[ciscn 2021 South China]rsa writeup

[CISCN 2021 South China ]rsa

Title Description

from flag import text,flag
import md5
from Crypto.Util.number import long_to_bytes,bytes_to_long,getPrime

assert md5.new(text).hexdigest() == flag[6:-1]

msg1 = text[:xx]
msg2 = text[xx:yy]
msg3 = text[yy:]

msg1 = bytes_to_long(msg1)
msg2 = bytes_to_long(msg2)
msg3 = bytes_to_long(msg3)

p1 = getPrime(512)
q1 = getPrime(512)
N1 = p1*q1
e1 = 3
print pow(msg1,e1,N1)
print (e1,N1)

p2 = getPrime(512)
q2 = getPrime(512)
N2 = p2*q2
e2 = 17
e3 = 65537
print pow(msg2,e2,N2)
print pow(msg2,e3,N2)
print (e2,N2)
print (e3,N2)

p3 = getPrime(512)
q3 = getPrime(512)
N3 = p3*q3
print pow(msg3,e3,N3)
print (e3,N3)
print p3>>200

We can know from the audit code ,flag == md5(text),text = msg1 + msg2 + msg3
So as long as we can solve msg1,2,3 Can solve flag

part1

p1 = getPrime(512)
q1 = getPrime(512)
N1 = p1*q1
e1 = 3
print pow(msg1,e1,N1)
print (e1,N1)

Small plaintext attack ,e Too small, resulting in the plaintext of the third power is still greater than n Small , Direct pair c1 Third power

part2

p2 = getPrime(512)
q2 = getPrime(512)
N2 = p2*q2
e2 = 17
e3 = 65537
print pow(msg2,e2,N2)
print pow(msg2,e3,N2)
print (e2,N2)
print (e3,N2)

Typical common mode attacks , The derivation process is as follows ：

First , Two encryption indices are coprime ：
gcd(e1,e2)=1

There is s1、s2 bring ：
s1 * e1+s2 * e2=1

Again because ：
c1≡m^e1 mod n
c2≡m mod n

Substitution and simplification can be obtained ：
c1^s1 * c2^s2 ≡ m mod n

You can find the plaintext

part3

p3 = getPrime(512)
q3 = getPrime(512)
N3 = p3*q3
print pow(msg3,e3,N3)
print (e3,N3)
print p3>>200

Title will p3 Move right 200 position , and p3 Originally 512 position , So we use Coppersmith partial information attack Algorithm for p3 The low order of can be solved p3,q3

among ,Coppersmith partial information attack be based on sage Realization .

complete exp

from Crypto.Util.number import *
import gmpy2
from hashlib import md5
import sys


sys.setrecursionlimit(1000000)


def egcd(a, b):
    if a == 0:
        return b, 0, 1
    else:
        g, y, x = egcd(b % a, a)
        return g, x - (b // a) * y, y


def modinv(a, m):
    g, x, y = egcd(a, m)
    if g != 1:
        raise Exception('modular inverse does not exist')
    else:
        return x % m


def CommonMode(e1, e2, c1, c2, n):
    s = egcd(e1, e2)
    s1 = s[1]
    s2 = s[2]

    if s1 < 0:
        s1 = - s1
        c1 = modinv(c1, n)
    elif s2 < 0:
        s2 = - s2
        c2 = modinv(c2, n)
    m = (pow(c1, s1, n) * pow(c2, s2, n)) % n
    return long_to_bytes(m)
c1 = 19105765285510667553313898813498220212421177527647187802549913914263968945493144633390670605116251064550364704789358830072133349108808799075021540479815182657667763617178044110939458834654922540704196330451979349353031578518479199454480458137984734402248011464467312753683234543319955893
e1 = 3
n1 = 123814470394550598363280518848914546938137731026777975885846733672494493975703069760053867471836249473290828799962586855892685902902050630018312939010564945676699712246249820341712155938398068732866646422826619477180434858148938235662092482058999079105450136181685141895955574548671667320167741641072330259009

msg1 = long_to_bytes(gmpy2.iroot(c1, e1)[0])
c2 = 54995751387258798791895413216172284653407054079765769704170763023830130981480272943338445245689293729308200574217959018462512790523622252479258419498858307898118907076773470253533344877959508766285730509067829684427375759345623701605997067135659404296663877453758701010726561824951602615501078818914410959610
c3 = 91290935267458356541959327381220067466104890455391103989639822855753797805354139741959957951983943146108552762756444475545250343766798220348240377590112854890482375744876016191773471853704014735936608436210153669829454288199838827646402742554134017280213707222338496271289894681312606239512924842845268366950
e2 = 17
e3 = 65537
n2 = 111381961169589927896512557754289420474877632607334685306667977794938824018345795836303161492076539375959731633270626091498843936401996648820451019811592594528673182109109991384472979198906744569181673282663323892346854520052840694924830064546269187849702880332522636682366270177489467478933966884097824069977

msg2 = CommonMode(e2, e3, c2, c3, n2)

c4 = 59213696442373765895948702611659756779813897653022080905635545636905434038306468935283962686059037461940227618715695875589055593696352594630107082714757036815875497138523738695066811985036315624927897081153190329636864005133757096991035607918106529151451834369442313673849563635248465014289409374291381429646
n3 = 113432930155033263769270712825121761080813952100666693606866355917116416984149165507231925180593860836255402950358327422447359200689537217528547623691586008952619063846801829802637448874451228957635707553980210685985215887107300416969549087293746310593988908287181025770739538992559714587375763131132963783147
p3_high = 7117286695925472918001071846973900342640107770214858928188419765628151478620236042882657992902

# This part of the code uses sage Go for a run 
# n=113432930155033263769270712825121761080813952100666693606866355917116416984149165507231925180593860836255402950358327422447359200689537217528547623691586008952619063846801829802637448874451228957635707553980210685985215887107300416969549087293746310593988908287181025770739538992559714587375763131132963783147
# p4=7117286695925472918001071846973900342640107770214858928188419765628151478620236042882657992902# It is known that P The high 
# e=65537
# pbits=512 #P Original digits 
# 
# kbits=pbits - p4.nbits()
# print (p4.nbits())
# p4 = p4 << kbits
# PR.<x> = PolynomialRing(Zmod(n))
# f = x + p4
# roots = f.small_roots(X=2^kbits,beta=0.4)
# #  After the above functions ,n and p Has been transformed into 10 Base number 
# if roots:
# p= p4 + int(roots([0]))
# print ("n",n)
# print ("p",p)
# print ("q",n/p)
p3 = 11437038763581010263116493983733546014403343859218003707512796706928880848035239990740428334091106443982769386517753703890002478698418549777553268906496423
q3 = 9918033198963879798362329507637256706010562962487329742400933192721549307087332482107381554368538995776396557446746866861247191248938339640876368268930589
d = gmpy2.invert(e3, (p3 - 1) * (q3 - 1))
msg3 = long_to_bytes(pow(c4, d, n3))
text = msg1 + msg2 + msg3
print(md5(text).hexdigest())

In the end, you get flag.