I tried to decrypt rsa using a private key that exists as a string, but it failed.
-----BEGIN RSA PRIVATE KEY-----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-----END RSA PRIVATE KEY-----
The private key is a string and exists as shown above.
from Crypto.PublicKey import RSA
import base64
from Crypto.Cipher import PKCS1_OAEP
Pkey ="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"
s="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"
code_bytes = s.encode('UTF-8')
by=base64.b64decode(code_bytes)
key = PKCS1_OAEP.new(Pkey)
rsadecrypt=key.decrypt(str(by))
Attempted to decode string 's' after base64 decoding.
However, an error was issued as follows:
Traceback (most recent call last):
File "/Users/isin-yeong/Desktop/element/code/rsadecrypt.py", line 14, in <module>
rsadecrypt=key.decrypt(str(by))
^^^^^^^^^^^^^^^^^^^^
File "/Library/Frameworks/Python.framework/Versions/3.11/lib/python3.11/site-packages/Crypto/Cipher/PKCS1_OAEP.py", line 161, in decrypt
modBits = Crypto.Util.number.size(self._key.n)
^^^^^^^^^^^
AttributeError: 'str' object has no attribute 'n'
How do I convert 'Pkey' that exists as a string into a private key and decode 's' decoded with base64?
The following methods are used for decryption:
"RSA/ECB/PKCS1Padding"
Try saving the private key you posted first in a file with the name 'pkey.pem', and instead of writing the string directly into your code, use the following to upload the key:
Pkey = RSA.importKey(open('pkey.pem').read())
For more information you can look here, the documentation of pycryptodome.
edited according to Topaco's comments
Your code should looks as follows:
from Crypto.PublicKey import RSA
import base64
from Crypto.Cipher import PKCS1_v1_5
from Crypto.Random import get_random_bytes
s="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"
code_bytes = s.encode('UTF-8')
by=base64.b64decode(code_bytes)
Pkey = RSA.importKey(open('pkey.pem').read())
key = PKCS1_v1_5.new(Pkey)
sentinel = get_random_bytes(16)
rsadecrypt=key.decrypt(by,sentinel)
Pay attention that you still going to get the following error:
ValueError: Ciphertext with incorrect length.
as Topaco explained in the comments
There are a few problems already noted in the comments by #Topaco:
Apart from the incorrect key import, the wrong padding is used. The
PyCryptodome counterpart to RSA/ECB/PKCS1Padding is PKCS1_v1_5 (and
not PKCS1_OAEP). Second, the ciphertext is apparently corrupted: The
posted (and thus compromised) private key has a length of 2048 bits =
256 bytes, i.e. the ciphertext must be of the same length. But the
posted (Base64 decoded) ciphertext is 380 bytes long (len(by)).
Furthermore, in decrypt() not str(by) but by must be passed.
You also have a typo in Pkey, the Base64 encoded body is ...boc61 and
not ...boc6 (i.e. the last character is missing). If this is fixed,
the key can be imported with RSA.importKey(base64.b64decode(Pkey))
When those are addressed we see that s, after base64 decoding, is too long to be the result of RSA encryption with a 2048 bit modulus. However by trying all offsets into the base64-decoded s and taking the next 256 bytes we get a successful decrypt at offset 3.
