David Evans cs.virginia/evans

1. Lecture 12: Non-secret Key Cryptosystems (How Euclid, Fermat and Euler Created E-Commerce) David Evans http://www.cs.virginia.edu/evans Real mathematics has no effects on war. No one has yet discovered any warlike purpose to be served by the theory of numbers. G. H. Hardy, The Mathematician’s Apology, 1940. CS588: Security and Privacy University of Virginia Computer Science

2. Applications of RSA • Privacy: • Bob encrypts message to Alice using EA • Only Alice knows DA • Signatures: • Alice encrypts a message to Alice using DA • Bob decrypts using EA • Knows it was from Alice, since only Alice knows DA • Things you use every day: ssh, SSL, DNS, ... CS588 Spring 2005

3. Public-Key Applications: Privacy Bob Alice • Alice encrypts message to Bob using Bob’s Private Key • Only Bob knows Bob’s Private Key only Bob can decrypt message Decrypt Ciphertext Encrypt Plaintext Plaintext Bob’s Public Key Bob’s Private Key CS588 Spring 2005

4. Signatures Bob Alice Signed Message Decrypt Encrypt • Bob knows it was from Alice, since only Alice knows Alice’s Private Key • Non-repudiation: Alice can’t deny signing message (except by claiming her key was stolen!) • Integrity: Bob can’t change message (doesn’t know Alice’s Private Key) Plaintext Plaintext Alice’s Public Key Alice’s Private Key CS588 Spring 2005

5. Public-Key Cryptography • Private procedure: E • Public procedure: D • Identity: E (D(m)) = D (E(m)) = m • Secure: cannot determine E from D • But didn’t know how to find suitable E and D CS588 Spring 2005

6. Properties of E and D Trap-door one way function: • D (E (M)) = M • E and D are easy to compute. • Revealing E doesn’t reveal an easy way to compute D Trap-door one way permutation: also • E (D (M)) = M CS588 Spring 2005

7. RSA E(M) = Me mod n D(C) = Cd mod n n = pqp, qare prime dis relatively prime to(p – 1)(q – 1) ed 1 (mod (p – 1)(q – 1)) (red things are secret) CS588 Spring 2005

8. Properties of E and D Trap-door one way function: • D (E (M)) = M • E and D are easy to compute. • Revealing E doesn’t reveal an easy way to compute D Trap-door one way permutation: also • E (D (M)) = M CS588 Spring 2005

9. Property 1: D (E (M)) = M E(M) = Me mod n D(E(M)) = (Me mod n)d mod n = Med mod n (as in D-H proof) Can we choosee, dandnwith this property: M  Med mod n equivalently:1  Med-1mod n CS588 Spring 2005

10. Finding e, d and n • We are looking for e, d and n such that:Med-11mod n • Euler’s Theorem: for a and n relatively prime: a(n)1mod n • Next: • What is (n) • Proof of Euler’s Theorem • How it works for arbitrary M • Given (n) how do we find e and d CS588 Spring 2005

11. Euler’s Totient Function  (n) = number of positive integers less than n that are relatively prime to n • If n is prime, (n) = n – 1 • Proof by contradiction • What if n = pq where p and q are prime? CS588 Spring 2005

12. Totient Products For primes,pandq:n = pq (n) = numbers < nnot relatively prime to pq = pq – 1 ; numbers less thanpq – (q – 1) ; size of p, 2p, …, (q –1)p – (p – 1) ; size of q, 2q, …, (p –1)q = pq – 1 – (q – 1) – (p – 1) = pq – (p + q) + 1 = (p – 1) (q – 1) = (p)(q) CS588 Spring 2005

13. Fermat’s Little Theorem If n is prime and a is not divisible by nan-11mod n CS588 Spring 2005

14. Fermat’s Little Theorem Proof If n is prime and a is not divisible by n: {a mod n, 2a mod n, … , (n-1)a mod n} = {1, 2, …, (n – 1) } Product of all elements in sets: a 2a  …  (n – 1) a (n – 1)! mod n (n – 1)!an-1 (n – 1)! mod n an-1  1 mod nQED. CS588 Spring 2005

15. Euler’s Theorem For a and n relatively prime: a(n)1mod n Partial Proof: If n is prime, (n) = n – 1 and an - 11mod n by Fermat’s Little Theorem What if n is not prime? CS588 Spring 2005

16. Euler’s Theorem, cont. For a and n relatively prime: a(n)1mod n (n) = number of numbers < nnot relatively prime ton We can write those numbers as: R = { x1, x2, … , x(n)} CS588 Spring 2005

17. Proving Euler’s Theorem R = { x1, x2, … , x(n)} multiply bya mod n: S = { ax1 mod n, ax2 mod n, …, ax (n) mod n} S is a permutation of R: • a is relatively prime to n • a is relatively prime to all xi • axi is relatively prime to n • Hence all elements of S are in R. • There are no duplicates in S. If aximod n = axjmod n then i = j. since a is relatively prime to n CS588 Spring 2005

18. Proving Euler’s Theorem x1 x2 … x(n) = ax1 mod n ax2 mod n … ax(n) mod n  (ax1 ax2 … ax(n)) mod n a(n) x1 x2 … x (n) mod n 1 a (n)mod n QED. CS588 Spring 2005

