KEQIN FENG AND MAOSHENG XIONG

Transcription

1 ON SELMER GROUPS AND TATE-SHAFAREVICH GROUPS FOR ELLIPTIC CURVES y = x 3 n 3 KEQIN FENG AND MAOSHENG XIONG Abstract. We study the distribution of the size of Selmer grous and Tate- Shafarevich grous arising from a -isogeny and its dual -isogeny for ellitic curves E n : y = x 3 n 3. We show that the -ranks of these grous all follow the same distribution. The result also imlies that the mean value of the -rank of the corresonding Tate-Shafarevich grous for square-free ositive integers n X is log log X as X. This is quite different from quadratic twists of ellitic curves with full -torsion oints over Q [9], where one Tate-Shafarevich grou is almost always trivial while the other is much larger.. Introduction In [9] one of the authors has studied asymtotic behavior of the size of Selmer grous arising from a -isogeny and its dual -isogeny for quadratic twists of the ellitic curve E a,b : y = xx + ax + b, where a, b are any integers satisfying aba b 0 and that ab is not a square. This result, combined with a result obtained by Yu [3], imlies exlicit information of the corresonding Tate-Shafarevich grous. Interested readers may refer to [9] for details. The curve E a,b can be characterized as ellitic curves with full -torsion oints over Q. It would be interesting to do a similar analysis for other family of ellitic curves, for examle for ellitic curves with one non-trivial -torsion oint over Q. It turns out, however, to be quite difficult in general. In this aer we focus on 000 Mathematics Subject Classification. G05, 4H5, L40, N45. Key words and hrases. Ellitic curves, Selmer grou, Tate-Shafarevich grou,distribution. The author is suorted by RGC grant number 606 and DAGSC0 from Hong Kong.

2 FENG AND XIONG the most interesting case, the ellitic curve E : y = x 3 and consider, for a square-free integer n, the quadratic twist E n given by the equation E n : y = x 3 n 3. E n hase comlex multilication 3, it various roerties have been studied extensively in the literature. Corresonding to the -torsion oint n, 0 there is a -isogeny φ : E n E n, where E n is another ellitic curve E n : Y = X 3 6nX 3n X, and the -isogeny φ is given by [5,. 74] y φx, y = x n, y3n x n x n Via Galois cohomology we obtain the short exact sequence. 0 E nq φe nq Sel φ E n /Q XE n /Q[φ] 0, where Sel φ E n /Q is the φ Selmer grou and XE n /Q[φ] is the φ-torsion of the Tate-Shafarevich grou. Denote by ˆφ the dual isogeny of φ. We also have the short exact sequence 0 E nq ˆφE n Q Sel ˆφ E n/q XE n/q[ ˆφ] 0. For X > 0 and integer h, define SX, h = { n X : n h mod and n is square-free}. We rove the following result. Theorem. Let E n be the ellitic curve given in. Put #Sel φ E n /Q = sn+, #Sel ˆφ E n/q = ŝn+, #XE n /Q[φ] = tn, #XE n/q[ ˆφ] = ˆtn,

3 ON ELLIPTIC CURVES y = x 3 n 3 3 where the quantities sn, ŝn, tn, ˆtn are called the -rank of those grous. For h = or 5 and X, as n varies in SX, h, the value γn = sn or ŝn, tn, ˆtn resectively follows the same distribution. More recisely, for any a > 0, we have lim X #SX, h # γn n SX, h : a log log n = a π e t dt. We remark that firstly, the distribution results may seem a little strange at the first glance, however, there is a very natural heuristics which we will exlain in the beginning of Section 4. Secondly, Theorem shows that the normalized - rank of Sel φ E n /Q and Sel ˆφ E n/q, i.e. the quantity γn/ log log n behaves asymtotically like the random variable max{0, Y }, where Y is a standard Gaussian distribution. This distribution is different from that obtained in [9], where one Selmer grou is much smaller, and the other is much larger. It is also different from results obtained in [8, 9, 3], where on average the -Selmer grous have intermediate size. Thirdly, for the sake of simlicity we only consider square-free integers modulo, there is no essential difficulty for general moduli under some congruence constraints as in [9]. Finally, the distribution of Selmer grous is with resect to the set SX, h as X. It would be interesting to study the statistics with the number of rime factors of n fixed, as was done by Swinnerton-Dyer for -Selmer grous in [8]. Theorem. Under the assumtions of Theorem, define #XE n /Q[φ] = tn, #XE n/q[ ˆφ] = ˆtn. Let γn = tn or ˆtn. Then for any integer k > 0, we have 3 #SX, h k/ log log X γn k λ k,

4 4 FENG AND XIONG and 4 #SX, h tn + ˆtn k/ k log log X λk, as X, where k k!/k/! : k is even, 5 λ k = k!/ π : k is odd. k Equation 3 can be derived from Theorem by comuting the k-th moment of the random variable max{0, X} see the remark after Theorem. Taking k =, we see that the mean value of tn and ˆtn is log log X. This behavior is again different from [9], where one Tate-Shafarevich grou is almost always trivial, and the other has a much larger mean value log log X. Since XE n/q[φ] XE n /Q[] and XE n/q[ ˆφ] XE n/q[], equation 3 shows that the -art of XE n /Q and XE n/q can be arbitrarily large. Finally, equation 4 imlies a strong correlation between tn and ˆtn, namely, tn and ˆtn can not be large at the same time, and this henomenon seems indeendent of Theorem. We also remark that while Selmer grous are relatively easy to handle, the Tate- Shafarevich grous are more mysterious. They aear in the Birch and Swinnerton- Dyer conjecture, and measure the degree of deviation from the Hasse rincile. Even the finiteness of such grous are not known in general. Various families of ellitic curves with large Tate-Shafarevich grous were identified by a number of authors see [],[3],[4],[5],[0],[],[],[3]. Moments [8], heuristic results [7], and uer bounds [4],[5] on the order of Tate-Shafarevitch Grous were also considered. There are three main ingredients in the roofs of the above results. First, to comute the Selmer grous, we use a grahical method, which lays an essential role in identifying the main contribution and reducing the comlexity of the roblem. Second, we emloy Heath-Brown s method based on character sums to obtain asymtotic formulas on the size of Selmer grous. Third, combining our results with

5 ON ELLIPTIC CURVES y = x 3 n 3 5 results obtained by Chang [6] and Stoll [7], we obtain information on the corresonding Tate-Shafarevich grous. These ingredients are similar to those in [9], however, the techniques involved here are more comlicated. The main reason is that the grahs constructed for the urose of comuting Selmer grous are more comlicated, which result in a character sum with 8 variables. By comarison, the character sum aearing in [9] has only 6 variables. Moreover, for each Selmer grou, one needs to deal with 4 such character sums, comaring with a single one in [9]. This increases the comlexity of the roblem. As it turns out, all the 4 character sums need to be combined together at the end so that extra savings can be exloited by combinatorial argument. Finally, the result is more difficult to analyze than that in [9]. Here we rely on the owerful machinery develoed by Granville and Soundararajan [6] on the Erdős-Kac theorem to obtain the distribution. Acknowledgement The authors want to exress their gratitude to the referee for careful reading and many useful suggestions.. Preliminaries.. -descent and Selmer grous. The -descent method is exlained in the last chater of Silverman s book [5]. For the articular curve E n in this aer, it can be secified as follows see also [3]. For a square-free ositive integer n, define a finite set S of rime divisors of the rational number field Q by S = { } { : 6n}. Let M be the multilicative subgrou of Q /Q generated by and the rime divisors of 6n. For each d M we have the homogeneous sace C d resectively C d defined by C d : dw = d t 4 6ndt z 3n z 4,

