简述

结合网上大量的文献，省略部分繁琐的过程和工具的使用，更方便快捷的解出RSA加密。

Kali终端 （注意带.pem文件）

openssl rsa -pubin -text -modulus -in warmup -in pub.pem

得到：
E= 65537
N= DD0289BC00E0C37B136A91B4071D0A398549F94CB885F756B73F9F3A14A1A56D

进制转换

将N值转换成10进制
N= 99965623838843374711411183391444104726307314029768628656811347707805304989037

Yafu工具

首先打开cmd，进入yafu所在路径
输入：

yafu-64.exe factor(30578675145816634962204467309994126955968568987449100734690153203822106214253)

得到：
P= 301421686937198008750983790559102741399
Q= 331647085034301039007512063728344459163

Python脚本

# 从公钥里面提取n 和 e
with open('rsa_public_key.pem', 'r') as f:
    key = RSA.import_key(f.read())
e = key.e
n = key.n
print('e = %d\nn = %d' % (e, n))

# p q e 得 d
p = 301421686937198008750983790559102741399
q = 331647085034301039007512063728344459163
phi = (p - 1) * (q - 1)
d = inverse(e, phi)
print('d =', d)

key = rsa.PrivateKey(n, e, d, q, p)
with open("venus.en", "rb") as f:
    f = f.read()
    print(rsa.decrypt(f, key))
# e = 65537
# n = 86934482296048119190666062003494800588905656017203025617216654058378322103517
# d = 81176168860169991027846870170527607562179635470395365333547868786951080991441
# b'flag{decrypt_256}\n'

填入P、Q
第一个open打开.pem
第二个open打开.enc