# https://stackoverflow.com/q/74840474/238704
import base64
from Cryptodome.Cipher import PKCS1_v1_5
from Cryptodome.PublicKey import RSA
from Cryptodome.Random import get_random_bytes
private_key_pem = '''-----BEGIN RSA PRIVATE KEY-----
MIIEvgIBADANBgkqhkiG9w0BAQEFAASCBKgwggSkAgEAAoIBAQCTiP1x58DboFjS
4WgzMm8tuY9VoHwANbOHNuuiElsZ4xIVFS+ZG7lu7Iz9gzmsno/YiqQXms8hXmUH
ouGgVJLJGPTw/NXAnLq6XlwB3C+zQMWpxvcMub4D6/IltP/PUpKNj9QzKKGhZF+6
s5B/QBzmSeNYlYYyGnO9GxyuHyR7P5xqF9AcpSskn9gjYy0koTLmvg/bwFx6jnci
np2qVHo5CFvSTAgrZKBCmKD2c2AGD9O+TSOrT0RDzvJosW4P1QJ9qsCjKFSPW/8K
IvB+AsSA+9xiVMDm/3YABBBM/R7wDcciTmKucZoEuDeWJg+cwLur8kKSVNwwkwCT
MqTIbHn5AgMBAAECggEABl5aemlDpPdl/ixmwBcEP5gL/OlBkQzAZCzVvRhHwHVR
2YEhnd1ZgtVJPMRGYBI0KWfKr44qNMmLWOoKDU65S179a82uOHNLiDH8jMQBdx5r
qemBzpXSAv9TY5dNl5h75Qp5YX/2gi4AB+IqcaPF25cC7lb+BPmpV1dtRILpozxJ
78mE0o0ddAJhg+0lIX6lZqjXbKJh4uiIqJIwezOM/B9U59qWEiJoHurkJIQoc00n
lp+XKUJfUWE5OHdy2DZNbAWEDqvFiD3DBd5baks/CPHy6IZLj7e5y+h/b65veDw6
MWpwXK30vroRmIzfNFcbONuQ3EjHBvCXzQYppUPnAQKBgQDD/k9NVQ8r2EnhjX0g
dQKobT7aCpY/y0QvDZ6aeuOVkfqnFB+rUUSf5WsXbANghqsHtzJa+cDUpP0C1PuE
hexGS+SsLywKI/33S014tvt5iF/MgEu08sLHtoxR/dH3H4k2LDGMNBrKILFnfDYP
40e3QKdogXMsEw4C0tb85YGrIQKBgQDAtJZp+onH0NSDXL0kbWZa8GFuVMGjTz2m
6E+j3NHK5GDGSAme755lDS61TTz2xNYaJFEwUdvYKQq1x/Q7jffKMPcHUJ5uy7bH
QjFho377rvN9bNSk+UdP8AkhvjaLNC0c5K5eNEUgumrd+yDLw2YNvjOe/RjE3QiW
zpGHLigL2QKBgQC2LUufMSJB+fBqlG6rXbgTwD/8wnx4XcNKDVnQc8vZenBHH0B0
qLyl98S8Z60X6vVM3a8TuzTPM3DuUfRccXN2wBKVLd+8qUnmtHsVatiDVgzd9J7q
WgBNTNRPXiPtlcWfsiJ4FPKV0R+1dlJ2ICfIIXO6gyD/5dJPM5WcSuRloQKBgQCq
Tu+gOgwKzEUE07FVwFn1hPyWxdFcw2t5V1cLOt0usyfUFVZrYNpb9IqTMO/wJ4YR
FIesbKwfHiaZAV6YQ5/60cuAa3+Z6/BdqeJ6qERRqw0GjGhiZOzheQKZD8KkxDga
kQCJwShXBGnuRUN20fofqzl0CbsaQT6WCXoUPnamAQKBgEY/ttlyLCrdHhamWlwN
ZkcgfsvQtLc//DGfEOSIiybs0eD0ORutxVQ2pwEgPI79xJTm0bdTHAFZVYziI9Lm
75k5WKxbBsYTgZm3i+LnnDtcHKPqkpXkq1v6WduTfaWpwMDJt0cJMOiTFzom76s1
mdFdmq2IXR+vXwl9f8fboc61
-----END RSA PRIVATE KEY-----'''
s = "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"
code_bytes = s.encode('UTF-8')
by = base64.b64decode(code_bytes)
private_key = RSA.import_key(private_key_pem)
cipher = PKCS1_v1_5.new(private_key)
sentinel = get_random_bytes(16)
rsadecrypt = cipher.decrypt(by[3: 3 + 256], sentinel)
if rsadecrypt == sentinel:
print('failure')
else:
print(f'success: {rsadecrypt.hex(" ")}')
output is:
success: 48 90 c1 c5 ed fd 67 84 ad 82 df d1 5b 22 40 6f
I don't know what the rest of the bytes of s are all about.
PKCS1 version 1.5 encryption padding is not all that great and is basically deprecated in favor of OAEP padding. One of the weaknesses is the unacceptably high probability that a decryption with the wrong key and/or corrupted ciphertext will succeed. It's unlikely in this case, but not unlikely enough to completely discard the possibility. Although you've provided no additional details about what the payload is supposed to be, the 16 random-looking bytes suggests a key of some sort, perhaps an AES-128 key.
