Math 324, Fall 2011 Assignment 6 Solutions

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1 Math 324, Fall 2011 Assignment 6 Solutions Exercise 1. (a) Find all positive integers n such that φ(n) = 12. (b) Show that there is no positive integer n such that φ(n) = 14. (c) Let be a positive integer. Show that if the equation φ(n) = has exactly one solution n then 36 divides n. Solutions :(a) If 1 pα is the prime factorization of n then 12 = φ(n) = p αj 1 (p j 1) j=1 If p j > 13 then p j 1 > 12, could not divide 12. It follows that the prime divisors of n must be less than or equal to 13. As 10 = 11 1 does not divide 12, we deduce that Since 12 is not divisible by 5, 7, nor 13, we see that n = 2 a 3 b 5 c 7 d 13 e. c 1, d 1, and e 1. In fact, we must have c = 0. For otherwise, we can write The multiplicativity of φ would lead us to n = 5n, gcd(5, n ) = = φ(n) = φ(5n ) = φ(5)φ(n ) = 4φ(n ), 3 = φ(n ). On the other hand, from class, φ(n ) is even if n > 2, while φ(1) = φ(2) = 1. Therefore there are no numbers n for which φ(n ) = 3. (I) e = 1 : In this case, one has n = 13l where gcd(13, l) = 1. The multiplicativity of φ yields 12 = φ(n) = φ(13l) = φ(13)φ(l) = 12φ(l) φ(l) = 1. The remars made above show l = 1 or 2, n = 13 or n = 13 2 = 26. (II) e = 0 : We have number of subcases. (i) d = 1 : In this case, n = 7l 1

2 where gcd(7, l) = 1. The multiplicativity of φ yields 12 = φ(n) = φ(7l) = φ(7)φ(l) = 6φ(l), Writing φ(l) = 2. l = 2 a 3 b, we deduce b 1, since 3 does not divide 2. If b = 1 then 2 = φ(l) = φ(2 a 3) = φ(2 a ) 2, and 1 = φ(2 a ). in which case a = 0 or 1. If b = 0 then a 1 ( since φ(1) = 1) and 2 = φ(l) = φ(2 a ) = 2 a 1, a = 2. In summary, if 7 is the largest prime dividing n and φ(n) = 12 then n = 7 3 = 21 or n = = 42 or n = 7 4 = 28. (ii) d = 0 : In this case, n = 2 a 3 b. Since 9 does not divide 12, b 2. On the other hand, b > 1, for otherwise φ(2 a 3 b ) is a power of 2. Thus the only posibilty is b = 2, in which case the multiplicativity of φ yields 12 = φ(n) = φ(2 a 9) = φ(2 a )φ(9) = φ(2 a ) 6, 2 = φ(2 a ). It was earlier observed that the last equation implies a = 2, n = 4 9 = 36. In summary, the integers n for which φ(n) = 12 are 13, 21, 26, 28, 36, and 42. (b) Observing φ(2 a ) is a power of 2, if φ(n) = 14 then n necessarily has an odd prime divisor p. Observing p 1 divides φ(n) = 14, we deduce p 1 {1, 2, 7, 14}, p {2, 3, 8, 15}. The fact p is an odd prime forces p = 3. Since 3 does not divide 14 = φ(n), we conclude n = 2 a 3 for some integer a. The multipcativity of φ thus yields 14 = φ(2 a 3) = φ(2 a )φ(3) = 2φ(2 a ), 2

