BUUCTF RSA题目全解3
1.[WUSTCTF2020]babyrsa
爆破n,就行了
from gmpy2 import*
from libnum import*
p = 189239861511125143212536989589123569301
q = 386123125371923651191219869811293586459
c = 28767758880940662779934612526152562406674613203406706867456395986985664083182
n = 73069886771625642807435783661014062604264768481735145873508846925735521695159
e = 65537
phi = (p-1)*(q-1)
d = invert(e,phi)
m =pow(c,d,n)
print(n2s(m))
flag{just_@_piece_0f_cak3}
2.RSA4
参考文章
CTF ——crypto ——RSA原理及各种题型总结
RAS4
from gmpy2 import *
from Crypto.Util.number import *
from functools import reduce
# 将5进制数转换为10进制数 int('',5)
N1 = int('331310324212000030020214312244232222400142410423413104441140203003243002104333214202031202212403400220031202142322434104143104244241214204444443323000244130122022422310201104411044030113302323014101331214303223312402430402404413033243132101010422240133122211400434023222214231402403403200012221023341333340042343122302113410210110221233241303024431330001303404020104442443120130000334110042432010203401440404010003442001223042211442001413004',5)
c1 = int('310020004234033304244200421414413320341301002123030311202340222410301423440312412440240244110200112141140201224032402232131204213012303204422003300004011434102141321223311243242010014140422411342304322201241112402132203101131221223004022003120002110230023341143201404311340311134230140231412201333333142402423134333211302102413111111424430032440123340034044314223400401224111323000242234420441240411021023100222003123214343030122032301042243',5)
N2 = int('302240000040421410144422133334143140011011044322223144412002220243001141141114123223331331304421113021231204322233120121444434210041232214144413244434424302311222143224402302432102242132244032010020113224011121043232143221203424243134044314022212024343100042342002432331144300214212414033414120004344211330224020301223033334324244031204240122301242232011303211220044222411134403012132420311110302442344021122101224411230002203344140143044114',5)
c2 = int('112200203404013430330214124004404423210041321043000303233141423344144222343401042200334033203124030011440014210112103234440312134032123400444344144233020130110134042102220302002413321102022414130443041144240310121020100310104334204234412411424420321211112232031121330310333414423433343322024400121200333330432223421433344122023012440013041401423202210124024431040013414313121123433424113113414422043330422002314144111134142044333404112240344',5)
N3 = int('332200324410041111434222123043121331442103233332422341041340412034230003314420311333101344231212130200312041044324431141033004333110021013020140020011222012300020041342040004002220210223122111314112124333211132230332124022423141214031303144444134403024420111423244424030030003340213032121303213343020401304243330001314023030121034113334404440421242240113103203013341231330004332040302440011324004130324034323430143102401440130242321424020323',5)
c3 = int('10013444120141130322433204124002242224332334011124210012440241402342100410331131441303242011002101323040403311120421304422222200324402244243322422444414043342130111111330022213203030324422101133032212042042243101434342203204121042113212104212423330331134311311114143200011240002111312122234340003403312040401043021433112031334324322123304112340014030132021432101130211241134422413442312013042141212003102211300321404043012124332013240431242',5)
N = [N1,N2,N3]
c = [c1,c2,c3]
