This time I created several challenges for TSJ CTF 2022, and I published my challenges and writeups in maple3142/My-CTF-Challenges. Due to GitHub's poor support of mathjax rendering in markdown, I decided to create this one for readers of Crypto writeups.

For other categories, please read it here

Futago#

• Category: Crypto, CSC (Cursed Shaman Challenges, guessy challenges in TSJ CTF)
• Score: 56/100 (CSC)
• Solves: 14/428

Description#

Who need source for RSA challenges?

Overview#

See the README of this challenge.

Solution#

Stage 1#

Guess that two public keys shares a prime, so you can factor it with gcd.

Stage 2#

Found out that $n_1=n_2$, but $e_1 \neq e_2$. Using common modulus attack with found that $\gcd(e_1,e_2)=3$, so you need to take the cube root over integers to decrypt the message.

Stage 3#

You got two seemingly normal 2048 bits RSA public key $n_1,n_2$, but it is easy to see $|n_1-n_2|$ is about 1024 bits only.

If you write them like this:

Then substracting them:

Suppose $p,q$ are balanced, this means $a,b$ needs to be really small to keep $|n_1-n_2|$ about 1024 bits only. So we have $p \approx p+a$ and $q \approx q+b$.

Then you can multiply them together:

It is easy to see $pq \approx (p+a)(q+b)$ and $p(q+b) \approx q(p+a)$ too. Running fermat factorization on $n_1 n_2$ will output these two pairs, and you can factor them with gcd.

from Crypto.PublicKey import RSA
from Crypto.Util.number import *


def read_stage(stage):
    with open(f"{stage}/key1.pub", "rb") as f:
        key1 = RSA.import_key(f.read())
    with open(f"{stage}/key2.pub", "rb") as f:
        key2 = RSA.import_key(f.read())
    with open(f"{stage}/flag.txt.key1.enc", "rb") as f:
        c1 = bytes_to_long(f.read())
    with open(f"{stage}/flag.txt.key2.enc", "rb") as f:
        c2 = bytes_to_long(f.read())
    ret = (key1.n, key1.e, c1, key2.n, key2.e, c2)
    return map(ZZ, ret)


def solve1():
    n1, e1, c1, n2, e2, c2 = read_stage("stage1")
    p = gcd(n1, n2)
    q = n1 // p
    d = inverse_mod(e1, (p - 1) * (q - 1))
    m = power_mod(c1, d, n1)
    return long_to_bytes(m)


def solve2():
    n1, e1, c1, n2, e2, c2 = read_stage("stage2")
    g, a, b = xgcd(e1, e2)
    mg = power_mod(c1, a, n1) * power_mod(c2, b, n1) % n1
    m = mg.nth_root(g)
    return long_to_bytes(m)


def solve3():
    n1, e1, c1, n2, e2, c2 = read_stage("stage3")

    def fermat(x, mx):
        a = floor(sqrt(x))
        b2 = a * a - x
        cnt = 0
        while True:
            if is_square(b2):
                b = floor(sqrt(b2))
                yield a + b, a - b
            a += 1
            cnt += 1
            if cnt == mx:
                return
            b2 = a * a - x

    ar = list(fermat(n1 * n2, 1000000))
    # print(ar)
    assert len(ar) == 2
    assert set(ar[1]) == set([n1, n2])
    p1 = gcd(ar[0][1], n1)
    # print(p1)
    p2 = gcd(ar[0][0], n2)
    # print(p2)

    q1 = n1 // p1
    d1 = inverse_mod(e1, (p1 - 1) * (q1 - 1))
    m = power_mod(c1, d1, n1)
    return long_to_bytes(m)


print(solve1() + solve2() + solve3())

RNG++#

• Category: Crypto
• Score: 213/500
• Solves: 21/428

This is actually a broken version of RNG+++.

Description#

I encrypted the flag and messages by xoring them with a random number generator.

Overview#

You got a simple LCG with $s_{i+1} = A s_i + C \pmod{M}$, where $A,C$ are prime and $M$ is a power of $2$. The flag are encrypted by xor it with initial state $s_0$. Some random messages are generated by randmsg are encrypted by the same way.

Solution#

The first thing to notice is randmsg generated messages are all decimal digits, and they are 0x30 to 0x39 in ASCII. It is obvious that their binary representation are all 0011????, so we can get serveral non continuous bits of states $s_i$.

