Q(173)Q(177)Q(188)Q(193)Q(203)

Transcription

1 MATH 313: SOLUTIONS HW3 Problem 1 (a) We use the Miller-Rabin test to check if it prime. We know that the smallest number which is a strong pseudoprime both base 2 and base 3 is ; hence, if passes the test for both base 2 and base 3 it will have to be prime since < Base 2: = (mod 30941) (mod 30941) pass the test. Base 3: (mod 30941) (mod 30941) pass the test. Therefore is prime and we are done! (b) Base 2: = (mod 33919) (mod 33919). Fail the test! so is composite. {2, 3, 5, 7, 17, 19, 31, 37, 43, 47, 53}. Also note that = 184 and we find that sieving between 170 and 203 is sufficient. Q(173)Q(177)Q(188)Q(193)Q(203) = ( )( )( )( )( ) =

2 gcd(( ) ( ), 33919) = 317 gcd(( ) + ( ), 33919) = 107 Hence, = (c) Base 2: = (mod 68569) (mod 68569) (mod 68569) (mod 68569). Fail the test! so is composite. {2, 3, 5, 7, 17, 19, 23, 29, 31, 53, 61}. Also note that = 262 and we find that sieving between 242 and 282 is sufficient. Q(262)Q(263)Q(264)Q(265) = (3 5 2 )( )(7 2 23)( ) = gcd(( ) ( ), 68569) = 191 gcd(( ) ( ), 68569) = 359 Hence, = (d) Base 2: = (mod ) (mod ). Pass the test! so can be a prime.

3 Base 3: (mod ) (mod ) (mod ). Fail the test! so is a composite. {2, 3, 5, 7, 11, 13, 17, 19, 29, 31, 37}. Also note that = 916 and we find that sieving between 896 and 936 is sufficient. Q(917)Q(923) = ( )( ) = gcd(( ) ( ), ) = 397 gcd(( ) + ( ), ) = 2113 Hence, = (e) Base 2: = (mod ). Pass the test! so can be a prime. Base 3: (mod ) (mod ) (mod ) (mod ) (mod ) (mod ). Fail the test! so is a composite. {2, 5, 7, 11, 19, 29, 31, 37, 41, 3, 47, 53}.

4 Also note that = 1008 and we find that sieving between 978 and 1038 is sufficient. Q(991)Q(999)Q(1011)Q(1022) = ( )( )( )( ) = gcd(( ) ( ), ) = 4051 gcd(( ) + ( ), ) = 251 Hence, = (f) Base 2: = (mod ) (mod ). Fail the test! so is a composite. {2, 3, 13, 17, 19, 23, 41, 61, 67, 71, 73}. Also note that = 1072 and we find that sieving between 1012 and 1131 is sufficient. Q(1077)Q(1085) = ( )( ) = gcd(( ) ( ), ) = 521 gcd(( ) + ( ), ) = 2207 Hence, =

5 Problem 2 Here are 6 possible factors: (1) Q(103) Q(107) = potential factors: gcd( , 10553) gcd( , 10553) (2) Q(103) Q(110) Q(111) = potential factors: gcd( , 10553) gcd( , 10553) (3) Q(107) Q(110) Q(111) = potential factors: gcd( , 10553) gcd( , 10553) Problem 3 Note that the form p 2 k 2047q2 k = ±Q k 1 p 2 k ±Q k 1 (mod 2047). In order to use the quadratic sieve, we want the product of some (±Q k 1 ) to be a perfect square. Observe that in the Q k+1 given, we already have a perfect square when p k = 181 and Q k+1 = 3 2. Therefore, we can use this one to try to factorize Following the observation above, we have: (mod 2047). Now, we find that gcd( , 2047) = 23 and 2047/23 = 89. Hence, we get 2047 = The period is at least 5. Problem 4 Let s look at the first few convergents of Note: You were supposed to compute the convergents using the algorithm used in class. I will omit this step in the solution, but you should be able to obtain the same 6 first convergents by going through the algorithm. [Steps of the algorithm: (1) compute α i s and a i s, (2) find p i s and q i s. You now have the convergents and are able to compute Q i s.]

6 First convergent: = 757. Not a perfect square. Second convergent: = 28 = Not a perfect square and cannot get a perfect square with a product of previous convergents. Third convergent: 11003/ = 55 = Not a perfect square and cannot get a perfect square with a product of previous convergents. Fourth convergent: / = 196 = = Perfect square. We can compute the gcd to find out if it gives us a factorization: gdc( , ) = 1 and gdc( , ) = So it doesn t give us a factorization! We continue looking at convergents... Fifth convergent: / = 565 = Not a perfect square and cannot get a perfect square with a product of previous convergents. Sixth convergent: / = 49 = 7 2. Perfect square. We can compute the gcd to find out if it gives us a factorization: gdc( , ) = 503 and /503 = 307. Hence we get =