Question

Can the Diffie-Hellman key exchange protocol be extended to three people, Alice, Bob and Carol to generate session keys? Describe your mathematical argument for why or why not.

Answer 1

Diffie–Hellman key agreement is not limited to a key shared by only two participants. Any number of users can take part in an agreement by performing iterations of the agreement protocol and exchanging intermediate data. For example, Alice, Bob, and Carol could participate in a Diffie–Hellman agreement as follows, with all operations taken to be modulo p:

step 1.The parties agree on the algorithm parameters p and g.
step 2.The parties generate their private keys, named a, b, and c respectively for Alice,Bob and Carol
step 3.Alice computes g^a and sends it to Bob.
step 4.Bob computes (g^a)^b = g^ab and sends it to Carol.
step 5:Carol computes (g^ab)^c = g^abc and uses it as her secret.
step 6:Bob computes g^b and sends it to Carol.
step 7:Carol computes (g^b)^c = g^bc and sends it to Alice.
step 8:Alice computes (g^bc)^a = g^abc and uses it as her secret.
step 9:Carol computes g^c and sends it to Alice.
step 10:Alice computes (g^c)^a = g^ca and sends it to Bob.
step 11:Bob computes (g^ca)^b = g^cab = g^abc and uses it as his secret.

An eavesdropper has been able to see g^a, g^b, g^c, g^ab, g^ac, and g^bc, but cannot use any combination of these to efficiently reproduce g^abc.