Rating:

# BSides Delhi CTF 2019 : BabyRSA

**category** : crypto

**points** : 930

## write-up

This is yet another RSA challenge
salt can be decrypt directly
Then, use wiener attack to factor n = p * q and get p1 * p2
Simply gcd p1 * p2 with n1 and n2 and get all the prime factors
Note that gcd(e1, (p1 - 1) * (q1 - 1)) != 1 and gcd(e2, (p2 - 1) * (q2 - 1)) != 1, we can't directly decrypt magic
Luckily, gcd(e1, q1 - 1) == 2 and gcd(e2, q2 - 1) == 2
We can get m ** 2 % q1 and m ** 2 % q2
Then use the same technique as in Rabin cryptosystem, which is modular square root and chinese remainder theorem

flag: bsides_delhi{JuG1iNg_WiTh_RS4}

# other write-ups and resources

Original writeup (https://github.com/OAlienO/CTF/tree/master/2019/BSides-Delhi-CTF/BabyRSA).