Professor Note about Network Security Fundamentals

Question: 1. In a public-key system using RSA, you intercept the ciphertext C =10 sent to a user whose public key is e=5, n=35. What is the plaintext M?2. In an RSA system, the public key of a given user is e=31, n=3599. What is the plaintext M?. Answer: 1. In a public-key system using RSA, you intercept the ciphertext C =10 sent to a user whose public key is e=5, n=35. What is the plaintext M? Cipher text C = 10 Public key PU = {e, n} = {5, 35} Therefore, f(n) = 24. It can be assume that two prime numbers are p = 5 and q = 7. Therefore, n = p*q = 5*7=35 f(n) = (p-1)(q-1)=(5-1)(7-1)=4*6=24 gcd(f(n), e) = gcd(24, 5) =1, 1 e f(n) Therefore, d e-1 mod f(n) equivalent to ed 1 mod f(n) or ed mod f(n) =1 mod 24 =1, and d = 5 because {e = 5, f(n) = 24. So, 5d} Now, the private key PR = {d, n} = {5, 35}. On the basis of RSA decryption algorithm, M = Cd mod n = 105 mod 35 = 5 Therefore, C = Me mod n = 55 mod 35 = 10 Therefore, we conclude that the plaintext M is 5. 2. In an RSA system, the public key of a given user is e=31, n=3599. What is the Private Key of the user? Since p and q must both be prime and n = 3599 = 15961, it can be assume that p=59 and q=61, and: (n) = (p-1)(q-1) = (59-1)(61-1) = 3480 d = (1 + k (n))/e = (1 + 3480k)/31 = -13921/31 = -449 (for k = -4) d = -449 (mod 3480) = 3031 (mod 3480) Bibliography Rani, S., Mittal, H. and Director, S.P.G.O.I., 2015. A Compound Algorithm Using Neural and AES for Encryption and Compare it with RSA and existing AES. Journal of Network Communications and Emerging Technologies (JNCET) www. jncet. org, 3(1). Gao, J., Cheng, H., Yang, Z. and Ding, Q., 2013, May. The research and design of embed RSA encryption algorithm network encryption card driver. In Sensor Network Security Technology and Privacy Communication System (SNS PCS), 2013 International Conference on (pp. 83-87). IEEE.