CSIS 5857: Encoding and Encryption
Assignment 3: Public Key Encryption
Due April 3
1. Problem 9.3 (page 280) in the textbook.
We first factor the modulus n into the primes 5 and 7.
This gives a totient value of 4 * 6 = 24.
We then find the private key d as the inverse of 5 mod 24. This is also 5, as 5 * 5 = 25
= 1 mod 24.
We then find the plaintext as 105 mod 35 = 5.
2. Problem 9.8 (page 280) in the textbook.
This is vulnerable to a short message attack. Since there are only 26 letters, all an
adversary has to do is to encrypt them all in advance using the public key, and then
compare the individual ciphertext letters to those values.
3. Problem 9.15 (page 282) in the textbook.
This is vulnerable to a form of man-in-the-middle attack, as all of the information
about the sender and recipient are outside of the encryption.
1) Darth intercepts the message [A, E(PUB, M), B] sent by A.
2) Darth then substitutes his own id to create [D, E(PUB, M), B], making B think the
message is actually from Darth. This would not be possible if the message was in the
form [D, E(PUB, [M, A]), B], as B would realize that D is not the same as A after
decrypting the message [M, A].
3) B then re-encrypts the message using Darth’s key PUD, sending [B, E(PUD, M),
D], which Darth can then decrypt to obtain the message.
4. Problem 10.1 (page 310) in the textbook.
a) 75 mod 71 = 51
b) 712 mod 71 = 4
c) 45 mod 71 = 5112 mod 71 = 30