8.78 Solutios to Problem Set 8. We ow that ( ) ( + x) x. Now we plug i x, ω, ω ad add the three equatios. If 3 the we ll get a cotributio of + ω + ω + ω + ω 0, whereas if 3 we ll get a cotributio of + + 3. So ( ) ( + ) + ( + ω) + ( + ω ) 3 3. We have + ( ω ) + ( ω) 3 ( + )/3 if 0 (mod 6) ( )/3 if 3 (mod 6). ( )/3 if, 4 (mod 6) ( + )/3 if, 5 (mod 6) d d (A(x)) x a dx dx! 0 x a! x a +,! 0 which is the expoetial geeratig fuctio of {a, a,... }. 3. Sice c is! times the coefficiet of x i A(x)B(x), c! a 0 b! ( )! ( ) a b 0 4. By part (a), d E(x) is the expoetial geeratig fuctio for the sequece {r, r, r 3 dx,... }. It follows that E (x) re(x). Sice E(0), solvig the differetial equatio, we get. r E( ) x x e rx.! 5. (a) I gp, x/(exp(x) ) gives the sequece of B /!, from which we deduce 0 0 3 4 5 6 7 8 9 0 B 0 5 6 30 0 4 0 0 30 66
(b) First, ote that O the other had, f(x) f( x) B x.! odd x x f(x) f( x) e x e x x xe x + e x e x x x( e ) ex So for 3 odd, B 0. x. (c) Multiplyig both sides of the defiig equatio by e x, we have ( ) x x x B.!! For, the coefficiet of x is (d) We have Therefore, 0 0 0 > 0 ( ) B. x S () ( + + + ) x!! 0 0 e x + e x + + e x x e x e e x e x x x e x ( ) ( ) l+ Bm x l ( ) m x m. (l + )! m! l0 m0 m Bm S()! ( ) ( m + )! m! m+ m0 + ) ( ) m B + 6. (a) If m a + b ad c + d, the m0 m m + m. m (a + b )(c + d ) (ac bd) + (ad bc). Now if p (mod 4) the p is a sum of two squares (show i class). If p 3 (mod 4) the q q + 0 is a sum of two squares. Fially, + is a sum of two squares. So ay iteger of the give form is a sum of two squares.
(b) We wat to use iductio o. Assume we have show that for all itegers less tha which are sums of two squares, every prime p 3 (mod 4) dividig such a iteger divides it to a eve power. Now suppose a + b ad let q 3 (mod 4) be a prime dividig (if there is o such prime, we are doe). We claim that q divides a ad b. Otherwise, say without loss of geerality that q b. Sice a + b 0 (mod q), we must have (ab ) (mod q), which is impossible. This shows that q a, b. Now write a a q ad b b q, so that q (a + b ). Lettig m a + b, by the iductive hypothesis it follows that m is divisible by primes cogruet to 3 mod 4 to eve powers. Sice q m, satisfies the same property. With the trivial base case, the iductio is complete. (c) Oe directio is obvious: if is a sum of two iteger squares, the it s a sum of two ratioal squares. Suppose ow that is a sum of two ratioal squares α ad β. Taig the commo deomiator, we write α a/d, β b/d. The a + b d. Now if we cosider ay prime q 3 (mod 4) the q divides a + b a eve umber of times. Obviously q also divides d a eve umber of times. Therefore, q divides a eve umber of times, so is of the form metioed i part (b), ad is thus a sum of two iteger squares. 7. (a) We have x 3 Φ 3 (x) x + x +. x Hece ω ω. Now for ay complex umber a + bω, a + bω (a + bω)(a + bω) (a + bω)(a + bω ) a + b + ab(ω + ω ) a ab + b. So if M a ab + b a + bω ad N c cd + d d + cω, the is of the same form. MN (a + bω)(d + cω) ad + bcω + (ac + bd)ω ad + bc( ω ) + (ac + bd)ω (ad bc) (ad bc)(ac + bd bc) + (ac + bd bc) (b) Suppose p (mod 3) ad p a ab + b. The p a or p b, sice otherwise p a ab + b would be divisible by p. I fact, if p a the p a ab + b implies p b, so p b as well. Thus, p divides either a or b. Ayway, (a b) + 3b 4(a ab + b ) 0 (mod p), so ( a b ) 3 (mod p). b Therefore, 3 is a square mod p. But we ve show before (usig quadratic reciprocity) that 3 is a square mod p if ad oly if p 3 or p (mod 3), cotradictio. 8. (a) For p 3, we have trivially 3 ()( ) + ( ). Now suppose p (mod 3). We ll prove by iductio o p that p is of the form a ab + b. Assume we have prove this statemet for primes less tha p. (We ca tae as our base case 7 3 (3)() +.) We ow 3 is a square mod p, so let x be the solutio to x 3 (mod p), ad write x y for some y. The y satisfies y y + 0 (mod p). We ca tae y < p/, so y p p y + < + + < p. 4 3
