Practice Assignment 2 Discussion 24/02/ /02/2018

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1 German University in Cairo Faculty of MET (CSEN 1001 Computer and Network Security Course) Dr. Amr El Mougy 1 RSA 1.1 RSA Encryption Practice Assignment 2 Discussion 24/02/ /02/2018 Perform encryption using the RSA algorithm, for the following: 1. p = 3; q = 11, e = 7; M = 5 Solution: n = p q = 3 11 = 33 C = M e mod n = 5 7 mod 33 = p = 5; q = 17, e = 3; M = 9 Solution: n = p q = 5 17 = 85 C = M e mod n = 9 3 mod 85 = p = 7; q = 5, e = 17; M = 8 Solution: n = p q = 7 5 = 35 C = M e mod n = 8 17 mod 35 = 8 2 RSA Decryption Perform decryption using the RSA algorithm, for the following: 1. p = 11; q = 13, e = 11; C = 106 Solution: n = p q = = 143 Φ(n) = (p 1)(q 1) = = 120 Now we have e = 11 and Φ(n) = 120, we know that 11 d = 1 + k 120,

2 120 = 11(10) = 10(1) = 11 10(1) Now substitute the first equation into 10: 1 = 11 (120 11(10)) Note that this is a linear combination of 120 and 11, after simplifying we get: We get k = 1 and d = 11 1 = (10) 1 = 11(1 + 10) = 11(11) = 11(11) M = C d mod n = mod 143 = (106 4 mod mod mod 143) mod 143 = ( ) mod 143 = 7 2. p = 17; q = 31, e = 7; C = 128 Solution: n = p q = = 527 Φ(n) = (p 1)(q 1) = = 480 Now we have e = 7 and Φ(n) = 480, we know that 7 d = 1 + k 480, 480 = 7(68) = 4(1) = 3(1) = 4 3(1) Now substitute the second equation into 3: 1 = 4 (7 4(1)) Page 2

3 Then we substitute the first equation into every instance of 4: 1 = (480 7(68)) (7 (480 7(68))(1)) Note that this is a linear combination of 480 and 7, after simplifying we get: 1 = 480 7(68) (68) 1 = 480(2) 7(137) 1 480(2) = 7(137) ( 2) = 7( 137) We get k = 2 and d = 137 which is in fact 343 mod 480 since = 343 so d = 343 M = C d mod n = mod 527 = (( mod 527) mod mod mod mod 527) mod 527 = ( ) mod 527 = 2 3. 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? Solution: By trial and error we try to find two prime numbers whose multiplication is equal to 35, we get; we then calculate Φ(n) p = 7 q = 5 Φ(n) = (p 1)(q 1) = 6 4 = 24 Now we have e = 5 and Φ(n) = 24, we know that 5 d = 1 + k 24, 24 = 5(4) = 4(1) = 5 4(1) Page 3

4 Now substitute the first equation into 4: 1 = 5 (24 5(4)) Note that this is a linear combination of 24 and 5, after simplifying we get: We get k = 1 and d = 5 1 = (4) 1 = 5(1 + 4) 24 1 = 5(5) = 5(5) M = C d mod n = 10 5 mod 35 = 5 4. p = 7; q = 11, e = 7; C = 59 Solution: n = p q = 7 11 = 77 Φ(n) = (p 1)(q 1) = 6 10 = 60 Now we have e = 7 and Φ(n) = 60, we know that 7 d = 1 + k 60, 60 = 7(8) = 4(1) = 3(1) = 4 3(1) Now substitute the second equation into 3: 1 = 4 (7 4(1))(1) Now substitute the first equation into every instance of 4: 1 = (60 7(8)) (7 (60 7(8))(1))(1) Note that this is a linear combination of 60 and 7, after simplifying we get: Page 4

5 1 = 60 7(8) (8) 1 = 60(2) 7( ) 1 = 60(2) 7(17) 1 60(20) = 7(17) ( 2) = 7( 17) We get k = 2 and d = 17 which is in fact 43 mod 60 since = 60 so d = 43 M = C d mod n = mod 77 = ((59 5 mod 77) mod 77) mod 77 = In an RSA system, Alices public key is e, n = 5, 851. Discover the corresponding private key. Solution: By trial and error we try to find two prime numbers whose multiplication is equal to 851, we get; we then calculate Φ(n) p = 23 q = 37 Φ(n) = (p 1)(q 1) = = 792 Now we have e = 5 and Φ(n) = 792, we know that 5 d = 1 + k 792, Now substitute the first equation into 2: 792 = 5(158) = 2(2) = 5 2(2) 1 = 5 2(792 5(158)) Note that this is a linear combination of 792 and 5, after simplifying we get: We get k = 2 and d = = 5 792(2) + 5(316) 1 = 5( ) 792(2) 1 = 5(317) 792(2) (2) = 5(317) Page 5