6 6 FENG AND XIONG C d : dw = d t 4 + 3dnt z + 3n z 4. The Selmer grou Sel φ E n /Q resectively Sel ˆφ E n/q measures the ossibility of C d or C d having non-trivial solutions in the local field Q v for all v S. Namely, Sel φ E n /Q = {d M : C d Q v for all v S}, Sel ˆφ E n/q = {d M : C dq v for all v S}, where C d Q v or C d Q v means that the homogeneous sace C d or C d has non-trivial solutions w, t, z 0, 0, 0 in Q v. We have {, 3} Sel φ E n /Q, {, 3} Sel ˆφ E n/q, since each of the homogeneous saces C, C 3, C, C 3 has non-trivial solution in Q. For the rank of the ellitic curve E n we obtain the formula see 86, [3] ranke n Q = dim F Sel φ E n /Q dim F XE n /Q[φ] + dim F Sel ˆφ E n/q dim F XE n/q[ ˆφ]. Thus we can calculate the rank from the dimensions of the Selmer grous and the Tate-Shafarevich grous... Even artitions. We use standard terminology in grah theory [7]. Let G = V, A be a simle directed grah where V = V G = {v,, v m } is the set of vertices, A = AG is the set of arcs. The adjacency matrix of G is defined by where MG = a ij i,j m,, if v i, v j A i j m, a ij = 0, otherwise. For the vertex v i, let d i = m j= a ij. The Lalace matrix of the grah G is defined by LG = diagd,, d m MG.

7 ON ELLIPTIC CURVES y = x 3 n 3 7 Definition. Let G = V, A be a directed grah. A artition of vertices V V = V is called even if for any v V, #{v V }, the total number of arcs from v to vertices in V is even, and for any v V, #{v V } is also even. Lemma. Let G = V, A be a directed grah, V = {v,..., v s+t } s, t 0. Then the number of even artition {V, V } of V such that {v s+,..., v s+t } V is equal to the number of vectors x,..., x s F s such that LG x,..., x s, 0,..., 0 T = 0. Lemma in various forms has been used by Xue and the authors in the study of new families of non-congruent numbers [],[],[3]. It was also used by Faulkner and James in [9]. The roof of Lemma is elementary. For examle, it can be derived easily by adating the roof of Lemma. in []..3. A theorem of Granville and Soundararajan. We use a result of Granville and Soundararajan [6], which rovides a most transarent account of the Erdös-Kac theorem. Recall that g is an additive function if g = 0, and gmn = gm + gn whenever gcdm, n =. If in addition g k = g for all k where is a rime, we say the function g is strongly additive. Let A = {a,..., a x } be a multi-set of x not necessarily distinct natural numbers. Let A d = #{n x : d a n }. We suose that there is a real valued, nonnegative multilicative function hd such that for square-free d we may write A d = hd d x + r d. Lemma Proosition 4, [6]. Let A be a multi-set of x integers, and let hd and r d be as above. Let P be a set of rimes, and let g be a real-valued, strongly additive function with g M for all P. Let µ P g = P g h, σ Pg = g h h. P

8 8 FENG AND XIONG Then, uniformly for all even natural numbers k σ P g/m /3, g µ P g a A a P k = C k xσ P g k + E, where E C k xσ P g k k 3 M + M k P k h r d, d D k P and for all odd natural numbers k σ P g/m /3, g µ P g a A a P k E, where E C k xσ P g k k 3/ M + M k P Here for any natural number k, k h r d. d D k P C k = Γk + k/ Γk/ +, and D k P denotes the set of square-free integers which are the roduct of at most k rimes all from the set P. 3. Exlicit formulas of the Selmer grous 3.. Solvability conditions. The roblem of finding the size of Selmer grous Sel φ E n /Q and Sel ˆφ E n/q is equivalent to the roblem of determining how many homogeneous saces C d resectively C d have non-trivial solutions over certain local fields. We collect solvability conditions for C d and C d in the following two lemmas.

9 ON ELLIPTIC CURVES y = x 3 n 3 9 Lemma 3. Let n be a square-free integer with gcdn, 6 =, and M Q /Q, the multilicative subgrou generated by and the rime divisors of 6n. Let denote an odd rime. For any d M, we have. d = C d Q =. For d we have C d Q = 3 = ; or 3 = 3 = 4 n/d =. 3. If 3 d, then C d Q 3 = d mod If n d, then C dq = mod 3 and d =. Proof.. For a rime, we denote by v the standard -adic exonential valuation. Let w, t, z 0, 0, 0 be a solution of C d in Q. From the equation of C d and d we know that at least two numbers in { + v w, 4v t, v t + v z, 4v z } reach the minimal value, which is denoted by min. It must be v t + v z = min since the other three numbers are distinct modulo 4. Then we must have v t + v z 4v t + 4v z, which is a contradiction. Therefore in this case C d Q =.. = : Let w, t, z 0, 0, 0 be a solution of C d in Q. We may assume that w, t, z Z 3 and 0 = min{ + v w, 4v t, v t + v z, 4v z}. It is easy to see that 4v t = v t + v z = 4v z < + v w so that 0 = v t = v z v w. Thus 0 t 4 6 n d t z 3 n d z 4 = t 3 n d z n d z mod. 3 3 Therefore = = C d Q =. Suose that = =, then 3 Q and t nz t 3 3 nz 0 mod, which means d d that

10 0 FENG AND XIONG A. t n d z mod, so that n d + 3 n/d 3 = = = or B. t 3 3 n d z mod, so that = 3 3 n d = 3 4 We have used the relation = Therefore C d Q = if 3 = = 3 4 n/d =. n/d =. n/d = : Taking w = 0 and z =, the equation of C d is reduced to t 4 6 n d t 3 n d = 3 0. If =, then 3 Q and the equation is t = 3 ± 3 n. If 3 = and d =, then 3+ 3n/d 3 3n/d 3 = = so that 3+ 3D/d = or 3 3n/d = which imlies by Hensel s lemma that there exists t = α Q such that α = n or d α = 3 3 n. Therefore w, z, t = 0,, α is d a solution of C d in Q. If = = =, then 3± 3n/d = and we also have C d Q by Hensel s lemma n/d 3. = : Suose that d mod 3. Choose z = 0, t =. Since d 3 =, by Hensel s lemma, the equation dw = has a solution in Q 3. Hence C d Q 3. = : Let w, t, z 0, 0, 0 be a solution of C d in Q 3. Then at least two terms in {v 3 w, 4v 3 t, + v 3 t + v 3 z, 4v 3 z + } reach the minimal value, which is denoted by min. The only ossibility is that v 3 w = 4v 3 t = min. We may assume that v 3 w = v 3 t = 0, hence v 3 z 0. The equation C d modulo 3 gives ;

11 ON ELLIPTIC CURVES y = x 3 n 3 us dw t 4 mod 3 and hence d 3 =, which contradicts d mod 3. This comletes the roof = : If mod 3, then =. Choose z = 0, t = we have dw d =, which is solvable in Q if = ; choose z =, t = 0 we have dw = 3 n, d which is solvable in Q if =. Since =, it is easy to see that one of 3d the two equations is always solvable. Hence C d Q. = : Let w, t, z 0, 0, 0 be a solution of C d in Q. Then at least two terms in {v w, 4v t, + v t + v z, 4v z + } reach the minimal value, which is denoted by min. There are two ossibilities. First, if v w = 4v t = min, we may assume that v w = v t = 0, hence v z 0. The equation C d modulo gives us dw t 4 d mod and hence =, which contradicts the assumtion that =. Second, if v w = 4v z + = min, we may assume that d v z = 0, v w =, hence v t. Let w = w, t = t, then v w = 0 v t. Dividing both sides of the equation C d by and then modulo we find that dw n 3 d z 4 3d mod. Hence =. Since mod 3, we 3 d have =, which imlies that =, which contradicts the assumtion that =. This concludes the roof. d Lemma 4. Under the same assumtion of lemma 3, we have. C d Q = d. If d then C d Q = 3 3 = ; 3 or = 3 = = 4 D/d 3. If 3 d, then C d Q 3 = d mod If n d, then C d Q = 3 = = d. 5. d < 0 C d Q =.