3 7 = φ(2 a ), a contradiction of the fact φ(2 a ) is a power of 2. Thus, there does not exist a positive integer n such that φ(n) = 14. (c) Note that it is sufficient to show that if φ(n) = has a unique solution then n is divisible by 4 and 9. For if the latter conditions hold, the fact gcd(4, 9) = 1 implies that that n is divisible by 4 9 = 36. If n is odd then the multiplicativity of φ yields φ(2n) = φ(2)φ(n) = 1 φ(n) = φ(n). The hypothesis that φ(n) = has a unique solution ensures that n must be even. Write n = 2n. If n were odd, the preceding argument shows that φ(n ) = φ(n), contradicting the uniqueness of the solution. Thus n is even, allowing us to conclude that 4 divides n. We can thus write n = 2 a m where m is odd and a 2. If gcd(3, m) = 1, consider Using the multiplicativity of φ, since a 1 1, l = 2 a 1 3 m. φ(l) = φ(2 a 1 )φ(3)φ(m) = 2 a 2 2φ(m) = 2 a 1 φ(m) = φ(2 a )φ(m) = φ(2 a m) = φ(n), a contradiction of the uniquness of the solution to φ(n) =. Thus, we have n = 2 a 3 m for some odd m. If m were not divisible by 3, the same argument would allow us to conclude φ(2 a+1 m ) = φ(2 a 3 m ) = φ(n) another contradiction of the uniqueness of the solution. We conclude that m is divisible by 9; since m divides n, it follows that n is divisible by 9. Exercise 2. (a) For which positive integers n is φ(n) divisible by 4? (b) For which positive integers n does φ(n) n? Solution : (a) If n is even, say n = 2 α0 n with n odd, then φ(n) = φ(2 α0 )φ(n ) = 2 α0 1 φ(n ). If α 0 3 then α 0 1 2, φ(n) is divisible by 4. If α 0 = 2 then φ(n) = 2φ(n ). If n > 1 then φ(n ) is nown to be even, in which case φ(n) will be divisible by 4. If n = 1 then n = 4 and φ(4) = 2 is not divisible by 4. The preceding two cases show that φ(n) is divisible by 4 if n = 4l with l > 1. If α 0 = 1 then φ(n) = φ(2n ) = φ(n ). In this case φ(n) is divisible 4 if and only if φ(n ) is divisible by 4. In the latter case, since φ(1) = 1, it would follow that n > 1 has a prime factorization n = p α1 1 pα 3

4 where 1 and the p j are distinct odd primes. The multiplicativity of φ thus yields φ(n ) = p αi 1 i (p i 1). Observing p i 1 is even for each i, we see that φ(n ) is divisible by 4 if > 1. If = 1 then φ(n ) = φ(p α1 1 ) = pαi 1 1 (p 1 1) is divisible by 4 if and only if p 1 1 mod 4. In summary, φ(n) is divisible by 4 precisely when n has one of the following forms. (i) n = 4l, l N, l > 1. (ii) n = 2n, n is odd with at least two distinct prime divisors or is a power of a prime p 1 mod 4. (iii) n is odd with at least two distinct prime divisors or is a power of a prime p 1 mod 4. (b) Let 1 pα. If φ(n) divides n then l = n φ(n) = p i p i 1 must be an integer. Formally, at most one factor of 2 can occur in the numerator of l, the denominator can have at most one factor of the form p i 1 with p i as an odd prime. It follows that n must have the form or 2 α, α 0, (i) 2 α p β, α 0, β 1, p odd prime. (ii) It is readily verified that φ(n) divides n for all n of the form (i). If n is of the form (ii) then the denominator of l must be even, which shows that the numerator must also be even, α > 0. In this case, l = n φ(n) = 2p p 1 Z shows that p 1 divides 2p. Since p and p 1 are relatively prime, p 1 must divide 2; in particular, p 1 2, p 3. On the other hand, p 3, being an odd prime, so p = 3. Thus, in case (ii), n must be of the form 2 α 3 β with α, β > 1. It is readily checed that all n of this form satisfy φ(n) n. In summary, if φ(n) divides n then either n = 2 α for some non-negative α or n = 2 α 3 β with α, β > 1. Exercise 3. (a) Which positive integers have exactly 4 positive divisors? Be sure to justify your answer. (b) Let be a positive integer. Show that the equation σ(n) = has at most a finite number of solutions. Solution :(a) We are required to find all integers n such that τ(n) = 4. (1) Since τ(1) = 1, any solution n of (1) must be greater than 1. it has a prime factorization of the form where 1 and each α > 0. In this case, 4 = τ(n) = 1 pα τ(p αi i ) = 4 (1 + α i ).

