Nullcon HackIM 2023 Crypto - Curvy Decryptor / deuterium's blog

Challenge Description

Curvy Decryptor 473 points Alice has hidden 2 flags in this challenge. And even though she is willing to decrypt most ciphers, she has some basic saveguards against stealing flags. Please submit flag1 here. nc 52.59.124.14 10005

Source Files

curvy_decryptor.py
ec.py
utils.py

Source Analysis

From curvy_decryptor.py

Python76 lines / static highlight

#!/usr/bin/env python3
import os
import sys
import string
from Crypto.Util import number
from Crypto.Util.number import bytes_to_long, long_to_bytes
from Crypto.Cipher import AES
from binascii import hexlify

from ec import *
from utils import *
from secret import flag1, flag2

#P-256 parameters
p = 0xffffffff00000001000000000000000000000000ffffffffffffffffffffffff
a = -3
b = 0x5ac635d8aa3a93e7b3ebbd55769886bc651d06b0cc53b0f63bce3c3e27d2604b
n = 0xffffffff00000000ffffffffffffffffbce6faada7179e84f3b9cac2fc632551
curve = EllipticCurve(p,a,b, order = n)
G = ECPoint(curve, 0x6b17d1f2e12c4247f8bce6e563a440f277037d812deb33a0f4a13945d898c296, 0x4fe342e2fe1a7f9b8ee7eb4a7c0f9e162bce33576b315ececbb6406837bf51f5)

d_a = bytes_to_long(os.urandom(32))
P_a = G * d_a

printable = [ord(char.encode()) for char in string.printable]

def encrypt(msg : bytes, pubkey : ECPoint):
    x = bytes_to_long(msg)
    y = modular_sqrt(x**3 + a*x + b, p)
    m = ECPoint(curve, x, y)
    d_b = number.getRandomRange(0,n)
    return (G * d_b, m + (pubkey * d_b))

def decrypt(B : ECPoint, c : ECPoint, d_a : int):
    if B.inf or c.inf: return b''
    return long_to_bytes((c - (B * d_a)).x)

def loop():
    print('I will decrypt anythin as long as it does not talk about flags.')
    balance = 1024
    while True:
        print('B:', end = '')
        sys.stdout.flush()
        B_input = sys.stdin.buffer.readline().strip().decode()
        print('c:', end = '')
        sys.stdout.flush()
        c_input = sys.stdin.buffer.readline().strip().decode()
        B = ECPoint(curve, *[int(_) for _ in B_input.split(',')])
        c = ECPoint(curve, *[int(_) for _ in c_input.split(',')])
        msg = decrypt(B, c, d_a)
        if b'ENO' in msg:
            balance = -1
        else:
            balance -= 1 + len([c for c in msg if c in printable])
        if balance >= 0:
            print(hexlify(msg))
            print('balance left: %d' % balance)
        else:
            print('You cannot afford any more decryptions.')
            return

if __name__ == '__main__':
    print('My public key is:')
    print(P_a)
    print('Good luck decrypting this cipher.')
    B,c = encrypt(flag1, P_a)
    print(B)
    print(c)
    key = long_to_bytes((d_a >> (8*16)) ^ (d_a & 0xffffffffffffffffffffffffffffffff))
    enc = AES.new(key, AES.MODE_ECB)
    cipher = enc.encrypt(flag2)
    print(hexlify(cipher).decode())
    try:
        loop()
    except Exception as err:
        print(repr(err))

Curvy Decryptor part 1

The solution of Curvy Decryptor part 1 is to find out flag1

Curvy Decryptor part 2

The solution of Curvy Decryptor part 1 is to find out flag2
The flag2 appears to be AES encrypted with a key which is

Plain text1 line / static highlight

key = long_to_bytes((d_a >> (8*16)) ^ (d_a & 0xffffffffffffffffffffffffffffffff))

The xor of most significant and least significant 64 bits of the 128 bits private key d_a
So in order to recover the flag2 we will need to break the ECC and recover d_a

