Solutions to Problem Set 6

Transcription

1 Solutios to Problem Set 6 1. Sice both sies are positive, it s eough to show their squares are the same. Now =, so. For a prime power p e, we have = = = = (). σ k (p e ) = 1 + p k + + p ek, which is o if a oly if p = or e is eve. So if = p e1 1 p r er, the σ k () = r i=1 σ k (p ei i ) is o if a oly if all the o primes iviig ivie it to a eve power, i.e., is a square or twice a square. 3. Suppose g = gc(a, b) > 1, a let S() = { N : } be the set of all positive ivisors of. We have a fuctio φ : S(a) S(b) S(ab) give by φ(, e) = e. The map φ is surjective, but ot ijective, because φ(g, 1) = φ(1, g). So σ k (ab) = x k x S(ab) < φ (t) k t S( a) S (b) = k e k a = σ k (a)σ k (b). Similarly, (ab) < (a)(b) just says S(ab) < S(a) S(b), which is obvious from the fact that φ is surjective but ot ijective, or from otig that () = σ 0 (). 4. (a) If m 1 is prime the σ( m 1) = m. Sice σ( m 1 ) = m 1 = m 1, if m 1 m = ( 1), the we have So is perfect. σ() = σ( m 1 )σ( m 1) = ( m 1) m =. e b 1

2 (b) Suppose is perfect a eve. Write = r s where r 1 a s is o. It s easy to rule out s = 1, so we ll assume s > 1. The σ() = ( r+1 1)σ(s) must equal = r+1 s. So r+1 1 ivies r+1 s, a sice gc( r+1 1, r+1 ) = 1, we have r+1 1 s. r+1 Now if s > 1, the s has at least the three ivisors 1, s, a s/( r+1 1), which are istict because r 1. Thus ( ) ( 1 r+1 ) σ(s) 1 + s 1 + > s, r+1 1 r+1 1 so ( r+1 1)σ(s) > r+1 s, cotraictio. Therefore we must have s = r+1 1, a σ(s) = r+1. But s = r+1 1 has at least the two ivisors 1 a s, which sum to r+1 alreay. So the oly possibility is that these are the oly two ivisors of s, i.e., s is prime. 5. The fuctio Ω() is the umber of primes iviig, with multiplicity, so Ω(m) = Ω(m) + Ω() for ay m,. Hece λ() is totally multiplicative, a λ() is multiplicative (but ot totally multiplicative). For prime powers p e, λ() = 1 + ( 1) ( 1) + + ( 1) e }{{} p e { 0 if e is o, = 1 if e is eve. (e+1) terms So for = p e1 1 per r, λ() will be 0 if ay of the e i are o, a 1 if all of the e i are eve (which occurs precisely whe is a perfect square). 6. Both sies are multiplicative, so it s eough to show the equality for prime powers p e. Sice ( 1) =, as ca be easily prove usig iuctio, 4 e () 3 = (p i ) 3 p e i=0 e = (i + 1) 3 i=0 e (e + 1) = 4 ( ) e+1 = i = i=1 (). p e 7. (a) This is just a multiplicative versio of the Möbius iversio formula. To prove it we use the fact that { 1 if = 1 µ() = ( ) 0 if > 1.

3 So ˆF () µ(/) = f(e) e = f(e) µ(/ef e,f ef = f(e) e ) µ(/) f m µ(m/f), where m = /e. By ( ), this expressio is simply f(). (b) Writig this equatio as a ( ) µ(/) =, (a,)=1! φ() we see that by part (a) it s eough to show that where! f() = f() = a=1 (a,)=1 is the left-ha sie. Now! is the prouct over x {1,..., } of x. For ay such x, let g = gc(x, ). The the fractio x is reuce to x/g. Coversely, for ay ivisor /g of a ay x { 1,..., } coprime x x / to, we have =, where x / {1,..., } has gc exactly g = / with. Therefore, which is what we set out to prove.! = ( a ) x x =1 (x, )=1 = f ( ), 8. (a) We have Z(f, s)z(g, s) = f(m) g() m s s m 1 = m, 1 f(m)g (), (m) s 1 which, whe recast as a sum over m = k, becomes m k f(m)g(k/m) = Z(f k s g, s). k 1 3

4 (b) First suppose f has a iverse g = f 1. The (f g)(1) = f(1)g(1) = 1, so f(1) = 0. Coversely, whe f(1) = 0, we will costruct a fuctio g such that f g = 1. First set g(1) = f(1) 1, which is force as above. Now we will efie g() for all, by iuctio o. The base case = 1 is oe. Suppose g() has bee efie for all less tha k. The we have 0 = 1(k) = (f g)(k) = f()g(k/) k = f(1)g(k) + k,>1 f()g(k/). All the g(k/) for > 1 have bee efie, by the iuctive hypothesis, so we ca solve this equatio uiquely for g(k). This completes the iuctio. By costructio, f g = 1, a by commutativity of we also have g f = (a) This is a staar proof by iuctio. (b) Splittig the itegers from 1 through by their gc with, we get j = j j=1 (j,)= = S(/) (c) Note that Therefore, a Möbius iversio gives = S(). S() ( + 1)( + 1) = j =. 6 j=1 ( ) S() 1 1 = + 3 +, 6 S() = ( ) 1 µ() () Sice the LHS a RHS are both multiplicative, it s eough to show equality for prime powers p e. I this case µ() = 1 p p e a e ( s ω(pe) φ(p e (p ) p 1) ) = ( p e 1)p e 1 (p 1) = 1 p e p. 4

5 (e) As show i part (c), S() 1 1 = 1 µ() + µ() + µ(). 3 6 Now, for ay, µ() = φ(). Also, µ() = 1(). So whe > 1, we get φ() ( 1) ω() φ()s() S() = (a) This follows from the Multiplicative versio of Möbius iversio, usig f() = Φ (x) a x 1 = Φ (x) = F (). (b) F () is the sum of the roots of the polyomial x 1, so it s equal to the egative of the coefficiet of x 1 i x 1. Therefore { 0 if > 1 F () = 1 if = 1. (c) Sice Φ = x (x) 1, the sum of the roots of the polyomial x 1 is F () = f(), where f() = e πia/ a=1 (a,)=1 is the sum of the roots of Φ (x). Therefore, f U = 1, so by Möbius iversio f = µ. 5

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