a2_2008sols

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

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 09/12/2009 for the course MATH 2068 taught by Professor Howlett during the One '08 term at University of Sydney.

### Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online