# 中国剩余定理算法
def chinese_remainder(modulus, remainders):
Sum = 0
prod = reduce(lambda a, b: a*b, modulus)
for m_i, r_i in zip(modulus, remainders):
p = prod // m_i
Sum += r_i * (inverse(p,m_i)*p)
return Sum % prod
e = 3
# print(chinese_remainder(N,c))
pow_m_e = chinese_remainder(N,c)
# pow_m_e = 17446992834638639179129969961058029457462398677361658450137832328330435503838651797276948890990069700515669656391607670623897280684064423087023742140145529356863469816868212911716782075239982647322703714504545802436551322108638975695013439206776300941300053940942685511792851350404139366581130688518772175108412341696958930756520037
m = iroot(pow_m_e,e)[0]
print(long_to_bytes(m))
flag{D4mn_y0u_h4s74d_wh47_4_b100dy_b4s74rd!}
3.[AFCTF2018]你能看出这是什么加密么
简单的RSA加密
from gmpy2 import*
from Crypto.Util.number import long_to_bytes
p=int('0x928fb6aa9d813b6c3270131818a7c54edb18e3806942b88670106c1821e0326364194a8c49392849432b37632f0abe3f3c52e909b939c91c50e41a7b8cd00c67d6743b4f',16)
q=int('0xec301417ccdffa679a8dcc4027dd0d75baf9d441625ed8930472165717f4732884c33f25d4ee6a6c9ae6c44aedad039b0b72cf42cab7f80d32b74061',16)
e=int('0x10001',16)
c=int('0x70c9133e1647e95c3cb99bd998a9028b5bf492929725a9e8e6d2e277fa0f37205580b196e5f121a2e83bc80a8204c99f5036a07c8cf6f96c420369b4161d2654a7eccbdaf583204b645e137b3bd15c5ce865298416fd5831cba0d947113ed5be5426b708b89451934d11f9aed9085b48b729449e461ff0863552149b965e22b6',16)
phi = (q-1)*(p-1)
n=p*q
d=invert(e,phi)
m=pow(c,d,n)
print(long_to_bytes(m))
flag{R54_|5_$0_$imp13}
4.[HDCTF2019]together（共模攻击）
先解析公钥
对于题目给的这两个flag文件，去直接base64解密，发现解密失败。将base64编码转换为unicode，再转数字。
python之将byte转换为int类型函数 int.from_bytes 详解与原码反码补码的简单介绍
# python3
import gmpy2, libnum, base64
def exgcd(a, b):
if b==0: return 1, 0
x, y = exgcd(b, a%b)
return y, x-a//b*y
n = 14853081277902411240991719582265437298941606850989432655928075747449227799832389574251190347654658701773951599098366248661597113015221566041305501996451638624389417055956926238595947885740084994809382932733556986107653499144588614105694518150594105711438983069306254763078820574239989253573144558449346681620784979079971559976102366527270867527423001083169127402157598183442923364480383742653117285643026319914244072975557200353546060352744263637867557162046429886176035616570590229646013789737629785488326501654202429466891022723268768841320111152381619260637023031430545168618446134188815113100443559425057634959299
e1 = 2333
c1 = int.from_bytes(base64.b64decode('R3Noy6r3WLItytAmb4FmHEygoilucEEZbO9ZYXx5JN03HNpBLDx7fXd2fl+UL5+11RCs/y0qlTGURWWDtG66eNLzGwNpAKiVj6I7RtUJl2Pcm3NvFeAFwI9UsVREyh7zIV6sI9ZP8l/2GVDorLAz5ULW+f0OINGhJmZm8FL/aDnlfTElhQ87LPicWpXYoMtyr6WrxjK6Ontn8BqCt0EjQ7TeXZhxIH9VTPWjDmFdmOqaqdVIT+LZemTgLNESwM5nn4g5S3aFDFwj1YiDYl0/+8etvKfOrfoKOwR0CxsRHagwdUUTES8EcHLmMGCxCkDZn3SzmmA6Nb3lgLeSgG8P1A=='), 'big')
e2 = 23333
c2 = int.from_bytes(base64.b64decode('O+rRCXI3aTB6P1rYIOPUdalUp6ujpwEq4I20CoWA+HIL8xxGtqY6N5gpr0guZv9ZgOEAMFnBxOqMdVNnB9GgnhmXtt1ZWydPqIcHvlfwpd/Lyd0XSjXnjaz3P3vOQvR71cD/uXyBA0XPzmnTIMgEhuGJVFm8min0L/2qI7wg/Z7w1+4mOmi655JIXeCiG23ukDv6l9bZuqfGvWCa1KKXWDP31nLbp0ZN2obUs6jEAa1qVTaX6M4My+sks+0VvHATrAUuCrmMwVEivqIJ/nS6ymGVERN6Ohnzyr168knEBKOVj0FAOx3YLfppMM+XbOGHeqdKJRLpMvqFXDMGQInT3w=='), 'big')
a, b = exgcd(e1, e2)
m = pow(c1, a, n)*pow(c2, b, n)%n
# m = gmpy2.powmod(c1, a, n) * gmpy2.powmod(c2, b, n) % n
print(libnum.n2s(m))
flag{23re_SDxF_y78hu_5rFgS}
4.[RoarCTF2019]babyRSA
