from gmpy2 import invert# 欧几里得算法def egcd(a, b): if a == 0: return (b, 0, 1) else: g, y, x = egcd(b % a, a) return (g, x - (b // a) * y, y)def main(): n = 22708078815885011462462049064339185898712439277226831073457888403129378547350292420267016551819052430779004755846649044001024141485283286483130702616057274698473611149508798869706347501931583117632710700787228016480127677393649929530416598686027354216422565934459015161927613607902831542857977859612596282353679327773303727004407262197231586324599181983572622404590354084541788062262164510140605868122410388090174420147752408554129789760902300898046273909007852818474030770699647647363015102118956737673941354217692696044969695308506436573142565573487583507037356944848039864382339216266670673567488871508925311154801 c1 = 22322035275663237041646893770451933509324701913484303338076210603542612758956262869640822486470121149424485571361007421293675516338822195280313794991136048140918842471219840263536338886250492682739436410013436651161720725855484866690084788721349555662019879081501113222996123305533009325964377798892703161521852805956811219563883312896330156298621674684353919547558127920925706842808914762199011054955816534977675267395009575347820387073483928425066536361482774892370969520740304287456555508933372782327506569010772537497541764311429052216291198932092617792645253901478910801592878203564861118912045464959832566051361 c2 = 18702010045187015556548691642394982835669262147230212731309938675226458555210425972429418449273410535387985931036711854265623905066805665751803269106880746769003478900791099590239513925449748814075904017471585572848473556490565450062664706449128415834787961947266259789785962922238701134079720414228414066193071495304612341052987455615930023536823801499269773357186087452747500840640419365011554421183037505653461286732740983702740822671148045619497667184586123657285604061875653909567822328914065337797733444640351518775487649819978262363617265797982843179630888729407238496650987720428708217115257989007867331698397 e1 = 11187289 e2 = 9647291 s = egcd(e1, e2) s1 = s[1] s2 = s[2] # 求模反元素 if s1<0: s1 = - s1 c1 = invert(c1, n) elif s2<0: s2 = - s2 c2 = invert(c2, n) m = pow(c1,s1,n)*pow(c2,s2,n) % n print (libnum.n2s(m))if __name__ == '__main__': main()

实现基于扩展欧几里得算法和模反元素计算的RSA共模攻击
为了实现基于扩展欧几里得算法和模反元素计算的RSA共模攻击，可以按照如下方式编写Python代码：
使用gmpy2库进行高效的大数运算
import gmpy2

def egcd(a, b):
"""Compute the greatest common divisor of a and b along with coefficients"""
if a == 0:
return (b, 0, 1)
else:
gcd, x1, y1 = egcd(b % a, a)
return (gcd, y1 - (b // a) * x1, x1)

def modinv(e, phi_n):
"""Calculate modular multiplicative inverse using Extended Euclidean Algorithm"""
gcd, x, _ = egcd(e, phi_n)
if gcd != 1:
raise Exception('Modular inverse does not exist')
else:
return x % phi_n

def rsa_common_modulus_attack(ciphertexts, public_keys):
"""
Perform RSA Common Modulus Attack given two ciphertexts encrypted under different exponents but same modulus.

:param ciphertexts: Tuple containing both ciphertext values c1 and c2 corresponding to each encryption exponent pair.
:param public_keys: List or tuple holding tuples of form (n,e), where n is shared between keys while e differs.
:return: Plaintext message recovered from provided inputs.
"""
# Unpack input parameters into variables for clarity
((c1, c2), [(n, e1), (_, e2)]) = zip(ciphertexts, public_keys)

# Compute Bezout's identity coefficient s,t such that se1-te2=1 via extended GCD function above
_, s, t = egcd(e1, e2)

# Calculate plaintext by raising respective powers according to CRT formula derived earlier
m1 = pow(gmpy2.invert(s*e1,phi(n)), abs(t), n)*pow(c1,abs(s),n)%n

return int(m1).to_bytes((int(m1).bit_length() + 7)//8,'big').decode()

# Helper functions used within main procedure
def phi(n):
"""Euler totient function φ(n)=n(1−p)(1−q)...for prime factors p,q,...of integer argument 'n'."""
result = n # Initialize result as number itself initially
p = 2 # Start checking divisibility starting at smallest possible factor i.e., 2
while p*p <= n:
if n%p==0:
while n%p==0:
n//=p
result-=result//p
p+=1
if n>1:
result -= result // n
return result
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此段代码实现了完整的RSA共模攻击流程[^1]。具体来说，这段程序定义了几项辅助函数用于执行必要的数学操作，并最终通过中国剩余定理(CRT)的形式恢复原始消息。
请注意，在实际应用中应当谨慎处理密钥材料的安全性和合法性验证等问题；此外，上述示例假设已知两个不同的公钥指数(e_1) 和 (e_2)[^2] ，以及相应的密文对 ((c_1,c_2))，这些条件共同构成了实施此类攻击的基础前提。