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Article #3663Question about choosing E & D in RSA encryption
From: Michael Mauldin
Date: Mon, 13 Jul 1987 05:08
Date: Mon, 13 Jul 1987 05:08
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1082 bytes
Does choosing E to be a power of 2 work for RSA encryption? If E = 2**k, then exp(M,E) mod N is just k squarings of M (mod N). This could reduce the number of multiplications necessary for encryption by an average of 33% (the decryption would still take as long, since Dalmost certainly not be a power of 2). Does this weaken the algorithm in any way? Has someone already suggested this? Michael L. Mauldin (Fuzzy) Department of Computer Science ARPA: Michael.Mauldin@NL.CS.CMU.EDU Carnegie-Mellon University Phone: (412) 268-3065 Pittsburgh, PA 15213-3890
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Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Path: arc1-ca.pugleaf.net!utzoo!mnetor!uunet!seismo!rochester!pt!nl.cs.cmu.edu!mlm From: mlm@nl.cs.cmu.edu (Michael Mauldin) Newsgroups: sci.crypt Subject: Question about choosing E & D in RSA encryption Message-ID: <15@nl.cs.cmu.edu> Date: Mon, 13-Jul-87 01:08:51 EDT Article-I.D.: nl.15 Posted: Mon Jul 13 01:08:51 1987 Date-Received: Tue, 14-Jul-87 00:48:06 EDT Organization: Carnegie-Mellon University, CS/RI Lines: 13 Keywords: RSA encryption fast