Abstract Algebra Theory and Applications - Computer Science ...

14.5 AN APPLICATION TO SOFTWARE DESIGN 247This is equivalent to showing thatk 1 m ≡ (b − a) (mod n)has a solution for k 1 . Since m and n are relatively prime, there exist integerss and t such that ms + nt = 1. Consequently,or(b − a)ms = (b − a) − (b − a)nt,[(b − a)s]m ≡ (b − a)(mod n).Now let k 1 = (b − a)s.To show that any two solutions are congruent modulo mn, let c 1 and c 2be two solutions of the system. That is,for i = 1, 2. Thenc i ≡ a (mod m)c i ≡ b (mod n)c 2 ≡ c 1 (mod m)c 2 ≡ c 1 (mod n).Therefore, both m and n divide c 1 − c 2 . Consequently, c 2 ≡ c 1 (mod mn).□Example 21. Let us solve the systemx ≡ 3 (mod 4)x ≡ 4 (mod 5).Using the Euclidean algorithm, we can find integers s and t such that 4s +5t = 1. Two such integers are s = −1 and t = 1. Consequently,x = a + k 1 m = 3 + 4k 1 = 3 + 4[(5 − 4)4] = 19.Theorem 14.19 (Chinese Remainder Theorem) Let n 1 , n 2 , . . . , n k bepositive integers such that gcd(n i , n j ) = 1 for i ≠ j. Then for any integersa 1 , . . . , a k , the systemx ≡ a 1 (mod n 1 )x ≡ a 2 (mod n 2 ).x ≡ a k (mod n k )