hw5-sol_Security

hw5-sol_Security - udaya shankar Page 1 of 4 10/12/2006 _...

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udaya shankar Page 1 of 4 10/12/2006 CMSC 414: HW 2 Solution and Grading ___________________________________________________________ 1. (text 5.1) Would it be reasonable to compute an RSA signature on a long message m by signing m mod-n (i.e., using (m mod-n) d mod-n as the signature). No. Recall that RSA restricts the message to be signed to be smaller than n. If m is larger than n, then message m and message (m mod-n) would have the same signature. So it would be easy to generate different messages that have the same signature. ___________________________________________________________ 2. (text 5.6) Why do MD4, MD5, and SHA-1 require padding of messages that are already a multiple of 512-bits? Otherwise it would be easiy to find two messages with the same hash. Let M' be any message that is not a multiple of 512 bits. Let M be M' padded as in MD4, so M is a multiple of 512 bits. If no padding is used for M (because it is a multiple of 512 bits) then MD4(M) would be the same as MD4(M'). ___________________________________________________________ 3. (text 6.3) In RSA, is it possible for more than one d to work with a given e, p, and q ? Because d is the multiplicative inverse of e mod-(p - 1) (q - 1), it is unique modulo (p - 1) (q - 1). [Recall e has a multiplicative inverse mod (p - 1) (q - 1) iff e is relatively prime to (p - 1) (q - 1). So multiplying the elements of Z (p - 1) (q - 1) by e results in a permutation of Z (p - 1) (q - 1) , so there is only one element in Z (p - 1) (q - 1) which yields 1 when multiplied by e.] ___________________________________________________________ 4. (text 6.8) Given your RSA signature on m
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hw5-sol_Security - udaya shankar Page 1 of 4 10/12/2006 _...

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