Mohak_Bheda_HW4.pdf - Homework 4 ECE/CS 5560 MOHAK BHEDA(906207830 Part 1 Chapter 9 Problem 9.3 Solution To find the Plaintext M We know M = cd mod 77 d

Mohak_Bheda_HW4.pdf - Homework 4 ECE/CS 5560 MOHAK...

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Homework 4 ECE/CS 5560 MOHAK BHEDA (906207830) Part 1 Chapter 9 Problem 9.3 Solution: To find the Plaintext M. We know, M = mod 77 c d d and e are multiplicative inverses in mod , (77) = 60 (77) Φ Φ Now, e = 13 D = 37 M = ( = 48 ) mod 77 20 37
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Problem 9.7 Solution: No, it’s not safe for Bob to generate new public and private key without generating a new modulus. Let’s see an example. We know that e*d = mod (n). Φ This way, the new keys can be easily extracted from the modulus, as the value of n is equal to n = pq. Additionally, consider the scenario as below, Let k= ed – 1. Then k is congruent to 0 mod φ(N) (where 'φ' is the Euler totient function). Select a random x in the multiplicative group Z(N). Then x^k≡ 1 mod N, which implies that x^(k/2) is a square root of 1 mod N. With 50% probability, this is a nontrivial square root of N, so that gcd(x^(k/2) – 1,N) will yield a prime factor of N. If x^(k/2) = 1 mod N, then try x^(k/4), x^(k/8), etc... This will fail if and only if x^(k/2^i)≡ –1 for some i.
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