Recall that by Theorems 10.3 and 10.4 together provide us the estimate o(n2 ), S(q) q 9, q=1

Transcription

1 Chaptr 11 Th singular sris Rcall that by Thorms 10 and 104 togthr provid us th stimat 9 4 n Rn = SnΓ 2 + on2, whr th singular sris Sn was dfind in Chaptr 10 as Sn = q=1 Sq q 9, with Sq = 1 a q gcda,q=1 Sq, a 9 an/q, Sq, a = q am /q m=1 Th dfinition of th Gamma function shows that Γ 4 > 0, hnc if w could prov that Sn > 0 thn th main rsult of our lcturs, Thorm 81, would b stablishd with c = Sn Γ Our aim in this chaptr is to prov Sn > 0 and furthrmor to provid a concptual dscription of Sn Dfin for ach q N, M n q := #{x 1,, x 9 Z [1, q] 9 : x x 9 nmod q}, 17

2 whr hr and blow, th x i dnot intgrs For prim powrs q = p k w might guss that for ach of th p 8k choics for th variabls 1 x 1,, x 8 p k thr xist at most solutions of th cubic quation in th variabl x 9, x x 9 nmod p k Hnc it is natural to considr th following limit for vry prim p, 112 σ p n := lim k M n p k p 8k Thorm 111 Th limit 112 xists and is positiv Furthrmor th infinit product p σ pn, takn ovr all prims, convrgs absolutly to th singular sris, Sn = p σ p n Th constants σ p n ar calld p-adic Hardy Littlwood dnsitis and, as 112 rvals, thy ar intimatly connctd to solving th quation x x 9 = n modulo positiv intgrs q Of cours, if thr is som q N such that x x 9 nmod q has no solutions for x i thn Rn = 0 On intrprtation of Thorm 111 is that it provids vidnc for th opposit; namly that if x x 9 = n is solubl modulo vry q thn it can b solvd in th intgrs This is not tru in gnral, a countrxampl is givn by 4x x 2 2 5x 2 = Rlating Sn to σ p n Lmma 112 Lt, q 2 b coprim intgrs and lt q := q 2 Thn for all a 1 Z [1, ], a 2 Z [1, q 2 ] w hav whr a := a 1 q 2 + a 2 S, a 1 Sq 2, a 2 = Sq, a, 174

3 Proof As th variabl m 1 rangs through all rsidu classs mod in th sum S, a 1 = a1 m 1 m 1 mod w s that, du to th coprimality of, q 2, th intgrs m 1 q 2 also covr all rsidu classs mod Hnc w may writ S, a 1 = a1 m 1 q 2, m 1 mod and th fact that for any positiv intgr q th function is priodic mod q, q allows us to writ S, a 1 = a1 m 1 q 2 = a1 m 1 q 2 + m 2 m 1 mod A similar argumnt shows that Thus w ar ld to S, a 1 Sq 2, a 2 = which quals Sq 2, a 2 = m 1 mod m 2 mod q 2 m 1 mod m 2 mod q 2 m 2 mod q 2 m 1 mod a2 m 1 q 2 + m 2 q 2 a1 m 1 q 2 + m 2 + a 2m 1 q 2 + m 2, q 2 a1 q 2 + a 2 m 1 q 2 + m 2 q 2 W can s that as th variabls m 1, m 2 rang through all availabl rsidu classs mod and mod q 2 rspctivly, thn th variabl m := m 1 q 2 + m 2 taks ach rsidu class mod q 2 onc Thrfor th last sum quals a1 q 2 + a 2 m, q 2 which concluds our proof mmod q 2 175

4 Lmma 11 Th function Sq is multiplicativ Proof Lt, q 2 b coprim positiv intgrs Thn th sts {a 1 Z [1, ] : gcda 1, = 1} {a 2 Z [1, q 2 ] : gcda 2, q 2 = 1} and {a Z [1, q] : gcda, q = 1} ar in 1 1 corrspondnc This can b sn by mapping a 1 mod, a 2 mod q 2 to amod q, whr a := a 1 q 2 + a 2 Hnc w may writ Sq = Sq, a 9 n a 1q 2 + a 2 q 1 a 1 1 a 2 q 2 gcda 1, =1 gcda 2,q 2 =1 Th idntity n a 1q 2 + a 2 = n a 1 n a 2 q q 2 and Lmma 112 allows us to dduc Sq = 1 a 1 gcda 1, =1 which is sufficint S, a 1 9 n a 1 1 a 2 q 2 gcda 2,q 2 =1 Sq 2, a 2 9 n a 2 q 2 Rcall that w hav provd in Chaptr 10 that Sn is an absolutly convrgnt sris, a fact which, whn combind with Lmma 11 shows that th Eulr product of Sn is 11 Sn = Sp m 1 + p 9m p m=1 and furthrmor that for ach prim p, k 114 lim 1 + k + m=1 Sp m p 9m Lmma 114 For ach prim p and k N w hav 1 + k m=1 xists Sp m p 9m = M np k p 8k 176,

