1. Week 4 - Friday

2.  What did we talk about last time?  Public key cryptography  A little number theory

7.  If p is prime and a is a positive integer not divisible by p, then: a p –1  1 (mod p)

8.  Assume a is positive and less than p  Consider the sequence a, 2a, 3a, …, (p – 1)a  If these are taken mod p, we will get:  1, 2, 3, …, p – 1  This bit is the least obvious part of the proof  However (because p is prime) if you add any non-zero element repeatedly, you will eventually get back to the starting point, covering all values (except 0) once  Multiplying this sequence together gives:  a ∙ 2a ∙ 3a ∙ … ∙ (p – 1)a  1 ∙ 2 ∙ 3 ∙ … ∙ (p – 1) (mod p)  a p – 1 (p – 1)!  (p – 1)! (mod p)  a p – 1  1 (mod p)

9.  Euler’s totient function  (n)   (n) = the number of positive integers less than n and relatively prime to n (including 1)  If p is prime, then  (p) = p – 1  If we have two primes p and q (which are different), then:  (pq) =  (p)∙  (q) = (p – 1)(q – 1)

10.  Euler’s Theorem: For every a and n that are relatively prime, a  (n)  1 (mod n)  This generalizes Fermat’s Theorem because  (p) = p – 1 if p is prime  Proof is messier

12.  Named for Rivest, Shamir, and Adleman  Take a plaintext M converted to an integer  Create an ciphertext C as follows: C = M e mod n  Decrypt C back into M as follows: M = C d mod n = (M e ) d mod n = M ed mod n

13. TermDetailsSource MMessage to be encryptedSender CEncrypted messageComputed by sender nModulus, n = pqKnown by everyone pPrime numberKnown by receiver qPrime numberKnown by receiver eEncryption exponentKnown by everyone dDecryption exponentComputed by receiver (n)(n) Totient of nKnown by receiver

14.  To encrypt: C = M e mod n  e could be 3 and is often 65537, but is always publically known  To decrypt: M = C d mod n = M ed mod n  We get d by finding the multiplicative inverse of e mod  (n)  So, ed  1 (mod  (n))

15.  We know that ed  1 (mod  (n))  This means that ed = k  (n) + 1 for some nonnegative integer k  M ed = M k  (n) + 1  M∙(M  (n) ) k (mod n)  By Euler’s Theorem M  (n)  1 (mod n)  So, M∙(M  (n) ) k  M (mod n)

16.  M = 26  p = 17, q = 11, n = 187, e = 3  C = M 3 mod 187 = 185   (n) = (p – 1)(q – 1) = 160  d = e -1 mod 160 = 107  C d = 185 107 mod 187 = 26  If you can trust my modular arithmetic

17.  You can’t compute the multiplicative inverse of e mod  (n) unless you know what  (n) is  If you know p and q, finding  (n) is easy  Finding  (n) is equivalent to finding p and q by factoring n  No one knows an efficient way to factor a large composite number

18.  Public key cryptography would come crashing down if  Advances in number theory could make RSA easy to break  Quantum computers could make it easy to factor large composites

19.  Choose your primes carefully  p < q < 2p  But, the primes can't be too close together either  Some standards insist that p and q are strong primes, meaning that p – 1 = 2m and p + 1 = 2n where m and n have large prime factors  There are ways to factor poorly chosen pairs of primes  Pad your data carefully  Take the example of a credit card number  If you know a credit card number is encrypted using RSA using a public n and an e of 3, how do you discover the credit card number?

21.  Once you have great cryptographic primitives, managing keys is still a problem  How do you distribute new keys?  When you have a new user  When old keys have been cracked or need to be replaced  How do you store keys?  As with the One Time Pad, if you could easily send secret keys confidentially, why not send messages the same way?

22.  We will refer to several schemes for sending data  Let X and Y be parties and Z be a message  { Z } k means message Z encrypted with key k  Thus, our standard notation will be:  X  Y: { Z } k  Which means, X sends message Z, encrypted with key k, to Y  X and Y will be participants like Alice and Bob and k will be a clearly labeled key  A || B means concatenate message A with B

23.  Typical to key exchanges is the idea of interchange keys and session keys  An interchange key is a key associated with a particular user over a (long) period of time  A session key is a key used for a particular set of communication events  Why have both kinds of keys?