I want to perform a sha256withRSA signature of an hexadecimal string (0xDEADBEEF in this example) using the PKCS1_PSS module in Python:
import Crypto
from Crypto.Signature import PKCS1_PSS
from Crypto.Hash import SHA256
from Crypto.PublicKey import RSA
from Crypto import Random
import binascii
message = 'DEADBEEF'
message = binascii.unhexlify(message)
h = SHA256.new(message) # sha256 hashing the message
my_hash = binascii.hexlify(h.digest())
I get the following hash :
b'5f78c33274e43fa9de5659265c1d917e25c03722dcb0b8d27db8d5feaa813953'
Then I sign it using the following Private Key :
-----BEGIN RSA PRIVATE KEY-----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-----END RSA PRIVATE KEY-----
Signing command :
key = RSA.import_key(open('private_key.pem').read())
signature = PKCS1_PSS.new(key).sign(h)
print(binascii.hexlify(signature))
The result :
b'67bcc8c0cd625a58272dc8808602beb630c0dc47622da153c6b3f7fcfdddd5e082beb9e73ed0e66f9751e68106b42ff71f8d291045ca7e9a5a265e885e19c016b6095e5f895801d3f735393e8cd3e4a18382a914487b46cf6c3ca3346c0b6f4bac923e491ca9933e12f826914b90955ce24d0203824dbb2c9cb7cb617af7cdef'
When I try to calculate the signature myself using the RSA encryption method (x = (m^d)%n ), I get a different value :
n = key.n
d = key.d
x = 0x5f78c33274e43fa9de5659265c1d917e25c03722dcb0b8d27db8d5feaa813953 # hash
calculated_signature = pow(x,d,n)
print (hex(calculated_signature))
I get this :
'0x4e3484dde9ca8987b77a52c696b6848e5980c858f635a62e10fe24d45bea52bac4873378b9612f47398cb73102243b73cfa2da87b487ec0d26f949e09e4edb299bd4acc4aa0eb43afa47b814b3430ff9b01b8e41bcf53ba310ae26c191cb516c07f3ebe272bcbb95acb9e7bbdc22f0c17997ba6c1884c177ba41d4e384184275'
I'm wondering why I get this difference? My understanding of the sign method previously used is that it encrypts the input (the hash) using the operation (x^d)%n.
Am I missing something ? Can anyone suggest me the reason of the difference ? Is it due to some bad manipulation/understanding from my side ?
I'm trying to understand the steps to take an OpenSSH public key like so:
ssh-rsa AAAAB3NzaC1yc2EAAAABIwAAAQEAqmEmDTNBC6O8HGCdu0MZ9zLCivDsYSttrrmlq87/YsEBpvwUTiF3UEQuFLaq5Gm+dtgxJewg/UwsZrDFxzpQhCHB6VmqrbKN2hEIkk/HJvCnAmR1ehXv8n2BWw3Jlw7Z+VgWwXAH50f2HWYqTaE4qP4Dxc4RlElxgNmlDPGXw/dYBvChYBG/RvIiTz1L+pYzPD4JR54IMmTOwjcGIJl7nk1VjKvl3D8Wgp6qejv4MfZ7Htdc99SUKcKWAeHYsjPXosSk3GlwKiS/sZi51Yca394GE7T4hZu6HTaXeZoD8+IZ7AijYn89H7EPjuu0iCAa/cjVzBsFHGszQYG+U5KfIw==
And then to convert it into an standard fingerprint like so:
2048 49:d3:cb:f6:00:d2:93:43:a6:27:07:ca:12:fd:5d:98 id_rsa.pub (RSA)
I have attempted to dive into the OpenSSH source to understand this, but it is over my head. My first guess was to do a simple MD5 on the key text, but the result does not match the above output.
