Cryptography

5. Answer #1 (cont.) x(k+1)(p–1)+1 mod p = xk(p–1)+(p-1)+1 mod p = xk(p–1)+1+(p-1) mod p = xk(p–1)+1x(p-1) mod p = xx(p-1) mod p (by inductive hypothesis) = xp mod p = xp mod p (by Fermat’s Little Theorem) Practical Aspects of Modern Cryptography

6. Problem #2 Show that for distinct primes p and q, x mod p = y mod p x mod q = y mod q together imply that x mod pq = y mod pq. Practical Aspects of Modern Cryptography

7. Answer #2 x mod p = y mod p  (x mod p) – (y mod p) = 0  (x – y) mod p = 0 (by first assignment)  (x – y) is a multiple ofp. Similarly x mod q = y mod q  (x – y) is a multiple ofq. Practical Aspects of Modern Cryptography

8. Answer #2 (cont.) Therefore, (x – y) is a multiple ofpq  (x – y) mod pq = 0  (x mod pq) – (y mod pq) = 0 x mod pq = y mod pq. Practical Aspects of Modern Cryptography

9. Problem #3 Put everything together to prove that xK(p–1)(q-1)+1 mod pq = x mod pq For K ≥ 0 and distinct primes p and q. Practical Aspects of Modern Cryptography

10. Answer #3 Let k1=K(q–1) and k2=K(p–1). xK(p–1)(q-1)+1 mod p = xk1(p–1)+1 mod p = x mod p and xK(p–1)(q-1)+1 mod q = xk1(q–1)+1 mod q = x mod q By Problem #1, and then by Problem #2 xK(p–1)(q-1)+1 mod pq = x mod pq. Practical Aspects of Modern Cryptography

11. Problem #4 E(x) = x43 mod 143 Find the inverse function D(x) = xd mod 143. Practical Aspects of Modern Cryptography

12. Answer #4 143 = 1113 We need to find d such that 43d mod (11–1)(13–1) = 1. Use the Extended Euclidean Algorithm to find a solution to find x and y such that 120x + 43y = 1. Practical Aspects of Modern Cryptography

13. Extended Euclidean Algorithm Given A,B > 0, set x1=1, x2=0, y1=0, y2=1, a1=A, b1=B, i=1. Repeat while bi>0: {i = i + 1; qi = ai-1 div bi-1; bi = ai-1-qibi-1; ai = bi-1; xi+1=xi-1-qixi; yi+1=yi-1-qiyi}. For alli: Axi + Byi = ai. Finalai = gcd(A,B). Practical Aspects of Modern Cryptography

14. Answer #4 (cont.) Practical Aspects of Modern Cryptography

15. Answer #4 (cont.) Practical Aspects of Modern Cryptography

16. Answer #4 (cont.) Practical Aspects of Modern Cryptography

17. Answer #4 (cont.) Practical Aspects of Modern Cryptography

18. Answer #4 (cont.) Practical Aspects of Modern Cryptography

19. Answer #4 (cont.) Practical Aspects of Modern Cryptography

20. Answer #4 (cont.) Practical Aspects of Modern Cryptography

21. Problem #5 Digital Signature Algorithm Public parameters: q = 11, p = 67, g = 9, y = 62 Private secret: x = 4 Message to be signed: M = 8 Selected random parameter: k = 2 Practical Aspects of Modern Cryptography

22. The Digital Signature Algorithm To sign a 160-bit message M, • Generate a random integer k with 0 < k < q, • Compute r = (gk mod p) mod q, • Compute s = ((M+xr)/k) mod q. The pair (r,s) is the signature on M. Practical Aspects of Modern Cryptography

23. Answer #5 • r = (gk mod p) mod q = (92 mod 67) mod 11 = (81 mod 67) mod 11 = 14 mod 11 = 3 • s = ((M+xr)/k) mod q = ((8+43)/2) mod 11 = (20/2) mod 11 = 10 mod 11 = 10 The pair (3,10) is the signature on 8. Practical Aspects of Modern Cryptography

24. The Digital Signature Algorithm A signature (r,s) on M is verified as follows: • Compute w = 1/s mod q, • Compute a = wM mod q, • Compute b = wr mod q, • Compute v = (gayb mod p) mod q. Accept the signature only if v = r. Practical Aspects of Modern Cryptography

25. Answer #5 (cont.) • w = 1/s mod q = 1/10 mod 11 = 10 • a = wM mod q = 108 mod 11 = 3 • b = wr mod q = 103 mod 11 = 8 • v = (93628 mod 67) mod 11 = (5915 mod 67) mod 11 = 14 mod 11 = 3 v = 3 andr = 3 so the signature is validated. Practical Aspects of Modern Cryptography