a2_2008sols - 2 The University of Sydney MATH2068 Number...

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The University of Sydney MATH2068 Number Theory and Cryptography (http://www.maths.usyd.edu.au/u/UG/IM/MATH2068/) Semester 2, 2008 Lecturer: R. Howlett Assignment 2 1. ( i ) Use the formula for φ ( n ) in terms of the factorization of n to compute φ (13), φ (21), φ (26), φ (28), φ (36) and φ (42). ( ii ) Find all positive integers n such that φ ( n ) is odd. ( iii ) Prove that if n is a positive integer and p a prime divisor of n then p φ ( n ) + 1, and classify all those cases for which equality holds. ( iv ) Find all positive integers n such that φ ( n ) = 2 or φ ( n ) = 6. Then find all positive integers n such that φ ( n ) = 12. [Hint: Let n = p k m where p is prime and m is not divisible by p , and use multiplicativity of φ .] Solution. ( i ) If n = p k 1 1 p k 2 2 ··· p k r r where the p i are pairwise distinct primes amd the k i are positive integers then (by the formulae on page 39 of the book of notes or Lecture 14) φ ( n ) = p k 1 - 1 1 ( p 1 - 1) p k 2 - 1 2 ( p 2 - 1) ··· p k r - 1 r ( p r - 1). Applying this with r = 1 and p k 1 1 = 13 1 gives φ (13) = 13 0 (13 - 1) = 12. To deal with 21 we need r = 2 and p k 1 1 p k 2 2 = 3 1 × 7 1 , and we get φ (21) = 3 0 (3 - 1)7 0 (7 - 1) = 2 × 6 = 12. The others are similar: φ (26) = φ (2 1 × 13 1 ) = 2 0 (2 - 1)13 0 (13 - 1) = 12 , φ (28) = φ (2 2 × 7) = 2 1 (2 - 1)7 0 (7 - 1) = 12 , φ (36) = φ (2 2 × 3 2 ) = 2 1 (2 - 1)3 1 (3 - 1) = 12 , φ (42) = φ (2 × 3 × 7) = 2 0 (2 - 1)3 0 (3 - 1)7 0 (7 - 1) = 12 . ( ii ) If n = 1 then by definition φ ( n ) = 1. (See the discussion of this point in Lecture 18.) So φ ( n ) is odd when n = 1. If n > 1 then by the Fundamental Theorem of Arithmetic we can express n in the form n = p k 1 1 p k 2 2 ··· p k r r , where the p i are distinct primes, the k i are positive integers, and r 1. As in Part ( i ) this gives φ ( n ) = p k 1 - 1 1 ( p 1 - 1) p k 2 - 1 2 ( p 2 - 1) ··· p k r - 1 r ( p r - 1) , ( * ) and if φ ( n ) is odd then all the factors in this expression must be odd. In partic- ular, p i - 1 must be odd for each i . But there is only one prime p such that p - 1 is odd, namely the prime p = 2. So 2 is the only prime divisor of n . That is, in the factorization n = p k 1 1 p k
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a2_2008sols - 2 The University of Sydney MATH2068 Number...

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