GordonOus

from Crypto.PulicKey import RSA

key = open('key.pub').read()
# STEP 1 => get the modulus and exponent
n = key.n
e = key.e

# STEP 2 => Since itis a weak rsa: we could factorise the modulus from using this site: http://factordb.com/
# after factorising we get p and q below

GordonOus

========== First =========
Lets compare both keys provided:

openssl rsa -pubin -inform PEM -text -noout < key1.pem
openssl rsa -pubin -inform PEM -text -noout < key2.pem

we can observe that both keys have SIMILAR modulus but DIFFERENT exponents: 
after googling online on RSA attacks i found one for Common Modulus

Exploiting it according to me was not very trivial especially without solid background in crypto (google to the rescue)

GordonOus

from Crypto.PublicKey import RSA
from Crypto.Util.number import inverse
from binascii import hexlify,unhexlify
from base64 import b64decode

# STEP 1 => import the public keys and get both moduli (n1,n2) and exponents (e1,e2)
key1 = RSA.import_key(open('key1.pem').read())
key2 = RSA.import_key(open('key2.pem').read())

n1,n2,e1,e2 = key1.n, key2.n, key1.e, key2.e

GordonOus

from binascii import unhexlify
import math

# STEP 1 => Get the bytes of the flag
encrypted_flag = unhexlify(open('output.txt','r').read().split(' ')[1].strip('\n'))
part_flag = b'HTB{'
# STEP 2 => Get the encryption key by xoring the known string with the similar placed output bytes
key = [chr(b ^ encrypted_flag[a]) for a,b in enumerate(part_flag)]

messages = []

GordonOus

import string
from binascii import unhexlify

encrypted = unhexlify(open('msg.enc','r').read())
plaintext = []
#brute force the encryption logic
for b in encrypted:
  for ch in string.printable:
    if ((123 * ord(ch) + 18) % 256 == b):
      plaintext.append(ch)

GordonOus

from binascii import unhexlify

enc_flag = 'Put the encrypted flag extracted from the pcap hear'
ct = 'the First Ciphertext we find'
ct2 = 'the second Ciphertext we find'
flag = ''
#Note the length of all the above texts are 36 bytes after decoding from ascii
for i,j in enumerate(ct):
  flag += chr(j ^ ct2[i] ^ enc_flag[i])

GordonOus

import time
import random
from pwn import *

def gen_seeds(s):
  return list(range((s - 20),(s + 20)))

possible_seeds = gen_seeds(int(time.time()))
conn = remote('IP ADDRESS',PORT)

GordonOus

import gmpy2
from Crypto.Util.number import long_to_bytes, bytes_to_long
from binascii import unhexlify

flag = open('output.txt','r').read().strip().split(' ')[1]
ciphertext = bytes_to_long(flag)
e = 3

#Fist method of getting cube root of ciphertext
gmpy2.get_context().precision = 4096

GordonOus

import numpy as np
from PIL import Image

img1 = Image.open('lemur.png')
img2 = Image.open('flag.png')

n1 = np.array(img1)*255
n2 = np.array(img2)*255

#our images have a mode of RGB which is assumed to be an 8-bit int

GordonOus

ssh-keygen -e -f 'bruce_rsa_6e7ecd53b443a97013397b1a1ea30e14.pub' -m 'PKCS8' > bruce.pub
openssl rsa -pubin -in foo.pub -noout -modulus | cut -d '=' -f2 | xargs echo "ibase=16; $1"| bc | tr -d '\' | tr -d '\n'