Proposition 3.99

Proposition 3.99 - q playing the role of p . There are...

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MATH 135 Fall 2008 Proposition 3.99 Let p,q be distinct primes, let M Z , and let k P . Then M k ( p - 1)( q - 1)+1 M (mod pq ). Note: This proposition does not appear in the textbook. It is closely related to Corollary 3.73, but the proof below is different. Proof: The strategy is to apply Proposition 3.64 (or equivalently, the Chinese Remainder Theorem). In other words, we consider the congruence separately modulo p and modulo q . First, consider the congruence modulo p . There are two cases: Suppose M 0 (mod p ). Then M k ( p - 1)( q - 1)+1 0 M (mod p ). Suppose M 6≡ 0 (mod p ). Then by Fermat’s Little Theorem, M p - 1 1 (mod p ), so M k ( p - 1)( q - 1)+1 ( M p - 1 ) k ( q - 1) · M 1 k ( q - 1) · M M (mod p ). In either case, we have shown that M k ( p - 1)( q - 1)+1 M (mod p ). Second, consider the congruence modulo q . We can apply the same logic with
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Unformatted text preview: q playing the role of p . There are again two cases: Suppose M 0 (mod q ). Then M k ( p-1)( q-1)+1 M (mod q ). Suppose M 6 0 (mod q ). Then by Fermats Little Theorem, M q-1 1 (mod q ), so M k ( p-1)( q-1)+1 ( M q-1 ) k ( p-1) M 1 k ( p-1) M M (mod q ). In either case, we have shown that M k ( p-1)( q-1)+1 M (mod q ). Finally, we apply Proposition 3.64. Since p and q are distinct primes, gcd( p,q ) = 1. Thus M k ( p-1)( q-1)+1 M (mod p ) and M k ( p-1)( q-1)+1 M (mod q ) implies M k ( p-1)( q-1)+1 M (mod pq ) as claimed....
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