Writeup "R u SAd?" from PlaidCTF 2019

by: dorianr (Dorian Rudolph)


Tears dripped from my face as I stood over the bathroom sink. Exposed again! The tears melted into thoughts, and an idea formed in my head. This will surely keep my secrets safe, once and for all. I crept back to my computer and began to type.

We are given a relatively long python script implementing RSA encryption, together with a public key and an encrypted file. They use a custom key format which is saved using the pickle module:

class Key:
    PRIVATE_INFO = ['P', 'Q', 'D', 'DmP1', 'DmQ1']

    def __init__(self, **kwargs):
        for k, v in kwargs.items():
            setattr(self, k, v)
        assert self.bits % 8 == 0

    def ispub(self):
        return all(not hasattr(self, key) for key in self.PRIVATE_INFO)

    def ispriv(self):
        return all(hasattr(self, key) for key in self.PRIVATE_INFO)

    def pub(self):
        p = deepcopy(self)
        for key in self.PRIVATE_INFO:
            if hasattr(p, key):
                delattr(p, key)
        return p

    def priv(self):
        raise NotImplementedError()

def genkey(bits):
    assert bits % 2 == 0
    while True:
        p = genprime(bits // 2)
        q = genprime(bits // 2)
        e = 65537
        d, _, g = egcd(e, (p - 1) * (q - 1))
        if g != 1: continue
        iQmP, iPmQ, _ = egcd(q, p)
        return Key(
            N=p * q, P=p, Q=q, E=e, D=d % ((p - 1) * (q - 1)), DmP1=d % (p - 1), DmQ1=d % (q - 1),
            iQmP=iQmP % p, iPmQ=iPmQ % q, bits=bits,

Notice that the values iPmQ and iQmP are not removed when constructing the public key. Let us call these values $a$ and $b$ in the following. If $a’, b’=egcd(p, q)$, then $a’p+b’q=gcd(p,q)$ by B├ęzout’s identity. Hence, \[ \quad (a+iq)p + (b+jp)q = 1\nl \Rightarrow ap+bq+(i+j)pq = 1\nl \Rightarrow ap+bq=1+zn=:c \] for small values $i,j,z\in \mathbb Z$.

Let $x, y = egcd(a, b)$. Since $gcd(a, b)=1$ (in our case), we have

\[ \quad ap+bq=1+zn=c\nl \quad ax+by=1\nl \Rightarrow a(p-xc)+b(q-yc) = 0\qquad \text{(subtract the second equation $c$ times from the first)}\nl \Rightarrow a(p-xc) = b(yc-q)\nl \Rightarrow (q-yc) \equiv 0 \mod a\nl \Rightarrow q-yc=ka \]

We can expect $q/a$ to be small. Hence, $k\approx -yc/a$. Then, $q=ka+yc$.


a = k.iPmQ
b = k.iQmP
n = k.N

x, y, _ = egcd(a, b)

for z in range(-10, 10):
    c = 1 + z*n
    for k in range(-y*c//a-10, -y*c//a+10):
        q = k*a + y*c
        if n % q == 0:
            print(q, n//q)