19. What if M is not relatively prime to n? Recap • We are looking for e, d and n such that: Med-11mod n • Euler’s Theorem: 1 a (n)mod n for a and n relatively prime • If n is prime,  (n) = n – 1. • For p and q prime, (pq) =  (p) (q) n = pq ed – 1 =  (n) = (p-1)(q-1) CS588 Spring 2005

20. M and n • Suppose M and n not relatively prime: gcd (M, n)  1 • Since n = pq and p and q are prime: gcd (M, p)  1 OR gcd (M, q)  1 Case 1: M = cp gcd (M, q) = 1 (otherwise M is multiple of both p and q, but M < pq). So, M(q) 1 mod q (by Euler’s theorem, since M and q are relatively prime) CS588 Spring 2005

21. M and n, cont Case 1: M = cp gcd (M, q) = 1 (otherwise M is multiple of both p and q, but M < pq). So, M (q) 1 mod q (by Euler’s theorem, since M and q are relatively prime) M(q) 1 mod q (M(q))(p) 1 mod q M (q)(p) 1 mod q M(n) 1 mod q CS588 Spring 2005

22. M and n M (n) 1 mod q M (n) = 1 + kq for some k M = cp recall gcd (M, p)  1 M M (n) = (1 + kq)cp M(n) + 1 =cp + kqcp = M + kcn M(n) + 1M mod n CS588 Spring 2005

23. Where’s ED? ed – 1 = (n) = (p-1)(q-1) • So, we need to choose e and d: ed =  (n) + 1 = n – (p + q) • Pick random d, relatively prime to  (n) gcd (d,  (n)) = 1 • Since d is relatively prime to  (n)it has a multiplicative inverse e: de  1 mod  (n) CS588 Spring 2005

24. Identity de  1 mod  (n) So,d * e = (k *  (n)) + 1 for some k. Hence, Med-1mod n = Mk * (n)mod n CS588 Spring 2005

25. D (E (M)) = M Med-1mod n = Mk * (n)mod n Euler says 1 M(n)mod n. So 1 Mk * (n)mod n 1  Med-1mod n M Med mod n QED. CS588 Spring 2005

26. Properties of E and D Trap-door one way function: • D (E (M)) = M • E and D are easy to compute. • Revealing E doesn’t reveal an easy way to compute D Trap-door one way permutation: also • E (D (M)) = M  CS588 Spring 2005

27. Movie Break Adam Glaser and Portman Wills CS588 Fall 2001 PS4

28. 1) "The reader is urged to  find a way to "break" the system. Once the method has withstood all attacks for a sufficient length of time it may be used with a reasonable amount of confidence." The authors appear to advocating the same method of validation that they called "fruitless" earlier in the paper (referring to the NBS certification). 2)  RSA seems to gloss over the whole PKI issue. They suggest either a single authority to hold all the keys or publishing a book to all the users. I'd also like to point out that RSA like to "excessively" to quotation marks for no apparent "purpose." 1. The problem is mentioned on page 4 and 6 of the paper: The trusted distribution of the public portion of the key. If one were to modify the public keys in transport or to attack a central repository, then it would be impossible to be sure of the authenticity of the keys. (The suggestion of having a telephone book is not even possibly applicable due to the need to securely deliver it to all users from trusted central source). This can be seen as present problem with the loss of keys by Microsoft (resulting in forced revocation) and the limited trust one can put in Verisign due to limited checks done. 2. The assumption in conclusion that a protocol is secure due to lack of success in attacks for some period of time. The recent attacks upon SHA/MD5 show that even 10 years could be insufficient time to prove security. Questionable Statements in RSA Paper: Finalists CS588 Spring 2005

29. Only Two Submissions • This is pathetic! • There will be a Short Quiz in class Tuesday • Closed book, closed notes • Covers material in RSA paper and new paper handed out today • Andrew and Aleks are exempt CS588 Spring 2005

30. Two “Questionable” Statements in RSA Paper • “The need for a courier between every pair of users has thus been replaced by the requirement for a single secure meeting between each user and the public file manager when the user joins the system.” (p. 6) CS588 Spring 2005

31. Two “Questionable” Statements in RSA Paper • “(The NBS scheme (DES) is probably somewhat faster if special-purposed hardware encryption devices are used; our scheme may be faster on a general-purpose computer since multiprecision arithmetic operations are simpler to implement than complicated bit manipulations.)” (p. 4) CS588 Spring 2005

32. Who really invented RSA? • General Communications Headquarters, Cheltenham (formed from Bletchley Park after WWII) • 1969 – James Ellis asked to work on key distribution problem • Secure telephone conversations by adding “noise” to line • Late 1969 – idea for PK, but function CS588 Spring 2005

33. RSA & Diffie-Hellman • Asks Clifford Cocks, Cambridge mathematics graduate, for help • He discovers RSA (four years early) • Then (with Malcolm Williamson) discovered Diffie-Hellman • Kept secret until 1997! • NSA claims they had it even earlier CS588 Spring 2005

34. Charge • Reread the parts of RSA paper you didn’t understand the first time • Work on your project! • Short Quiz on RSA material and Encrypted Searches paper in class Tuesday • Closed-book, closed-notes, open-T-shirt • Next time: RSA Properties 2, 3 and 4 CS588 Spring 2005