The most simple unintended solution is to use z3. You just let the state be $\log_2{M}$ bits unsigned bitvector, and write LCG and specify the constraints, and it will magically found the state, so you can get the flag too.

Another more mathematical unintended solution by @Kuruwa: Since $M$ is a power of $2$, reducing it to $\bmod{2^{8i}}$ works too. And just do some bruteforce searching over the digits to see if it holds.

The intended solution will be in the writeup of RNG+++.

babyRSA#

• Category: Crypto
• Score: 276/500
• Solves: 13/428

Description#

Overview#

We have $n=pq$, where $p$ is 1024 bit prime and $q$ is 512 bit prime. Define an elliptic curve $E: y^2=x^3+px+q$ over $\mathbb{Z}_n$, flag (with some random padding) are encoded as x coordinate of a point $P$ on curve. The ciphertext is another point $C=eP$, where $e=65537$.

Solution#

By substituting the coorinates of $C$ back to the original curve, we have $ap+q \equiv r \pmod{n}$, where $a, r$ are known. If we change this equation to $\bmod{p}$, we get $q \equiv r \pmod{p}$.

Define another polynomial $f(x)=r-x$, it is easy to see $f(q)=0 \pmod{p}$. Since $q$ is small compared to $n$, we use Coppersmith's method to solve for the root $q$ and factor $n$.

For more details: Extensions of Coppersmith algorithm

Once you got $p,q$, the remaining part should be easy. Change the curve to $\mathbb{F}_p$ and $\mathbb{F}_q$ and compute the order separately, and invert the scalar multiplication to get the flag.

from Crypto.Util.number import *

proof.arithmetic(False)

n = 1084688440161525456565761297723021343753253859795834242323030221791996428064155741632924019882056914573754134213933081812831553364457966850480783858044755351020146309359045120079375683828540222710035876926280456195986410270835982861232693029200103036191096111928833090012465092747472907628385292492824489792241681880212163064150211815610372913101079146216940331740232522884290993565482822803814551730856710106385508489039042473394392081462669609250933566332939789
xx, yy = (
    1079311510414830031139310538989364057627185699077021276018232243092942690870213059161389825534830969580365943449482350229248945906866520819967957236255440270989833744079711900768144840591483525815244585394421988274792758875782239418100536145352175259508289748680619234207733291893262219468921233103016818320457126934347062355978211746913204921678806713434052571635091703300179193823668800062505275903102987517403501907477305095029634601150501028521316347448735695,
    950119069222078086234887613499964523979451201727533569872219684563725731563439980545934017421736344519710579407356386725248959120187745206708940002584577645674737496282710258024067317510208074379116954056479277393224317887065763453906737739693144134777069382325155341867799398498938089764441925428778931400322389280512595265528512337796182736811112959040864126090875929813217718688941914085732678521954674134000433727451972397192521253852342394169735042490836886,
)

e = 65537

r = ZZ((yy ^ 2 - xx ^ 3) % n)
P = PolynomialRing(Zmod(n), "q")
q = P.gen()
f = r - q
q = ZZ(f.monic().small_roots(X=2 ^ 512, epsilon=0.8, beta=0.66)[0])
p = n // q
assert p * q == n

print((log(p) / log(n)).n())

E = EllipticCurve(Zmod(n), [p, q])

phi = E.change_ring(GF(p)).order() * E.change_ring(GF(q)).order()
d = inverse_mod(e, phi)
x, y = (d * E(xx, yy)).xy()
print(long_to_bytes(x))

Another way to factor it by @Kuruwa:

Since $r \equiv q \pmod{p} \implies r=q+xp$, the vector $v=(q^2,2^{512}q)$ will be in lattice spanned by the following basis $B$.

Since $q$ is just 512 bits, $v$ is probably a short vector in it. Run LLL or Lagrange-Gauss algorithm could produce the vector $v$.

Top Secret#

• Category: Crypro
• Score: 325/500
• Solves: 9/428

Description#

In year 2087, the TSJ corporation invented a new patented stream cipher to protect their communication. One day, another member stole an important file from the CEO's computer, but it turns out to be encrypted. Fortunately, the script used to encrypt the file were stolen too. You, as a member of Nantou Resistance, accept the challenge to decrypt it.

