This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Leo Reyzin. Notes for BU CAS CS 538. 1 4 Working with composite moduli and the BlumBlumShub generator 4.1 Chinese Remainder Theorem Let p = q be two primes. The Chinese Remainder Theorem (CRT) says that working modulo n = pq is essentially the same as working modulo p and modulo q at the same time: more formally (for those comfortable with abstract algebra), that the ring Z n is isomorphic to the product ring Z p Z q . (Actually, this is the light version of CRT, which is all we need for this course. The fullfledged version says that working modulo a 1 a 2 . . . a k , where a i are pairwise relatively prime, is the same as working simultaneously modulo a 1 , a 2 , . . . , a k .) Here is an example. Consider all the values modulo 35. They are in onetoone correspondence with values modulo 5 and modulo 7. 1 2 3 4 5 6 15 30 10 25 5 20 1 21 1 16 31 11 26 6 2 7 22 2 17 32 12 27 3 28 8 23 3 18 33 13 4 14 29 9 24 4 19 34 Observe that if you want to add, say, 17 and 29 (underlined in the table), is the same as adding 3 (which is 17 mod 7) and 1 (which is 29 mod 7) modulo 7 to get 4; adding 2 (which is 17 mod 5) and 4 (which is 29 mod 5) modulo 5 to get 1; and then looking up the value corresponding to coordinates 4 and 1 in the table to get 11 (in a box in the table). Thus, we can do addition coordinatewise. Same for multiplication. We now formally state and prove the observations above, generalized to p and q instead of 5 and 7. Theorem 1. Let p = q be primes, n = pq . For each a Z p , b Z q , there is unique c , c < n such that c a (mod p ) and c b (mod q ) . Proof. Let r = p 1 mod q and s = q 1 mod p . Let c = rpb + sqa . Then c rpb + sqa r b + 1 a a (mod p ), and c rpb + sqa 1 b + s a b (mod q ). Let c = c mod pq . Then pq  ( c c ), so p  ( c c ), so c c (mod p ). Similarly, c c (mod q ). Hence, c satisfies all the conditions: 0 c < n , and c a (mod p ) (because c c a (mod p )), and c b (mod q ) (because c c b (mod q )). Thus, for every pair ( a, b ) there is a c . There are pq...
View
Full
Document
This document was uploaded on 02/27/2012.
 Spring '09

Click to edit the document details