12 FENG AND XIONG Proof. The roof is very similar to that of Lemma 3, so we omit the details. 3.. The size of the Selmer grous. For any secific value n we may determine if C d Q v and C d Q v for v S, d M by Lemma 3 and Lemma 4, so that we can determine the Selmer grous Sel φ E n /Q and Sel ˆφ E n/q. However when the number of rime factors of n gets larger, the comutation becomes more comlicated and it seems difficult to exress the size of Sel φ E n /Q and Sel ˆφ E n/q exlicitly with certain knowledge of n. We find a reasonable way to calculate the size of the Selmer grous in terms of some secific grahs associated to n. This grahtheoretical method has been used to find rank zero ellitic curves y = x 3 n x [],[],[3] so we find new non-congruent numbers n. This grahical method also works nicely for the curve E n, and the final result can be stated in a very simle way. For a square-free ositive integer n with gcdn, 6 =, we factor it as a roduct of rimes n = P P t Q Q s t q q s, where P i, Q j 5, λ 7, q µ mod, t, s, t, s 0, t + s + t + s. Put P = {P,..., P t }, Q = {Q,..., Q s }, = {,..., t }, q = {q,..., q s }. Then S = {,, 3} P Q q, and M =,, 3, P,..., P t, Q,..., Q s,,..., t, q,..., q s Q /Q.

13 ON ELLIPTIC CURVES y = x 3 n 3 3 We construct a grah G n = V, E as follows: V = P Q q {3}, E = { ππ π π =, π P, π P Q q} { π3 3 π 4 =, π P} { ππ π π =, π q, π P }. and 5 of Lemma 4 imly that if d or d < 0, then d / Sel ˆφ E n/q. of Lemma 4 imlies that if d and 3 = which means that Q q, then d Sel ˆφ E n/q. Hence Sel ˆφ E n/q 3, P,..., P t,,..., t. Since 3 Sel ˆφ E n/q, we may restrict our attention to the subgrou M = P,..., P t,,..., t. Each d M is associated to a artition V = V d V d given by V d = {π P : π d}, V d = V V d, and this artition satisfies the roerty that V d Q q {3}. There is a oneto-one corresondence between the elements of M and the artitions V = V V of G n with V Q q {3}, and it is easy to see from Lemma 4 that for any d M, d Sel ˆφ E n/q if and only if the artition V = V d V d is an even artition with V d Q q {3}. To comute this cardinality ŝn, we order the vertices of V as 6 V = {P,..., P t,,..., t, q,..., q s, Q,..., Q s, 3} = {v,..., v m }, where m = t + t + s + s +. Let LG n be the Lalace matrix of the grah G n. By Lemma, the quantity ŝn equals the number of vectors x,..., x t, y,..., y t, 0,..., 0 F m such that 7 LG n x,..., x t, y,..., y t, 0,..., 0 T = 0.

14 4 FENG AND XIONG The matrix LG n can be written exlicitly in the form LG n = A PP A P A qp A q , where A PP is a symmetric t t matrix, A P is a t t matrix, A qp is a s t matrix, and A q is a s t matrix. Define 8 Γn = A PP A qp A P A q. and a = rank F Γn. Then the number of solutions of the equation 7 is t+t a. Hence ŝn = t + t a. For any integer b with gcdb, =, let ω b n = n b mod. It is worth noting that a ω n + min{ω 7 n, ω n}, hence 9 ŝn max {0, ω 7 n ω n}.

15 ON ELLIPTIC CURVES y = x 3 n 3 5 To comute #Sel φ E n /Q, we use a similar method. Gn = V, E as follows: We construct a grah V = P Q q {3} E = { ππ π π =, π P, π P Q q} { π3 3 π 4 =, π P} { ππ π π π, π P} { ππ π =, π π, π q}. Since 3 Sel φ E n /Q, by Lemma 3, it is enough to consider the subgrou T = {d, P,..., P t, q,..., q s : d mod 3}. Each d T is associated to a artition V = V d V d where V d = { P q : d}, V d = V V d, and this artition satisfies the roerty that V d Q {3}. There is a oneto-one corresondence between the elements of T and the artitions V = V V with Q {3} V, and it is also easy to see from Lemma 3 that for any d T, d Sel φ E n /Q if and only if V = V d V d is an even artition in the grah Gn with Q {3} V. To comute this cardinality sn, similarly we order the vertices as in 6. Let LGn be the Lalace matrix of the grah Gn. By Lemma, the quantity sn equals the number of vectors x,..., x t, 0,..., 0, y,..., y s, 0,..., 0 F m such that 0 LGn x,..., x t, 0,..., 0, y,..., y s, 0,..., 0 T = 0.

16 6 FENG AND XIONG Comaring with LG n, the matrix LGn can be written exlicitly as A PP A Pq A P A q 0 0 LGn = , where A PP is symmetric, A P = A T P, A Pq = A T qp, and A q = A T q. Using the same matrix Γn in 8, and a = rank F Γn, we find that the number of solutions of the equation 0 is t+s a. Hence sn = t + s a. It is also worth noting that sn max {0, ω n ω 7 n}. The matrices A PP, A P, A qp and A q deend on the factorization of the integer n and can be written down exlicitly. We choose not to do that for the sake of simlicity. As alications we resent three interesting consequences, which can be derived immediately from above. Corollary. i. Sel φ E n /Q = {, 3} and Sel ˆφ E n/q = {, 3} hence ranke n Q = 0 if and only if s = t and Γn is nonsingular over F. ii. If n is a roduct of distinct rimes which are all congruent to 5 modulo, then Sel φ E n /Q = {, 3} and Sel ˆφ E n/q = {, 3}, hence ranke n Q = 0. iii For the following cases ranke n Q = 0 in what follows, q are distinct odd rimes: n =, 5 mod ; or mod, 3 = ; 4

17 ON ELLIPTIC CURVES y = x 3 n 3 7 n = q,, q 5, 5 mod ;, q, mod,, q, mod,, q, 5 mod,, q 7, mod, 3 = = ; q = = ; q 4 4 q = 4 ; = ; 3 q 3 q We oint out that iii of Corollary was first obtained by the first author in [0]. 4. Averaging #Sel φ E n /Q Let n be a square-free ositive integer such that gcdn, 6 =. Our task now is to study how the -rank sn and ŝn is distributed as n varies in the set SX, h. We can give a heuristic argument first. The matrix Γn in 8 deends on the factorization of n, and can be considered as a random matrix when n varies in SX, h. Its F -rank tends to be as large as ossible, if there are no obvious obstructions, with high robability. Considering the width and height of Γn, its largest rank is obviously min{ω n + ω n, ω n + ω 7 n}. By this argument and, the quantity sn should be distributed as max{0, ω n ω 7 n}, which behaves like a Gaussian random variable by the famous Erdős-Kac theorem. The argument for ŝn is the same. We will rove below that it is indeed the case. From Lemma 3, we have sn = F n + F n,

