| 編號 | 繳交日期 | 內容 | 解答 |
|---|---|---|---|
| 1 | 10/31 | 2.13.5, 2.13.8, 2.13.17, 2.14.7 | |
| 2 | 11/07 | Prove that if gcd(a, n)!=1 then the inverse of a does not exist. | |
| 3 | 11/28 | 1. Let p = 2 p' + 1, q = 2 q' + 1, p, q, p', q' are all large primes, n = p q, show that Zn* is not a cyclic group 2. 3.13.16, 3.13.35 |
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| 4 | 12/05 | 4.9.11 | |
| midterm solution | |||
| 5 | 12/12 | 4.9.10 | |
| 6 | 12/19 | Show that if you use an additional MC step at the 10-th round of the encryption of Rijndael encryption algorithm, you would not be able to use the same hardware to decrypt the ciphertext. | |
| 7 | 01/09 | Part 1. Let p = 37, q = 79, so that n = 37 * 79,
find the secret decryption key d for e = 11 Part 2. Suppose that Bob uses an RSA cryptosystem which is hard to factorize. Alice sends an encrypted message to Bob, obtained by encrypting it, letter-by-letter, with each one of the 26 letters of the alphabet encrypted using the RSA crytposystem. So A --> 0, B --> 1, C --> 2, etc, with appropriate padding for each number. Describe how Oscar can easily decrypt this message. Part 3. Suppose that three users, Bob, Bart and Bert, use an RSA cryptosystem with moduli n1 < n2 < n3, respeectively, that are hard to factorize. Suppose that all three have the same public encryption exponent e = 3. Alice wants to send to each one of them the same plaintext x < n1. So she sends y1 = x3 (mod n1) to Bob, y2 = x3 (mod n2) to Bart, and y3 = x3 (mod n3) to Bert. Describe how Oscar can obtain the plaintext x from the ciphertexts y1, y2, y3; |
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