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QUESTION

QUESTION. Question 10 (Pg. 160) Supposed two users Alice & Bob have the same RSA modulus n and suppose that their encryption exponents eA and eB are relatively prime. Charles wants to send the message m to Alice and Bob, so he encrypts to get cA ≡ m eA and cB ≡ m eB (mod n).

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QUESTION

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  1. QUESTION Question 10 (Pg. 160) Supposed two users Alice & Bob have the same RSA modulus n and suppose that their encryption exponents eA and eB are relatively prime. Charles wants to send the message m to Alice and Bob, so he encrypts to get cA ≡ meA and cB ≡ meB (mod n). Show how Eve can find m if she intercepts cA and cB.

  2. Given Information • Alice use <n, eA> • Bob use <n, eB> • Gcd(eA, eB) = 1 (Means eA and eB are relatively prime) Encryption Exponents Common RSA modulus

  3. Objective cA ≡ meA (mod n) cB ≡ meB (mod n) Objective : Get rid of the eA & eB.

  4. Getting rid of the eA & eB.(1) • Use the Euclidean Algorithm to compute x & y, where: x•eA + y•eB = 1 x & y can be found. (one of them will be negative.)

  5. Getting rid of the eA & eB.(2) x•eA + y•eB = 1, x & y found. cA ≡ meA (mod n) cB ≡ meB (mod n) (cA)x•(cB)y = meA•x•meB•y (mod n) = meA•x + eB•y (mod n) = m (mod n) (m is obtained by Eve) (Since cA and cB have a common modulus n.)

  6. Example • p=23, q=37 n = 851 • M = 88 • eA = 13 • eB = 11 • cA = 88^13 (mod 851) = 421 • cB = 88^11 (mod 851) = 45

  7. Example • n = 851 • eA = 13 Public • eB = 11 • cA = 421 Ciphertexts • cB = 45

  8. Example • n = 851 13x + 11y = 1 • eA = 13 x = -5 • eB = 11 y = 6 • cA = 421 • cB = 45 Use Euclidean Algorithm ((4216)*(455))(mod 851) = 88 = M Note: Since x is negative, use –x in place of x, and swap x & y.

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