RSA 攻击

在RSA中e也称为加密指数。由于e是可以随意选取的,选取小一点的e可以缩短加密时间,但是选取不当的话,就会造成安全问题。

e=3时的小明文攻击
介绍:
当e=3时,如果明文过小,导致明文的三次方仍然小于n,那么通过直接对密文三次开方,即可得到明文。

这里写图片描述
当e=3,暴力是个好方法
m^e=kn+c
m=(kn+c)^1/e
m是整数时跳出循环

#! /usr/bin/env python
# -*- coding: utf-8 -*-
import gmpy2
gmpy2.get_context().precision = 1000 //设置精度
c=431396049519259356426983102577521801906916650819409770125821662319298730692378063287943809162107163618549043548748362517694341497565980142708852098826686158246523270988062866178454564393347346790109724455155942667492571325721344535616869
N=69125924513738898062336798085735284919075483446361453239926516926518012932962624644852640880359045605575998083615990230652355588280160345758926109633882386877041312214620631872182423375717799704396440136980182986299545351418528973531328797438870658736243635425250389995560895709088731225183252726124183169549
# d=c**(1.0/3)
i=0
# c=100
# N=9
while(1):
    a=gmpy2.root(c+i*N, 3)
    # print a
    # print a-(int)(a)
    if a-(int)(a)==0.00000:
        print a
    #   break
    i+=1
    # print 1
    # d=i*n+c**(1.0/3)
    # print  ('a=%.100f' %(d))
    # i+=1
    if(i==1000000000):
        break

猜你喜欢

转载自blog.csdn.net/qq_38677814/article/details/80139606
RSA