5 Observing 1 + α i 2, it follows immediately that 2. If = 1 then Therefore, n = p 3 for some prime p. If = 2 then 4 = 1 + α 1, α 1 = 3. 4 = (1 + α 1 )(1 + α 2 ) with both 1 + α 1 and 1 + α 2 both > 1. The prime factorization of 4 allows us to conclude 1 + α 1 = 1 + α 2 = 2, α 1 = α 2 = 1. Therefore, n = pq for distinct primes p and q. In summary, if τ(n) = 4 then either n = p 3 for a prime p or there exists distinct primes p and q such that n = pq. On the other hand, it is readily checed that if n has either of the two forms described then it has exactly 4 positive divisors. (b) We recall that if n has prime factorization then Observing if σ(n) = then It follows that the equation has at most solutions, as required. σ(n) = 1 pα l l l p αi+1 i 1 p i 1. p αi+1 i 1 = p αi i + + p i + 1 p αi i, p i 1 = σ(n) l σ(n) = p αi i = n. Exercise 4. Let be a positive integer. Given a positive integer n, let σ (n) be the sum of the -th powers of the (positive) divisors of n, that is σ (n) = d. (a) Find a closed formula for σ (p α ) where p is prime and α is a positive integer. (b) Prove that σ (n) is multiplicative. (c) Find a closed formula for σ (n). Solution : (a) If p is prime then the positive divisors of p α are precisely p i, 0 i α. Therefore, σ (p α ) = α (p α ) = i=0 α i=0 5 (p ) α = (p ) α+1 1 p 1.

6 (b) If n and m are positive integers then the exponentiation rules yield (mn) = m n. This shows that the function n n is a (completely) multiplicative function f. The multiplicativity of σ thus follows from the fact that it is the summatory function of the multiplicative function f. (c) In light of (a) and (b), if the integer n has the prime factorization 1 pα l l then σ (n) = σ (p α1 1 pα l l ) = σ (p α1 1 ) σ (p α l l ) l ( (p ) αi+1 ) 1 = p. 1 Exercise 5. (a) Let n be a positive integer. Show that τ(2 n 1) τ(n). (b) An integer n is said to be -perfect if σ(n) = n. Find all 3-perfect numbers of the form n = 2 3 p where p is an odd prime. Solution : (a) We recall that if then 2 d 1 divides 2 n 1. Furthermore, if d 1 and d 2 are divisors of n with 2 d1 1 = 2 d2 1 then It follows that the map d 1 = d 2. d 2 d 1, provides an injective map from the set of divisors of n to the set of divisors of 2 n 1. We deduce that, τ(2 n 1) = { divisors of 2 n 1 } { divisors of n } = τ(n). (b) Let n = 2 3 p is a 3-perfect number. We first observe that p must be different from 3. Indeed, if p = 3 then = 3n = σ(2 3 2 ) = σ(2 ) Multiplying both sides by 2, = 26σ(2 ) = 2 13 (2 +1 1), a contradiction of the Fundamental Theorem of Arithmetic, as the prime 13 appears on the right but not on the left. Since p is an odd prime different from 3, we have p = 3n = σ(2 3 p) = σ(2 ) 4 (p + 1). (1) Observing that the right hand side is divisible by 8, the Fundamental Theorem of Arithmetic allows us to conclude that 3. In this case, σ(2 ) = , so (1) yields p = (2 +1 1)(p + 1). (2) Observing gcd(2 2, ) = gcd(p, p + 1) = 1, the identity (2) allows us to conclude 2 2 p + 1 and p

7 Writing the equation (2) allows us to conclude There are three cases to consider. (i) s = 1, t = 9 : In this case, It follows that = 5 and thus p + 1 = 2 2 s and = pt, 3 2 = st. p + 1 = 2 2 and 9p = , = 9p = 9(2 2 1). 8 = = 2 2 (9 2 3 ) = 2 2, p = = = 7. In this case, we see n = = 672. (ii) s = t = 3 : In this case, p + 1 = and 3p = From the first of these equations p = Substituting into the second equation, We deduce = 3, In this case, we see n = = 120. (iii) s = 9, t = 1 : In this case, Substituting for p in the first equation, an obvious contradiction = , 2 = = 2 2 (9 8) = 2 2. p = = = 5. p + 1 = and = p = 9 2 2, In summary, if n is a 3-perfect number of the form 2 3 p where p is an odd prime then n = 120 or 672. Exercise 6. (a) If n is a positive integer, show µ(n)µ(n + 1)µ(n + 2)µ(n + 3) = 0. (b) Find simple formulae for the following expressions. (i) dµ(d). (ii) τ(d)µ(d). Solution : (a) One of the four consecutive numbers n, n + 1, n + 2, and n + 3 7

