Cryptography Basics I Assignment

Running head: CRYPTOGRAPHY BASICS
CRYPTOGRAPHY BASICS
Name of the Student
Name of the University
Author Note

CRYPTOGRAPHY BASICS
1
Table of Contents
RSA..................................................................................................................................................2
Feige-Fiat-Shamir scheme...............................................................................................................4
Diffie-Hellman protocol..................................................................................................................5

CRYPTOGRAPHY BASICS
2
RSA
1. Solution
Given public key (143, 11)
Here n = 143, e = 11
We know n = p x q
Hence the factor of the 143 = 11, 13
Now ∅ ( n ) =( p−1)×( q−1) [11, 13 one of them is p or q]
=> ∅ ( n ) =10 ×12=120
Now for private key
=> d ×e=1 mod ∅ ( n )
=> d ×11=1 mod 120
Using Chinese remainder theorem
Multiply both side by 11
=> d (11 ×11)=1 1mod 120
=> d=11mod 120
Hence the d = 11, and private key pair is (143,11) Ans.
2. Solution
Key is given (143, 11)