Question

1. (a) Explain the terms “data encryption, authentication, and message integrity,” often used in the networks security literature. (3 Points) (b) Lorenzo likes to send to his close friend Art a secret market data related to their business using public key cryptography (RSA algorithm). He chooses two prime numbers 7 and 11, and a public key e = 13 to encrypt the data. Art uses d=37 to decrypt the data. Indicate why (e, 77) and (d, 77) are valid public and private keys. (2 Points) (c) Show how the number 5 is encrypted and decrypted using the above keys. (3 Points)

Answer 1

1.

a)

Data Encryption - In computer networks, when the sender sends data to the receiver, there can be a chance that the data can be exposed or tampered by elements on the path of the data transfer. So, in order to safely transfer data from the sender to the receiver, the data is transformed to a different format, so that the attacker cannot obtain the actual data. This method is called data encryption. In data encrpytion, the source data is called the plaintext and the encrypted data is called the ciphertext.

Authentication - When a user tries to access a server , the communication is authenticated with the help of username and password. The method of identifying an user by a server or the concerned body is called authentication. Suppose, a person A wants to access his mail account. Another person B knows the mailid of the person A but not the password. When, person B tries to login, he can't access the mail account because he does not know the password. In order to have authentication of the user A, the user requires to fill the mail id as well as the password. With the help of this, the server becomes sure that the mail is accessed from the authenticated user.

Integrity - It is the characteristic of network communication which ensures that the data transmitted from the sender to the receiver is not altered or tampered in the middle of the transmission.Suppose, a teacher sends a mail to the student that there will be a cultural programme and everyone has to be present. Now, if someone tampers the message in the middle and modifies it by mentioning that the particular day is a holiday, then the student won't go to see the programme. This violates the integrity property as the message was altered in the middle of the communication.

b)

The 2 prime numbers used in RSA are :

p = 7 and q = 11.

n = p*q = 7*11 = 77

Φ(n) = (p-1)*(q-1) = 6*10 = 60.

The public key e = 13 and the private key d = 37.

These 2 are valid public and private key because according to the RSA algorithm,

( e*d ) mod Φ(n) should be equal to 1.

After putting the values of e, d and Φ(n), the respective value is :

(13*37) mod 60 = 481 mod 60.

This leaves the remainder as 1 and the quotient as 8 because 481 = 8*60 + 1.

So, it is proved that (13*37) mod 60 = 481 mod 60 = 1.

So, (e, 77) and (d,77) that is (13, 77) and (37, 77) are valid public and private keys.

c)

Given,

For encryption,

The plaintext is P = 5.

e = 13, d = 37 and n = 77.

Let the Ciphertext be C.

C = (P^e) mod n

= (5^13 ) mod 77

Using the method of modular exponentiation :

i	( 5^i ) mod 77	Output
1	5 mod 77	5
2	(5^2) mod 77	25
4	(5^4) mod 77 = ((5^2) mod 77 ) * (5^2) mod 77 )) mod 77	9
8	(5^8) mod 77 = ((5^4) mod 77 ) * (5^4) mod 77 )) mod 77	4
12	(5^12) mod 77 = ((5^8) mod 77 ) * (5^4) mod 77 )) mod 77	36
13	(5^13) mod 77 = ((5^1) mod 77 ) * (5^12) mod 77 )) mod 77	26

So,

Ciphertext C = (5^13) mod 77

= 26.

So, the number 5 is encrypted as the number 26.

For decryption,

The ciphertext is C = 5.

e = 13, d = 37 and n = 77.

Let the plaintext be P.

P = (C^d) mod 77

= (5^37) mod 77

Using the method of modular exponentiation :

i	( 5^i ) mod 77	Output
1	5 mod 77	5
2	(5^2) mod 77	25
4	(5^4) mod 77 = ((5^2) mod 77 ) * (5^2) mod 77 )) mod 77	9
8	(5^8) mod 77 = ((5^4) mod 77 ) * (5^4) mod 77 )) mod 77	4
16	(5^16) mod 77 = ((5^8) mod 77 ) * (5^8) mod 77 )) mod 77	16
32	(5^32) mod 77 = ((5^16) mod 77 ) * (5^16) mod 77 )) mod 77	25
36	(5^36) mod 77 = ((5^32) mod 77 ) * (5^4) mod 77 )) mod 77	71
37	(5^37) mod 77 = ((5^36) mod 77 ) * (5^1) mod 77 )) mod 77	47

So, the plaintext P = (5^37) mod 77

= 47.

So, the number 5 is decrypted as the number 47.