Analysing curvy_decryptor.py

Python10 lines / static highlight

#P-256 parameters
p = 0xffffffff00000001000000000000000000000000ffffffffffffffffffffffff
a = -3
b = 0x5ac635d8aa3a93e7b3ebbd55769886bc651d06b0cc53b0f63bce3c3e27d2604b
n = 0xffffffff00000000ffffffffffffffffbce6faada7179e84f3b9cac2fc632551
curve = EllipticCurve(p,a,b, order = n)
G = ECPoint(curve, 0x6b17d1f2e12c4247f8bce6e563a440f277037d812deb33a0f4a13945d898c296, 0x4fe342e2fe1a7f9b8ee7eb4a7c0f9e162bce33576b315ececbb6406837bf51f5)

d_a = bytes_to_long(os.urandom(32))
P_a = G * d_a

indeed look like a standard curve P-256 (note that a = -3 is equivalent to a = p - 3)
So there doesnt appear to be any standard weakness based on weak curve parameters.
The private key d_a is initialized to be per-instance system-random 128 bit number
And the Public key P_a is simply G*d_a

Encryption

Python6 lines / static highlight

def encrypt(msg : bytes, pubkey : ECPoint):
    x = bytes_to_long(msg)
    y = modular_sqrt(x**3 + a*x + b, p)
    m = ECPoint(curve, x, y)
    d_b = number.getRandomRange(0,n)
    return (G * d_b, m + (pubkey * d_b))

The msg is encoded as the x coordinate of the point, the corresponding y is found so as to find the point on the curve to generate the point m
A secure random number d_b in range (0 - curve order) is generated as the nonce, The points G*d_b and m + (pubkey * d_b) are returned

Decryption

Python3 lines / static highlight

def decrypt(B : ECPoint, c : ECPoint, d_a : int):
    if B.inf or c.inf: return b''
    return long_to_bytes((c - (B * d_a)).x)

It simply reverses the encrypt function if correct d_a is provided
i.e if we do

Python2 lines / static highlight

B,c = encrypt(flag1, P_a)
xx = decrypt(B,c)

We will get,

\text{decrypt}((G * d_b * d_a), \text{flag}_m + (P_a * d_b))

= (\text{flag}_m + (P_a * d_b) - G * d_b * d_a).x

= (\text{flag}_m + G * d_a * d_b - G * d_b * d_a).x

= (\text{flag}_m).x = \text{flag}

Main Loop

The main loop of the program just repeatedly asks for input of two EC points B and c and tries to decrypt it with the servers private key d_a
BUT
if the decryption contains b'ENO' i.e the start of the flag, it exits.
Otherwise it decreases the balance proportionate to the number of printable characters in the decryption.

So If we directly input the B and c corresponding to the flag1, it will abort and hence no flags for us :’(

But it doesnt care what points we ask it to decrypt.
So what if we try to decrypt the points B, c + A

We will get,

\text{decrypt}((G * d_b * d_a), \text{flag}_m + (P_a * d_b) + A)

= (\text{flag}_m + (P_a * d_b) + A - G * d_b * d_a).x

= (\text{flag}_m + A + G * d_a * d_b - G * d_b * d_a).x

= (\text{flag}_m + A).x

As long as we know A, we can always get the point from the x coordinate, and subtract A from it to get the original point from the curve for the sake of simplicity, we can even pick it to be G

or we can even try decrypting the points B + A, c which will lead to

\text{decrypt}((G * d_b * d_a) + A, \text{flag}_m + (P_a * d_b))

= (\text{flag}_m + (P_a * d_b) - (G * d_b + A) * d_a).x

= (\text{flag}_m + G * d_a * d_b - G * d_b * d_a - A * d_a).x

= (\text{flag}_m - A * d_a).x

if we choose A to be G or -G, we will end up with the point flag - P_a or flag + P_a and since we even know P_a, it will also work
With a high probability, we wont observe any b'ENO' in the resulting point, and if we do, we can always pick countless possibilities of A to make it work

Python45 lines / static highlight

from ec import *
from utils import *
import pwn

p = 0xffffffff00000001000000000000000000000000ffffffffffffffffffffffff
a = -3
b = 0x5ac635d8aa3a93e7b3ebbd55769886bc651d06b0cc53b0f63bce3c3e27d2604b
n = 0xffffffff00000000ffffffffffffffffbce6faada7179e84f3b9cac2fc632551
curve = EllipticCurve(p,a,b, order = n)
G = ECPoint(curve, 0x6b17d1f2e12c4247f8bce6e563a440f277037d812deb33a0f4a13945d898c296, 0x4fe342e2fe1a7f9b8ee7eb4a7c0f9e162bce33576b315ececbb6406837bf51f5)

HOST, PORT = "52.59.124.14", 10005
REM = pwn.remote(HOST, PORT)