8 is divisible by 4; in particular at least one of them is not square-free. It follows that at least one of µ(n), µ(n + 1), µ(n + 2), and µ(n + 3) is zero, µ(n)µ(n + 1)µ(n + 2)µ(n + 3) = 0. (b) Note that if f is a non-zero multiplicative function then so is the product µf. General theory then asserts that the function G(n) = µ(d)f(d) is multiplicative. Observing G(p α ) = α µ(p i )f(p i ) = 1 f(p) if p is prime, multiplicativity of G allows us to conclude that µ(d)f(d) = G(n) = G(p α1 1 ) G(pα ) = (1 f(p 1)) (1 f(p )) where 1 pα is the prime factorization of n (i) Applying the above to f(n) = n, if is the prime factorization of n then 1 pα dµ(d) = (1 p 1 ) (1 p ). (ii) Recalling that τ(p) = 2 if p is prime, if is the prime factorization of n then 1 pα µ(d)τ(d) = (1 τ(p 1 )) (1 τ(p )) = ( 1). Exercise 7. The Margoldt function Λ is defined for all positive integers n by { log p, if n = p Λ(n) = where p is prime and is a positive integer; 0, otherwise. (a) Show that if n is a positive integer then Λ(d) = log n. (b) Prove Λ(n) = µ(d)log d. Solution : (a) Let 1 pα 8

9 be the prime factorization of n. If then Λ(d) = 0 unless d = p βj j Λ(d) = = = α j j=1 β j=1 α j j=1 β j=1 Λ(p βj j ) log p j α j log p j j=1 = log j=1 p αj j = log n. for some β j, 1 β j α j. We deduce (b) By (a), log n is the summatory function of Λ(n). Therefore, Möbius Inversion yields Λ(n) = µ(d)log(n/d) = µ(d)(log n log d) = µ(d) log n µ(d)log d = logµ(d)log d, since µ(d) = 0 if n > 1 and log n = 0 if n = 1. Bonus Question. Suppose n is an odd perfect number. (a) Show n = p α m 2 where p is an odd prime, p α 1 mod 4, and m is an odd integer. (b) Deduce n 1 mod 4. Solution : (a) Since σ(1) = 1, n > 1. Therefore, if 1 pα is the prime factorization of n then 1 and each prime p i is odd. Furthermore, since n is perfect, 2n = σ(n) = σ (p αi i ). Since the left side is even, one of the factors appearing on the right must be even. On the other hand, the left-side is not divisible by 4, so at most one of the factors on the right is even, in which case it is precisely divisible by 4. By rearranging the factors, we can assume that σ(p α1 1 ) is even and the remaining factors σ(p αj j ), j 2, are all odd. Let q be an odd prime. Recalling α σ(q α ) = q i, 9 i=0

10 σ(q α ) is a sum of α + 1 odd numbers. In particular, σ(q α ) is odd if and only if α + 1 is odd, that is α is even. Applying to the case at hand, we deduce that α j is even for all j 2, where 1 pα2 2 pα = p α1 1 (pα2/2 2 p α /2 ) 2 = p α1 m = p α2/2 1 p α /2 1 m2, is an odd integer. On the other hand, the fact σ(p α1 1 ) is even forces α 1 to be odd. Writing p 1 = p and α 1 = α, there exist β 0 such that α = 2β + 1. In this case, we have σ(p α ) = α p i = i=0 = = 2β+1 i=0 p i β (p 2l + p 2l+1 ) l=0 β β p 2l (1 + p) = (p + 1) p 2l. l=0 Since σ(p α ) is divisible by 2 but not divisible by 4, the even number p + 1 must be congruent to 2 modulo 4, p 2 1 = 1 mod 4. Furthemore, β p 2l = l=0 β (p 2 ) l must be odd, β must be even (p 2 is odd), which shows that l=0 α = 2β mod 4. l=0 (b) Since m = ±1 mod 4, Thus, m 2 1 mod 4. n = p α m 2 1 α 1 = 1 mod 4. 10