# Therefore the prime p is not among p 1 p k which

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1, which is impossible. Therefore, the prime p is not among p 1 , . . . , p k , which contradicts our assumption that these are the only primes. 4

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Chapter 2 Congruences This chapter reviews the notion of congruences. 2.1 Definitions and Basic Properties For positive integer n and for a, b Z , we say that a is congruent to b modulo n if n | ( a - b ), and we write a b (mod n ). If n - ( a - b ), then we write a 6≡ b (mod n ). The number n appearing in such congruences is called the modulus . A trivial observation is that a b (mod n ) if and only if there exists an integer c such that a = b + cn . Another trivial observation is that if a b (mod n ) and n 0 | n , then a b (mod n 0 ). A key property of congruences is that they are “compatible” with integer addition and multi- plication, in the following sense: Theorem 2.1 For all positive integers n , and all a, a 0 , b, b 0 Z , if a a 0 (mod n ) and b b 0 (mod n ) , then a + b a 0 + b 0 (mod n ) and a · b a 0 · b 0 (mod n ) . Proof. Suppose that a a 0 (mod n ) and b b 0 (mod n ). This means that there exist integers c and d such that a 0 = a + cn and b 0 = b + dn . Therefore, a 0 + b 0 = a + b + ( c + d ) n, which proves the first equality of the theorem, and a 0 b 0 = ( a + cn )( b + dn ) = ab + ( ad + bc + cdn ) n, which proves the second equality. 2 2.2 Solving Linear Congruences For a positive integer n , and a Z , we say that a is a unit modulo n if there exists a 0 Z such that aa 0 1 (mod n ), in which case we say that a 0 is a multiplicative inverse of a modulo n . Theorem 2.2 An integer a is a unit modulo n if and only if a and n are relatively prime. 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 0 (mod n ) if and only if x x 0 (mod n ) . More generally, if d = gcd( a, n ) , then ax ax 0 (mod n ) if and only if x x 0 (mod n/d ) . Proof. For the first statement, assume that gcd( a, n ) = 1, and let a 0 be a multiplicative inverse of a modulo n . Then, ax ax 0 (mod n ) implies a 0 ax a 0 ax 0 (mod n ), which implies x x 0 (mod n ), since a 0 a 1 (mod n ). Conversely, if x x 0 (mod n ), then trivially ax ax 0 (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 0 (mod n ) holds if and only if ( a/d ) x ( a/d ) x 0 (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 0 (mod n ) holds if and only if x x 0 (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 0 is a solution if and only if x x 0 (mod n ) .

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• Spring '13
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