# 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}`
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