REM.recvline() # My public key is:
pubkey = REM.recvline().strip()[6:-1].split(b',')
P_a = ECPoint(curve, int(pubkey[0]), int(pubkey[1]))

REM.recvline() # Good luck decrypting this cipher.
B_text = REM.recvline().strip()[6:-1].split(b',')
B = ECPoint(curve, int(B_text[0]), int(B_text[1]))

c_text = REM.recvline().strip()[6:-1].split(b',')
c = ECPoint(curve, int(c_text[0]), int(c_text[1]))

flag2_enc = bytes.fromhex(REM.recvline().strip().decode())

REM.recvline() # I will decrypt anythin as long as it does not talk about flags.

def get_decryption(B,c):
    REM.sendline("{},{}".format(B.x, B.y))
    REM.sendline("{},{}".format(c.x, c.y))
    status = REM.recvline()
    if b'cannot afford' in status:
        return -1, None
    balance = int(REM.recvline().strip().split(b': ')[-1])
    return balance, bytes.fromhex(status.strip()[6:-1].decode())

bal, BG = get_decryption(B, c+G)
BG_int = int.from_bytes(BG)
y = modular_sqrt(BG_int**3+a*BG_int+b,p)  # getting the valid y coordinate for the x
point_BG1 = ECPoint(curve, BG_int, y)
point_BG2 = -point_BG1
print(int.to_bytes((point_BG1-G).x, 32,'big'))
print(int.to_bytes((point_BG2-G).x, 32,'big'))

Note that we only get the x coordinate of the given point lifting the point would result in two points and we will need to try with both of them (x,y) and (x,-y)

The last part will also work with

Python7 lines / static highlight

bal, BG = get_decryption(B-G, c)
BG_int = int.from_bytes(BG)
y = modular_sqrt(BG_int**3+a*BG_int+b,p)  # getting the valid y coordinate for the x
point_BG1 = ECPoint(curve, BG_int, y)
point_BG2 = -point_BG1
print(int.to_bytes((point_BG1-P_a).x, 32,'big'))
print(int.to_bytes((point_BG2-P_a).x, 32,'big'))

And we get the flag to part 1 b'\x00\x00ENO{ElGam4l_1s_mult1pl1cativ3}'

Analysing ec.py

Python66 lines / static highlight

from Crypto.Util.number import inverse

class EllipticCurve(object):
        def __init__(self, p, a, b, order = None):
                self.p = p
                self.a = a
                self.b = b
                self.n = order

        def __str__(self):
                return 'y^2 = x^3 + %dx + %d modulo %d' % (self.a, self.b, self.p)

        def __eq__(self, other):
                return (self.a, self.b, self.p) == (other.a, other.b, other.p)

class ECPoint(object):
        def __init__(self, curve, x, y, inf = False):
                self.x = x % curve.p
                self.y = y % curve.p
                self.curve = curve
                self.inf = inf
                if x == 0 and y == 0: self.inf = True

        def copy(self):
                return ECPoint(self.curve, self.x, self.y)

        def __neg__(self):
                return ECPoint(self.curve, self.x, -self.y, self.inf)

        def __add__(self, point):
                p = self.curve.p
                if self.inf:
                        return point.copy()
                if point.inf:
                        return self.copy()
                if self.x == point.x and (self.y + point.y) % p == 0:
                        return ECPoint(self.curve, 0, 0, True)
                if self.x == point.x:
                        lamb = (3*self.x**2 + self.curve.a) * inverse(2 * self.y, p) % p
                else:
                        lamb = (point.y - self.y) * inverse(point.x - self.x, p) % p
                x = (lamb**2 - self.x - point.x) % p
                y = (lamb * (self.x - x) - self.y) % p
                return ECPoint(self.curve,x,y)

        def __sub__(self, point):
                return self + (-point)

        def __str__(self):
                if self.inf: return 'Point(inf)'
                return 'Point(%d, %d)' % (self.x, self.y)

        def __mul__(self, k):
                k = int(k)
                base = self.copy()
                res = ECPoint(self.curve, 0,0,True)
                while k > 0:
                        if k & 1:
                                res = res + base
                        base = base + base
                        k >>= 1
                return res

        def __eq__(self, point):
                return (self.inf and point.inf) or (self.x == point.x and self.y == point.y)

Looking closely at the ECPoint class, one would note that on the initialization of the point with arbitrary x, y coordinates, it works as usual and doesnt check whether the supplied x, y satisfy the curve equation $y^2 = x^3 + ax + b \mod p$