Overview#

This is a weird stream cipher that has an internal state s, and it will update its state using the forward/fast_forward function:

def forward(s: int, n: int, k: int) -> int:
    for _ in range(n):
        s = (s >> 1) ^ ((s & 1) * k)
    return s

The k is a known constant, and the n is the key of the cipher. The initialize state of the cipher is fixed too. The objective is to decrypt a PNG encrypted by it.

Solution#

First, it is necessary to understand the meaning of s = (s >> 1) ^ ((s & 1) * k). If you ever tried to read the implementation of AES, you may find that it is pretty similar to the xtime operation. (An example in C)

xtime operation means multiplying an $GF(2^8)$ element by $x$, and each element is represented by a polynomial of $x$, modulo $x^8+x^4+x^3+x+1$. In AES, the lowest bit represents $x^0$, and the highest bit represents $x^7$.

In this challenge, the order of bits are swapped, so the bit shifting direction are different. And the constant k obvious represents a 128 degree polynomial...?

Actually, key = randbelow(2 ** 128) is meant to confuse people. If you try to construct the polynomial by f'{k:0128b}1' you will see it is not a prime polynomial, because its leftmost bit is 0. The correct degree is 127, so you can construct the field in Sage and implements fast_forward like this:

from sage.all import GF, var


def fast_forward(s, n, k):
    x = var("x")
    coefs = [int(x) for x in (f"{k:0127b}1")]
    poly = sum(c * x ** i for i, c in enumerate(coefs))
    F = GF(2 ** 127, "a", modulus=poly)
    a = F.gen()
    s_coefs = [int(x) for x in f"{s:0127b}"]
    ss = sum(c * a ** i for i, c in enumerate(s_coefs))
    sn = ss * a ** n
    return int("".join(map(str, sn.polynomial().list())).ljust(127, "0"), 2)

The next step is to observe that the first 16 bytes of PNG is known, not just 8 bytes, because of the IHDR chunk. Xor the first chunk of ciphertext and the first 16 bytes of PNG gives second state of the cipher. This can be written as:

So this is a discrete log problem in $GF(2^{127})$. Unfortunately, $2^{127}-1$ is a mersenne prime, so you can't use Pohlig–Hellman to compute $key$.

The intended way is to use Fast Evaluation of Logarithms in Fields of Characteristic Two , which can solve discrete log in this size easily, and it is implemented in pari too.

In Sage, you can just use pari.fflog(e, g) to find x such that g^x == e. In newer version of Sage (9.5 or above), e.log(g) works too.

from sage.all import var, GF, pari

s = 0x6BF1B9BAE2CA5BC9C7EF4BCD5AADBC47
k = 0x5C2B76970103D4EEFCD4A2C681CC400D

x = var("x")
coefs = [int(x) for x in (f"{k:0127b}1")]
poly = sum(c * x ** i for i, c in enumerate(coefs))
F = GF(2 ** 127, "a", modulus=poly)
alpha = F.gen()


def int2el(r):
    return sum(c * alpha ** i for i, c in enumerate(int(x) for x in f"{r:0127b}"))


with open("flag.png.enc", "rb") as f:
    data = f.read()

pngheader = bytes.fromhex("89504E470D0A1A0A0000000d49484452")

ks = bytes(x ^ y for x, y in zip(data, pngheader))
A = int2el(s)
B = int2el(int.from_bytes(ks, "big"))
key = int(pari.fflog(B / A, alpha))  # https://trac.sagemath.org/ticket/32842

from challenge import Cipher

with open("flag.png", "wb") as f:
    f.write(Cipher(key).decrypt(data))

By the way, @Utaha tells me that it is not necessary to compute the discrete log to solve this challenge. Because the key stream is just $x^{key}s_0, (x^{key})^2 s_0, (x^{key})^3 s_0 \cdots$ like a geometric progression. And $x^{key}$ can be simply computed by $s_0 s_1^{-1}$, so it is enough to decrypt the flag actually.

Cipher Switching Service#

• Category: Crypto
• Score: 416/500
• Solves: 4/428

Description#

You can freely swap the encrypted flag between two cryptosystem, but you still don't know what is the flag.

