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<H1>[pycrypto] PyCrypto AND Crypt_RSA integration</H1>
<B>Dwayne C. Litzenberger</B>
<A HREF="mailto:pycrypto%40lists.dlitz.net?Subject=%5Bpycrypto%5D%20PyCrypto%20AND%20Crypt_RSA%20integration&In-Reply-To=3c5f192d0902090436r3e9c905n2edd78019033118%40mail.gmail.com"
TITLE="[pycrypto] PyCrypto AND Crypt_RSA integration">dlitz at dlitz.net
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<I>Tue Feb 10 18:32:53 CST 2009</I>
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<PRE>On Mon, Feb 09, 2009 at 10:36:40AM -0200, Mauricio Arozi wrote:
><i>Am I helpless?
</I>
I think the problem is that you're asking the mailing list for the *Python*
Cryptography Toolkit about how to use an obscure *PHP* library.
We can help with the Python side of things. I wouldn't expect the people
here to know and/or care much about PHP.
><i> According to this site: <A HREF="http://pajhome.org.uk/crypt/rsa/rsa.html,">http://pajhome.org.uk/crypt/rsa/rsa.html,</A> and
</I>><i> yet others, the e(exponent?) is used for the public key, and d for the
</I>><i> private key.
</I>
The notation I've seen most often is something like this:
n - modulus (public)
e - public exponent
d - private exponent
(n, e) - public key
(n, d) - private key
(p, q) - the (private) primes from which the keypair is derived.
PyCrypto uses a similar notation:
from Crypto.PublicKey import RSA
import os
# DO NOT USE RandomPool (see below)
keypair = RSA.generate(2048, os.urandom)
print "PRIVATE KEYPAIR:"
print "n:", keypair.n # modulus (public)
print "e:", keypair.e # public exponent
print "d:", keypair.d # private exponent
print "p:", keypair.p # prime (private)
print "q:", keypair.q # other prime (private)
print "u:", keypair.u # I forget what this for (but it's private)
pub = keypair.publickey()
print ""
print "PUBLIC KEY:"
print "n (pub):", pub.n # modulus (public)
print "e (pub):", pub.e # public exponent
print "d (pub):", pub.d # raises an exception
print "p (pub):", pub.p # raises an exception
print "q (pub):", pub.q # raises an exception
print "u (pub):", pub.u # raises an exception
This outputs the following:
PRIVATE KEYPAIR:
n: 277...[truncated]
e: 65537
d: 232...[truncated]
p: 159...[truncated]
q: 174...[truncated]
u: 125...[truncated]
PUBLIC KEY:
n (pub): 277...[truncated]
e (pub): 65537
d (pub):
Traceback (most recent call last):
File "x.py", line 21, in ?
print "d (pub):", pub.d
File "/usr/lib/python2.4/site-packages/Crypto/PublicKey/RSA.py", line 154, in __getattr__
return getattr(self.key, attr)
AttributeError: rsaKey instance has no attribute 'd'
><i> My problem is that while using PyCrypto to generate both public and
</I>><i> private keys, the e(exponent?) is always the same.
</I>
Mads Kiilerich already talked a bit about this, but I won't go into detail.
What you're describing here is normal, and it really helps improve the
performance of encryption/verification.
If you're concerned about the security of using RSA, I suggest reading Dan
Boneh's 1999 article, "Twenty years of attacks on the RSA cryptosystem":
<A HREF="http://crypto.stanford.edu/~dabo/abstracts/RSAattack-survey.html">http://crypto.stanford.edu/~dabo/abstracts/RSAattack-survey.html</A>
><i>So in simple words, I only need to be able to encrypt/decrypt sign and
</I>><i>verify signs on php and python, simultaneously, if possible, using RSA
</I>><i>algo.
</I>
PyCrypto's PublicKey package is very low-level, so people shouldn't use it
directly unless they REALLY know what they are doing. Mere mortals should
use a separate library in addition to PyCrypto for that. You should never
do anything like this:
><i>privkeyA = RSA.generate(512, rpool.get_bytes)
</I>><i>pubkeyA = privkeyA.publickey()
</I>><i>
</I>><i>msg = 'This is the secret phrase testing.'
</I>><i>msgc = pubkeyA.encrypt(msg, '')
</I>
That is called "textbook RSA", and it's insecure. (Also, it uses a 512-bit
key, which is way too short, but I assume that's just for demonstration.)
I strongly recommend looking at PKCS#1v2 (also known as RSAES-OAEP).
PyCrypto doesn't include an implementation yet, but Sergey Chernov
mentioned that he is working on one.
Also, I noticed in your code that you used RandomPool. Don't. RandomPool
is a security disaster, and it will be removed from future versions. See
the following messages:
<A HREF="http://lists.dlitz.net/pipermail/pycrypto/2008q3/000000.html">http://lists.dlitz.net/pipermail/pycrypto/2008q3/000000.html</A>
<A HREF="http://lists.dlitz.net/pipermail/pycrypto/2008q3/000020.html">http://lists.dlitz.net/pipermail/pycrypto/2008q3/000020.html</A>
I hope you find the above information helpful.
Cheers,
- Dwayne
--
Dwayne C. Litzenberger <<A HREF="http://lists.dlitz.net/cgi-bin/mailman/listinfo/pycrypto">dlitz at dlitz.net</A>>
Key-signing key - 19E1 1FE8 B3CF F273 ED17 4A24 928C EC13 39C2 5CF7
Annual key (2008) - 4B2A FD82 FC7D 9E38 38D9 179F 1C11 B877 E780 4B45
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