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Entries from April 2019.

Midnight Sun CTF 2019 EZDSA Writeup
7th April 2019

This is FAUST playing CTF again, this time midnightsun.

Crew: siccegge

OK so we're looking at the EZDSA service. This is a signature service and the task is essentially to recover the signing key. Code is reproduced below.

from hashlib import sha1
from Crypto import Random
from flag import FLAG

class PrivateSigningKey:

    def __init__(self):
        self.gen = 0x44120dc98545c6d3d81bfc7898983e7b7f6ac8e08d3943af0be7f5d52264abb3775a905e003151ed0631376165b65c8ef72d0b6880da7e4b5e7b833377bb50fde65846426a5bfdc182673b6b2504ebfe0d6bca36338b3a3be334689c1afb17869baeb2b0380351b61555df31f0cda3445bba4023be72a494588d640a9da7bd16L
        self.q = 0x926c99d24bd4d5b47adb75bd9933de8be5932f4bL
        self.p = 0x80000000000001cda6f403d8a752a4e7976173ebfcd2acf69a29f4bada1ca3178b56131c2c1f00cf7875a2e7c497b10fea66b26436e40b7b73952081319e26603810a558f871d6d256fddbec5933b77fa7d1d0d75267dcae1f24ea7cc57b3a30f8ea09310772440f016c13e08b56b1196a687d6a5e5de864068f3fd936a361c5L
        self.key = int(FLAG.encode("hex"), 16)

    def sign(self, m):

        def bytes_to_long(b):
            return long(b.encode("hex"), 16)

        h = bytes_to_long(sha1(m).digest())
        u = bytes_to_long(
        assert(bytes_to_long(m) % (self.q - 1) != 0)

        k = pow(self.gen, u * bytes_to_long(m), self.q)
        r = pow(self.gen, k, self.p) % self.q
        s = pow(k, self.q - 2, self.q) * (h + self.key * r) % self.q
        assert(s != 0)

        return r, s

The outer service was not provided but you could pass in base64 encoded byte arrays and got back r and s as already indicated. Looking at the final computation for s we notice that given \((h + k * r)\) and \(h, r\) we can easily recover \(k\). For this to work it would be convenient if the first term ends up being 1. Unfortunately, the easiest way to get there is prevented: \(g^{q-1} = 1\). Fortunately this is not the only exponent where this works and a good candidate is \((q-1 / 2)\).

pow(gen, (q-1)//2, q)

From there the only thing left is solving \(s = (h + k * r)\). Fortunately gmpy has the solution prepackaged again: divm. So we proceed by getting a valid "signature" on \((q-1 / 2)\). The rest is simple calculation:

sha1(binascii.unhexlify("%x" % ((q-1)//2))).hexdigest()

base64.b64encode(binascii.unhexlify("%x" % ((q-1)//2)))

gmpy2.divm(s-h, r, q)

binascii.unhexlify("%x" % 39611266634150218411162254052999901308991)

OK so why does \((q-1 / 2)\) work? Essentially, the field defined \(F_q\) -- calculations mod q -- has q elements additively and \(q-1\) elements multiplicatively(and we're considering exponentiation as repeated multiplication). Therefore it contains cyclic subgroups for all factors of \(q-1\) and for every element \(e\), \(e^o = 1\) where o is the order of the subgroup that element belongs to. as the generator is trivially not \(-1\) -- the subgroup of size 2 -- \((q-1 / 2)\) must be a multiple of the generated group's order.

Tags: crypto, ctf, faust, python, security, writeup.

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