Overview#

The server prints one 1024 bits RSA public key and one 1024 bits ElGamal public key on connection, and gives an RSA encrypted randomly padded flag (padded length is 96 bytes). The length of the flag is 20.

Server supports two operations, allowing you to decrypt a RSA ciphertext and encrypt it using ElGamal, and vice versa.

Solution#

Note: There is a much more simpler unintended solution idea from @Mystiz, scroll to the bottom to see

Using the homomorphic property of RSA, we can get any ciphertext of $am \bmod{n}$. ElGamal is encrypted as $c_1=g^r, c_2=y^r m$ for some random number $r$ and public key $y=g^x$.

I use $L(x,y)$ to denote Legendre Symbol here because it is hard to type in Latex. Obviously, $L(c_2,p)=L(y^r,p) L(m,p)=L(g,p)^{rx} L(m,p)$. Also, by how $g$ is generated, it is easy to see $L(g,p)=1$, so we have $L(m,p)=L(c_2,p)$.

When it decrypts $a^e m^e$ with RSA private, it actually gives $am \bmod{n}$, so it is $L(c_2,p)=L(am \bmod{n},p)$ actually.

Suppose $am \lt n$, $am \bmod{n}$ is just $am$, so $L(c_2,p)=L(a,p) L(m,p)$ should hold. If $am \ge n$, $L(c_2,p)=L(a,p) L(m,p)$ might not hold. From this, it is possible find $a$ such that $am \approx n$ with binary search, so the flag $m$ is approximately $\lfloor\frac{n}{a}\rfloor$.

Since the oracle is not really robust, might need to use some heuristic to make it more stable.

from pwn import *
import ast
from Crypto.Util.number import *
from gmpy2 import powmod as pow


def to_elgamal(c):
    io.sendlineafter(b"> ", b"1")
    io.sendlineafter(b"c = ", str(c).encode())
    return ast.literal_eval(io.recvlineS())


def to_rsa(c1, c2):
    io.sendlineafter(b"> ", b"2")
    io.sendlineafter(b"c1 = ", str(c1).encode())
    io.sendlineafter(b"c2 = ", str(c2).encode())
    return int(io.recvlineS())


def legendre(a, p):
    return pow(a, (p - 1) // 2, p)


io = remote("localhost", 8763)
io.recvuntil(b"RSA")
n, e = ast.literal_eval(io.recvlineS())
io.recvuntil(b"ElGamal")
p, g, y = ast.literal_eval(io.recvlineS())
io.recvuntil(b"len(flag) = ")
flagln = ast.literal_eval(io.recvlineS())
io.recvuntil(b"flagenc = ")
c = ast.literal_eval(io.recvlineS())
assert legendre(g, p) == 1


def legendre(a, p):
    r = pow(a, (p - 1) // 2, p)
    return r if r != p - 1 else -1


cnt = 0


def get_legendre(a):
    # get legendre((a * m) % n, p)
    global cnt
    cnt += 1
    ca = pow(a, e, n)
    cc = (ca * c) % n
    c1, c2 = to_elgamal(cc)
    lc2 = legendre(c2, p)
    return lc2


ln = 96
k_lb = int.from_bytes(b"TSJ{" + b"\x00" * (ln - 4), "big")
k_ub = int.from_bytes(b"TSJ{" + b"\xff" * (ln - 4), "big")
a_lb = n // k_ub
a_ub = n // k_lb

expected = get_legendre(1)


def oracle(a, b):
    # return True if a * k < n
    # no guarantee
    for i in range(a - b, a + b):
        if get_legendre(i) * legendre(i, p) != expected:
            return False
    return True


b = 3  # 3~4 is the best
t = 10
until = (a_ub - a_lb).bit_length() - (flagln - 4) * 8 - 5


def search(l, r, hist):
    # dfs with some pruning
    global b
    if (r - l).bit_length() < until:
        for aa in range(l, r):
            f = long_to_bytes(n // aa)
            if f.startswith(b"TSJ{"):
                print(f)
                break
        print(f"{cnt = }")
        exit()
    if sum(hist[-t:]) >= t or sum(hist[-t:]) <= -t:
        # because oracle may have false positive
        # discard current branch if it is search on a single direction
        print("bad", f"{t = }")
        b = min(b + 1, 10)  # increase bruteforcing window
        return t  # rewind t recursive call
    print((r - l).bit_length(), b)
    m = (r + l) // 2
    while True:
        if oracle(m, b):
            r = search(m, r, hist + [1])
            if r is not None and r > 0:
                return r - 1
        else:
            r = search(l, m, hist + [-1])
            if r is not None and r > 0:
                return r - 1
        if r != 0:
            break


search(a_lb, a_ub, [])

