'2019/11/12'에 해당되는 글 2건

암호문과 pem 형식으로 된 두 개의 공개키를 제공해 준다. pem 파일에서 n을 각각 추출한 뒤 최대 공약수를 확인해보면 공개키 생성 과정에서  공통된 소수를 사용해 p 값을 구할 수 있다.

python RsaCtfTool.py --key ../rsa/seccon/pub1 --dumpkey

python RsaCtfTool.py --key ../rsa/seccon/pub2 --dumpkey

solve.py

from gmpy2 import *
from Crypto.PublicKey import RSA

n1 = 763718912475160487902804749669814117361530270298225094625871588939939773892509348006077810445741086683427253000695920011673348476297973798322806091336777405584801442639626925406721932533140226556519019440300864340670199686368307155860493615065198319490060598587202051942638792919648596288576294804549738135969737494734307362891313864027749187674251692407867312885251279302785352318725391842117065840058358065676707016006124478822206302825992616559261930620693061673348139416033418864269248381876692676529410115745518353146254670349865568255213560376368953292931958006941630719304442332912813624543600126197554727832190226632919876204677667384620275620336951964833888599634720101911051166398907898747710391394105753614253527704990658698844796442515669670816004761855554187277637871343595647487793209271354240148469085627742503649786300484610102224828274384484809539697728008552925590472129497180290668277790132130824141651399551803499770513576176720509094833332201946177880267399933460994496277590932311628302240154240967341858152145815276163397709272690500041597393678630136932574450837982593210399370333578887450410911663220219423601973078501237709613593311133945501455828164291429228495931943107997137587307522565029820756690578833

n2 = 847796795638781450678718708664542960446354226336422534142899480441478781747168340722544245493739168125415291376063352480076469305992008517388366298914970810765149321160596112226917023146371080685458239747986992197343482255681414590678694753885521068656675164266739274608505094927725285868801825144058835115051061857233824816176186349372737400492228548130718029270482313906040935333159681273527671359469173939861889228505791626525994180917670701909816457560622922219797109618405842358250940062127841301193206255543580550667149741357342408703680375118252445296600962244008043713708516004025217199954286478272550961283413053789743740774140439284662596273452124779590753836094450154459078034659421366584354005421566290862244000178569163732971816333961711759000171156293724203334889496589376907683142167883476184246287660764821622298961378903818077158234683927321823293466245584486980446609569799420950073036635602004293246754252079757527610246223280774592566063422181158810677919909523144033543195999214800559998598003329294711989868113423468536273984117507008756542731210465222475943122324672467669623079721727526348599823126409108190030551584092950358235465140166515452377475799157617227362838246912839086733412861436353117994501342653

p = gcd(n1,n2)

n = n2

q = n/p
e = 65537

phi = (p-1)*(q-1)

d = invert(e,phi)
f = open("./cipher","rb")
c = f.read()
c = int(c.encode("hex"),16)

print("%x"%pow(c,d,n))[1:].decode("hex")

문제 파일은 아래와 같다.

revolutional-secure-angou-de97106aa248a41a40fdd001fc5f7b4b4f28a39eb6bcabf8401b108b7a8961c5.7z

압축을 풀어보면 암호화된 flag, pem 형식의 공개키, 문제 소스가 존재한다.

generator.rb

require 'openssl'

e = 65537

while true

  p = OpenSSL::BN.generate_prime(1024, false)

  q = OpenSSL::BN.new(e).mod_inverse(p)

  next unless q.prime?

  key = OpenSSL::PKey::RSA.new

  key.set_key(p.to_i * q.to_i, e, nil)

  File.write('publickey.pem', key.to_pem)

  File.binwrite('flag.encrypted', key.public_encrypt(File.binread('flag')))

  break

end

소스는 루비를 통해 pem 형식의 공개키를 만들고 해당 공개키로 flag를 암호화하는 구조이다.

pem 파일에 존재하는 공개키를 구하기 위해 아래 툴을 사용하였다.

ssrinc@ssrinc-virtual-machine:~/RsaCtfTool$ ./RsaCtfTool.py --key ./publickey.pem --dumpkey

[*] n: 16809924442712290290403972268146404729136337398387543585587922385691232205208904952456166894756423463681417301476531768597525526095592145907599331332888256802856883222089636138597763209373618772218321592840374842334044137335907260797472710869521753591357268215122104298868917562185292900513866206744431640042086483729385911318269030906569639399362889194207326479627835332258695805485714124959985930862377523511276514446771151440627624648692470758438999548140726103882523526460632932758848850419784646449190855119546581907152400013892131830430363417922752725911748860326944837167427691071306540321213837143845664837111

[*] e: 65537