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Proof the integer x ba where a is a multiplicative

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Proof. The integer x = ba 0 , where a 0 is a multiplicative inverse of a modulo n , is clearly a solution. For any integer x 0 , we have ax 0 b (mod n ) if and only if ax 0 ax (mod n ), which by Theorem 2.3 holds if and only if x x 0 (mod n ). 2 In particular, this theorem implies that multiplicative inverses are uniquely determined modulo n . More generally, we have: Theorem 2.5 Let n be a positive integer and let a, b Z . Let d = gcd( a, n ) . If d | b , then the congruence ax b (mod n ) has a solution x , and any integer x 0 is also a solution if and only if x x 0 (mod n/d ) . If d - b , then the congruence ax b (mod n ) has no solution x . Proof. Let n, a, b, d be as defined above. For the first statement, suppose that d | b . In this case, by Theorem 2.3, we have ax b (mod n ) if and only if ( a/d ) x ( b/d ) (mod n/d ), and so the statement follows immediately from Theorem 2.4. For the second statement, assume that ax b (mod n ) for some integer x . Then since d | n , we have ax b (mod d ). However, ax 0 (mod d ), since d | a , and hence b 0 (mod d ), i.e., d | b . 2 Next, we consider systems of congruences with respect to moduli that that are relatively prime in pairs. The result we state here is known as the Chinese Remainder Theorem, and is extremely useful in a number of contexts. 6
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Theorem 2.6 (Chinese Remainder Theorem) Let k > 0 , and let a 1 , . . . , a k Z , and let n 1 , . . . , n k be positive integers such that gcd( n i , n j ) = 1 for all 1 i < j k . Then there ex- ists an integer x such that x a i (mod n i ) ( i = 1 , . . . , k ) . Moreover, any other integer x 0 is also a solution of these congruences if and only if x x 0 (mod n ) , where n := Q k i =1 n i . Proof. Let n := Q k i =1 n i , as in the statement of the theorem. Let us also define n 0 i := n/n i ( i = 1 , . . . , k ) . It is clear that gcd( n i , n 0 i ) = 1 for 1 i k , and so let m i be a multiplicative inverse of n 0 i modulo n i for 1 i k , and define z i := n 0 i m i ( i = 1 , . . . , k ) . By construction, one sees that for 1 i k , we have z i 1 (mod n i ) and z i 0 (mod n j ) for 1 j k with j 6 = i . That is to say, for 1 i, j k , z i δ ij (mod n j ), where δ ij := 1 for i = j and δ ij := 0 for i 6 = j . Now define x := k X i =1 z i a i . One then sees that for 1 j k , x k X i =1 z i a i k X i =1 δ ij a i a j (mod n j ) . Therefore, this x solves the given system of congruences. Moreover, if x 0 x (mod n ), then since n i | n for 1 i k , we see that x 0 x a i (mod n i ) for 1 i k , and so x 0 also solves the system of congruences. Finally, if x 0 solves the system of congruences, then x 0 x (mod n i ) for 1 i k . That is, n i | ( x 0 - x ) for 1 i k . Since gcd( n i , n j ) = 1 for i 6 = j , this implies that n | ( x 0 - x ), i.e., x 0 x (mod n ). 2 2.3 Residue Classes It is easy to see that for a fixed value of n , the relation · ≡ · (mod n ) is an equivalence relation on the set Z ; that is, for all a, b, c Z , we have a a (mod n ) , a b (mod n ) implies b a (mod n ) , and a b (mod n ) and b c (mod n ) implies a c (mod n ) .
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