5 Proof W bgin by dtcting solutions x i of th quation α=1 x x 9 nmod p k using crtain xponntial functions To this nd obsrv that for ach intgr x w hav p 1 k α xp { 1 if x 0mod p k, = p k k 0 othrwis Thus writing M n p k = 1 x 1,,x 9 p k a n x 1, x 9 with a n x 1, x 9 = { 1 if x x 9 n 0mod p k, 0 othrwis and invrting th ordr of summation, w s that M n p k quals 1 p k p k α=1 1 x 1,,x 9 p k α x x 9 n = 1p p k p k k α=1 Sp k, α 9 αn/p k Lt ν p α := t whr t is th intgr such that p t divids α but p t+1 dos not divid α Thn ach intgr α in th last sum can b factorisd uniquly as α = p k m a, whr m := k ν p α and a is coprim to p Not that 1 α p k, hnc th only possibl valus for m and a ar 0 m k, 1 a p m Not that th idntity α = p k m a implis that Sp k, α = ax = p k m Sp m, a, 1 x p k hnc w obtain that p k α=1 Spk, α 9 αn/p k is qual to p 9k k m=0 p 9m 1 a p m gcda,p m =1 This is sufficint for our lmma p m Sp m, a 9 an/p m = p 9k 177 k p 9m Sp m m=0

6 Combining 114 and Lmma 114 shows that th limit 112, that dfins σ p n xists In addition, Lmma 114 and 11 show that Sn = p σ p n, hnc th only rmaining part rgarding th vrification of Thorm 111 is th positivity of ach σ p n This is th aim of th last sction Rmark 115 Th absolut convrgnc of th sris dfining Sn guarants that th infinit product in Thorm 111 is absolutly convrgnt As such, it has a strictly positiv valu if and only if ach of th p-adic factors is strictly positiv Thrfor th positivity of ach σ p n guarants that th constant c in Thorm 81 dos not vanish, which, in turn, implis that for all larg nough intgrs n th function Rn is positiv, i thr xists at last on rprsntation of n as a sum of xactly 9 positiv intgr cubs 112 Positivity of th p-adic dnsitis For prims p dfin th quantity γ p := { 2 if p = 2,, 1 if p > Lmma 116 For ach prim p, vry lmnt in Z/p γp Z is th sum of at most 9 cubs of lmnts of Z/p γp Z, at last on of which is coprim to p Proof Th statmnt is obvious whn p = 2 or, sinc on can add 1 svral tims Assum that p >, so that γ p = 1 W hav that 0modp quals modp, hnc it is sufficint to prov that ach lmnt of Z/pZ := Z/pZ \ {0} is a sum of at most 9 cubs W know that this st forms a cyclic group undr multiplication Pick a gnrator g and considr th subgroup Γ p := {g m modp : m N}, which has ordr p 1 gcdp 1, 178

7 If p 2mod thn Γ p = Z/pZ, hnc our lmma holds In th rmaining cas, p 1mod, th st Γ p has p 1/ lmnts Lt C 1 := Γ p and for ach m N with m 2 dnot by C m th lmnts of Z/pZ that ar a sum of m lmnts of Γ m but not a sum of m 1 lmnts of Γ m Fix j 1 and considr th minimum lmnt x Z/pZ that is not in any of C 1, C 2,, C j Thn x 1 or x 2 is also in Z/pZ and must thrfor b a sum of at most j cubs Owing to x = x and x = x , w s that x C j+1 or x C j+2 Applying this for j = 1 and j = w infr that at last of C 1,, C 5 must b non-mpty Also not that for ach j w hav Γ p C j C j, hnc if C j is not mpty thn it must contain at last #Γ p = p 1 lmnts Assum that Thn p 1 > 5 #C j = j=1 Z/pZ 5 i=1c i 1 j 5 #C j 0 #C j p 1 1 j 5 #C j 0 1 p 1, which is a contradiction This provs that ach lmnt of x Z/pZ is a sum of at most 5 cubs, all of which ar coprim to p W dduc that for ach n N and prim p, thr is at last on solution of x x 9 n mod p γp with p x j for som j For ach i j and k > γ p thr ar p k γp lmnts y i mod p k with y i x i mod p γp For any of thos p k γp 8 choics w not that n i j y i n i j x i x jmod p, hnc µ := n i j y i is an intgr coprim to p for which th quation x µmod p has a solution Hnsl s lmma allows us to lift this solution to a solution mod p k, thrby giving ris to a solution of 9 x i n mod p k i=1 This implis that M n p k p k γp 8, hnc σ p n p 8γp > 0, thus concluding th proof of Thorm