dp in RSA

最新推荐文章于 2024-04-10 21:31:10 发布

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最新推荐文章于 2024-04-10 21:31:10 发布

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分类专栏： CTF writeup 文章标签：信息安全 rsa python

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dp leaking

0x01 Calculation idea

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0x02 What about “p”

ergodic x

got (p-1)

got p

d = gmpy2.invert(e,phi)%phi

0x03 Examples

e = 65537
n = 156808343598578774957375696815188980682166740609302831099696492068246337198792510898818496239166339015207305102101431634283168544492984586566799996471150252382144148257236707247267506165670877506370253127695314163987084076462560095456635833650720606337852199362362120808707925913897956527780930423574343287847
c = 108542078809057774666748066235473292495343753790443966020636060807418393737258696352569345621488958094856305865603100885838672591764072157183336139243588435583104423268921439473113244493821692560960443688048994557463526099985303667243623711454841573922233051289561865599722004107134302070301237345400354257869
dp = 734763139918837027274765680404546851353356952885439663987181004382601658386317353877499122276686150509151221546249750373865024485652349719427182780275825

#!/usr/bin/python
# -*- coding:utf-8 -*-

import  gmpy2
import  rsa
import  binascii

e = 65537
n = 156808343598578774957375696815188980682166740609302831099696492068246337198792510898818496239166339015207305102101431634283168544492984586566799996471150252382144148257236707247267506165670877506370253127695314163987084076462560095456635833650720606337852199362362120808707925913897956527780930423574343287847
c = 108542078809057774666748066235473292495343753790443966020636060807418393737258696352569345621488958094856305865603100885838672591764072157183336139243588435583104423268921439473113244493821692560960443688048994557463526099985303667243623711454841573922233051289561865599722004107134302070301237345400354257869
dp = 734763139918837027274765680404546851353356952885439663987181004382601658386317353877499122276686150509151221546249750373865024485652349719427182780275825

temp = dp*e
for i in range(1,e) :
    if (temp-1)%i==0:
        x = (temp-1)//i+1
        y = n%x
        if y==0:
            p=x
            break
q = n//p #'//'代表向下取整,'/'得到的是浮点数
d = gmpy2.invert(e,(p-1)*(q-1))
key = rsa.PrivateKey(n,e,d,p,q)
flag = pow(c,d,n)
print(binascii.unhexlify(hex(flag)[2:]))  #unhexlify()的作用是返回16进制数对应的字符串

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