How do you verify an RSA SHA1 signature in Python?

Use M2Crypto. Here's how to verify for RSA and any other algorithm supported by OpenSSL:

pem = """-----BEGIN PUBLIC KEY-----MIGfMA0GCSqGSIb3DQEBAQUAA4GNADCBiQKBgQDfG4IuFO2h/LdDNmonwGNw5srWnUEWzoBrPRF1NM8LqpOMD45FAPtZ1NmPtHGo0BAS1UsyJEGXx0NPJ8Gw1z+huLrlXnAVX5B4ec6cJfKKmpL/l94WhP2v8F3OGWrnaEX1mLMoxe124Pcfamt0SPCGkealVvXw13PLINE/YptjkQIDAQAB-----END PUBLIC KEY-----""" # your example keyfrom M2Crypto import BIO, RSA, EVPbio = 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

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:

30819f300d06092a864886f70d010101050003818d0030818902818100df1b822e14eda1fcb743366a27c06370e6cad69d4116ce806b3d117534cf0baa938c0f8e4500fb59d4d98fb471a8d01012d54b32244197c7434f27c1b0d73fa1b8bae55e70155f907879ce9c25f28a9a92ff97de1684fdaff05dce196ae76845f598b328c5ed76e0f71f6a6b7448f08691e6a556f5f0d773cb20d13f629b63910203010001

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.

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.