# 6.8 sol - y d = x ed(mod p-1 = x(mod p 6.8.8 In order to...

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4th Cryptography Homework 6.8.2 Suppose your RSA modulus is n = 55 = 5 × 11 and your encryption exponent is e = 3 6.8.2.a Find the decryption modulus d . n = 5 × 11, φ ( n ) = 4 × 10 = 40 We obtain that e = 3 , φ ( n ) = 40 , and d = 27. 6.8.2.b Assume that gcd( m, 55) = 1. Show that if c = m 3 (mod 55) is the ciphertext, then the plaintext is m = c d (mod 55). Do not quote the fact that RSA decryption works. That is what you are showing in this speciﬁc case. m = c 27 = ( m 3 ) 27 = m 81 (mod φ (55)) = m 81 (mod 40) = m 1 (mod 55) 6.8.4 Suppose you encrypt messages m by computing c = m 3 (mod 101). How do you decrypt? Since 101 is a prime, φ ( n ) = 100 , e = 3 , d = 67. The decryption is as following: m = c 67 (mod 101) = m 3 · 67 (mod 100) (mod 101) . 6.8.5 Let p be a large prime. Suppose you encrypt a message x by computing y = x e (mod p ) for some (suitably chosen) encryption exponent e . How do you ﬁnd a decryption d such that y d = x (mod p )? Note that φ ( p ) = p - 1. Hence we ﬁnd d s.t. ed = 1 (mod p - 1) and
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Unformatted text preview: y d = x ed (mod p-1) = 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 ) = ( p-1)( q-1), where p, q are large primes. Hence gcd( e, φ ( n )) 6 = 1....
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## This note was uploaded on 10/01/2009 for the course SCIENCE 111 taught by Professor Mckessh during the Spring '09 term at École Normale Supérieure.

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