Math 4400/6400 Homework #7 solutions

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1 MATH 4400 problems. Math 4400/6400 Homewor #7 solutios 1. Let p be a prime umber. Show that the order of 1 + p modulo p 2 is exactly p. Hit: Expad (1 + p) p by the biomial theorem, ad recall from MATH 4000 that ( p ) is a multiple of p for all with 0 < < p. Solutio. First, we show that (1 + p) p 1 (mod p 2 ). We have (1 + p) p = p ( ) p 1 p ( ) p 1 p p = 1 + p + p p. =0 For all 1 < < p, we have that p ( ) p ad that p p, so that p 2 ( p ) p. Sice p 2, we also have that p 2 p p. Thus, looig modulo p 2, we see that (1+p) p (mod p 2 ). Hece, the order of 1 + p is a divisor of p, so is either 1 or p. But it is ot 1, sice 1 < 1 + p < p Let p be a prime umber. Let a be a primitive root mod p. (a) Show that the order of a modulo p 2 is either p 1 or p(p 1). Proof. Let j be the order of a modulo p 2. The j φ(p 2 ) = p(p 1). Moreover, it is ot hard to show that p 1 j. To see this, otice that sice a j 1 (mod p 2 ), we have that p p 2 a j 1, ad so that a j 1 (mod p). Sice a has order p 1 modulo p, it follows that p 1 j. So j is a divisor of p(p 1) that is also a multiple of p 1. Write j = (p 1)q, where q Z. Sice j divides p(p 1), we ca write p(p 1) = jq = (p 1)qq, with q Z. Cacelig p 1 from both sides gives p = qq, ad so either q = 1 or q = p (sice p is prime). If q = 1, the j = p 1, while if q = p, the j = p(p 1). (b) Show that if the order of a modulo p 2 is p 1, the the order of a(1+p) is p(p 1). Hit: Use the previous exercise i cojuctio with oe of the lemmas we used to prove U(Z p ) is cyclic. Proof. Suppose that the order of a modulo p 2 is p 1. We ow from the previous exercise that the order of 1 + p is p. Sice p ad p 1 are relatively prime, oe of the group-theoretic lemmas from class gives that a(1 + p) has order p(p 1). (c) Deduce that U(Z p 2) is cyclic for every prime p. Proof. We have to show that there is a iteger whose order modulo p 2 is φ(p 2 ) = p(p 1). By (a), the order of a modulo p 2 is either p 1 or p(p 1). If it is p(p 1), the we are doe; a the iteger we were looig for. If ot, by part (b), we ca tae the iteger a(1 + p). 3. Recall that the th Fermat umber F is defied by F = For example, F 3 = = 257. Show that if p is a prime dividig F, the p 1 (mod 2 +1 ). =1

2 Proof. If p F, the (mod p). Squarig, ( 1) 2 1 (mod p). Thus, the order of 2 modulo p divides 2 +1 but does ot divide 2. But the oly divisor of 2 +1 that is ot a divisor of 2 is 2 +1 itself. Hece, 2 has order 2 +1 modulo p. Sice the order of 2 modulo p divides p 1, it follows that p 1 (mod 2 +1 ). 4. Fid ad prove simple formulas for each of the fuctios µ(d)τ(/d), µ(d)τ(d), µ(d) 2 φ(d). Express your aswers for the 2d ad 3rd of these i terms of the prime factorizatio of. Proof. All three fuctios are multiplicative. Ideed, we proved i class that f(d)h(/d) is a multiplicative fuctio of wheever f ad h are. [I the first example, we ca apply this fact with f = µ ad h = τ. I the secod, we tae f() = µ()τ() ad h idetically 1, ad i the third, we tae f() = µ() 2 φ() ad h idetically 1.] A multiplicative fuctio is determied by its behavior at prime powers. So let us evaluate all three fuctios at prime powers p e. For the first, we fid d p e µ(d)τ(p e /d) = µ(1)τ(p e ) + µ(p)τ(p e 1 ) + µ(p 2 )τ(p e 2 ) + + µ(p e )τ(1) For the secod, we fid that = 1 (e + 1) 1 e = (e + 1) e = 1. d p e µ(d)τ(d) = µ(1)τ(1) + µ(p)τ(p) + µ(p 2 )τ(p 2 ) + + µ(p e )τ(p e ) For the third, we fid = ( 1) = 1. d p e µ(d) 2 φ(d) = µ(1) 2 φ(1) + µ(p) 2 φ(p) + µ(p 2 ) 2 φ(p 2 ) + + µ(p e ) 2 φ(p e ) 2 = ( 1) 2 (p 1) = p. So if we write = p e 1 1 p e 2 2 p e, we get from multiplicativity that µ(d)τ(/d) = 1 = 1, p that µ(d)τ(d) = ( 1) = ( 1), p 2