6 3 Diffie-Hellman 1. Users A and B use the Diffie-Hellman key exchange technique with a common prime q = 71 and a primitive root a = 7. (i) If user A has private key X A = 5, what is A s public key Y A? (ii) If user B has private key X B = 12, what is B s public key Y B? (iii) What is the shared secret key? Solution: (i) Y A = a X A mod q = 7 5 mod 71 = 51 (ii) Y B = a X B mod q = 7 12 mod 71 = 4 X (iii) K = Y A B mod q = 4 5 mod 71 = This problem illustrates the point that the Diffie-Hellman protocol is not secure without the step where you take the modulus; i.e. the Indiscrete Log Problem is not a hard problem! You are Eve and have captured Alice and Bob and imprisoned them. You overhear the following dialog. Bob: Oh, lets not bother with the prime in the Diffie-Hellman protocol, it will make things easier. Alice: Okay, but we still need a base a to raise things to. How about a = 3? Bob: All right, then my result is 27. Alice: And mine is 243. What is Bob s secret X B and Alice s secret X A? What is their secret combined key? Solution: Since there is no modulus operation involved here. This simply reduces to getting the log directly. So Bob s secret can be computed as follows: Similarly for Alice s: Then the secret key K is: X B = log a Y B = log 3 27 = 3 X A = log a Y A = log = 5 Y B X A = Y A X B = 27 5 = = Alice and Bob wish to exchange a secret key using the Diffie-Hellman algorithm. They agree to use the prime number 71 and its primitive root 7. Alice chooses the private key 6 while Bob chooses 12. To their misfortune, Eve, intercepts the public keys sent by Alice and Bob and executes a Man In The Middle (or rather Woman In the Middle) attack to trick them into believing they have exchanged a key with each other, while in reality they would have exchanged two keys with Eve. Execute Eves Page 6

7 attack and obtain the secret keys shared with Alice and Bob. You may assume Eve chooses the private keys 8 and 14. Solution: Having q = 71, a = 7, X A = 6 and X B = 12, we calculate the public keys of both Alice and Bob as follows; Y A = a X A mod q = 7 6 mod 71 = 2 Y B = a X B mod q = 7 12 mod 71 = (7 11 mod 71 7 mod 71) mod 71 = (31 7) mod 71 = 4 We can also calculate Eves public key with Alice and Eves public key with Bob knowing that her private key with Alice is X C = 8 and her private key with Bob is X D = 14; Y C = a X C mod q = 7 8 mod 71 = 27 Y D = a X D mod q = 7 14 mod 71 = (7 11 mod mod 71) mod 71 = (31 59) mod 71 = 54 We now can use these public keys to calculate Eves common key with Alice K AC and her common key with Bob K BD ; K AC = Y A X C mod q = 2 8 mod 71 = 43 K BD = Y B X D mod q = 4 14 mod 71 = 5 4. In the man-in-the-middle attack on the Diffie-Hellman key exchange protocol the adversary generates two public private key pairs for the attack. Could the same attack be accomplished with one pair? Explain. Solution: Yes, it is possible. Consider the following scenario. A and B want to share a secret key using the Diffie-Hellman protocol, while Eve is the attacker. q and a are known as usual to the three entities. 1. Eve chooses X E and generates Y E = a X E mod q 2. A chooses X A and generates Y A = a X A mod q 3. B chooses X B and generates Y B = a X B mod q 4. A sends B his public key Y A. However, Eve intercepts the message and sends B her public key Y E instead. 5. B receives Y E and computes the presumed common key with A; X K BE = Y B E mod q 6. B now sends A his public key Y B. However, Eve intercepts the message and sends A her public key Y E instead. Page 7

8 7. A receives Y E and computes the presumed common key with B; X K AE = Y A E mod q 8. Eve has both public keys for A and B; Y A and Y B respectively. So she computes both shared keys K AE = Y A X E mod q and K BE = Y B X E mod q Now K AE is the common shared key between Eve and A, while K BE is the common shared key between Eve and B. Page 8