It is the MD5 sum of the base64-encoded key:
import base64
import hashlib
def lineToFingerprint(line):
key = base64.b64decode(line.strip().split()[1].encode('ascii'))
fp_plain = hashlib.md5(key).hexdigest()
return ':'.join(a+b for a,b in zip(fp_plain[::2], fp_plain[1::2]))
https://github.com/ojarva/sshpubkeys
pip install sshpubkeys
Usage:
import sshpubkeys
key = sshpubkeys.SSHKey("ssh-rsa AAAAB3NzaC1yc2EAAAABIwAAAQEAqmEmDTNBC6O8H" +
"GCdu0MZ9zLCivDsYSttrrmlq87/YsEBpvwUTiF3UEQuFLaq5Gm+dtgxJewg/UwsZrDFxz" +
"pQhCHB6VmqrbKN2hEIkk/HJvCnAmR1ehXv8n2BWw3Jlw7Z+VgWwXAH50f2HWYqTaE4qP4" +
"Dxc4RlElxgNmlDPGXw/dYBvChYBG/RvIiTz1L+pYzPD4JR54IMmTOwjcGIJl7nk1VjKvl" +
"3D8Wgp6qejv4MfZ7Htdc99SUKcKWAeHYsjPXosSk3GlwKiS/sZi51Yca394GE7T4hZu6H" +
"TaXeZoD8+IZ7AijYn89H7EPjuu0iCAa/cjVzBsFHGszQYG+U5KfIw== user#host")
print(key.bits) # 2048
print(key.hash()) # '49:d3:cb:f6:00:d2:93:43:a6:27:07:ca:12:fd:5d:98'
In python I am trying to import a public key as follows (omitting a lot of characters):
public = "MIGfMA0G...."
RSA.importKey(public)
but I get the error
ValueError: RSA key format is not supported
How to create a RSA object with the public key when I have the key in a string?
In order to import the key, if it has the header, is imported correctly:
from Crypto.PublicKey import RSA
key = "-----BEGIN PUBLIC KEY-----\nMIGfMA0GCS{...}QVAwwIDAQAB\n-----END PUBLIC KEY-----"
RSA.importKey(key)
So the only thing you need is to place the beginning and the end.
I'm attempting to write a script to generate SSH Identity key pairs for me.
from M2Crypto import RSA
key = RSA.gen_key(1024, 65337)
key.save_key("/tmp/my.key", cipher=None)
The file /tmp/my.key looks great now.
By running ssh-keygen -y -f /tmp/my.key > /tmp/my.key.pub I can extract the public key.
My question is how can I extract the public key from python? Using key.save_pub_key("/tmp/my.key.pub") saves something like:
-----BEGIN PUBLIC KEY-----
MFwwDQYJKoZIhvcNAQEBBQADASDASDASDASDBarYRsmMazM1hd7a+u3QeMP
...
FZQ7Ic+BmmeWHvvVP4Yjyu1t6vAut7mKkaDeKbT3yiGVUgAEUaWMXqECAwEAAQ==
-----END PUBLIC KEY-----
When I'm looking for something like:
ssh-rsa AAAABCASDDBM$%3WEAv/3%$F ..... OSDFKJSL43$%^DFg==
Use cryptography! pycrypto is not in active development anymore and if possible you should be using cryptography. Since June it's possible to generate SSH public keys as well:
from cryptography.hazmat.primitives import serialization as crypto_serialization
from cryptography.hazmat.primitives.asymmetric import rsa
from cryptography.hazmat.backends import default_backend as crypto_default_backend
key = rsa.generate_private_key(
backend=crypto_default_backend(),
public_exponent=65537,
key_size=2048
)
private_key = key.private_bytes(
crypto_serialization.Encoding.PEM,
crypto_serialization.PrivateFormat.PKCS8,
crypto_serialization.NoEncryption()
)
public_key = key.public_key().public_bytes(
crypto_serialization.Encoding.OpenSSH,
crypto_serialization.PublicFormat.OpenSSH
)
Note: You need at least version 1.4.0.
Note: If your SSH client does not understand this private key format, replace PKCS8 with TraditionalOpenSSL.
Just in case there are any future travellers looking to do this. The RSA module support writing out the public key in OpenSSH format now (possibly didn't at the time of earlier posts). So I think you can do what you need with:
from os import chmod
from Crypto.PublicKey import RSA
key = RSA.generate(2048)
with open("/tmp/private.key", 'wb') as content_file:
chmod("/tmp/private.key", 0600)
content_file.write(key.exportKey('PEM'))
pubkey = key.publickey()
with open("/tmp/public.key", 'wb') as content_file:
content_file.write(pubkey.exportKey('OpenSSH'))
The files are opened with a 'wb' as the keys must be written in binary mode.
