66 68 Exercises 68 Exercises 1 Show that d n is odd if and only if n is a

66 68 exercises 68 exercises 1 show that d n is odd

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6.8. Exercises 6.8 Exercises 1. Show that d ( n ) is odd if and only if n is a square. 2. Show that for a multiplicative function f , we have f (1) = 0 or f (1) = 1. Also, show that if f (1) = 0 then f is identically the zero function. 3. Show that d ( n ) is divisible by 3 if and only if in the prime factorization n = p α 1 1 · · · p α r r of n , at least one of the α i is 2 (mod 3). 4. Show that a multiplicative function f is completely multiplica- tive if and only if for all primes p and all α 1, we have f ( p α ) = f ( p ) α . 5. (Shoup, Exercise 2.23) Show that φ ( mn ) = gcd ( m, n ) φ ( lcm ( m, n )). 6. Show that if n is divisible by r distinct odd primes, then 2 r | φ ( n ). 7. Prove the identities X r | n μ ( r ) d ( n/r ) = 1 and X r | n μ ( r ) d ( r ) = ( - 1) ω ( n ) where ω ( n ) is the number of distinct prime divisors of n . 8. Prove that X r | n μ ( r ) σ ( n/r ) = n and X r | n μ ( r ) σ ( r ) = ( - 1) ω ( n ) p 1 · · · p r where p 1 , · · · , p r are the distinct prime divisors of n . 9. (Shoup, Exercise 2.51) Show that n is squarefree if and only if X d | n μ ( d ) 2 φ ( d ) = n. 67
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6. Arithmetic Functions 10. Show that if f is an arithmetic function with f (1) 6 = 0, there is a unique arithmetic function g so that f g = δ . 6.9 Solutions to Exercises 1. If n = p α 1 1 · · · p α r r then d ( n ) = ( α 1 + 1) · · · ( α r + 1) and for this expression to be odd, all the α i must be even. Let β i = α ) i/ 2 and m = p β 1 1 · · · p β r r . Then n = m 2 . Conversely, if n is a square, all the α i are even and so the expression for d ( n ) shows that it is odd. 2. Again, using the formula for d ( n ), we see that 3 | d ( n ) if and only if 3 | ( α i + 1) for some i . (This is because 3 is prime.) Hence, for this i , we have α i 2 (mod 3). 3. Since f is multiplicative, f (1) = f (1 · 1) = f (1) 2 and the result follows. 4. If f is the zero function, there is nothing to prove, so sup- pose that f is not identically zero. If f is completely mul- tiplicative, then this can be proved inductively. It holds for α = 1. If it holds for smaller values of α , then f ( p α ) = f ( p · p α - 1 ) = f ( p ) f ( p α - 1 ) (since f is completely multiplica- tive). This last quantity is equal to f ( p ) f ( p ) α - 1 (using the induction hypothesis) and this is equal to f ( p ) α . Conversely, suppose f ( p α ) = f ( p ) α for all primes p and all α 1. Notice that this holds for α = 0 as well, since by the previous exercise f (1) = 1. Write n = p α 1 1 · · · p α r r and m = p β 1 1 · · · p β r r where some of the α i or β j could be 0. Then mn = p α 1 + β 1 1 · · · p α r + β r r . Since we are given that f is multiplicative, f ( mn ) = f ( p α 1 + β 1 1 ) · · · f ( p α r + β r r ) . By the supposition, the right hand side is f ( p 1 ) α 1 + β 1 · · · f ( p r ) α r + β r and this is f ( p 1 ) α 1 f ( p 1 ) β 1 · · · f ( p r ) α r f ( p r ) β r = f ( p α 1 1 ) · · · f ( p α r r ) · f ( p β 1 1 ) · · · f ( p β r r ) and again using the fact that f is multiplicative, this is equal to f ( m ) f ( n ) as required. 68
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6.9. Solutions to Exercises 5. The left hand side is equal to mn Y p | mn 1 - 1 p .
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