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Unformatted text preview: 5 Proof. This follows immediately from the fact that a and n are relatively prime if and only if there exist s,t ∈ Z such that as + bt = 1. 2 We now prove a simple a “cancellation law” for congruences: Theorem 2.3 If a is relatively prime to n , then ax ≡ ax (mod n ) if and only if x ≡ x (mod n ) . More generally, if d = gcd( a,n ) , then ax ≡ ax (mod n ) if and only if x ≡ x (mod n/d ) . Proof. For the first statement, assume that gcd( a,n ) = 1, and let a be a multiplicative inverse of a modulo n . Then, ax ≡ ax (mod n ) implies a ax ≡ a ax (mod n ), which implies x ≡ x (mod n ), since a a ≡ 1 (mod n ). Conversely, if x ≡ x (mod n ), then trivially ax ≡ ax (mod n ). That proves the first statement. For the second statement, let d = gcd( a,n ). Simply from the definition of congruences, one sees that in general, ax ≡ ax (mod n ) holds if and only if ( a/d ) x ≡ ( a/d ) x (mod n/d ). Moreover, since a/d and n/d are relatively prime, the first statement of the theorem implies that ( a/d ) x ≡ ( a/d ) x (mod n ) holds if and only if x ≡ x (mod n/d ). That proves the second statement. 2 We next look at solutions x to congruences of the form ax ≡ b (mod n ), for given integers n,a,b . Theorem 2.4 Let n be a positive integer and let a,b ∈ Z . If a is relatively prime to n , then the congruence ax ≡ b (mod n ) has a solution x ; moreover, any integer x is a solution if and only if x ≡ x (mod n ) . Proof. The integer x = ba , where a is a multiplicative inverse of a modulo n , is clearly a solution. For any integer x , we have ax ≡ b (mod n ) if and only if ax ≡ ax (mod n ), which by Theorem 2.3 holds if and only if x ≡ x (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 is also a solution if and only if x ≡ x (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....
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 Spring '13
 MRR
 Math, Algebra, Number Theory

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