RSA密钥生成与读取

需要安装pycryptodome模块

公钥生成：

from Crypto.PublicKey import RSA

p= 787228223375328491232514653709
q= 814212346998672554509751911073
n= 640970939378021470187479083920100737340912672709639557619757
d= 590103645243332826117029128695341159496883001869370080307201
e= 65537


rsa_components = (n, e)
keypair = RSA.construct(rsa_components)
with open('pubckey.pem', 'wb') as f :
    f.write(keypair.exportKey())

私钥生成：

from Crypto.PublicKey import RSA

p= 787228223375328491232514653709
q= 814212346998672554509751911073
n= 640970939378021470187479083920100737340912672709639557619757
d= 590103645243332826117029128695341159496883001869370080307201
e= 65537


rsa_components = (n,e,d,p,q)
keypair = RSA.construct(rsa_components)
with open('private1.pem', 'wb') as f :
    f.write(keypair.exportKey()) 

公钥读取：

from Crypto.PublicKey import RSA
with open("pubckey.pem","rb") as f:
    key = RSA.import_key(f.read())
    print('n = %d' % key.n)
    print('e = %d' % key.e) 

私钥读取：

from Crypto.PublicKey import RSA
with open("private1.pem","rb") as f:
    key = RSA.import_key(f.read())
    print('n = %d' % key.n)
    print('e = %d' % key.e)
    print('d = %d' % key.d)
    print('p = %d' % key.p)
    print('q = %d' % key.q) 

出题脚本 -基于N分解的题目

import libnum
import gmpy2
from Crypto.PublicKey import RSA

p=libnum.generate_prime(1024)
#下一个素数
q=int(gmpy2.next_prime(p))
e=65537
m="flag{a272722c1db834353ea3ce1d9c71feca}"
m=libnum.s2n(m)
n=p*q
c=pow(m,e,n)
flag_c=libnum.n2s(c)
rsa_components = (n, e)
keypair = RSA.construct(rsa_components)
with open('pubckey1.pem', 'wb') as f :
    f.write(keypair.exportKey())
with open("flag.txt","wb") as f:
    f.write(flag_c) 
    

解题脚本：

import libnum
import gmpy2
from Crypto.PublicKey import RSA


def isqrt(n):
  x = n
  y = (x + n // x) // 2
  while y < x:
    x = y
    y = (x + n // x) // 2
  return x

def fermat(n, verbose=True):
    a = isqrt(n) # int(ceil(n**0.5))
    b2 = a*a - n
    b = isqrt(n) # int(b2**0.5)
    count = 0
    while b*b != b2:
        # if verbose:
        #     print('Trying: a=%s b2=%s b=%s' % (a, b2, b))
        a = a + 1
        b2 = a*a - n
        b = isqrt(b2) # int(b2**0.5)
        count += 1
    p=a+b
    q=a-b
    assert n == p * q
    # print('a=',a)
    # print('b=',b)
    # print('p=',p)
    # print('q=',q)
    # print('pq=',p*q)
    return p, q

with open("pubckey1.pem","rb") as f:
    key = RSA.import_key(f.read())
    n=key.n
    e=key.e

with open("flag.txt","rb") as f:
    c=f.read()
    c=libnum.s2n(c)

#费马分解,
n1=fermat(n)
p=n1[0]
q=n1[1]
phi_n=(p-1)*(q-1)
#求逆元
d=libnum.invmod(e,phi_n)
m=pow(c,d,n)
print(m)
print(libnum.n2s(int(m)).decode())