已知n,e,c,分解n,得p,q,求m

题目:

from Crypto.Util.number import *

flag = b'NSSCTF{******}'

p = getPrime(256)
q = getPrime(256)
n = p*q
e = 65537
phi = (p-1)*(q-1)

m = bytes_to_long(flag)

c = pow(m, e, n)

print(f'n = {n}')
print(f'e = {e}')
print(f'c = {c}')

'''
n = 7382582015733895208810490097582153009797420348201515356767397357174775587237553842395468027650317457503579404097373070312978350435795210286224491315941881
e = 65537
c = 6511001389892474870028836129813814173158254564777610289284056550272120510686249909340499673868720839756059423749304765055919251717618117507007046973023557
'''

解题思路:

分解合数n得到质数p和q

解答:

• 分解n

from Crypto.Util.number import *
n = 7382582015733895208810490097582153009797420348201515356767397357174775587237553842395468027650317457503579404097373070312978350435795210286224491315941881
p = 70538125404512947763739093348083497980212021962975762144416432920656660487657
q = 104660876276442216612517835199819767034152013287345576481899196023866133215633
e = 65537
c = 6511001389892474870028836129813814173158254564777610289284056550272120510686249909340499673868720839756059423749304765055919251717618117507007046973023557

n=p*q
phi = (p-1)*(q-1)
d = inverse(e, phi)
m = pow(c, d, n)

print(long_to_bytes(m))
#NSSCTF{factordb_is_useful}