24.  If only a single key (instead of interchange and session keys) were used, participants are more vulnerable to:  Known plaintext attacks (and potentially chosen plaintext attacks)  Attacks requiring many copies of encrypted material for comparison  Replay attacks in which old encrypted data is sent again from a malicious party  Forward search attacks in which a user computes many likely messages using a public key and thereby learns the contents of such a message when it is sent

25.  To be secure, a key exchange whose goal is to allow secret communication from Alice to Bob must meet this criteria: 1. Alice and Bob cannot transmit their key unencrypted 2. Alice and Bob may decide to trust a third party (Cathy or Trent) 3. Cryptosystems and protocols must be public, only the keys are secret

26.  If Bob and Alice have no prior arrangements, classical cryptosystems require a trusted third party Trent  Trent and Alice share a secret key k Alice and Trent and Bob share a secret key k Bob  Here is the protocol: 1. Alice  Trent: {request session key to Bob} k Alice 2. Trent  Alice: { k session } k Alice || { k session } k Bob 3. Alice  Bob: { k session } k Bob

27.  Unfortunately, this protocol is vulnerable to a replay attack  (Evil user) Eve records { k session } k Bob sent in step 3 and also some message enciphered with k session (such as "Deposit $500 in Dan's bank account")  Eve can send the session key to Bob and then send the replayed message  Maybe Eve is in cahoots with Dan to get him paid twice  Eve may or may not know the contents of the message she is sending  The real problem is no authentication

28.  We modify the protocol to add random numbers (called nonces) and user names for authentication 1. Alice  Trent: { Alice || Bob || rand 1 } k Alice 2. Trent  Alice: { Alice || Bob || rand 1 || k session || {Alice || k session }k Bob } k Alice 3. Alice  Bob: { Alice || k session }k Bob 4. Bob  Alice: { rand 2 } k session 5. Alice  Bob: { rand 2 – 1 } k session

29.  Needham-Schroeder assumes that all keys are secure  Session keys may be less secure since they are generated with some kind of (possibly predictable) pseudorandom generator  If Eve can recover a session key (maybe after a great deal of computational work), she can trick Bob into thinking she's Alice as follows: 1. Eve  Bob: { Alice || k session }k Bob 2. Bob  Alice: { rand 3 } k session [intercepted by Eve] 3. Eve  Bob: { rand 3 – 1 } k session

30.  Denning and Sacco use timestamps (T) to let Bob detect the replay 1. Alice  Trent: { Alice || Bob || rand 1 } k Alice 2. Trent  Alice: { Alice || Bob || rand 1 || k session || {Alice || T || k session }k Bob } k Alice 3. Alice  Bob: { Alice || T || k session }k Bob 4. Bob  Alice: { rand 2 } k session 5. Alice  Bob: { rand 2 – 1 } k session  Unfortunately, this system requires synchronized clocks and a useful definition of when timestamp T is "too old"

31.  The Otway-Rees protocol fixes these problem by using a unique integer num to label each session 1. Alice  Bob: num || Alice || Bob || { rand 1 || num || Alice || Bob } k Alice 2. Bob  Trent: num || Alice || Bob || {rand 1 || num || Alice || Bob } k Alice || { rand 2 || num || Alice || Bob } k Bob 3. Trent  Bob: num || { rand 1 || k session } k Alice || { rand 2 || k session } k Bob 4. Bob  Alice: num || { rand 1 || k session } k Alice

32.  Strange as it seems, these key exchange protocols are actually used  Kerberos was created at MIT as a modified Needham- Schroeder protocol (with timestamps)  Originally used to control access to network services for MIT students and staff  Current versions of Windows use a modified version of Kerberos for authentication  Many Linux and Unix implementations have an implementation of Kerberos  Kerberos uses a central server that issues tickets to users which give them the authority to access a service on some other server

34.  Suddenly, the sun comes out!  Public key exchanges should be really easy  The basic outline is: 1. Alice  Bob: { k session } e Bob  e Bob is Bob's public key  Only Bob can read it, everything's perfect!  Except…  There is still no authentication

35.  Alice only needs to encrypt the session key with her private key  That way, Bob will be able to decrypt it with her public key when it arrives  New protocol: 1. Alice  Bob: {{ k session } d Alice }e Bob  Any problems now?

36.  A vulnerability arises if Alice needs to fetch Bob's public key from a public server Peter  Then, Eve can cause problems  Attack: 1. Alice  Peter: Send me Bob's key [intercepted by Eve] 2. Eve  Peter: Send me Bob's key 3. Peter  Eve: e Bob 4. Eve  Alice: e Eve 5. Alice  Bob: { k session } e Eve [intercepted by Eve] 6. Eve  Bob { k session } e Bob

38.  The previous man in the middle attack shows a significant problem  How do we know whose public key is whose?  We could sign a public key with a private key, but then…  We would still be dependent on knowing the public key matching the private key used for signing  It's a massive chicken and egg or bootstrapping problem

39.  A typical approach is to create a long chain of individuals you trust  Then, you can get the public key from someone you trust who trusts someone else who… etc.  This can be arranged in a tree layout, with a central root certificate everyone knows and trusts  This system is used by X.509  Alternatively, it can be arranged haphazardly, with an arbitrary web of trust  This system is used by PGP, which incorporates different levels of trust

41.  Finish key management  Hash functions  David Gallop presents

42.  Read section 12.4  Finish Project 1  Due tonight by midnight!