X-Loop: openpgp.net
From: "Frank Andrew Stevenson" <[EMAIL PROTECTED]>

> 2) Calculate MD5( passphrase ) and pads it repeadedly to get the number
>    of bits in p less one.
> 3) calculates B = g ** md5md5...  mod p , and publishes this as public key

I was under the impression that the exponents has to be a prime number,
too... anyone? [Well, if that's the case, I guess you could search for the
first prime number after the result of MD5 padding.]

Mark


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