Tom Rollins writes:
If this is the file that the Co$ is trying to crack, then what the is being asked for is a pass phrase that can be handed to the Co$ that will pass the PGP valid key check and still not decrypt the data to anything usefull.
Well, I don't have the PGP 'conventional' encryption format memorized, but there is probably a constant after the IV that is prepended to the data. The constant is used to determine if the key is correct. Since the conventional encryption runs in CFB mode and there is a full block of random IV at the beginning of the file, it is extremely unlikely that a key could be found that would properly decrypt only the first two blocks while leaving the rest unreadable...
If Larry Wollersheim does have the valid key. It would be a simpler process to know what fake key to use and work it backwards through the MD5 to arrive at an ascii string to produce the fake key.
Not really. Even if you could find an IDEA key that would produce the desired output it would be hard to find a passphrase that would produce that key when hashed. One of the properties of one-way hash functions is that it is difficult to find a plaintext that produces a given hash. Hence the term 'one-way'.... Even if you did find a passphrase (which, if MD5 is strong, would require something like 2^64 operations), it would likely be long, have 8-bit chars, and would be impossible to type in. It would be tough to convince anyone that it was the real passphrase.
andrew
There was a hack to pgp ui published a while back that would allow someone decrypting a RSA encrypted file to print out the idea key.
Another feature of the hack allowed someone with the idea key to decrypt an RSA PGP encrypted file ignoring the RSA headers and using the IDEA key directly.
Using this software should allow the reciever of an RSA PGP encrypted file to allow someone else to decrypt it (by giving them the IDEA key) without exposing the secret key. The IV block check will allow them to check that they are using the correct idea key.
Looking at the source code showes that all that is needed to pass the PGP key check is for the first two blocks to decode in such a way that the last 2 bytes of the IV match the 2 check bytes before the actual message. Thus the first 6 bytes of the IV and the last 6 bytes of the next block need not match the actual message. There was a bug in the older versions of PGP that set the IV to a constant instead of a random value when encrypting with the "-c" option. I made a mistake thinking that knowledge of the correct key would help in creating a fake key.