1. CSCE 715: Network Systems Security Chin-Tser Huang huangct@cse.sc.edu University of South Carolina

2. 02/04/20092 RSA Invented by Rivest, Shamir & Adleman of MIT in 1977 Best known and widely used public-key scheme Based on exponentiation in a finite (Galois) field over integers modulo a prime exponentiation takes O((log n) 3 ) operations (easy) Use large integers (e.g. 1024 bits) Security due to cost of factoring large numbers factorization takes O(e log n log log n ) operations (hard)

3. 02/04/20093 RSA Key Setup Each user generates a public/private key pair by select two large primes at random: p, q compute their system modulus n=p·q note ø(n)=(p-1)(q-1) select at random the encryption key e where 1<e<ø(n), gcd(e,ø(n))=1 solve following equation to find decryption key d e·d=1 mod ø(n) and 0≤d≤n publish their public encryption key: KU= {e,n} keep secret private decryption key: KR= {d,n}

4. 02/04/20094 RSA Usage To encrypt a message M: sender obtains public key of receiver KU={e,n} computes: C=M e mod n, where 0≤M<n To decrypt the ciphertext C: receiver uses its private key KR={d,n} computes: M=C d mod n Message M must be smaller than the modulus n (cut into blocks if needed)

5. 02/04/20095 Why RSA Works Euler's Theorem: a ø(n) mod n = 1 where gcd(a,n)=1 In RSA, we have n=p·q ø(n)=(p-1)(q-1) carefully chosen e and d to be inverses mod ø(n) hence e·d=1+k·ø(n) for some k Hence : C d = (M e ) d = M 1+k·ø(n) = M 1 ·(M ø(n) ) k = M 1 ·(1) k = M 1 = M mod n

6. 02/04/20096 RSA Example: Computing Keys 1. Select primes: p=17, q=11 2. Compute n=pq=17×11=187 3. Compute ø(n)=(p–1)(q-1)=16×10=160 4. Select e: gcd(e,160)=1 and e<160  choose e=7 5. Determine d: de=1 mod 160 and d<160  d=23 since 23×7=161=1×160+1 6. Publish public key KU={7,187} 7. Keep secret private key KR={23,187}

7. 02/04/20097 RSA Example: Encryption and Decryption Given message M = 88 ( 88<187 ) Encryption: C = 88 7 mod 187 = 11 Decryption: M = 11 23 mod 187 = 88

8. 02/04/20098 Exponentiation Use a property of modular arithmetic [(a mod n)  (b mod n)]mod n = (a  b)mod n Use the Square and Multiply Algorithm to multiply the ones that are needed to compute the result Look at binary representation of exponent Only take O(log 2 n) multiples for number n e.g. 7 5 = 7 4 ·7 1 = 3·7 = 10 (mod 11) e.g. 3 129 = 3 128 ·3 1 = 5·3 = 4 (mod 11)

9. 02/04/20099 RSA Key Generation Users of RSA must: determine two primes at random - p,q select either e or d and compute the other Primes p,q must not be easily derived from modulus n=p·q means p,q must be sufficiently large typically guess and use probabilistic test Exponents e, d are multiplicative inverses, so use Inverse algorithm to compute the other

10. 02/04/200910 Security of RSA Four approaches to attacking RSA brute force key search (infeasible given size of numbers) mathematical attacks (based on difficulty of computing ø(n), by factoring modulus n) timing attacks (on running of decryption) chosen ciphertext attacks (given properties of RSA)

11. 02/04/200911 Factoring Problem Mathematical approach takes 3 forms: factor n=p·q, hence find ø(n) and then d determine ø(n) directly and find d find d directly Currently believe all equivalent to factoring have seen slow improvements over the years as of May-05 best is 200 decimal digits (663 bits) with LS biggest improvement comes from improved algorithm cf “Quadratic Sieve” to “Generalized Number Field Sieve” to “Lattice Sieve” 1024+ bit RSA is secure barring dramatic breakthrough ensure p, q of similar size and matching other constraints

12. 02/04/200912 Timing Attacks Developed in mid-1990’s Exploit timing variations in operations e.g. multiplying by small vs large number Infer operand size based on time taken RSA exploits time taken in exponentiation Countermeasures use constant exponentiation time add random delays blind values used in calculations

13. 02/04/200913 Chosen Ciphertext Attacks RSA is vulnerable to a Chosen Ciphertext Attack (CCA) attackers chooses ciphertexts and gets decrypted plaintext back choose ciphertext to exploit properties of RSA to provide info to help cryptanalysis can counter with random pad of plaintext or use Optimal Asymmetric Encryption Padding (OAEP)

14. 02/04/200914 Key Management Asymmetric encryption helps address key distribution problems Two aspects distribution of public keys use of public-key encryption to distribute secret keys

15. 02/04/200915 Distribution of Public Keys Four alternatives of public key distribution Public announcement Publicly available directory Public-key authority Public-key certificates

16. 02/04/200916 Public Announcement Users distribute public keys to recipients or broadcast to community at large E.g. append PGP keys to email messages or post to news groups or email list Major weakness is forgery anyone can create a key claiming to be someone else’s and broadcast it can masquerade as claimed user before forgery is discovered

17. 02/04/200917 Publicly Available Directory Achieve greater security by registering keys with a public directory Directory must be trusted with properties: contains {name, public-key} entries participants register securely with directory participants can replace key at any time directory is periodically published directory can be accessed electronically Still vulnerable to tampering or forgery

18. 02/04/200918 Public-Key Authority Improve security by tightening control over distribution of keys from directory Has properties of directory Require users to know public key for the directory Users can interact with directory to obtain any desired public key securely require real-time access to directory when keys are needed

19. 02/04/200919 Public-Key Authority

20. 02/04/200920 Public-Key Certificates Certificates allow key exchange without real- time access to public-key authority A certificate binds identity to public key usually with other info such as period of validity, authorized rights, etc With all contents signed by a trusted Public- Key or Certificate Authority (CA) Can be verified by anyone who knows the CA’s public key

21. 02/04/200921 Public-Key Certificates

22. 02/04/200922 Distribute Secret Keys Using Asymmetric Encryption Can use previous methods to obtain public key of other party Although public key can be used for confidentiality or authentication, asymmetric encryption algorithms are too slow So usually want to use symmetric encryption to protect message contents Can use asymmetric encryption to set up a session key

23. 02/04/200923 Simple Secret Key Distribution Proposed by Merkle in 1979 A generates a new temporary public key pair A sends B the public key and A’s identity B generates a session key K s and sends encrypted K s (using A’s public key) to A A decrypts message to recover K s and both use

24. 02/04/200924 Problem with Simple Secret Key Distribution An adversary can intercept and impersonate both parties of protocol A generates a new temporary public key pair {KU a, KR a } and sends KU a || ID a to B Adversary E intercepts this message and sends KU e || ID a to B B generates a session key K s and sends encrypted K s (using E’s public key) E intercepts message, recovers K s and sends encrypted K s (using A’s public key) to A A decrypts message to recover K s and both A and B unaware of existence of E

25. 02/04/200925 Next Class Key exchange Diffie-Hellman key exchange protocol Elliptic curve cryptography Read Chapters 11 and 12