This leads to an interesting vulnerability aka Invalid Curve Attack

Which can be noted by the facts that

The point addtion of two points $P$ and $Q$ over the curve $y^2 = x^3 + ax + b \mod p$ if $P \ne Q$ is independent of both the curve parameters $a$ and $b$
The point doubling i.e $P = Q$ is just dependent on $P$ and $Q$ and $a$ but not on $b$ again

This means that the group addition operation is independent of the parameter $b$ , the number of points in the group of some point $P$ is just dependent on $P$ and $a$ but independent of $b$

So if we choose a curve $C'$ with parameter $b'$ and pick a valid point $P'$ on it, and run the point addition over the original curve $C$ , the order of point $P'$ when used on $C$ will be the same as order of point $P'$ when used on $C'$

This implies that we are not stuck with the original prime order of P-256, but we can vary $b = -3$ such that the order of the curve $C'$ has small factors. We can then easily find points $P'$ with those small factors as their order by using the fact that -
If $G'$ is the generator of $C'$ with order $o = f_1f_2f_3...f_n$ , the point $G' * (o/f_1)$ will have the order $f_1$

Once we have sufficient number of small orders, we can take a chinese remainder theorem over them to recover the original private key d_a

To find the order of curve $C'$ and the corresponding generators, we can utilize the greate library of sagemath

Python19 lines / static highlight

def get_invalid_curves(a,b,n,cutoff=10**5):
    factors, total, i = {}, 0, 0
    while total < n*2:
        i += 1
        try:
            E = EllipticCurve(GF(p), [a,i])
            order = E.order()
            n_facs = order.factor()
        except ArithmeticError: #the parameter i defines a singular curve
            continue
        for prime, power in n_facs:
            if prime > cutoff: # dont take any factors bigger than it
                break
            gen = E.gen(0)*(order//prime)
            factors[prime] = [int(gen[0]), int(gen[1]),i]
            total *= prime
        print(i, total)
    # key : [gen_x, gen_y, b']
    return factors

And we can easily bruteforce a given point

Python4 lines / static highlight

def bruteforce(point, generator, order):
    for i in tqdm(range(order),desc=f"bruteforcing {order=}"):
        if point.y == (generator*i).y:
            return i

We get the following invalid curves and we just searched over 12 values of $b$ !