5.[ACTF新生赛2020]crypto-rsa3
由题目可以得到值 n,c,e，已知p,q接近，所以可以yafu爆破获取到pq
from flag import FLAG
from Cryptodome.Util.number import *
import gmpy2
import random
e=65537
p = getPrime(512)
q = int(gmpy2.next_prime(p))
n = p*q
m = bytes_to_long(FLAG)
c = pow(m,e,n)
print(n)
print(c)
Python3脚本
from libnum import *
from gmpy2 import *
c = 1457390378511382354771000540945361168984775052693073641682375071407490851289703070905749525830483035988737117653971428424612332020925926617395558868160380601912498299922825914229510166957910451841730028919883807634489834128830801407228447221775264711349928156290102782374379406719292116047581560530382210049
p = 13326909050357447643526585836833969378078147057723054701432842192988717649385731430095055622303549577233495793715580004801634268505725255565021519817179293
q = 13326909050357447643526585836833969378078147057723054701432842192988717649385731430095055622303549577233495793715580004801634268505725255565021519817179231
e=65537
n = p*q
phi = (p-1)*(q-1)
d=invert(e,phi)
m =pow(c,d,n)
print(n2s(m))
flag{p_and_q_should_not_be_so_close_in_value}
7.[RoarCTF2019]RSA
题目
A=(((y%x)**5)%(x%y))**2019+y**316+(y+1)/x
p=next_prime(z*x*y)
q=next_prime(z)
A = 2683349182678714524247469512793476009861014781004924905484127480308161377768192868061561886577048646432382128960881487463427414176114486885830693959404989743229103516924432512724195654425703453612710310587164417035878308390676612592848750287387318129424195208623440294647817367740878211949147526287091298307480502897462279102572556822231669438279317474828479089719046386411971105448723910594710418093977044179949800373224354729179833393219827789389078869290217569511230868967647963089430594258815146362187250855166897553056073744582946148472068334167445499314471518357535261186318756327890016183228412253724
n = 117930806043507374325982291823027285148807239117987369609583515353889814856088099671454394340816761242974462268435911765045576377767711593100416932019831889059333166946263184861287975722954992219766493089630810876984781113645362450398009234556085330943125568377741065242183073882558834603430862598066786475299918395341014877416901185392905676043795425126968745185649565106322336954427505104906770493155723995382318346714944184577894150229037758434597242564815299174950147754426950251419204917376517360505024549691723683358170823416757973059354784142601436519500811159036795034676360028928301979780528294114933347127
c = 41971850275428383625653350824107291609587853887037624239544762751558838294718672159979929266922528917912189124713273673948051464226519605803745171340724343705832198554680196798623263806617998072496026019940476324971696928551159371970207365741517064295956376809297272541800647747885170905737868568000101029143923792003486793278197051326716680212726111099439262589341050943913401067673851885114314709706016622157285023272496793595281054074260451116213815934843317894898883215362289599366101018081513215120728297131352439066930452281829446586562062242527329672575620261776042653626411730955819001674118193293313612128
尝试分解一下n，factordb
顺利获取到p,q的值
发现题目并没有给出e的值，可以采取分析爆破的方式计算出e，但会很麻烦。尝试用一下e的一般值 65537（0x10001） 成功解出flag
from gmpy2 import*
from libnum import*
