230hw4sol - Math 230 E Homework 4 Fall 2011 Drew Armstrong...

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Math 230 E Fall 2011 Homework 4 Drew Armstrong Problem 1. (a) Consider a,b,d Z with gcd( a,d ) = 1. Prove: If d | ab then d | b . (b) Consider a,b,m Z with gcd( a,b ) = 1. Prove: If a | m and b | m then ab | m . (c) Consider a,b,c,n Z with gcd( c,n ) = 1. Prove: If ca cb mod n then a b mod n . Proof. First we prove (a). Consider a,b,d Z with a,d coprime, and suppose that d | ab . We will show that d | b . Since d | ab , there exists k Z such that dk = ab . Then since a,d are coprime, B´ ezout’s Lemma implies that there exist x,y Z such that ax + dy = 1. Finally, we multiply both sides of this equality by b to get b ( ax + dy ) = b · 1 , bax + bdy = b, dkx + dby = b, d ( kx + by ) = b. We conclude that d | b . To show (b), consider a,b,m Z with a,b coprime. Suppose further that a | m and b | m . We will show that ab | m . There are two possible solutions. Solution 1. (From Scratch) There exist k,‘ Z such that ak = m and b‘ = m . Then since a,b are coprime, B´ ezout’s Lemma gives us x,y Z such that ax + by = 1. Multiply both sides by m to get m ( ax + by ) = m · 1 , max + mby = m, b‘ax + akby = m, ab ( ‘x + ky ) = m. Hence ab | m . Solution 2. (Quote Part (a)) There exist k,‘ Z such that ak = m = b‘ , hence ak = b‘ . Since b | ak with a,b coprime, part (a) implies that b | k . That is, there exists x Z such that k = bx . Substitute this into ak = m to get abx = m . We conclude that ab | m . Finally, we will prove (c). Consider a,b,c,n Z with c,n coprime and ca cb mod n . We will show that
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This note was uploaded on 01/08/2012 for the course MATH 461 taught by Professor Armstrong during the Fall '11 term at University of Miami.

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230hw4sol - Math 230 E Homework 4 Fall 2011 Drew Armstrong...

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