18 8 FENG AND XIONG where d and d { + 4 F n = 3 n=dd F n = { n=dd + + d { 3 4 d d { 3 4 d d + + } 3 } d, 3 4 d Here F n accounts for d Sel φ E n /Q with d > 0 and F n accounts for d Sel φ E n /Q with d < 0. We deal with F n first. Exanding the roduct, we obtain where F n = F,n + F, n, } 3 } d. F, n = F, n = g, ɛ, n=ɛ ɛ ɛ 8 g, ɛ, n=ɛ ɛ ɛ 8 g, ɛ = ωɛ ωɛ 3ωɛ 3 ωɛ ɛ 3 ɛ 7 3 ɛ3 ɛ 4 ɛ 7 3 ɛ ɛ 3 ɛ 4 ɛ 6 ɛ 7 ɛ 9 ɛ ɛ ɛ 4 ɛ 5 ɛ 7 ɛ 8 3 ɛ 5 ɛ 6 ɛ 7 ɛ 0 4 ɛ4 ɛ 6 ɛ 7 ɛ 8 ɛ ɛ ɛ 3 ɛ 4 ɛ 8 ɛ 9 ɛ ɛ ɛ3 ɛ 5 ɛ5 ɛ 6 ɛ 7 ɛ 0, ɛ 5 ɛ 6 ɛ 7 ɛ 0 ɛ 3 ɛ 5 ɛ 5 ɛ 6 ɛ 7 ɛ 0 ɛ 3 ɛ 5

19 ON ELLIPTIC CURVES y = x 3 n 3 9 and g, ɛ = ωɛ ωɛ 3ωɛ 3 ωɛ ɛ 3 ɛ 7 ɛ ɛ 3 3 ɛ3 ɛ 4 ɛ 7 3 ɛ ɛ 3 ɛ 4 ɛ 6 ɛ 7 ɛ 9 ɛ ɛ ɛ 4 ɛ 5 ɛ 7 ɛ 8 3 ɛ 5 ɛ 6 ɛ 7 ɛ 0 4 ɛ4 ɛ 6 ɛ 7 ɛ 8 ɛ ɛ ɛ 3 ɛ 4 ɛ 8 ɛ 9 ɛ ɛ ɛ3 ɛ 5 ɛ5 ɛ 6 ɛ 7 ɛ 0. ɛ 5 ɛ 6 ɛ 7 ɛ 0 ɛ 3 ɛ 5 ɛ 5 ɛ 6 ɛ 7 ɛ 0 ɛ 3 ɛ 5 Here the sums in F, n and F, n are over all ositive integers ɛ, ɛ,..., ɛ 8 such that n = ɛ ɛ ɛ 8, and we ut ɛ = ɛ 7 ɛ 8, ɛ = ɛ ɛ ɛ 3 ɛ ɛ 3 ɛ 4 ɛ 5 ɛ 6, ɛ 3 = ɛ 4 ɛ 5 ɛ 6 ɛ 7 ɛ 8 ɛ 9 ɛ 0 ɛ, and the function ωn counts the number of distinct rime factors of n. Our goal is to estimate We shall first estimate sn = F n + F, n. F n. We sum over the 8 variables ɛ i, subject to the conditions that each ɛ i is square-free, ositive, that they are relatively rime in airs, and that their roduct n satisfies n X, n h mod, where h = or 5. We divide the range of each variable ɛ i into intervals [A i, A i, where A i runs over owers of. For such A i s, we define A = A,..., A 8 and SA = ɛ i A i i 8 g, ɛ, where ɛ i A i means A i ɛ i < A i, and the sum is under the restriction that ɛ ɛ 8 X, ɛ i s are relatively rime in airs and ɛ ɛ 8 h mod. We

20 0 FENG AND XIONG collect all such admissible A s together and we call it the set L. It is clear that #L = Olog 8 X. We may assume that for any A L, Now we estimate the sum 8 i= A i X. SA. A L Following Heath-Brown [8, 9], we shall describe the variables ɛ i and ɛ j as being linked if exactly one of the Jacobi symbols ɛi, ɛ j occurs in the exression for g, ɛ. It is easy to see that the airs ɛ i, ɛ j with i {4, 6, 7, 8}, j {5, 6, 7, 0} are linked. The airs ɛ i, ɛ j with i {3, 5}, j {,, 3, 4, 8, 9,, } are also linked. 4.. Case one. Given A L, suose that A i, A j log X 640 and that ɛ i, ɛ j are ɛj linked variables. We may write g, ɛ in the form ɛi g, ɛ = aɛ i bɛ j, ɛ j where the function aɛ i deends all the variables ɛ k excet ɛ j, and similarly the function bɛ j deends on all the variables ɛ k excet ɛ i. Moreover we have ɛ i aɛ i, bɛ j. We can now write SA ɛ k, k 8 k i,j ɛ,ɛ j ɛi ɛ j aɛ i bɛ j. We need the following result.

21 ON ELLIPTIC CURVES y = x 3 n 3 Lemma 5 Lemma 4 in [8] and Lemma 4. in [30]. Suose ɛ > 0 is any fixed number, X, M and N are sufficiently large real numbers, and {a m }, {b n } are two comlex sequences, suorted on odd integers, satisfying a m, b n. Fix ositive integers h, q satisfying gcdh, q = and q {minm, N} ɛ/3. Let S := m,n a m b n m n, where the summation is subject to M m < M, N n < N, mn X and mn h mod q. Then S MN 5/6+ɛ + M 5/6+ɛ N, where the constant involved in the symbol deends on ɛ only. As a consequence of this lemma one finds that SA Therefore 8 k= A k {mina i, A j } /3 Xlog X 0. Lemma 6. We have SA Xlog X 0 whenever there is a air of linked variables ɛ i, ɛ j with A i, A j log X Case two. We now examine the case that A i log X 640, A j < log X 640 while ɛ i, ɛ j is a air of linked variables. Using the quadratic recirocity law we ut g, ɛ in the form g, ɛ = r ωɛ i ɛi ɛ j χɛ i c,

22 FENG AND XIONG where χ is a character modulo 8, which may deend on the variables ɛ k s other than ɛ i, and the factor c is indeendent of ɛ i and satisfies c, and r = ±, ±4 or ±8, deending on the choice of i. It follows that 3 SA ɛ k,k i k 8 ɛ i r ωɛi ɛi ɛ j χɛ i, where the inner sum is restricted by the conditions that ɛ i must be square-free and corime to all the other variables ɛ k, k i. Next, we emloy the following result, which generalizes Lemma 4 in [8]. Lemma 7. Lemma 4., [30] Suose s is a fixed rational number. Let N be sufficiently large. Then for arbitrary ositive integers q, r and any nonrincial character χ mod q, we have n X,gcdn,r= µ ns ωn χn Xτr ex η log X with a ositive constant η = η s,n, uniformly for q log N X. Here τ is the usual divisor function and µ is the Möbius function. To use this result we remove the condition ɛ i h mod from the inner sum on the right side of 3 and insert instead a factor 4 ψ mod ψɛ i ψh.

