国城杯2024-Crypto

crypto

babyRSA

from secret import flag
from Crypto.Util.number import*
from gmpy2 import*

flag = b'D0g3xGC{****************}'

def gen_key(p, q):
    public_key = p*p*q
    e = public_key
    n = p*q
    phi_n = (p-1)*(q-1)
    private_key = inverse(e,phi_n)
    return public_key,private_key,e

p = getPrime(512)
q = getPrime(512)

N,d,e = gen_key(p,q)

c = gmpy2.powmod(bytes_to_long(flag),e,N)

print(N)
print(d)
print(c)

参考：blog.csdn.net/MikeCoke/article/details/113915715

Schmidt-Samoa密码:$N=p^2*q,e=N$

解密：

$$m=c^d\mod p*q\\$$

获取p*q：

$$e*d=1\mod (p-1)*(q-1)\\ N*d=1+k*(p-1)*(q-1)\\ 由欧拉定理得：2^{N*d}\equiv2^{1+k*(p-1)*(q-1)}\equiv=2\mod p*q=2+k*pq\\ 与N求gcd即可得到p*q$$

from gmpy2 import*
from libnum import*
N = 539403894871945779827202174061302970341082455928364137444962844359039924160163196863639732747261316352083923762760392277536591121706270680734175544093484423564223679628430671167864783270170316881238613070741410367403388936640139281272357761773388084534717028640788227350254140821128908338938211038299089224967666902522698905762169859839320277939509727532793553875254243396522340305880944219886874086251872580220405893975158782585205038779055706441633392356197489
d = 58169755386408729394668831947856757060407423126014928705447058468355548861569452522734305188388017764321018770435192767746145932739423507387500606563617116764196418533748380893094448060562081543927295828007016873588530479985728135015510171217414380395169021607415979109815455365309760152218352878885075237009
c = 82363935080688828403687816407414245190197520763274791336321809938555352729292372511750720874636733170318783864904860402219217916275532026726988967173244517058861515301795651235356589935260088896862597321759820481288634232602161279508285376396160040216717452399727353343286840178630019331762024227868572613111538565515895048015318352044475799556833174329418774012639769680007774968870455333386419199820213165698948819857171366903857477182306178673924861370469175

pq = gcd(pow(2,d*N,N)-2,N)

m = pow(c,d,pq)
print(n2s(m))
#b'D0g3xGC{W1sh_Y0u_Go0d_L@ucK-111}'

Curve

#sagemath
from Crypto.Util.number import *

def add(P, Q):
    (x1, y1) = P
    (x2, y2) = Q

    x3 = (x1*y2 + y1*x2) * inverse(1 + d*x1*x2*y1*y2, p) % p
    y3 = (y1*y2 - a*x1*x2) * inverse(1 - d*x1*x2*y1*y2, p) % p
    return (x3, y3)

def mul(x, P):
    Q = (0, 1)
    while x > 0:
        if x % 2 == 1:
            Q = add(Q, P)
        P = add(P, P)
        x = x >> 1
    return Q

p = 64141017538026690847507665744072764126523219720088055136531450296140542176327
a = 362
d = 7
e=0x10001

gx=bytes_to_long(b'D0g3xGC{*****************}')

PR.<y>=PolynomialRing(Zmod(p))
f=(d*gx^2-1)*y^2+(1-a*gx^2)
gy=int(f.roots()[0][0])

assert (a*gx^2+gy^2)%p==(1+d*gx^2*gy^2)%p

G=(gx,gy)

eG = mul(e, G)
print(eG)

#eG = (34120664973166619886120801966861368419497948422807175421202190709822232354059, 11301243831592615312624457443883283529467532390028216735072818875052648928463)

直接套鸡块师傅的脚本就行了：https://tangcuxiaojikuai.xyz/post/187210a7.html

from Crypto.Util.number import *
p = 64141017538026690847507665744072764126523219720088055136531450296140542176327
a = 362
d = 7
e=0x10001
eG = (34120664973166619886120801966861368419497948422807175421202190709822232354059, 11301243831592615312624457443883283529467532390028216735072818875052648928463)
c=1
#part2 map to ECC
F = GF(p)
dd = F(d*c^4)
A = F(2) * F(a+dd) / F(a-dd)
B = F(4) / F(a-dd)
a = F(3-A^2) / F(3*B^2)
b = F(2*A^3-9*A) / F(27*B^3)

def edwards_to_ECC(x,y):
    x1 = F(x) / F(c)
    y1 = F(y) / F(c)
    #now curve is a*x^2+y^2 = 1+dd*x^2*y^2

    x2 = F(1+y1) / F(1-y1)
    y2 = F(x2) / F(x1)
    #now curve is By^2 = x^3 + Ax^2 + x

    x3 = (F(3*x2) + F(A)) / F(3*B)
    y3 = F(y2) / F(B)
    #now curve is y^2 = x^3 + ax + b

    return (x3,y3)
 
def ECC_to_edwards(x,y):
    x2 = (F(x) * F(3*B) - F(A)) / F(3)
    y2 = F(y) * F(B)
    #now curve is By^2 = x^3 + Ax^2 + x

    x1 = F(x2) / F(y2)
    y1 = F(1) - (F(2) / F(x2+1))
    #now curve is a*x^2+y^2 = 1+dd*x^2*y^2

    x_ = F(x1) * F(c)
    y_ = F(y1) * F(c)
    #now curve is a*x^2+y^2 = c^2(1+d*x^2*y^2)
    
    return (x_,y_)
 
