Public-Key Cryptography

This challenge is simulating the infamous Logjam attack on many internet protocols like HTTPS, SSH, IPsec, SMTPS and protocols rely on TLS that uses Diffie-Hellman key exchange. The Logjam attack allows a man-in-the-middle attacker to downgrade vulnerable TLS connections to 512-bit export-grade cryptography, as there is an option for clients back when the paper is published to use DHE_EXPORT level of security. There is no indication of the cipher suites the server has chosen, so a MiTM can easily modify the client’s ciphersuite to be DHE_EXPORT....

The task is to find the generator of the finite field. There are multiple ways to do this: Naive implementation (brute-force). Credits to Landryl @ Cryptohack. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 ''' Rather than using a set and checking if every element of Fp has been generated, we can also rapidly disregard a number from being a generator by checking if the cycle it generates is smaller in size than p....

The challenge is straightforward - just factor out the modulus used N. There are two approaches, one using FactorDB and one using the intended way of using Sage. Credits to pdro solution on Cryptohack: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 n = 580642391898843192929563856870897799650883152718761762932292482252152591279871421569162037190419036435041797739880389529593674485555792234900969402019055601781662044515999210032698275981631376651117318677368742867687180140048715627160641771118040372573575479330830092989800730105573700557717146251860588802509310534792310748898504394966263819959963273509119791037525504422606634640173277598774814099540555569257179715908642917355365791447508751401889724095964924513196281345665480688029639999472649549163147599540142367575413885729653166517595719991872223011969856259344396899748662101941230745601719730556631637 ''' Key observation: the number is 2033 bits and it has 30+ factors. The smallest factor will be ~2033/30 = 68 bits in the worst case (i....

A interesting challenge, we are given the modulus n, public exponent e and ciphertext ct. 1 2 3 n = 17258212916191948536348548470938004244269544560039009244721959293554822498047075403658429865201816363311805874117705688359853941515579440852166618074161313773416434156467811969628473425365608002907061241714688204565170146117869742910273064909154666642642308154422770994836108669814632309362483307560217924183202838588431342622551598499747369771295105890359290073146330677383341121242366368309126850094371525078749496850520075015636716490087482193603562501577348571256210991732071282478547626856068209192987351212490642903450263288650415552403935705444809043563866466823492258216747445926536608548665086042098252335883 e = 3 ct = 243251053617903760309941844835411292373350655973075480264001352919865180151222189820473358411037759381328642957324889519192337152355302808400638052620580409813222660643570085177957 From this, denote $m$ as the plaintext, we need to solve the equation $$ m^e \equiv ct \mod N $$ The above equation is equivalent to: $$ m ^ e - ct = 0 \mod N $$ The idea that I had is to use Coppersmith attack for low-exponent 3, we can find the solution to the equation quickly....

In this challenge, we are acting as the MiTM which will intercept the key exchange messages between Alice and Bob. We are able to modify the A and B - each of the shared secret by doing g^a and g^b of Alice and Bob. The flag is sent from Alice to Bob, hence we only need to care about the response of the key exchange message from Bob to Alice. Recall that when Bob’s secret B is sent over to Alice, Alice will do B^a on her side, where a is the secret of Alice....