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【zer0pts CTF 2022】 Anti-Fermat

2022-07-27 21:07:00 【[email protected]】

1.题目

from Crypto.Util.number import isPrime, getStrongPrime
from gmpy import next_prime
from secret import flag
 
# Anti-Fermat Key Generation
p = getStrongPrime(1024)
q = next_prime(p ^ ((1<<1024)-1))
n = p * q
e = 65537
 
# Encryption
m = int.from_bytes(flag, 'big')
assert m < n
c = pow(m, e, n)
 
print('n = {}'.format(hex(n)))
print('c = {}'.format(hex(c)))
 
#n = 0x1ffc7dc6b9667b0dcd00d6ae92fb34ed0f3d84285364c73fbf6a572c9081931be0b0610464152de7e0468ca7452c738611656f1f9217a944e64ca2b3a89d889ffc06e6503cfec3ccb491e9b6176ec468687bf4763c6591f89e750bf1e4f9d6855752c19de4289d1a7cea33b077bdcda3c84f6f3762dc9d96d2853f94cc688b3c9d8e67386a147524a2b23b1092f0be1aa286f2aa13aafba62604435acbaa79f4e53dea93ae8a22655287f4d2fa95269877991c57da6fdeeb3d46270cd69b6bfa537bfd14c926cf39b94d0f06228313d21ec6be2311f526e6515069dbb1b06fe3cf1f62c0962da2bc98fa4808c201e4efe7a252f9f823e710d6ad2fb974949751
#c = 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

2.复现

重点看p,q的生成方式，p是随机的1024位素数，q = next_prime(p ^ ((1<<1024)-1))。

1<<1024-1=2**1024-1化为2进制就是1024个1,由于一个二进制数异或1就是取它的反，意思就是1变成0,0变成1。所以p^2**1024-1的结果x是把p的每一位取反，所以x+p=2**1024-1。

由于q是x的下一个素数所以q=x+r，带进去p+q=2**1024-1+r，所以p+q就约等于2**1024。

 又由于n=(\frac{p+q}{2})^{2}-(\frac{p-q}{2})^{2} 把p+q带进去n=(\frac{2^{1024}}{2})^{2}-(\frac{p-q}{2})^{2}=\frac{2^{2048}}{4}-\frac{(p-q)^{2}}{4}

所以4n=2**2048-(p-q)**2,p-q=\sqrt{2^{2048}-4n},因为p+q= 2^{1024},所以p=\frac{\sqrt{2^{2048}-4n}+2^{1024}}{2}

然后再向后面爆破p就行了。

import gmpy2
import sympy
import libnum
n = int(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)
c = int(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)
e = 65537
p=(gmpy2.iroot(pow(2,2048)-4*n,2)[0]+pow(2,1024))//2
# p=(2**1024+gmpy2.iroot((2**1024)**2-4*n,2)[0])//2
p=int(p)
while(1):
    p=sympy.nextprime(p)
    if(n%p==0):
        print(p)
        break
q=n//p
phi=(p-1)*(q-1)
d=gmpy2.invert(e,phi)
m=pow(c,d,n)
flag=libnum.n2s(int(m))
print(flag)
# b'Good job! Here is the flag:\n+-----------------------------------------------------------+\n| zer0pts{F3rm4t,y0ur_m3th0d_n0_l0ng3r_w0rks.y0u_4r3_f1r3d} |\n+-----------------------------------------------------------+'

 

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版权声明
本文为[[email protected]]所创，转载请带上原文链接，感谢
https://blog.csdn.net/qq_61774705/article/details/124987392

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