Obviously don't store you're private key in /tmp...
Edit 05/09/2012:
I just realized that pycrypto already has this:
import os
from Crypto.PublicKey import RSA
key = RSA.generate(2048, os.urandom)
print key.exportKey('OpenSSH')
This code works for me:
import os
from Crypto.PublicKey import RSA
key = RSA.generate(2048, os.urandom)
# Create public key.
ssh_rsa = '00000007' + base64.b16encode('ssh-rsa')
# Exponent.
exponent = '%x' % (key.e, )
if len(exponent) % 2:
exponent = '0' + exponent
ssh_rsa += '%08x' % (len(exponent) / 2, )
ssh_rsa += exponent
modulus = '%x' % (key.n, )
if len(modulus) % 2:
modulus = '0' + modulus
if modulus[0] in '89abcdef':
modulus = '00' + modulus
ssh_rsa += '%08x' % (len(modulus) / 2, )
ssh_rsa += modulus
public_key = 'ssh-rsa %s' % (
base64.b64encode(base64.b16decode(ssh_rsa.upper())), )
The key used by ssh is just base64 encoded, i don't know M2Crypto very much, but after a quick overview it seems you could do what you want this way:
import os
from base64 import b64encode
from M2Crypto import RSA
key = RSA.gen_key(1024, 65537)
raw_key = key.pub()[1]
b64key = b64encode(raw_key)
username = os.getlogin()
hostname = os.uname()[1]
keystring = 'ssh-rsa %s %s#%s' % (b64key, username, hostname)
with open(os.getenv('HOME')+'/.ssh/id_rsa.pub') as keyfile:
keyfile.write(keystring)
I didn't test the generated key with SSH, so please let me know if it works (it should i think)
The base64 decoded version of ssh-keygen output to the contents of key.pub() the format of the keyfile is
b64encode('\x00\x00\x00\x07ssh-rsa%s%s' % (key.pub()[0], key.pub()[1]))
If you want, you could just also use ssh-keygen itself.
You can extend this to also create your file, and just use open to read the content later, but i focused on creating a .pub key from an already existing key here.
from subprocess import Popen, PIPE
import os
home = f'{os.path.expanduser("~")}'
cert_pos = f'{home}/.ssh/my_key'
your_key_pw = ''
cmd = ['ssh-keygen', '-y', '-f', cert_pos]
if your_key_pw:
cmd.append('-P')
cmd.append(your_key_pw)
p = Popen(cmd, stdout=PIPE)
p.wait()
res, err = p.communicate()
cert_content = res.decode('utf-8')
Here is an example using the Twisted Conch library which leverages PyCrypto under the covers. You can find the API documentation at http://twistedmatrix.com/documents/current/api/twisted.conch.ssh.keys.html:
from twisted.conch.ssh import keys
# one-time use key
k="""-----BEGIN RSA PRIVATE KEY-----
PRIVATE KEY STUFF
-----END RSA PRIVATE KEY-----"""
# create pycrypto RSA object
rsa = keys.RSA.importKey(k)
# create `twisted.conch.ssh.keys.Key` instance which has some nice helpers
key = keys.Key(rsa)
# pull the public part of the key and export an openssh version
ssh_public = key.public().toString("openssh")
print ssh_public
You can use pycryptodome as described in documentation:
from Crypto.PublicKey import RSA
key = RSA.generate(2048)
private_key = key.export_key()
file_out = open("private.pem", "wb")
file_out.write(private_key)
public_key = key.publickey().export_key()
file_out = open("receiver.pem", "wb")
file_out.write(public_key)
Just guessing... but have you tried something like this?:
print "ssh-rsa " + "".join([ l.strip() for l in open('/tmp/my.key.pub') if not l.startswith('-----')])
Can you get the AAAA...Dfg== string out of it while it's an object? If so, you could simply open a file yourself and save that instead of using the built in save_pub_key function.
I don't know of such a library that comes standard with Python.
If you want to look to third-party libraries, you might find the paramiko library useful (also available from PyPI). It implements the SSH protocol, and has functionality for handling existing keys, but not generating them.
Generation of keys might be a useful addition to that library (you could work with the developers to incorporate it into the Paramiko library), and an easier start than doing it from scratch.