I made new Messenger with Identity Based Cryptography.
The flag is in admin's messages. Can you read it?
IBM.py - (logging 부분 제외하고 소스 공개)
Server (pk, sk) = (n, g) with RSA public key n
Client (pk, sk) = h(id), g^h(id) % n
Key agreement btw id1, id2 : sk1 ^ pk2 % n = sk2 ^ pk1 % n = g ^ (pk1 * pk2) % n
(1, a, b) = egcd(h(id1), h(id2)) then g = sk1 ^ a * sk2 ^ b % n