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so5 - Math 55 Homework 5 Solutions Friday 1 For a positive...

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Math 55 - Homework 5 Solutions Friday, July 15, 2011 1. For a positive integer n , define Z * n to be the integers from 1 to n - 1 which are relatively prime to n . That is Z * n = { k Z : 0 k < n and gcd( k, n ) = 1 } . Defined φ ( n ) to be the size of Z * n . For example, Z * 12 = { 1 , 5 , 7 , 11 } and φ (12) = 4. (a) If p is a prime, and i is a positive integer, prove that φ ( p i ) = p i - 1 ( p - 1). Since the prime factorization of p i is unique, the only prime which divides p i is p . So the only integers from 0 to p i - 1 not relatively prime with p i are multiples of p : 0 , p, 2 p, 3 p, . . . ( p i - 1 - ) p . There are p i - 1 of those. So φ ( p i ) = Z * p i = p i - p i - 1 = p i - 1 ( p - 1) . (b) If m and are relatively prime, prove that φ ( m‘ ) = φ ( m ) · φ ( ). k Z * m‘ ⇐⇒ k coprime to m‘ ⇐⇒ k coprime to m and ⇐⇒ k MOD m coprime to m and k MOD coprime to ⇐⇒ ( k MOD m, k MOD ) Z * m × Z * By the Chinese Remainder Theorem, the map k ( k MOD m, k MOD ) is a bijection, so φ ( m‘ ) = | Z * m‘ | = | Z * m | · | Z * | = φ ( m ) φ ( ) (c) If n = p e 1 1 p e 2 2 . . . p e j j , where the p ’s are distinct primes and the e ’s are positive, prove that φ ( n ) = n 1 - 1 p 1 1 - 1 p 2 . . . 1 - 1 p j . The various p e i i are relatively prime, so by applying (b) and then (a), we get: φ ( n ) = φ ( p e 1 1 p e 2 2 . . . p e j j ) = φ ( p e 1 1 ) φ ( p e 2 2 ) . . . φ ( p e j j ) = ( p e 1 1 - p e 1 - 1 1 ) ( p e 2 2 - p e 2 - 1 2 ) . . . p e j j - p e j - 1 j = p e 1 1 1 - 1 p 1 p e 2 2 1 - 1 p 2 p e 2 1 1 - 1 p 2 = n 1 - 1 p 1 1 -
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