n = 117930806043507374325982291823027285148807239117987369609583515353889814856088099671454394340816761242974462268435911765045576377767711593100416932019831889059333166946263184861287975722954992219766493089630810876984781113645362450398009234556085330943125568377741065242183073882558834603430862598066786475299918395341014877416901185392905676043795425126968745185649565106322336954427505104906770493155723995382318346714944184577894150229037758434597242564815299174950147754426950251419204917376517360505024549691723683358170823416757973059354784142601436519500811159036795034676360028928301979780528294114933347127
c = 41971850275428383625653350824107291609587853887037624239544762751558838294718672159979929266922528917912189124713273673948051464226519605803745171340724343705832198554680196798623263806617998072496026019940476324971696928551159371970207365741517064295956376809297272541800647747885170905737868568000101029143923792003486793278197051326716680212726111099439262589341050943913401067673851885114314709706016622157285023272496793595281054074260451116213815934843317894898883215362289599366101018081513215120728297131352439066930452281829446586562062242527329672575620261776042653626411730955819001674118193293313612128
p = 842868045681390934539739959201847552284980179958879667933078453950968566151662147267006293571765463137270594151138695778986165111380428806545593588078365331313084230014618714412959584843421586674162688321942889369912392031882620994944241987153078156389470370195514285850736541078623854327959382156753458569
q = 139916095583110895133596833227506693679306709873174024876891023355860781981175916446323044732913066880786918629089023499311703408489151181886568535621008644997971982182426706592551291084007983387911006261442519635405457077292515085160744169867410973960652081452455371451222265819051559818441257438021073941183
e = 65537
phi = (p-1)*(q-1)
d = invert(e,phi)
m= pow(c,d,n)
print(n2s(m))
flag{wm-l1l1ll1l1l1l111ll}
8.[AFCTF2018]可怜的RSA
9.RSA & what
10.[网鼎杯 2020 青龙组]you_raise_me_up（离散对数）
涉及知识点：离散对数，可以通过sage来解决
也可以通过 python 第三方sympy库来解决。
这个考点以后可以写一篇专门的文章来解析离散对数及解决代码。
可以参考一下的文章：
1.[网鼎杯2020青龙组] you_raise_me_up
2.离散对数概念
3.Sympy常用函数总结
from sympy import*
求离散对数（如下15 = 7**3 mod 41）：
discrete_log(41, 15, 7)
来分析一下这道题
#!/usr/bin/env python
# -*- coding: utf-8 -*-
from Crypto.Util.number import *
import random
n = 2 ** 512
m = random.randint(2, n-1) | 1
c = pow(m, bytes_to_long(flag), n)
print 'm = ' + str(m)
print 'c = ' + str(c)
# m = 391190709124527428959489662565274039318305952172936859403855079581402770986890308469084735451207885386318986881041563704825943945069343345307381099559075
# c = 6665851394203214245856789450723658632520816791621796775909766895233000234023642878786025644953797995373211308485605397024123180085924117610802485972584499
从题目可以看出来，要解出flag，只需要一个公式即可。
c = m ** flag mod n (这里的等号是同余号)。
意思为 c同余m的flag次方 模 n
我们要解flag的值，用函数
discrete_log(n,c,m) #(模数，同余数，同余数)
from gmpy2 import*
from sympy import*
from libnum import*
n = 2 ** 512
# n = 13407807929942597099574024998205846127479365820592393377723561443721764030073546976801874298166903427690031858186486050853753882811946569946433649006084096
m = 391190709124527428959489662565274039318305952172936859403855079581402770986890308469084735451207885386318986881041563704825943945069343345307381099559075
c = 6665851394203214245856789450723658632520816791621796775909766895233000234023642878786025644953797995373211308485605397024123180085924117610802485972584499
flag = discrete_log(n,c,m)
print(n2s(flag))
flag{5f95ca93-1594-762d-ed0b-a9139692cb4a}