The algorithm is very simple: 1. Choose a big random value x from some very broad range (say, {1,2,..,N^2}). 2. Pick a random element g (mod N). 3. Compute y = g^x (mod N). 4. Ask for the discrete log of y to the base g, and get back some answer x' such that y = g^x' (mod N). 5. Compute x-x'. Note that x-x' is a multiple of phi(N), and it is highly likely that x-x' is non-zero. It is well-known that given a non-zero multiple of phi(N), you can factor N in polynomial time.

`Not exactly. Consider N = 3*7 = 21, phi(N) = 12, g = 4, x = 2, x' = 5.`

`You'll only get a multiple of phi(N) if g was a generator of the`

`multiplicative group Z_N^*.`

Peter -- [Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278 --------------------------------------------------------------------- The Cryptography Mailing List Unsubscribe by sending "unsubscribe cryptography" to [EMAIL PROTECTED]