An unintended solution by @Mystiz is to find a $k$ such that $km$ is just slightly above $p$, this means $km \bmod{p}$ is $km-p$. By using homomorphic property of ElGamal you can get the RSA ciphertext of $km \bmod{p}$, which is just $(km-p)^e$. And you can just do $\gcd(m^e-c,(km-p)^e-c')$ to find the plaintext.

Signature#

• Category: Crypto
• Score: 469/500
• Solves: 2/428

Description#

Another boring crypto challenge about signatures.

Overview#

This challenge signs 6 signatures using ECDSA on Secp256k1. The nonce $k$ is generated by xoring hash $z$ and private key $d$. The flag is encrypted with AES-CTR, key are derived from ECDSA private and IV are unknown.

Solution#

This challenge is based on jonasnick/ecdsaPredictableNonce.

Need to use two basic facts:

• $a \oplus b = a + b - 2 (a \land b)$
• For two n bits integer $a,b$, $a \land b = \sum_{i=0}^{n-1} 2^i a_i b_i$, where $a_i$ and $b_i$ is their n-th bit.

So $k_i=d \oplus z_i$ can be written as $k_i=d+z-2 \sum_{j=0}^{n-1} d_i z_{ij}$. And it can be substituted back to $sk \equiv z+rd \pmod{n}$ to get a bunch of equations that have $d_i$ as their roots.

In Secp256k1, $d$ is a 256 bit number. So in the jonasnick/ecdsaPredictableNonce, you can collect 256 signatures and solve $d_i$ with linear algebra.

But it forgot an important fact: $0 \le d_i \le 1$. This problem can be transform into finding a short vector in lattice, and use LLL to solve it.

Details are in the solution script

After recovering $d$, we still need IV to decrypt the flag with AES-CTR. Notice there is a prefix Congrats! This is your flag: in the plaintext, we can take the first block and xor it with the first block of ciphertext, they decrypt it with the key. This is how AES-CTR works.

from fastecdsa.curve import secp256k1 as CURVE
from Crypto.Cipher import AES
from hashlib import sha256
from operator import xor
import ast

with open("output.txt") as f:
    sigs = []
    for _ in range(6):
        z, r, s = map(int, f.readline().strip().split())
        sigs.append((z, r, s))
    msgct = ast.literal_eval(f.readline())


def recover_d(sigs):
    P = PolynomialRing(Zmod(CURVE.q), "d", 256)
    ds = P.gens()
    dd = sum([2 ** i * di for i, di in enumerate(ds)])
    polys = []
    for z, r, s in sigs:
        d_and_z = sum([2 ** i * ((z & (1 << i)) >> i) * di for i, di in enumerate(ds)])
        # fact: (a xor b) = a + b - 2 * (a and b)
        k = dd + z - 2 * d_and_z
        polys.append((s * k) - (z + r * dd))
    M, v = Sequence(polys).coefficient_matrix()
    print(v.T)
    M = M.T.dense_matrix()
    a, b = M.dimensions()
    B = block_matrix(
        ZZ, [[matrix.identity(b) * CURVE.q, matrix.zero(b, a)], [M, matrix.identity(a)]]
    )
    B[:, :b] *= 2 ^ 64
    print("LLL", B.dimensions())
    for row in B.LLL():
        if row[:b] == 0 and row[-1] == 1 and all(0 <= x <= 1 for x in row[b:-1]):
            dbits = row[b:-1]
            d = int("".join(map(str, dbits[::-1])), 2)
            return d


d = recover_d(sigs)
print(f"{d = }")
key = sha256(str(d).encode()).digest()[:16]
nonce = AES.new(key, AES.MODE_ECB).decrypt(
    bytes(map(xor, b"Congrats! This is your flag: "[:16], msgct[:16]))
)[:8]
cipher = AES.new(key, AES.MODE_CTR, nonce=nonce)
print(cipher.decrypt(msgct))

RNG+++#

• Category: Crypto
• Score: 469/500
• Solves: 2/428

This upgraded version of RNG++, so you might want to read it before reading this.

