Why SRP is not plaintext-equivalent?

Question

About the SRP Protocol: http://en.wikipedia.org/wiki/Secure_remote_password_protocol

I can see that the generation of the session key (K) is perfectly safe, but in the last step the user sends proof of K (M). If the network is insecure and the attacker in the midlle captures M, he would be able to authenticate without having K. right?

Answer 1

A little background

Well known values (established beforehand):

  n    A large prime number. All computations are performed modulo n.
  g    A primitive root modulo n (often called a generator).

The users password is established as:

x = H(s, P)
v = g^x 

  H()  One-way hash function
  s    A random string used as the user's salt
  P    The user's password
  x    A private key derived from the password and salt
  v    The host's password verifier

The authentication:

+---+------------------------+--------------+----------------------+
|   | Alice                  | Public Wire  | Bob                  |
+---+------------------------+--------------+----------------------+
| 1 |                        |        C --> | (lookup s, v)        |
| 2 | x = H(s, P)            | <-- s        |                      |
| 3 | A = g^a                |        A --> |                      |
| 4 |                        | <-- B, u     | B = v + g^b          |
| 5 | S = (B - g^x)^(a + ux) |              | S = (A · v^u)^b      |
| 6 | K = H(S)               |              | K = H(S)             |
| 7 | M[1] = H(A, B, K)      |     M[1] --> | (verify M[1])        |
| 8 | (verify M[2])          | <-- M[2]     | M[2] = H(A, M[1], K) |
+---+------------------------+--------------+----------------------+

    u    Random scrambling parameter, publicly revealed
  a,b    Ephemeral private keys, generated randomly and not publicly revealed
  A,B    Corresponding public keys
  m,n    The two quantities (strings) m and n concatenated
    S    Calculated exponential value 
    K    Session key

The answer to your question:

As you can see, both parties calculate K (=the session key) separately, based upon the values available to each of them.
If Alice's password P entered in Step 2 matches the one she originally used to generate v, then both values of S will match.

The actual session key K is however never send over the wire, only the proof that both parties have successfully calculated the same session key. So a man-in-the middle could resend the proof, but since he does not have the actual session key, he would not be able to do anything with the intercepted data.

Answer 2

The proof is only valid for a certain K.

Without MITM:

Alice <-K-> Bob

Alice produces a proof for K and Bob accepts it

With MITM:

Alice <-K1-> Eve <-K2-> Bob

Alice produces a proof for K1 but when Eve presents it to Bob he doesn't accept it because it doesn't fit K2.