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intsu11h3s

# intsu11h3s - Introductory Number Theory Homework 3 Due...

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Introductory Number Theory. Homework 3 Due: Tuesday, June 7, 2011, 4:45PM Note: My words to work on your own seem to continue to be ignored. At least, don’t make it so obvious! At least, UNDERSTAND WHAT YOU ARE COPYING!!! I know you are copying without really understanding when I see meaningless statements, or incomplete statements, that make sense when one sees the complete form in someone else’s paper. I may not fail you for it, but rest assured that you I am aware of it. My apologies to the few of you who work on their own and don’t allow others to copy from them. 1. Let m, n be integers, not both 0. Consider the set A = { mx + ny : x, y Z } . Show that this set coincides with the set of all multiples of the greatest common divisor of m, n . That is, if d = gcd ( m, n ), show A = { kd : k Z } . Solution. By Bezout, there exist integers a, b such that d = am + bn . For every k Z we have kd = kam + kbn A , proving that the set { kd : k Z } ⊂ A . But one also has to prove the converse inclusion! Assume now x, y Z . Since d | m and d | n , there exists integers j, ‘ such that m = jd , n = ‘d , hence mx + ny = jdx + ‘dy = ( jx + ‘y ) d = kd with k = jx + ‘y . The converse inclusion follows. 2. If a, m N and gcd ( a, m ) = 1, then the order of an integer a modulo m is the smallest positive integer k such that a k 1 (mod m ). Prove: If gcd ( a, m ) = 1 and if k is the order of a modulo m , then k | φ ( m ). Hint: Divide φ ( m ) by k ; what can you say about the remainder? Solution. Since a φ ( m ) 1 (mod m ) by Euler’s theorem, we must have 1 k φ ( m ). By the division algorithm, we can write φ ( m ) = qk + r for unique integers q, r , 0 r < k . Now, once again by Euler’s theorem, and because a k 1 (mod m ), 1 a φ ( m ) = a qk a r = ( a k ) q a r 1 q a r = a r (mod m ); i.e., a r 1 (mod m ). If r > 0 then 1 r < k , contradicting that k

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intsu11h3s - Introductory Number Theory Homework 3 Due...

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