The RSA Public-Key Encryption Algorithm

1. The RSA Public-Key Encryption Algorithm Network Security

2. The RSA Public-Key Encryption Algorithm Objectives of the Topic • After completing this topic, a student will be able to • explain working ofthe RSA Public-Key encryption algorithm.

3. The RSA Public-Key Encryption Algorithm Figures and material in this topic have been adapted from • “Network Security Essentials : Applications and Standards”, 2014, by William Stallings.

4. The RSA Public-Key Encryption Algorithm • RSA is the best known, and widely used general public key encryption scheme. • It was first published by Rivest, Shamir & Adleman of MIT in 1978.

5. The RSA Public-Key Encryption Algorithm • The RSA scheme is a cipher in which the plaintext and ciphertext are integers between 0 and n - 1 for some n. • A typical size for n is 1024 bits, or 309 decimal digits. That is, n is less than 21024.

6. The RSA Public-Key Encryption Algorithm • RSA is based on exponentiation in a finite (Galois) field over integers modulo a prime. • Security due to cost of factoring large numbers.

7. The RSA Public-Key Encryption Algorithm Description • Plaintext is encrypted in blocks, with each block having a binary value less than some number n. • The block size must be less than or equal to log2(n) + 1; in practice, the block size is i bits, where 2i< n ≤2i+1.

8. The RSA Public-Key Encryption Algorithm • Encryption and decryption are of the following form, for some plaintext block M and ciphertext block C. • C = Me mod n • M = Cd mod n = (Me)d mod n = Med mod n

9. The RSA Public-Key Encryption Algorithm • Both sender and receiver must know the value of n. • The sender knows the value of e, and only the receiver knows the value of d.

10. The RSA Public-Key Encryption Algorithm • Here, the public key of PU = {e, n} and the private key of PR = {d, n}. • For this algorithm to be satisfactory for public-key encryption, following requirements must be met:

11. The RSA Public-Key Encryption Algorithm • 1. It is possible to find values of e, d, and n such that Med mod n = M for all M < n. • 2. It is relatively easy to calculate Me mod n and Cd mod n for all values of M < n. • 3. It is infeasible to determine d given e and n.

12. The RSA Public-Key Encryption Algorithm Summary

13. The RSA Public-Key Encryption Algorithm Example • Key Generation • 1. Select two prime numbers, p= 17 and q= 11. • 2. Calculate n = pq = 17 × 11 = 187. • 3. Calculate ⱷ(n) = (p-1)(q-1)= 16 × 10 =160.

14. The RSA Public-Key Encryption Algorithm • 4. Select e such that e is relatively prime to ⱷ(n) = 160 and less than ⱷ(n); we choose e=7. • 5. Determine d such that de mod 160 =1 and d < 160.The correct value is d=23, because 23 × 7=161= (1 × 160)+1.

15. The RSA Public-Key Encryption Algorithm • The resulting keys are public key PU={7, 187} and private key PR= {23, 187}. • Encryption • Lets use these keys for a plaintext input of M = 88. • Here, we need to calculate C= 887mod 187.

16. The RSA Public-Key Encryption Algorithm • Decryption • We need to calculate M = 1123mod 187 • Above expressions can be evaluated by exploiting the properties of modular arithmetic.

17. The RSA Public-Key Encryption Algorithm The Security of RSA • There are two possible approaches to defeating the RSA algorithm. • The first is the brute-force approach: Try all possible private keys.

18. The RSA Public-Key Encryption Algorithm • Thus, the larger the number of bits in e and d, the more secure the algorithm. • However, because the calculations involved are complex, the larger the size of the key, the slower the system will run.

19. The RSA Public-Key Encryption Algorithm • Most discussions of the cryptanalysis of RSA have focused on the task of factoring n into its two prime factors. • For a large n with large prime factors, factoring is a hard problem.

20. The RSA Public-Key Encryption Algorithm • A large key size such as a 1024-bit key size (about 300 decimal digits) is considered strong enough for virtually all applications. End