Description#

I encrypted the flag and messages by xoring them with a random number generator again. But it should be harder to break this time.

Overview#

Basically same as RNG++, but $M$ is a prime this time.

Solution#

Since we only know some bits in each state $s_i$, it can be written as:

$Y_i$ is a known part of each state and $0 \leq u_{ij} \leq 15$ are unknowns.This is equivalent to this binary representation:

0011????0011????0011????0011????0011????...

It is easy to see $A s_i + C - s_{i+1} \equiv 0 \pmod{M}$, and trying to substitute $s_i$ with $u_{ij}$ will find out that it is just some linear combination of $u_{ij}$. And the fact $0 \leq u_{ij} \leq 15$ tells us we need to find something small, so it isn't too hard to think about lattice.

Using coefficient matrix of those polynomials ($A s_i + C - s_{i+1}$), we can transform this problem into finding closest vector of a lattice. And using Babai Nearest Plane algorithm on a BKZ reduced basis (LLL reduced basis doesn't work well here) allows you to find the correct $u_{ij}$.

from Crypto.Util.number import *
from operator import xor


with open("output.txt") as f:
    lines = f.readlines()
    M, A, C = [ZZ(x) for x in lines[0].strip().split()]
    sz = round(M.log(2))
    cts = [bytes_to_long(bytes.fromhex(line.strip())) for line in lines[1:]]

flagct = cts[0]
cts = cts[1:1+4]
n = len(cts)

mask1 = sum(
    [
        (1 << (7 + 8 * i))
        + (1 << (6 + 8 * i))
        + (1 << (5 + 8 * i))
        + (1 << (4 + 8 * i))
        for i in range(sz // 8)
    ]
)
mask2 = sum([+(1 << (5 + 8 * i)) + (1 << (4 + 8 * i)) for i in range(sz // 8)])
ys = [xor((mask1 & s), mask2) for s in cts]

F = Zmod(M)
# t = F(C / (1 - A))
unames = ",".join(sum([[f"u_{i}_{j}" for j in range(sz // 8)] for i in range(n)], []))
P = PolynomialRing(F, unames)
U = matrix(n, sz // 8, P.gens())
rs = [ys[i] + sum([2 ^ (8 * j) * U[i, j] for j in range(sz // 8)]) for i in range(n)]
fs = [A * r + C - rr for r, rr in zip(rs, rs[1:])]
# rs = [r - t for r in rs]  # substitution
# fs = [A ^ 1 * r - rr for r, rr in zip(rs, rs[1:])]

B, v = Sequence(fs).coefficient_matrix()
print(vector(v))
B = B.T.dense_matrix().change_ring(ZZ)
target = (-B[-1]).list()
B = B[:-1]
a, b = B.dimensions()
print(a, b)
I = matrix.identity(a)
B = block_matrix([[B, I], [M * matrix.identity(b), matrix.zero(b, a)]])
print(B.dimensions())


def Babai_CVP(mat, target):
    # M = mat.LLL()
    M = mat.BKZ(algorithm="NTL", prune=5)
    G = M.gram_schmidt()[0]
    diff = target
    for _ in range(1):
        for i in reversed(range(G.nrows())):
            diff -= M[i] * ((diff * G[i]) / (G[i] * G[i])).round()
    return target - diff


cvp_target = vector(target + [8] * (n * sz // 8))
rr = Babai_CVP(B, cvp_target)
xx = B.solve_left(rr)
print("CVP")
print(rr)
print(xx)
print((rr - cvp_target).norm().n())
print()
found = xx[: n * (sz // 8)]
print(found)

if all(0 <= x < 16 for x in found):
    print("good")
    mat = matrix(n, sz // 8, found)
    ss = [
        ys[i] + sum([2 ^ (8 * j) * mat[i, j] for j in range(sz // 8)]) for i in range(n)
    ]
    print(ss)
    assert (A * ss[0] + C) % M == ss[1]
    flag = long_to_bytes(xor(ZZ(F(ss[0] - C) / A), flagct)).decode()
    print(f"TSJ{{{flag}}}")