3 ad that µ(d) 2 φ(d) = p. p If = 1, we caot factor ito prime powers, but all three formulas still remai valid, provided that the empty product is iterpreted as Let λ() be the arithmetic fuctio defied as follows. If = p e 1 1 p e 2 2 p e, the λ() = ( 1) e 1+e 2 + +e. For example, sice 12 = 2 2 3, we have λ(12) = ( 1) 2+1 = 1. Show that if is ot a perfect square, the λ(d) = 0. What happes if is a square? Proof. Let f() = λ(d). I claim that f is multiplicative. To prove the claim, it suffices (i view of the theorems proved i class) to show that λ is multiplicative. Let ad m be ay two positive itegers. Let p 1, p 2,..., p be a list of the primes dividig either or m. The we ca write = p e 1 1 p e ad m = p f 1 1 p f, for some oegative itegers e 1,..., e ad f 1,..., f. The ad so m = p e 1+f 1 1 p e +f, λ(m) = ( 1) (e 1+f 1 )+(e 2 +f 2 )+ +(e +f ) = ( ( 1) e 1+ +e ) ( ( 1) f 1 + +f ) = λ()λ(m). This proves that λ is multiplicative. (I fact, we ve proved eve more, sice we did t eed to assume that ad m were coprime to mae this argumet wor.) At this poit we ow that f is multiplicative. Let us determie its values o prime powers. We have f(p e ) = d p e λ(d) = λ(1) + λ(p) + λ(p 2 ) + λ(p 3 ) + + λ(p e ) = ( 1) e. We see that f(p e ) = 1 if e is eve, ad f(p e ) = 0 otherwise. Suppose that is ot a perfect square. Factorig = p e 1 1 p e, oe of the e i must be odd, say (without loss of geerality) that it is e 1. The f() = f(p e 1 1 ) f(p e ) = 0 f(p 2 ) e2 f(p e ) = 0. So f() is 0 uless is a perfect square. O the other had, if is a perfect square, the each e i is eve, ad so each f(p e i i ) = 1 always. Thus, f() = = 1. 3

4 6. What is wrog with the followig proof that all perfect umbers are eve? If is a perfect umber, the σ() = 2. I other words, 2 = d. So if f ad g are the arithmetic fuctios defied by f() = 2 ad g() =, the f() = g(d). So by Möbius iversio, = g() = f(d)µ(/d) = 2dµ(/d) = 2 dµ(/d). The fial parethesized expressio is a iteger, ad so is eve. Proof. I Möbius iversio, you assume that f() = g(d) for every value of. I the cotext of this problem, that relatio holds oly for perfect umbers. So oe caot ivert. MATH 6400 problems. Do ay two of the followig. G1. Let N be a positive iteger. (a) Show that µ() N d m µ(d), ad the reverse the order of sum- Hit: Start by evaluatig m N matio. µ(d) = d N m N d m N = 1. Proof. I class, we showed that d m µ(d) = 1 if m = 1, while d m µ(d) = 0 otherwise. Thus, m N d m µ(d) = = 1. O the other had, reversig the order of summatio gives 1 = N µ(d). d d N µ(d) m N d m Comparig these two evaluatios gives the result i (a). (b) Deduce from (a) that N µ() 1. Proof. We write N/ = N/ {N/} ad use the result of (a) to fid that N N µ() = 1 + N µ() {N/}. Whe = 1, we have {N/} = 0, ad so the sum o the right is uchaged if we add the coditio 2. Sice both µ ad {N/} are bouded i 1 by absolute value, the sum o the right is the bouded i absolute value by N 1. It follows by the triagle iequality that N N N. Dividig by N gives the desired result. 4 µ()

5 G2. Let p be a prime, ad let be a atural umber dividig p 1. Show that the umber of icogruet ozero th powers modulo p is exactly p 1. We did this i class i the special case = 2. Hit: Use the fact that U(Z p ) is cyclic, ad so isomorphic to Z p 1 uder additio. Proof. Raisig to the th power i U(Z p ) correspods, uder the isomorphism, to multiplyig by i Z p 1. So this problem is equivalet to showig that there are exactly p 1 distict multiples of modulo p 1. I fact, these distict multiples are exactly 1, 2,..., p 1 p 1, of which there are. This follows from the fact that a b (mod p 1) if ad oly if a b (mod p 1 ), ad the fact that 1, 2,..., p 1 represet all of the elemets modulo p 1 exactly oce. G3. Let p be a prime umber. Show that for ay iteger a, oe ca fid itegers x ad y with a x 2 +y 2 (mod p). I other words, every elemet of Z p is a sum of two squares. Proof. If p = 2, this is obvious, sice every elemet of Z p is itself a square. So suppose i what follows that p is odd. Let S be the set of elemets of Z p represeted by a iteger of the form x 2, ad let T be the set of elemets of Z p represeted by a iteger of the form a y 2. Sice there are p+1 squares i Z 2 p (icludig 0), both S ad T have size p+1. Sice the total 2 umber of elemets i Z p is p, while the sizes of S ad T add to p + 1, the sets S ad T caot be disjoit. Hece, there are itegers x ad y with x 2 a y 2 (mod p). But the a x 2 + y 2 (mod p), as desired. 5

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