Python3 Simple RSA CTF Solver

rsa.py

# python3 -m pip install pycryptodome==3.4.3
from Crypto.Util.number import inverse, long_to_bytes
import decimal, binascii

# If you have the id_rsa.pub or equivalent
# ssh-keygen -f id_rsa.pub -e -m PKCS8 > id_rsa.pem

# If you have a private key (pem) file
# - http://certificate.fyicenter.com/2145_FYIcenter_Public_Private_Key_Decoder_and_Viewer.html
# -- Fill in n, e, d, p, q (Note: numeric values of n,d,p,q - Not displayed hex values)
# -- Convert hex to decimal over at: https://www.binaryhexconverter.com/hex-to-decimal-converter

n = 0 # Enter given n value here - Leave as 0 if not available
e = 0 # Enter given e value here - Required
c = 0 # Enter given c value here - Leave as 0 if not available (If 0, will assume you have a cipher file named cipher)
d = 0 # Enter given d value here - Leave as 0 if not available

# If you have n
# - Go to http://factordb.com/ and enter in n
# - Under "Number", p = left, q = right. If they have ^2 on them, add **2 to the end of the number

# If you have p and q, just enter them

p = 0
q = 0

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if (n == 0):
  n = p * q

# Cubed Root Attack (Thanks John!)
if e == 3:
  decimal.getcontext().prec = 3000
  c_decimal = decimal.Decimal(str(c))
  m = c_decimal ** (decimal.Decimal('1') / 3)
  m = int(m) + 1
  result = long_to_bytes(m)
  print("Possible Solution: " + result.decode("utf-8"))

if p != 0 and q != 0:
  # They match - Different calc
  if p == q:
    phi = p * (p - 1)
  else:
    phi = (p - 1) * (q - 1)
  # Either solving for c or d
  if d == 0:
    d = inverse(e,phi)
  if c == 0:
    with open("cipher", "rb") as f:
      data = f.read()
    c = int(binascii.hexlify(data), 16)

  # We should have everything now
  m = pow( c, d, n )
  print(long_to_bytes(m))

Generate private key from n, p, q, and e

python3 rsatool.py -n CHANGEME -p CHANGEME -q CHANGEME -e CHANGEME(Probs 65537) -f PEM -o priv.key

chmod 600 priv.key