23 ON ELLIPTIC CURVES y = x 3 n 3 3 One has SA ɛ k,k i k 8 A i ex η log A i k i = A i ex η log A i k i ɛ k τɛ k A i ex η log A i k i A k log X τɛ k Xlog X 7 ex η log A i, rovided that ɛ j and 8 ɛ j log N A i for some N > 0. We summarize the above results as follows. Lemma 8. For any constant κ with 0 < κ < one has SA Xlog X 9 whenever there are linked variables ɛ i, ɛ j for which A i ex{log X κ } and ɛ j > Case Three. Put I = {7, 8}, I = {,, 3,, 3, 4, 5, 6}, and I 3 = {4, 5, 6, 7, 8, 9, 0, }. For κ > 0 to be a sufficiently small real number, let 4 C = ex {log X κ },

24 4 FENG AND XIONG For any ositive integers λ, λ, λ 3, we define #{i I : A i > C} = λ L λ, λ, λ 3 = A L : #{i I : A i > C} = λ #{i I 3 : A i > C} = λ 3. For each j =,, 3, write m j = ɛ i, n j = ɛ i <C i I ɛ i. ɛ i C i I j For A L λ, λ, λ 3, fix m j and n j, the number of ɛ i such that i I j,ɛ i <C ɛ i = m j is at most #I j ωm j, and the number of ɛ i such that i I j,ɛ i >C ɛ i = n j is at most #Ij ωn λ j λ j j. Hence SA A L λ,λ,λ 3 For j =,, 3, define ωm ωm 3ωm3 ωm+3ωm+3ωm3 m,m,m 3 C 8 ωn ωn 3ωn 3 λ ωn λ ωn λ ωn 3 3. n n n 3 X/m m m 3 where µ is the Möbius function, and Then f j n n X f j n = α ωn j µ n, α = λ, α = λ 4, α 3 = λ 3 8. X log X n X By Mertens estimate, this gives us 5 f j n n X X log X ex α ωn j µ n n X α j X log X X Xlog X α j. + α j.

25 ON ELLIPTIC CURVES y = x 3 n 3 5 We also have n n X f n f n n X Alying 5 we obtain that Similarly n n n 3 X Now we can comute SA A L λ,λ,λ 3 Using the bound n n X n X f n n X/n f n + n X f n f n Xlog X α +α. f n f n f n f 3 n 3 Xlog X α +α +α 3. m,m,m 3 C 8 ωm m,m,m 3 C 8 ωm γ ωn n X n n n 3 X/m m m 3 n X/n f n. f n f n f 3 n 3 X log X α +α +α 3. m m m 3 + γ log X γ, which is valid for any fixed γ > 0, we find that SA Xlog X α +α +α 3 +4κ, where A L λ,λ,λ 3 α = λ, α = λ 4, α 3 = λ 3 8. If λ + λ 4 + λ 3 8 <, we may choose κ > 0 sufficiently small so that λ + λ 4 + λ κ < 0. Then in this case the total error is bounded by O Xlog X /0.

26 6 FENG AND XIONG 5. The remaining cases The remaining cases which are not treated in Case one, Case two and Case three in Section 4 must satisfy the condition that We list all ossibilities as follows. λ = ; λ = 0;. A 3 > C or A 5 > C;.. A 4 > C or A 6 > C;.. A 4 C and A 6 C;. A 3 C and A 5 C; 3 λ = ;.. A 4 > C or A 6 > C;.. A 4 C and A 6 C; 3. A 8 > C and A 7 C; 3.. A 3 > C or A 5 > C; 3.. A 3 C and A 5 C; 3. A 7 > C and A 8 C; 3.. A 3 > C or A 5 > C; 3.. A 3 C and A 5 C; We deal with each case individually. λ + λ 4 + λ Case. If λ =, so that A 7 > C and A 8 > C. This imlies that A 5 = A 6 = A 7 = A 0 =. For such A, we can obtain SA ωɛ7 ωɛ8 ɛ 7,ɛ 8 ɛ i A i i 7,8 ɛ7 3.

27 ON ELLIPTIC CURVES y = x 3 n 3 7 The next lemma can be used to get a desired uer bound. Lemma 9. Let s, s be non-zero rational numbers. Let C be a ositive integer, and let A > 0 be any fixed number. For X >, let T ex log X and M, N T be given. There exists some constant η > 0 and B > 0 such that, for any ositive integer r, any integer h corime to C, and any distinct characters χ, χ mod q, where Cq log X A, we have m,n µ mµ ns ωm s ωn χ mχ n Xτr ex η log T log X B, where the sum is over corime variables satisfying the conditions M < m M, N < n N, mn X, mn h mod C, gcdmn, r =, and the constant involved in the -symbol deends on s, s and C only. We remark that Lemma 9 slightly generalizes Lemma.4 of [3], which roves the result when s = s is a ositive integer. Equied with Lemma 7, however, the roof of Lemma.4 of [3] can be adated easily to obtain Lemma 9. The roof of Lemma 9 also follows the same line as that of Lemma 0 in [8]. We omit the details here. By Lemma 9, we find that SA Xlog X 9. The total contribution from this case is bounded by O Xlog X. 5.. Case... This is the case that λ = 0, A 3 > C or A 5 > C, and A 4 > C or A 6 > C. The conditions imly that A 7, A 8 C and A = A = A 3 = A 4 = A 5 = A 6 = A 7 = A 8 = A 9 = A 0 = A = A =. Hence λ = λ 3 = 0.

28 8 FENG AND XIONG Since λ + λ 4 + λ 3 8, we must have λ = 4, which imlies that A 3, A 5 > C. However, in this case we have SA ɛ i A i i 3,5 3 i 8 ωɛ3 ωɛ5 ɛ 3,ɛ 5 ɛ3 By Lemma 9 again, SA is bounded by O Xlog X 9. So the total contribution from this case is O Xlog X Case... This is the case that λ = 0, A 3 > C or A 5 > C, and A 4 C and A 6 C. The conditions imly that A 7, A 8 C and A = A = A 3 = A 4 = A 8 = A 9 = A = A =. Hence λ = 0, λ and λ 3 4. Since λ + λ 4 + λ 3 8, we must have λ = and λ 3 = 4, which imlies that A 3, A 5 > C. By Lemma 9, similar to the argument of Case.., we find that SA is bounded by O Xlog X 9. So the total contribution from this case is O Xlog X Case... This is the case that λ = 0, A 3 C and A 5 C, and A 4 > C or A 6 > C. The conditions imly that A 7, A 8 C and A 5 = A 6 = A 7 = A 0 =. If A 4 > C and A 6 > C, by Lemma 9, similar to the argument of Case.., SA is bounded by O Xlog X 9. So we can assume that exactly one element of {A 4, A 6 } is larger than C. If at least two elements of {A, A, A 3, A } are larger than C, then A 3 = A 5 =, and by Lemma 9, similar to the argument of Case.., SA is bounded by O Xlog X 9. So we can assume that at most one element of {A, A, A 3, A } is larger than C. If at least two elements of {A 4, A 8, A 9, A } are larger than C, then A 3 = A 5 =, and by Lemma 9, similar to the argument of Case.., SA is bounded by

29 ON ELLIPTIC CURVES y = x 3 n 3 9 O Xlog X 9. So we can assume that at most one element of {A 4, A 8, A 9, A } is larger than C. We are left with λ = 0, λ, λ 3, and This has been dealt with in Case three. λ + λ 4 + λ 3 8 < Case... This is the case that λ = 0, A 3 C and A 5 C, and A 4 C and A 6 C. The conditions imly that A 7, A 8 C. If at least two elements of {A, A, A 3, A } are larger than C, then A 3 = A 5 =, and by Lemma 9, similar to the argument of Case.., SA is bounded by O Xlog X 9. So we can assume that at most one element of {A, A, A 3, A } is larger than C. Hence λ. Similarly, if at least two elements of {A 4, A 8, A 9, A } are larger than C, then A 3 = A 5 =, and by Lemma 9, SA is bounded by O Xlog X 9. So we can assume that at most one element of {A 4, A 8, A 9, A } is larger than C. Hence λ 3 5. We are left with λ = 0, λ, λ 3 5, and This has been dealt with in Case three. λ + λ 4 + λ 3 8 < Case 3... This is the case that A 8 > C, A 7 C, A 3 > C or A 5 > C. The conditions imly that A = A = A 3 = A 4 = A 5 = A 6 = A 7 = A 8 = A 9 = A 0 = A = A =. Hence λ =, λ 3 = 0. If A 3 > C, since A 8 > C, by Lemma 9, similar to the argument of Case.., SA is bounded by O Xlog X 9. So we can assume that A 3 C, and hence A 5 > C. If A 4 > C, since A 8 > C, by Lemma 9 again, SA is bounded by O Xlog X 9. So we can assume that A 4 C.

