Hw4 - (1.1) (1.1) p. 109 #27. a) Since there are only two...

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Unformatted text preview: (1.1) (1.1) p. 109 #27. a) Since there are only two primes in the factorization, then there are at most 4 possible outputs, by the CRT. Hence, the probability of getting a wrong answer n times in a row is , so that there is only a 1% chance of not having a correct answer after 16 random tries. (0.75)^16; 0.01002259576 b) Knowing the message is essentially the same as being able to factor n. Here is the argument. If we can factor n, then we can find the square roots of n and read the message using the CRT. On the other hand, if we can find solutions y mod n for which mod n. If we can do that, then mod n, so n divides the product, but because factor, so a proper factor of n is in each gcd( x - y , n ) and gcd( x + y , n ), so we can factor n. c) Given access to a decoding machine, we run it until we find such x and y. We then use the Euclidean algorithm to find the gcd, which yields the factorization. p. 109 #30. a) If and gcd( n , a ) = 1, then gcd( x , n ) = 1 and = = +1. Now take the contrapositive. b) Since 35 is composite, we are dealing with a Jacobi symbol. ( ) = = since 3 is odd, gcd(3,5) = 1 and 5 = 1 mod 4 = since 5 = 2 mod 3 = (-1) since 3 = 3 mod 8 = since 3 is odd and both 3 and 7 are 3 mod 4 (3.1) (3.1) = since 7 = 1 mod 3 = +1 since = 1 If , then 35 | ( ) so 5 | ( ) so would be 1, but = since 3 is odd and gcd(3,5) = 1, and 5 = 1 mod 4. But = = -1, since 3 = 3 mod 8. A Legendre symbol is +1 iff the top number is a quadratic residue, but the same CANNOT be said for Jacobi symbols. p. 109 #31. a) The Jacobi symbol = +1 since 15 mod 8 = 7. On the other hand, 2&^((15-1)/2) mod 15; 8 so the property p. 89 #2, which is true for Legendre symbols, is NOT true for Jacobi symbols....
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This note was uploaded on 02/14/2012 for the course MATH 470 taught by Professor Staff during the Spring '08 term at Texas A&M.

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Hw4 - (1.1) (1.1) p. 109 #27. a) Since there are only two...

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