概述

HackTheBox 网站CTF靶场密码学相关题目RSAisEasy，题目地址https://app.hackthebox.eu/challenges/rsaiseasy，主要考点为RSA模不互素的攻击。

题目

下载附件并解压缩得到output.txt和RSAisEasy.py

output.txt内容为

n1: 101302608234750530215072272904674037076286246679691423280860345380727387460347553585319149306846617895151397345134725469568034944362725840889803514170441153452816738520513986621545456486260186057658467757935510362350710672577390455772286945685838373154626020209228183673388592030449624410459900543470481715269
c1: 92506893588979548794790672542461288412902813248116064711808481112865246689691740816363092933206841082369015763989265012104504500670878633324061404374817814507356553697459987468562146726510492528932139036063681327547916073034377647100888763559498314765496171327071015998871821569774481702484239056959316014064
c2: 46096854429474193473315622000700040188659289972305530955007054362815555622172000229584906225161285873027049199121215251038480738839915061587734141659589689176363962259066462128434796823277974789556411556028716349578708536050061871052948425521408788256153194537438422533790942307426802114531079426322801866673
(n1 * E) + n2: 601613204734044874510382122719388369424704454445440856955212747733856646787417730534645761871794607755794569926160226856377491672497901427125762773794612714954548970049734347216746397532291215057264241745928752782099454036635249993278807842576939476615587990343335792606509594080976599605315657632227121700808996847129758656266941422227113386647519604149159248887809688029519252391934671647670787874483702292498358573950359909165677642135389614863992438265717898239252246163

RSAisEasy.py的内容为

#!/usr/bin/env python3
from Crypto.Util.number import bytes_to_long, getPrime
from secrets import flag1, flag2
from os import urandomflag1 = bytes_to_long(flag1)
flag2 = bytes_to_long(flag2)p, q, z = [getPrime(512) for i in range(3)]e = 0x10001n1 = p * q
n2 = q * zc1 = pow(flag1, e, n1)
c2 = pow(flag2, e, n2)E = bytes_to_long(urandom(69))print(f'n1: {n1}')
print(f'c1: {c1}')
print(f'c2: {c2}')
print(f'(n1 * E) + n2: {n1 * E + n2}')

其中getPrime()的意思是

即得到1个512 bit的质数

解题思路

通过查看代码，n1 = p * q，n2 = q * z，则他们有相同的公约数q，可知模不互素，由代码和output.txt可知：

n1 = p * q
sum = n1 * e + n2 = p * q * e + q * z = q * (p * e + z)

通过gmpy2.gcd(n1,sum)得到公约数q的值，p = n1 / q
而n2 = sum % n1，这是关键的一步，注意p,q,z为512bit的限制
z = n2 / q

解题代码

import gmpy2
from Crypto.Util.number import long_to_bytes # n1 = p * q
n1 = 101302608234750530215072272904674037076286246679691423280860345380727387460347553585319149306846617895151397345134725469568034944362725840889803514170441153452816738520513986621545456486260186057658467757935510362350710672577390455772286945685838373154626020209228183673388592030449624410459900543470481715269c1 = 92506893588979548794790672542461288412902813248116064711808481112865246689691740816363092933206841082369015763989265012104504500670878633324061404374817814507356553697459987468562146726510492528932139036063681327547916073034377647100888763559498314765496171327071015998871821569774481702484239056959316014064c2 = 46096854429474193473315622000700040188659289972305530955007054362815555622172000229584906225161285873027049199121215251038480738839915061587734141659589689176363962259066462128434796823277974789556411556028716349578708536050061871052948425521408788256153194537438422533790942307426802114531079426322801866673e = int('0x10001',16)# sum = (n1 * e) + n2 = p * q * e + q * z = q * (p * e + z)
sum = 601613204734044874510382122719388369424704454445440856955212747733856646787417730534645761871794607755794569926160226856377491672497901427125762773794612714954548970049734347216746397532291215057264241745928752782099454036635249993278807842576939476615587990343335792606509594080976599605315657632227121700808996847129758656266941422227113386647519604149159248887809688029519252391934671647670787874483702292498358573950359909165677642135389614863992438265717898239252246163# 关键的一步
n2 = sum % n1# 求公约数q
q = gmpy2.gcd(n1, sum)
assert (q != 1)
p = n1 / q
z = n2 / q d1 = gmpy2.invert(e, (q - 1) * (p - 1))
d2 = gmpy2.invert(e, (q - 1) * (z - 1))
m1 = pow(c1, d1, n1)
m2 = pow(c2, d2, n2)
print long_to_bytes(m1) + long_to_bytes(m2)

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HackTheBox-RSAisEasy

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