30 30 FENG AND XIONG If A 6 C, then λ =, and λ + λ 4 + λ , which has been dealt with in Case three. We are left with the case that A = A = A 3 = A 4 = A 5 = A 6 = A 7 = A 8 = A 9 = A 0 = A = A =, A 3, A 4, A 7 C and A 5, A 6, A 8 > C. Collecting this case together we define L = A L : For any subset B L, let A = A =... = A =, A 3, A 4, A 7 C, A 5, A 6, A 8 > C. SB = A B SA.. We obtain for L 6 SL = A L ωɛ 7ɛ 8 ωɛ 3 ɛ 4 ɛ 5 ɛ 6 ωɛ 7 ɛ3 ɛ 4 ɛ Case 3... This is the case that A 8 > C, A 7 C, A 3 C and A 5 C. The conditions imly that A 5 = A 6 = A 7 = A 0 =. Fist, since A 8 > C, by Lemma 9, using the same argument as in Case 3.., we may assume that A 4 C. If max{a, A, A 3, A 4, A 8, A 9, A, A } C, then λ =, λ and λ 3 = 0, and λ + λ 4 + λ , which has been dealt with in Case three. Now assume that max{a, A, A 3, A 4, A 8, A 9, A, A } > C. Then A 3 = A 5 =. Since A 8 > C, by Lemma 9 again, using the same argument as in Case 3.., we may assume that A, A, A 3, A 4, A 8, A 9 C.

31 ON ELLIPTIC CURVES y = x 3 n 3 3 If either A C or A 6 C, then λ =, λ and λ 3, and λ + λ 4 + λ , which has been dealt with in Case three. So we may assume that A, A 6 > C. Collecting this case which is left together we define A, A, A 3, A 4, A 8, A 9, A 4, A 7 C, L = A L : A 5 = A 6 = A 7 = A 0 =,. A, A 6, A 8 > C. We obtain 7 SL = A L ωɛ ɛ ɛ 3 ɛ ɛ 4 ɛ 6 3ωɛ 4 ɛ 8 ɛ 9 ɛ ωɛ 7 ɛ 8 ωɛ ɛ 3 ɛ 7 3 ɛ4 ɛ 7 ɛ ɛ 3 ɛ 4 ɛ 9 ɛ ɛ ɛ 4 ɛ Case 3... This is the case that A 7 > C, A 8 C, A 3 > C or A 5 > C. The conditions imly that A = A = A 3 = A 4 = A 5 = A 6 = A 7 = A 8 = A 9 = A 0 = A = A =. Hence λ =, λ 3 = 0. If A 5 > C or A 6 > C, since A 7 > C, by Lemma 9, similar to the argument of Case.., SA is bounded by O Xlog X 9. So we can assume that A 5 C and A 6 C. Hence A 3 > C. If A 4 C, then λ =, and λ + λ 4 + λ 3 8 = 3 4, which has been dealt with in Case three. So A 4 > C. We are left with the case that A = A = A 3 = A 4 = A 5 = A 6 = A 7 = A 8 = A 9 = A 0 = A = A =, A 3, A 4, A 7 > C and A 5, A 6, A 8 C. Collecting this case together we define L 3 = A L : A = A =... = A =, A 3, A 4, A 7 > C, A 5, A 6, A 8 C..

32 3 FENG AND XIONG We obtain 8 SL 3 = A L 3 ωɛ 7ɛ 8 ωɛ 3 ɛ 4 ɛ 5 ɛ 6 ωɛ 7 ɛ3 ɛ 4 ɛ Case 3... This is the case that A 7 > C, A 8 C, A 3 C and A 5 C. The conditions imly that A 5 = A 6 = A 7 = A 0 =. Fist, since A 7 > C, by Lemma 9, using the same argument as in Case 3.., we may assume that A 6 C. If max{a, A, A 3, A 4, A 8, A 9, A, A } C, then λ =, λ and λ 3 = 0, and λ + λ 4 + λ , which has been dealt with in Case three. Now assume that max{a, A, A 3, A 4, A 8, A 9, A, A } > C. Then A 3 = A 5 =. Since A 7 > C, by Lemma 9 again, using the same argument as in Case 3.., we may assume that A, A 3, A 8, A 9, A, A C. If either A C or A 4 C, then λ =, λ and λ 3, and λ + λ 4 + λ , which has been dealt with in Case three. So we may assume that A, A 4 > C. Collecting this case which is left together we define A, A 3, A 8, A 9, A, A, A 6, A 8 C, L 4 = A L : A 5 = A 6 = A 7 = A 0 =,. A, A 4, A 7 > C. We obtain 9 SL 4 = A L 4 ωɛ ɛ ɛ 3 ɛ ɛ 4 ɛ 6 3ωɛ 4 ɛ 8 ɛ 9 ɛ ωɛ 7 ɛ 8 ωɛ ɛ 3 ɛ 7 3 ɛ4 ɛ 7 ɛ ɛ 3 ɛ 4 ɛ 9 ɛ ɛ ɛ 4 ɛ 8 3.

33 ON ELLIPTIC CURVES y = x 3 n Conclusions Combining all results in Sections 4 and 5, we conclude that F, n = SL + SL + SL 3 + SL 4 + O Xlog X /0. We can also conclude by similar analysis that F, n = SL + SL + SL 3 + SL 4 + O Xlog X /0, where and L = L 4 = A L : A L : A, A 3, A 8, A 9, A, A, A 4, A 7 C, A 5 = A 6 = A 7 = A 0 =, A, A 6, A 8 > C. A, A, A 3, A 4, A 8, A 9, A 6, A 8 C, A 5 = A 6 = A 7 = A 0 =, A, A 4, A 7 > C. As for F n, we can also exand the roduct to obtain F n = F,n + F, n,,. where F, n = F, n = g, ɛ, n=ɛ ɛ ɛ 8 g, ɛ, n=ɛ ɛ ɛ 8 g, ɛ = ωɛ ωɛ 3ωɛ 3 ωɛ ɛ 3 ɛ 7 3 ɛ3 ɛ 4 ɛ 7 3 ɛ ɛ 3 ɛ 4 ɛ 5 ɛ 9 ɛ 0 ɛ 3 ɛ 5 ɛ ɛ ɛ 4 ɛ 5 ɛ 7 ɛ 8 3 ɛ 5 ɛ 6 ɛ 7 ɛ 0 4 ɛ4 ɛ 6 ɛ 7 ɛ 8 ɛ ɛ ɛ 3 ɛ 4 ɛ 8 ɛ 9 ɛ ɛ ɛ3 ɛ 5 ɛ5 ɛ 6 ɛ 7 ɛ 0, ɛ 5 ɛ 6 ɛ 7 ɛ 0 ɛ 3 ɛ 5 ɛ 5 ɛ 6 ɛ 7 ɛ 0 ɛ 3 ɛ 5

