303 CSCI Homework 6
Problem 1 (31.1-1): Prove that there are innitely many primes. Solution 1: Assume that there are only nitely many primes, p1 , p2 , . . . , pk . Let
k
n=1+
i=1
pi
Then for all i {1, 2, . . . , k }, if you divide n by pi the remainder is 1. By the fundamental theorem of arithmatic, all numbers can be written as a product of prime factors. Since none of the primes divide n, it cannot be the product of prime factors, and we have a contradiction. Therefore our assumption that there are only nitely many primes must be false, so there are innitely many primes. Problem 2 (31.1-7): For any integer k > 0, we say that an integer n is a kth power if there exists an integer a such that ak = n. We say that n > 1 is a nontrivial power if it is a k th power for some integer k > 1. Show how to determine if a given -bit integer n is a nontrivial power in time polynomial in . Solution 2: Given an n of bits, we wish to check whether there exist integers a and k such that ak = n. Note that a 2 because 1k = 1 for all k , and n must be greater than 1 to be a nontrivial power. This implies that k is at most lg n , because a lg n +1 > n for all a 2. To determine if n is a nontrivial power, all we need to do is check whether any of n1/2 , n1/3 , . . . , n1/ lg n are integers. Checking whether the mth root of n is an integer can be done in time polynomial in , and we need to check at most roots (since lg n ), so determining whether n is a nontrivial power can be done in time polynomial in . Problem 3 (Derived from 31.2-2): Use the extended-Euclids algorithm to compute gcd(899, 493) and numbers x and y such that 899x + 493y = gcd(899, 493). Show your work. Solution 3: (eq 1) (eq 2) (eq 3) (eq 4) (eq 5) (eq 6) 899 = (1) 899 + (0) 493 493 = (0) 899 + (1) 493 406 = (1) 899 + (1)493 = (eq 1) 1 (eq 2) 87 = (1)899 + (2) 493 = (eq 2) 1 (eq 3) 58 (5) = 899 + (9)493 = (eq 3) 4 (eq 4) 29 = (6)899 + (11) 493 = (eq 4) 1 (eq 5) (since (since (since (since
899 493 493 406 406 87 87 58
= 1) = 1) = 4) = 1)
Finally, we see that 58 = 2 29, so gcd(899, 493) = 29, and 29 = (6)899 + (11)493. Problem 4 (Not in book): Your RSA private key is 7, 33 . Digitally sign the message M = 9. Show your work.
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Solution 4: To digitally sign the message, we need to compute 97 mod 33. 97 = 9 92 94 = 9 81 812 Here, we need to compute 81 mod 33, and we nd that 81 = (2)33 + 15, so 81 15 (mod 33).
97 9 15 152 9 15 225
(mod 33) (mod 33) (mod 33).
Next, we need to compute 225 mod 33, and we nd that 225 = (6)33 + 27, so 225 27
97 9 15 27 135 27
(mod 33) (mod 33) (mod 33).
Next, we need to compute 135 mod 33, and we nd that 225 = (4)33 + 3, so 135 3
97 3 27 81 15 Because we found 81 15 (mod 33) earlier.
(mod 33) (mod 33) (mod 33)
Therefore, the digital signature of the message M = 9 is Sig = 15. Problem 5 (31.7-1): Consider an RSA key set with p = 11, q = 29, n = 319, and e = 3. What value of d should be used in the secret key? What is the encryption of the message M = 100? Show your work. Solution 5: p = 11 n=pq = 319 e=3 We need to compute d. To do so, we use the extended-Euclids algorithm with inputs 280 and 3. 280 = (1)280 + (0)3 3 = (0)280 + (1)3 1 = (1)280 + (93)3
2
q = 29 (n) = (p 1) (q 1) = 10 28 = 280
So d = 93 + 280 = 187. Lets check it. 3(187) = 561 561 1 (mod 280) The secret key is 187, 319 . Now we wish to encrypt M = 100 using the public key 3, 319 . We need to nd 1003 mod 319. 1003 = 100 1002 = 100 10 000 We need to compute 10 000 mod 319, and we nd that 10 000 = (31)319 + 111, so 10 000 111 (mod 319). 1003 100 111 11 100
(mod 319) (mod 319) (mod 319).
Similarly, we nd that 11 100 = (34)319 + 254, so 11 100 254 1003 254 So the cyphertext is C = 254.
(mod 319)
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