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RSA Encryption simple proof

  1. Dec 1, 2009 #1
    b7nfig.jpg

    I started by saying

    cince c = me mod n (1)

    then me ≡ c mod n

    Since this is true then

    mek ≡ cek-1 mod n

    and cek-1 = (me)ek-1 mod n, by (1)

    = mek mod n

    => k is a generator of e

    and so both are proved? Not sure..

    Also I haven't a clue is it's dangerous or not for RSA.

    Thanks for any help.
     
  2. jcsd
  3. Dec 2, 2009 #2
    Bump for desperate help!
     
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