Python58 lines / static highlight

invalid_curves = {
3: [79692280239272980873245387831874823476097665365069163558817570386218657526967,
   15885657487155030912288031888128427124936813080859472376494663130798867119982, 12],
5: [30463586456259052716174121724723788478797318939762291523651966151233767925799,
   14521026652335616630611219515390291411023400641948957680811406721043438902186, 12],
7: [7494160963166719022445789448670075468300539216220596044581361034311676798234,
   98511591492536111584332615786483658886888663084020823615343609813895225923287, 0],
13: [44238399751822344629155927349410921734336660036385908812849527496419061724190,
   111209137730801733774021088162408683888753865848469665403753663663899601005809, 3],
17: [9274144687945784364291903707116312659963917031850121885976057784793297477861,
   86301980488975426887521079169756244075123784201450393961147458432303011672794, 10],
19: [108657251488837839710095894743866739052486880271258033613510419634191398226376,
   8593905934316229092193387452437731577526088690676465668457131094758391852209, 4],
37: [6829338390266237482283310246665103308891228336319477318479644522260556056309,
   74701668267551028200304338837410580676774474001240882947774298643661074111881, 3],
71: [47689891150662520216418276050802771367044708366511766907543187521601344242001,
   76986723505258444611874235435887405018513758524921757765033540214285112109625, 1],
97: [109352438132789597676269849271161933029115963700376783044214805643475162939438,
   40494199582133551395560104592591896448854412368997470369685720658455096277720, 3],
113: [24758423058742208238204864443231318968571918830166822957638906079202832915346,
   64680694216633390865014362020707904150333340051904806580803858267985332824228, 3],
179: [83115631016490504822328655777895864162660782325660359674792065332260812135544,
   35707141486916353816358123900356888673488260709581875628819622187261352405569, 4],
251: [22589597796365257246296758128505770638799961769310824687868041010311103597978,
   22730375842099404560129412560882492093998724011749582473003682871994848593450, 10],
389: [61737418306809996908630595437832052272700263892021361415640028314169193468679,
   38812622012907358702971098652989910931710935180589252493039839244464470874161, 12],
653: [96946680343613920300091880607027973891460464096741803273170797219756170839615,
   82843786165425711709725308210080928839669646246950072002603300248219042690214, 9],
823: [101591078169875595753109991494905506967978136240961172388509275621325253165256,
   10303457253039461887297603877250453486351456285884301634668455036581356870325, 1],
1151: [4942947285962241518079147671001480777229821084370946279070977308149340420785,
   16048516228456466259745940658231903287363041924322695524263376011778714624389, 7],
1229: [91709248688928381574970306879959143256779911355970659117798234761550615769703,
   85856020107303390216619790339131383987393998107943546966635616939179964220668, 1],
2447: [107091037109612570995136294213336682923913717986054179094643922074841981090569,
   38297847735446351346601186761335949464902974429727652825128988635682228100545, 5],
4003: [69634612360547639692978050736475584000001950346963254134893659331303767659709,
   7267690154676021708711188497795027916155791784399213931355851351510639163175, 6],
7103: [110323527740892356276833844768860449554291010208201255792825053403232044044793,
   34702628194678317337541067016532288256370093140368313590333192034888987762792, 7],
7489: [69497610789705595174058737106242513100950130190920702467431032172354669590563,
   43000989776377667520933328800675765150040604546037676698173382008099239610730, 1],
13003: [69994388431307856080322572731970917270151067511018517619530568914812259046195,
   51645889020375608054366335957352074257130320976341249224010343992676177045239, 4],
16033: [80150849770701280770379260802876332257245651607220436873841708569583336291111,
   67454443034144602807761039494179913732280388550455172486599878108995578176702, 6],
19423: [86763316696116146207846209443089376095966542281990071872698734124275764832625,
   1497373281188841342082112917519408391664673991593710756225816313602284346637, 11],
30203: [27140306769124364253212826889951250714782929180685455599284687702513066987645,
   27465668374540052358785326933904597047991904030378513297677516281690992773738, 1],
52183: [20786893006200668135980517481305198967871522130773700571327256180224225598537,
   38476450712159672989047119873673988596095516096648184067210103163599625447149, 12],
72337: [16864673136043278693040185572303485743677125999233419976437302471094264721938,
   39982110747848740588957884598316010802865483814669576940304708444368796141014, 9],
81173: [41965847134675666863089670621412699297207446259915277832939899605119013001686,
   97595102592346869875749873612676528534971198259624427507680148771098555985918, 8]
}

Attack procedure

For each $b'$ and $G'$ of order $o'$
Send $B = -G'$ and $c = G'$ we will get back x coordinate of $G' * (d_a + 1)$

Note that if we send 0, it will just return b'' so no use
Lift the x coordinate Over C’ to get $G' * (d_a + 1)$ But this wont help us to figure out of the two possible points. For this
Send another $B = -G'$ and $c = G' * 2$ to get back x coordinate of $G' * (d_a + 2)$ To figure out the correct lifting of the x coordinate to get $G' * (d_a + 1)$
Bruteforce and recover the ECDLP to get $d_a' = (d_a + 1) \mod o'$
Combine all $d_a'$ using chinese remainder theorem to get back original $d_a$
PROFIT ????

Python27 lines / static highlight

recovered_order = {}
for order, (gen_x, gen_y, b_i) in invalid_curves.items():
    gen = ECPoint(curve, gen_x, gen_y)
    bal, BG = get_decryption(-gen, gen) # gen*(da+1)
    BG_int = int.from_bytes(BG)
    y = modular_sqrt(BG_int**3+a*BG_int+b_i,p)
    point_BG = ECPoint(curve, BG_int, y)

    bal, BG2 = get_decryption(-gen, gen*2) # gen*(da+2)
    BG_int2 = int.from_bytes(BG2)
    if (point_BG+gen).x == BG_int2:
        recovered_order[order] = bruteforce(point_BG, gen, order)
    elif (point_BG-gen).x == BG_int2:
        recovered_order[order] = bruteforce(-point_BG, gen, order)
    else:
        print("something went wrong")

mods, values = [],[]
for i,v in recovered_order.items():
    mods.append(i)
    values.append((v-1)%i)

d_a = crt(mods,values)[0]

key_int = (d_a >> (8*16)) ^ (d_a & 0xffffffffffffffffffffffffffffffff)
key = key_int.to_bytes(16, 'big')
print(AES.new(key, AES.MODE_ECB).decrypt(flag2_enc))

ENO{be5t_th1nk_out_0f_th3_curv3}

Solve script

solve.py