E = EllipticCurve(GF(p), [a, b])
order = E.order()
eG = E(edwards_to_ECC(eG[0],eG[1]))
t = inverse(e,order)
G = t*eG
G = ECC_to_edwards(G[0],G[1])
print(long_to_bytes(int(G[0])))
#b'D0g3xGC{SOlvE_The_Edcurv3}'

EZ_sign

from Crypto.Util.number import *
from gmpy2 import *
from hashlib import*
import random,os

flag = b'D0g3xGA{***************}'
msg = b'e = ?'

def sign(pub, pri, k):
    (p,q,g,y) = pub
    x = pri
    r = int(powmod(g, k, p) % q)
    H = bytes_to_long(sha1(os.urandom(20)).digest())
    s = int((H + r * x) * invert(k, q) % q)
    return (H,r,s)

k1 = getPrime(64)
k2 = k1 ** 2
pri = bytes_to_long(msg)
a = 149328490045436942604988875802116489621328828898285420947715311349436861817490291824444921097051302371708542907256342876547658101870212721747647670430302669064864905380294108258544172347364992433926644937979367545128905469215614628012983692577094048505556341118385280805187867314256525730071844236934151633203
b = 829396411171540475587755762866203184101195238207
g = 87036604306839610565326489540582721363203007549199721259441400754982765368067012246281187432501490614633302696667034188357108387643921907247964850741525797183732941221335215366182266284004953589251764575162228404140768536534167491117433689878845912406615227673100755350290475167413701005196853054828541680397
y = 97644672217092534422903769459190836176879315123054001151977789291649564201120414036287557280431608390741595834467632108397663276781265601024889217654490419259208919898180195586714790127650244788782155032615116944102113736041131315531765220891253274685646444667344472175149252120261958868249193192444916098238

pub = (a, b, g, y)

H1, r1, s1 = sign(pub, pri, k1)

H2, r2, s2 = sign(pub, pri, k2)

p = getPrime(128)
q = getPrime(128)
n = p * q
c = powmod(bytes_to_long(flag), e, n)

C = p**2 + q**2

print(f'(H1, r1, s1) = {H1}, {r1}, {s1}')
print(f'(H2, r2, s2) = {H2}, {r2}, {s2}')
print(c)
print(C)

前面按照线性k的dsa正常分析

$$k_2=a*k_1+b\\ r_1=(g^{k_1}\mod p) \mod q\quad\quad s_1=k_1^{-1}*(H_1+x*r_1)\mod q\\ r_2=(g^{k_2}\mod p) \mod q\quad\quad s_2=k_2^{-1}*(H_2+x*r_2)\mod q\\ 签名对(r,s_1),(r,s_2)$$

$$s_1的式子乘以r_2:s_1*k_1*r_2=H_1*r_2+x*r_1*r_2\mod q\\\\ s_2的式子把k_2=的式子带进去:\\ s_2=(a*k_1+b)^{-1}*(H_2+x*r_2)\mod q\\ s_2*(a*k_1+b)=H_2+x*r_2\mod q\\ s_2*a*k_1+s_2*b=H_2+x*r_2\mod q\\ s_2*a*k_1*r_1+s_2*b*r_1=H_2*r_1+x*r_2*r_1\mod q\\\\ 两式相减，得：\\ k_1*(r_2*s_1-a*r_1*s_2)=H_1*r_2-H_2*r_1+b*r_1*s_2\mod q\\ 到这一步后，我们知道a=k,b=0,所以有:\\ s_1r_2k_1-r_1s_2k_1^2=H_1r_2-H_2r_1\mod q$$

之后利用small_roots求解出k，然后正常解密出x就得到了msg中得e

之后得RSA需要利用C求出p和q，但是直接用sagemath中得two_squares得求出来得不是我们需要的，之后就找了个这个网站可以多个解，然后慢慢试：

Generic two integer variable equation solver

from Crypto.Util.number import *
(H1, r1, s1) = 659787401883545685817457221852854226644541324571, 334878452864978819061930997065061937449464345411, 282119793273156214497433603026823910474682900640
(H2, r2, s2) = 156467414524100313878421798396433081456201599833, 584114556699509111695337565541829205336940360354, 827371522240921066790477048569787834877112159142

p = 149328490045436942604988875802116489621328828898285420947715311349436861817490291824444921097051302371708542907256342876547658101870212721747647670430302669064864905380294108258544172347364992433926644937979367545128905469215614628012983692577094048505556341118385280805187867314256525730071844236934151633203
q = 829396411171540475587755762866203184101195238207
g = 87036604306839610565326489540582721363203007549199721259441400754982765368067012246281187432501490614633302696667034188357108387643921907247964850741525797183732941221335215366182266284004953589251764575162228404140768536534167491117433689878845912406615227673100755350290475167413701005196853054828541680397

x=(H1*r2-H2*r1)%q

PR.<k>=PolynomialRing(Zmod(q))
f=s1*r2*k-s2*r1*k^2-x
f=f.monic()
k=int(f.small_roots(X=2**64, beta=0.4,epsilon=0.01)[0])

print(long_to_bytes(int(((s1*k-H1)*inverse(r1,q))%q)))
#b'e = 44519'
c = 18947793008364154366082991046877977562448549186943043756326365751169362247521
C = 179093209181929149953346613617854206675976823277412565868079070299728290913658

p=302951519846417861008714825074296492447
q=295488723650623654106370451762393175957
e = 44519
n=p*q
phi=(p-1)*(q-1)
d=inverse_mod(e,phi)
m=int(pow(c,d,n))
print(long_to_bytes(m))
#b'D0g3xGC{EZ_DSA_@nd_C0mplex_QAQ}'