A Question For The Techys ?

[timetrex]

The information needed to c/s may also be held in the card encryted in some tier.. But really for most peeps u need to dump the flash which for the newer boxes with no jtag points will be difficult and even render the box as a doorstopper and prevent future updates to the tiers.

[timetrex]

Are there random delays between the IRD and card when computing the new key?

[leachy070667]

Originally Posted by TheCoder

It is actually truly random !

Sequence of events is -

1/ Card informs box that it has no session key or session expired
2/ Box asks card to generate a random session key
3/ Card signals that it has a new random session key
4/ Box asks for the new random session key
5/ Card returns the session key in an RSA encrypted form using a "shared" rsa modulus and private/public keyset. Only the card has the private key.
6/ Box receives the new session key and decrypts it using the shared public key. It then modifies the key in a certain way and re-encrypts using the public key.
7/ Box returns the newly encrypted key back to the card
8/ Card verifies that the key has been modified in the correct way
9/ Card and box now use the new session key for all ECM CW transfers.
10 After 1 hour - start again !

So, the session key really is truly randon and changes every 1 hour. This is not something VM/Kudelski initiate - they dont know the resultant key. Its just a process between between stb and card.

I don’t know how (or even if) it will ever be done, but you can never say never.

Personally I am convinced the way forward will be by somehow deciphering or diverting the data from a paired card and box (from a basic subscription).

Although looking at the communication process you have posted it looks fool proof.

[TheCoder]

Originally Posted by leachy070667

Although looking at the communication process you have posted it looks fool proof.

It isn't !

Its defeated routinely enough in both Europe and the US. If you study how its done in those places then the solution will become relatively obvious.

[leachy070667]

Originally Posted by TheCoder

It isn't !

Its defeated routinely enough in both Europe and the US. If you study how its done in those places then the solution will become relatively obvious.

So let me get this right (or wrong), what you are saying is, that if you have said box with paired card in your possession and a basic subscription, then your half way there ?

I am intrigued ?

[TheCoder]

Originally Posted by leachy070667

So let me get this right (or wrong), what you are saying is, that if you have said box with paired card in your possession and a basic subscription, then your half way there ?

I am intrigued ?

You will only ever get what the card is subscribed to - the interesting thing is that if you defeat the pairing/session management then you have the ability to share the card to other boxes at the same time (you still only get what the card is subscribed to on the other boxes of course).

This is what cardsharing is all about and its been occuring for quite a few years now on various encryption systems. There's nothing technically different about cable in the UK which precludes exactly the same techniques being used.