Hece y y + p for some < p, ad we have i additio that > 0 sice y y + (y /) + 3/4 > 0. Now let m be the smallest positive iteger such that mp ca be writte i the form a ab + b. Note that by the above proof m < p, ad if m the we are doe. Assume, for the sae of cotradictio, that m >. Let mp a ab + b. We may assume that g gcd(a, b), else g m ad thus we ca divide a ad b by g to reduce m to m/g. Now let l be a prime dividig m. The l a or l b; say l b. As i Problem 7, we have ( a b ) 3 (mod l), b so l 3 or l (mod 3). First, suppose l 3. The we have a ab + b 0 (mod 3). Sice 3 caot divide both a ad b, it ca be easily checed that the oly possibility is that a (mod 3) ad b (mod 3) (or vice versa). The ( ) ( ) ( ) a + b ) ( a + b a b a b b a ab + ( m ) + p, 3 3 3 3 3 3 so we have a smaller multiple of p, cotradictio. Therefore we must have l > 3. The x x + 0 (mod l) for x ab (mod l). Also, sice l m < p, by the iductive hypothesis l is of the form l c cd + d. Agai, we ca assume that l d, so y y + 0 (mod l) for y cd. Now x x + y y + (mod l), so (x y)(x + y ) 0 (mod l). Thus either x y (mod l) or x y (mod l). I the secod case, replacig (c, d) by (d c, d), we ote that (d c) (d c)d + d d cd + c l ad (d c)d cd y, so we may assume that x y (mod l). It follows that ab cd (mod l), so l ad bc. Now we showed i Problem 7 that (a ab + b )(c cd + d ) (ad bc) (ad bc)(ac + bd bc) + (ac + bd bc). The LHS ad the first two terms of the RHS are divisible by l. Thus, l ad + bd bc. Writig ad bc xl ad ac + bd bc yl, we ow have (mp)(l) x l xyl + y l. So ( m ) p x xy + y, l showig that m is ot miimal, cotradictio. Therefore every prime p (mod 3) ca be writte i the form a ab + b. (b) Oe directio is easy: suppose is positive ad every prime q (mod 3) divides to a eve power. We showed that 3 ad primes p (mod 3) are of the form a ab + b. Ad for q (mod 3), we have trivially that q q q 0 + 0 is also of this form. Sice the set of umbers of the form a ab + b is closed uder multiplicatio, it follows that is of the form a ab + b for some itegers a, b. To prove the coverse, we first ote that if a ab + b the ( b a ) + 4 ( ) b > 0.
(We will exclude the case a b 0.) We ow proceed with iductio o. The base case 0 + 0 is obvious. Suppose q (mod 3) divides 4. We claim that q a, b. Otherwise, without loss of geerality, assume that q b. The ( ) a b 3 (mod q), b 9. Computig, showig that 3 is a square mod q, which is impossible. So we ca write a a q, b b q, ad thus q (a a b + b ). By the iductive hypothesis, q divides a a b + b to a eve power, so it divides to a eve power as well. This completes the iductio. 657 676 7 + 783 783 7 + 783/676 7 + 07 + 676 7 + + 676/07 7 + 7 + 7 + 7 + + + + + 34 6 + 07 6 + 07/34 6 + 6 + 5 3 + 34 [7,, 6, 3, 34/5] [7,, 6, 3, 6, 5/4] [7,, 6, 3, 6,, 4]. 3 + 34/5 5
Next, 5 3 + 5 3 6 3 + 5 + 3 3 + (3 + 5)/6 3 + 5 3 + 6 3 + + 6/( 5 3) 3 + 3 + + 5 + 3 + 6 + 5 3 [3,, 6, [ ],... ] 3,, 6. 0. Taig the log of both sides, Differetiatig, so log si z log z + ( z ) log π. z cot z + π, z z π z cot z + z z π ( ) z π z π ( ) z z π π 0 z. π O the other had, we have x x r e x B r, r! r 0 6
ad pluggig i x iz, Taig the real part of this equatio, we get (iz) r iz B r r! e iz ize iz e iz e iz z cot z r 0 r eve iz(cos z i si z) i si z z cot z iz. B r (i) r r! z r ( ) B z ()! 0 ( ) B z. ()! Equatig the two expressios, ad taig the coefficiet of z, B ( ). ()! π So we coclude that ζ() ( ) B π. ()! 7
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