[sci.crypt] Question about choosing E & D in RSA

mlm@nl.cs.cmu.edu (Michael Mauldin) (07/13/87)

In article <15@nl.cs.cmu.edu>, mlm@nl.cs.cmu.edu (Michael Mauldin) writes:
> 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).

I meant 2**k + 1.  Obviously if p & q are odd, then phi(p*q) = 
(p-1)(q-1) = product of two even numbers has 4 as a factor, and
then gcd (2**k, phi(N)) >= 4.

If E = 2**k + 1, then exp(M,E) mod N takes k+1 multiplications.

Using 2**k + 1, you can even choose a relatively small k to further
reduce the number of multiplications.  With 100 digit p and q and 200
digit N, using E = 2**32 + 1 reduced encryption time from 169.4
seconds (on a microvax) to 6.8 seconds (25 times faster).

The question still is, does this weaken the encryption?  Is it easier
to break RSA using E = 2**k+1 for relatively small k?

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