34 34 FENG AND XIONG and g, ɛ = ωɛ ωɛ 3ωɛ 3 ωɛ ɛ 3 ɛ 7 ɛ ɛ 3 3 ɛ3 ɛ 4 ɛ 7 3 3ɛ ɛ 3 ɛ 4 ɛ 5 ɛ 9 ɛ 0 ɛ 3 ɛ 5 ɛ ɛ ɛ 4 ɛ 5 ɛ 7 ɛ 8 3 ɛ 5 ɛ 6 ɛ 7 ɛ 0 ɛ4 ɛ 6 ɛ 7 ɛ 8 ɛ ɛ ɛ 3 ɛ 4 ɛ 8 ɛ 9 ɛ ɛ ɛ3 ɛ 5 ɛ5 ɛ 6 ɛ 7 ɛ 0. ɛ 5 ɛ 6 ɛ 7 ɛ 0 ɛ 3 ɛ 5 ɛ 5 ɛ 6 ɛ 7 ɛ 0 ɛ 3 ɛ 5 Here the sums in F, n and F, n are over all ositive integers ɛ, ɛ,..., ɛ 8 such that n = ɛ ɛ ɛ 8, ɛ = ɛ 7 ɛ 8, ɛ = ɛ ɛ ɛ 3 ɛ ɛ 3 ɛ 4 ɛ 5 ɛ 6, ɛ 3 = ɛ 4 ɛ 5 ɛ 6 ɛ 7 ɛ 8 ɛ 9 ɛ 0 ɛ, and the function ω counts the number of distinct rime factors. The functions g, ɛ and g, ɛ are very similar to g, ɛ and g, ɛ and can be treated in the same way. By careful analysis and a little bit of extra work we also obtain that and Recall that we obtain F, n = SL + SL 4 + O Xlog X /0, F, n = SL SL 4 + O Xlog X /0. sn = F, n + F, n + F, n + F, n, sn = SL + SL + SL 3 + SL 4 + O Xlog X /0. Our goal now is to identify the main terms. 4

35 ON ELLIPTIC CURVES y = x 3 n Let HX = SL + SL 3. Since L L 3 = A L : A =... = A =, A 3, A 4, A 7 C, A 5, A 6, A 8 > C; or A 3, A 4, A 7 > C, A 5, A 6, A 8 C.. For A L L 3, the summand can be simlified as ωɛ 7ɛ 8 ωɛ 3 ɛ 4 ɛ 5 ɛ 6 ωɛ 7 ɛ3 ɛ 4 ɛ 7. 3 Hence HX = A L L 3 ωɛ7ɛ8 ωɛ3ɛ4ɛ5ɛ6 ωɛ7 ɛ3 ɛ 4 ɛ 7. 3 Removing the constraints on A 3,..., A 8, we consider the set H = {A L : A =... = A = }. It is easy to see, following the argument in the revious two sections that We can rewrite it as SH = SH = HX + O Xlog X /0. ɛ 3 ɛ 4 ɛ 5 ɛ 6 ɛ 7 ɛ 8 =n ωɛ 7ɛ 8 ωɛ 3 ɛ 4 ɛ 5 ɛ 6 ωɛ 7 ɛ3 ɛ 4 ɛ 7. 3 Since the function in the inner sum is multilicative, we can comute directly that Therefore SH = n = = #SX, h. HX = #SX, h + O Xlog X /0.

36 36 FENG AND XIONG 6.. On the other hand, let Then Here and GX = SL + SL 4. GX = A L L 4 ωɛ ɛ ɛ 3 ɛ ɛ 4 ɛ 6 3ωɛ 4 ɛ 8 ɛ 9 ɛ ωɛ 7 ɛ 8 ωɛ ɛ 3 ɛ 7 3 ɛ4 ɛ 7 ɛ ɛ 3 ɛ 4 ɛ 9 ɛ ɛ ɛ 4 ɛ 8 3. L = A L : L 4 = A L : A, A, A 3, A 4, A 8, A 9, A 4, A 7 C, A 5 = A 6 = A 7 = A 0 = A 3 = A 5 =, A, A 6, A 8 > C. A, A 3, A 8, A 9, A, A, A 6, A 8 C, A 5 = A 6 = A 7 = A 0 = A 3 = A 5 =, A, A 4, A 7 > C. Removing the constraints on A, A, A 3, A 4, A 8, A 9, A, A, A 4, A 6, A 7, A 8, we consider the set H = {A L : A 5 = A 6 = A 7 = A 0 = A 3 = A 5 = }.,. It is easy to see, following the argument in the revious two sections that Moreover, we can rewrite it as SH = SH = GX + O Xlog X /0. ɛ ɛ ɛ 3 ɛ 4 ɛ 8 ɛ 9 ɛ ɛ ɛ 4 ɛ 6 ɛ 7 ɛ 8 =n ωɛ ɛ ɛ 3 ɛ ɛ 4 ɛ 6 3ωɛ 4ɛ 8 ɛ 9 ɛ ωɛ 7 ɛ 8 ωɛ ɛ 3 ɛ 7 3 ɛ ɛ 3 ɛ 4 ɛ 9 ɛ ɛ ɛ 4 ɛ 8 ɛ4 ɛ 7. 3 Since the function in the inner sum is multilicative, we can comute directly that SH = ω n ω 7 n.

37 ON ELLIPTIC CURVES y = x 3 n 3 37 Here for any integer a with gcda, =, we define Therefore GX = We finally conclude that For each n, let ω a n = sn = #SX, h + n a mod. ω n ω 7 n + O Xlog X /0. ω n ω 7 n + O Xlog X /0. 7. Distribution of sn gn = ω n ω 7 n, λn = max{0, gn}. To find the distribution of sn for n SX, h as X, we first write the asymtotic equation in the form 0 sn = gn + O X. It is known from that sn λn gn. We shall study the distribution of the right hand side first. 7.. Moments of λn. We first aly Lemma for the strongly additive function gn, and A = SX, h, P is the set of rimes not exceeding X. From the roof of [9, Lemma 4] we find that #A = 3 4π X + O X ex η log X,

38 38 FENG AND XIONG and for a square-free d with d, =, if A d = #{n A : d n}, then A d #A = d + d + O τd ex η log X, where η is a sufficiently small ositive constant, τ is the divisor function. Hence in alication of Lemma, hd = d +, and Since we can choose M =. We obtain µ P g = X r d τd d + X ex η log X. d mod, g = 7 mod, 0 7, mod, g h = mod X This in turn, by Merten s estimate, gives us On the other hand, σ P g = X = g h mod X By Mertens estimate, we obtain µ P g = O. h mod X σ P g = log log X + O. 7 mod X

39 ON ELLIPTIC CURVES y = x 3 n 3 39 For a fixed ositive integer k, it is also easy to see that r d X ex η log X µ dτd X ex η log X. d d D k P d X k Then by Lemma, for any fixed even integer k, we have #SX, h k/ gn O k = C k log log X + O +O log log X k, and for any fixed odd integer k, we have #SX, h gn O k C k k 3/ log log X + O k /. We can conclude that for any fixed ositive integer k, #SX, h k/ gn k = C k δ log log X k/ + O log log X /, where for any α R, δ α = if α Z, and δ α = 0 if α Z. gn Noticing that, all the moments of, as n varies in the set SX, h with log log X X, are the same as the moments of a standard Gaussian distribution, where the odd moments vanish and the even moments are x r e x / dx = C r, π gn this imlies that asymtotically is a standard Gaussian distribution. log log X Let X be a standard Gaussian distribution and Y = max{0, X}. We can comute easily that for any ositive integer k, EY k = λ k, where λ k s are the constants defined in 5. Since λn = max{0, gn}, λn Y, log log X we find that for any ositive integer k, as X, #SX, h λn k = λ k + o k/ log log X.

