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

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 02/09/2009

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 Ks and sends encrypted Ks (using A’s public key) to A A decrypts message to recover Ks and both use 02/09/2009

Problem with Simple Secret Key Distribution An adversary can intercept and impersonate both parties of protocol A generates a new temporary public key pair {KUa, KRa} and sends KUa || IDa to B Adversary E intercepts this message and sends KUe || IDa to B B generates a session key Ks and sends encrypted Ks (using E’s public key) E intercepts message, recovers Ks and sends encrypted Ks (using A’s public key) to A A decrypts message to recover Ks and both A and B unaware of existence of E 02/09/2009

Distribute Secret Keys Using Asymmetric Encryption if A and B have securely exchanged public-keys ? 02/09/2009

Problem with Previous Scenario Message (4) is not protected by N2 An adversary can intercept message (4) and replay an old message or insert a fabricated message 02/09/2009

Order of Encryption Matters What can be wrong with the following protocol? AB: N BA: EKUa[EKRb[Ks||N]] An adversary sitting between A and B can get a copy of secret key Ks without being caught by A and B! 02/09/2009

Diffie-Hellman Key Exchange First publicly proposed public-key type scheme By Diffie and Hellman in 1976 along with advent of public key concepts A practical method for public exchange of secret key Used in a number of commercial products 02/09/2009

Diffie-Hellman Key Exchange Use to set up a secret key that can be used for symmetric encryption cannot be used to exchange an arbitrary message Value of key depends on the participants (and their private and public key information) Based on exponentiation in a finite (Galois) field (modulo a prime or a polynomial) – easy Security relies on the difficulty of computing discrete logarithms (similar to factoring) – hard 02/09/2009

Primitive Roots From Euler’s theorem: aø(n) mod n=1 Consider am mod n=1, GCD(a,n)=1 must exist for m= ø(n) but may be smaller once powers reach m, cycle will repeat If smallest is m= ø(n) then a is called a primitive root if p is prime and a is a primitive root of p, then successive powers of a “generate” the group mod p Not every integer has primitive roots 02/09/2009

Primitive Root Example: Power of Integers Modulo 19 02/09/2009

Discrete Logarithms Inverse problem to exponentiation is to find the discrete logarithm of a number modulo p Namely find x where ax = b mod p Written as x=loga b mod p or x=dloga,p(b) If a is a primitive root of p then discrete logarithm always exists, otherwise may not 3x = 4 mod 13 has no answer 2x = 3 mod 13 has an answer 4 While exponentiation is relatively easy, finding discrete logarithms is generally a hard problem 02/09/2009

Diffie-Hellman Setup All users agree on global parameters large prime integer or polynomial q α which is a primitive root mod q Each user (e.g. A) generates its key choose a private key (number): xA < q compute its public key: yA = αxA mod q Each user publishes its public key 02/09/2009

Diffie-Hellman Key Exchange Shared session key for users A and B is KAB: KAB = αxA.xB mod q = yAxB mod q (which B can compute) = yBxA mod q (which A can compute) KAB is used as session key in symmetric encryption scheme between A and B Attacker needs xA or xB, which requires solving discrete log 02/09/2009

Diffie-Hellman Example Given Alice and Bob who wish to swap keys Agree on prime q=353 and α=3 Select random secret keys: A chooses xA=97, B chooses xB=233 Compute public keys: yA=397 mod 353 = 40 (Alice) yB=3233 mod 353 = 248 (Bob) Compute shared session key as: KAB= yBxA mod 353 = 24897 = 160 (Alice) KAB= yAxB mod 353 = 40233 = 160 (Bob) 02/09/2009

Elliptic Curve Cryptography Majority of public-key crypto (RSA, D-H) use either integer or polynomial arithmetic with very large numbers/polynomials Imposes a significant load in storing and processing keys and messages An alternative is to use elliptic curves Offers same security with smaller bit sizes 02/09/2009

Real Elliptic Curves An elliptic curve is defined by an equation in two variables x and y, with coefficients Consider a cubic elliptic curve of form y2 = x3 + ax + b where x, y, a, b are all real numbers also define zero point O Have addition operation for elliptic curve geometrically, sum of P+Q is reflection of intersection R 02/09/2009

Real Elliptic Curve Example 02/09/2009

Finite Elliptic Curves Elliptic curve cryptography uses curves whose variables and coefficients are finite Two families are commonly used prime curves Ep(a,b) defined over Zp use integers modulo a prime best in software binary curves E2m(a,b) defined over GF(2m) use polynomials with binary coefficients best in hardware 02/09/2009

Elliptic Curve Cryptography ECC addition is analog of modulo multiply ECC repeated addition is analog of modulo exponentiation Need a “hard” problem equivalent to discrete logarithm Q=kP, where Q, P belong to a prime curve is “easy” to compute Q given k, P but “hard” to find k given Q, P known as the elliptic curve logarithm problem Certicom example: E23(9,17) 02/09/2009

ECC Diffie-Hellman Can do key exchange analogous to D-H Users select a suitable curve Ep(a,b) Select base point G=(x1, y1) with large order n s.t. nG=O A and B select private keys nA<n, nB<n Compute public keys: PA=nA×G, PB=nB×G Compute shared key: K=nA×PB, K=nB×PA same since K=nA×nB×G 02/09/2009

ECC Encryption/Decryption Must first encode any message M as a point on the elliptic curve Pm Select suitable curve and point G as in D-H Each user chooses private key nA<n and computes public key PA=nA×G To encrypt Pm: Cm={kG, Pm+kPB}, k random To decrypt Cm: Pm+kPB–nB(kG) = Pm+k(nBG)–nB(kG) = Pm 02/09/2009

ECC Security Relies on elliptic curve logarithm problem Fastest method is “Pollard rho method” Compared to factoring, ECC can use much smaller key sizes than with RSA For equivalent key lengths computations are roughly equivalent Hence for similar security ECC offers significant computational advantages 02/09/2009

Comparable Key Sizes 1 02/09/2009

Next Class Message authentication Hashing functions Message digests Read Chapters 11 and 12 02/09/2009