/*
  Author:  Pate Williams (c) 1997

  Exercise
  "6.12 Suppose Bob is using the Pedersen-van Heyst
  Fail-stop Signature scheme, where p = 3467,
  alpha = 4, a0 = 1567 and beta = 514 (of course,
  the value of a0 is not known to Bob).
  (a) Using the fact that a0 = 1567, determine all
      possible keys
      K = (gamma1, gamma2, a1, a2, b1, b2)
      such that sig(42) = (1118, 1449).
  (b) Suppose that sig(420 = (1118, 1449) and
      sig(969) = (899, 471). Without using the
      fact that a0 = 1567, determine the value
      of K (this shows that the scheme is a
      one-time scheme)." -Douglas R. Stinson-
  See "Cryptography: Theory and Practice" by
  Douglas R. Stinson page 232.
*/