40 40 FENG AND XIONG 7.. Moments of sn. For each n SX, h, write sn = λn + δn, then δn Z and δn 0. Since λn gn, from 0 we obtain X λn+δn λn sn gn. Since we obtain λn δn δn δn λn+δn X. λn+δn, For each ositive integer r, ut a r = # {n SX, h : λn + δn = r, δn }. The asymtotic formula can be written as r a r X. r N Hence a r X. r Then for any fixed ositive integer k, δn λn + δn k = r k a r r= r= r k X r X. For such a fixed ositive integer k, we exand #SX, h sn k = #SX, h λn + δn k = #SX, h λn k +E,

41 ON ELLIPTIC CURVES y = x 3 n 3 4 where E = The right hand side is #SX, h δn k r=0 Hence we can bound E by #SX, h k r=0 k λn r δn k r r 0 E #SX, h δn k λn r δn k r. r #SX, h δn k r=0 λn + δn k. We deduce from that for any fixed ositive integer k, #SX, h sn k = λ k + o sn k/ log log X. k λn r δn k r. r Noticing that, all the moments of, as n varies in the set SX, h with log log X X, are the same as the moments of the random variable Y, this imlies that asymtotically sn Y. log log X This comletes the roof of Theorem for sn. 8. Distribution of tn 8.. Stoll s formula. Chang [6] has used Stoll s formula [6] to obtain uer bound for average rank of quadratic twists of the ellitic curve y = x 3 A for any integer A under some congruence conditions. For the secial ellitic curve E n : y = x 3 n 3, we use [7], which imroves uon [6] by Stoll himself, in order to treat as many E n as ossible. We follow Chang s ideas and focus on E n with n ositive, square-free and n, 5 mod. Interested readers may refer to [6] and [7] for general setting and details.

42 4 FENG AND XIONG Let ζ Q be a rimitive third root of unity and K = Qζ. The ellitic curve E n : y = x 3 n 3 can be considered as an ellitic curve over K. We denote simly by ζ the endomorhism on E given by x, y ζx, y which is defined over K. Denote by λ this endomorhism ζ. Via Galois cohomology, we have the short exact sequence 0 E nk λe nk Sel λ E n /K XE n /K[λ] 0. If n is square-free, then E n K ranke n Q rank F3 λe n K rank F 3 Sel λ E n /K. For n, 5 mod and n > 0, Stoll s formula Corollary. and Theorem. of [7] states that 3 rank F3 Sel λ E n /K = dim F3 Cl Q n [3], where Cl Q n [3] denotes the 3-torsion art of the class grou of the number field Q n, whose discriminant equals 4n. Let h 3 denote #ClF [3], where F is the quadratic extension of Q with discriminant. For ositive integers N and m, denote by N X, m, N the set of fundamental discriminants such that X < < 0 and m mod N. For h =, 5, recall the following secial case of a general theorem roved by Nakagawa and Horie [4] 4 lim X #N 4X, 4h, 48 N 4X, 4h,48 h 3 =. The set N 4X, 4h, 48 can be identified with SX, h by the maing 4n n. Define #Sel λ E n /K = 3 s n.

43 ON ELLIPTIC CURVES y = x 3 n 3 43 The equations 3 and 4 can be restated as lim X #SX, h 3 s n =. Let ranke n Q = rn. Since rn s n, from above we obtain 5 3 rn X. 8.. Moments of tn. Recall that φ : E n E n is a -isogeny and ˆφ : E n E n is the dual -isogeny. Hence ˆφ φ = [ ], and one has the following commutative diagrams see 97, []: E n Q φe nq Sel φ E n /Q XE n /Q[φ] 0 0 E nq E nq Sel E n /Q XE n /Q[] 0 0 E nq ˆφE n Q Sel ˆφ E n/q XE n/q[ ˆφ] 0 0 Ĉ Ĉ Define #Sel φ E n /Q = sn+, #Sel ˆφ E n/q = ŝn+, #XE n /Q[φ] = tn, #XE n/q[ ˆφ] = ˆtn. From the commutative diagrams, we have rn = sn tn + ŝn ˆtn,

44 44 FENG AND XIONG where rn is the rank of E n over Q. Let Using tn = sn r n, we exand #SX, h r n = sn tn. tn k = #SX, h sn k + E, where Since E = #SX, h k i=0 0 r n rn, k i sn i r n k i. i for each fixed ositive integer k, we deduce from 5 that Recalling we find that E max 0 i k #SX, h #SX, h #SX, h r n k #SX, h rn k. k/ sn k = λ k + o log log X, sn i / #SX, h rn k i / log log X k. We conclude that as X, #SX, h k/ tn k = λ k + o log log X.

45 ON ELLIPTIC CURVES y = x 3 n 3 45 Noticing again that, all the moments of, as n varies in the set SX, h log log X with X, are the same as the moments of the random variable Y described at tn the end of Section 7, this imlies that asymtotically tn Y. log log X This comletes the roof of Theorem for tn. 9. Distribution of ŝn and ˆtn 9.. ŝn and ˆtn. Let n be a square-free ositive integer such that gcdn, 6 =. From Lemma 4, we have where d { + 4 F n = 3 n=dd + Exanding the roduct, we obtain ŝn = F n, d { 3 4 d F n = F n + F n, d } 3 } d. where Here F n = g ɛ, F n = n=ɛ ɛ ɛ 8 n=ɛ ɛ ɛ 8 g ɛ. g ɛ = ωɛ ωɛ 3ωɛ 3 ωɛ ɛ 3 ɛ ɛ ɛ 3 ɛ 5 ɛ 6 ɛ 8 ɛ 9 ɛ ɛ ɛ 4 ɛ 5 ɛ 7 ɛ 8 ɛ 3 ɛ 4 ɛ 7 ɛ 5 ɛ 6 ɛ 7 ɛ 0 ɛ3 ɛ 4 ɛ 5 ɛ 6 ɛ 7 ɛ 8 ɛ ɛ... ɛ, ɛ 5 ɛ 6 ɛ 7 ɛ 0 ɛ 3 ɛ 5 4

46 46 FENG AND XIONG and g ɛ = ωɛ ωɛ 3ωɛ 3 ωɛ ɛ 3 ɛ 7 ɛ ɛ... ɛ ɛ ɛ 3 ɛ 5 ɛ 6 ɛ 8 ɛ 9 ɛ ɛ ɛ 4 ɛ 5 ɛ 7 ɛ 8 ɛ 3 ɛ 4 ɛ 7 ɛ 5 ɛ 6 ɛ 7 ɛ 0 ɛ3 ɛ 4 ɛ 5 ɛ 6 ɛ 7 ɛ 8 ɛ ɛ... ɛ. ɛ 5 ɛ 6 ɛ 7 ɛ 0 ɛ 3 ɛ 5 The sums in F n and F n are over all ositive integers ɛ, ɛ,..., ɛ 8 such that n = ