【zer0pts CTF 2022】 Anti-Fermat

1. subject

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 = 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
#c = 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

2. Reappear

The key to see p,q How to generate ,p Is random 1024 Bitprime number ,q = next_prime(p ^ ((1<<1024)-1)).

1<<1024-1=2**1024-1 Into 2 Hexadecimal is 1024 individual 1, Because of a binary XOR 1 Is to take its inverse , It means 1 become 0,0 become 1. therefore p^2**1024-1 Result x It's a p Every bit of is reversed , therefore x+p=2**1024-1.

because q yes x So q=x+r, Take it in p+q=2**1024-1+r, therefore p+q It's about equal to 2**1024.

  And because of   hold p+q Take it in

therefore 4n=2**2048-(p-q)**2,p-q=, because p+q= , therefore p=

Then blast back p That's it .

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+-----------------------------------------------------------+'