How do you verify an RSA SHA1 signature in Python?
Solution 1:
Use M2Crypto. Here's how to verify for RSA and any other algorithm supported by OpenSSL:
pem = """-----BEGIN PUBLIC KEY-----
MIGfMA0GCSqGSIb3DQEBAQUAA4GNADCBiQKBgQDfG4IuFO2h/LdDNmonwGNw5srW
nUEWzoBrPRF1NM8LqpOMD45FAPtZ1NmPtHGo0BAS1UsyJEGXx0NPJ8Gw1z+huLrl
XnAVX5B4ec6cJfKKmpL/l94WhP2v8F3OGWrnaEX1mLMoxe124Pcfamt0SPCGkeal
VvXw13PLINE/YptjkQIDAQAB
-----END PUBLIC KEY-----""" # your example key
from M2Crypto import BIO, RSA, EVP
bio = BIO.MemoryBuffer(pem)
rsa = RSA.load_pub_key_bio(bio)
pubkey = EVP.PKey()
pubkey.assign_rsa(rsa)
# if you need a different digest than the default 'sha1':
pubkey.reset_context(md='sha1')
pubkey.verify_init()
pubkey.verify_update('test message')
assert pubkey.verify_final(signature) == 1
Solution 2:
The data between the markers is the base64 encoding of the ASN.1 DER-encoding of a PKCS#8 PublicKeyInfo containing an PKCS#1 RSAPublicKey.
That is a lot of standards, and you will be best served with using a crypto-library to decode it (such as M2Crypto as suggested by joeforker). Treat the following as some fun info about the format:
If you want to, you can decode it like this:
Base64-decode the string:
30819f300d06092a864886f70d010101050003818d0030818902818100df1b822e14eda1fcb74336
6a27c06370e6cad69d4116ce806b3d117534cf0baa938c0f8e4500fb59d4d98fb471a8d01012d54b
32244197c7434f27c1b0d73fa1b8bae55e70155f907879ce9c25f28a9a92ff97de1684fdaff05dce
196ae76845f598b328c5ed76e0f71f6a6b7448f08691e6a556f5f0d773cb20d13f629b6391020301
0001
This is the DER-encoding of:
0 30 159: SEQUENCE {
3 30 13: SEQUENCE {
5 06 9: OBJECT IDENTIFIER rsaEncryption (1 2 840 113549 1 1 1)
16 05 0: NULL
: }
18 03 141: BIT STRING 0 unused bits, encapsulates {
22 30 137: SEQUENCE {
25 02 129: INTEGER
: 00 DF 1B 82 2E 14 ED A1 FC B7 43 36 6A 27 C0 63
: 70 E6 CA D6 9D 41 16 CE 80 6B 3D 11 75 34 CF 0B
: AA 93 8C 0F 8E 45 00 FB 59 D4 D9 8F B4 71 A8 D0
: 10 12 D5 4B 32 24 41 97 C7 43 4F 27 C1 B0 D7 3F
: A1 B8 BA E5 5E 70 15 5F 90 78 79 CE 9C 25 F2 8A
: 9A 92 FF 97 DE 16 84 FD AF F0 5D CE 19 6A E7 68
: 45 F5 98 B3 28 C5 ED 76 E0 F7 1F 6A 6B 74 48 F0
: 86 91 E6 A5 56 F5 F0 D7 73 CB 20 D1 3F 62 9B 63
: 91
157 02 3: INTEGER 65537
: }
: }
: }
For a 1024 bit RSA key, you can treat "30819f300d06092a864886f70d010101050003818d00308189028181"
as a constant header, followed by a 00-byte, followed by the 128 bytes of the RSA modulus. After that 95% of the time you will get 0203010001
, which signifies a RSA public exponent of 0x10001 = 65537.
You can use those two values as n
and e
in a tuple to construct a RSAobj.
Solution 3:
A public key contains both a modulus(very long number, can be 1024bit, 2058bit, 4096bit) and a public key exponent(much smaller number, usually equals one more than a two to some power). You need to find out how to split up that public key into the two components before you can do anything with it.
I don't know much about pycrypto but to verify a signature, take the hash of the string. Now we must decrypt the signature. Read up on modular exponentiation; the formula to decrypt a signature is message^public exponent % modulus
. The last step is to check if the hash you made and the decrypted signature you got are the same.