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Cryptography Fundamentals: Fermat's Little Theorem and Digital Signatures

Explore solutions to cryptographic problems using Fermat's Little Theorem and induction, uncovering practical aspects and demonstrating techniques like the Digital Signature Algorithm. Learn to find inverses using modular arithmetic and verify signatures with key parameters.

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Cryptography Fundamentals: Fermat's Little Theorem and Digital Signatures

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  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

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