This preview shows page 1. Sign up to view the full content.
Unformatted text preview: y d = x ed (mod p1) = x (mod p ). 6.8.8 In order to increase security, Bob chooses n and two encryption exponents e 1 , e 2 . He asks Alice to encrypt her message m to him by ﬁrst computing c 1 = m e 1 (mod n ), then encrypting c 1 to get c 2 = c 1 e 2 (mod n ). Alice then sends c 2 to Bob. Does this double encryption increase security over single encryption? Why or why not? No, Since c 2 = c e 2 1 = m e 1 e 2 1 (mod n ). Put E = e 1 e 2 and we see that is equals the single encryption. 6.8.10 The exponents e = 1 and e = 2 should not be used in RSA. Why? If we use e = 1 in RSA and compute c = m 1 (mod n ), then c is the plaintext! In the case of e = 2, We know that φ ( n ) = ( p1)( q1), where p, q are large primes. Hence gcd( e, φ ( n )) 6 = 1....
View Full
Document
 Spring '09
 Mckessh
 Biology, pH, Work

Click to edit the document details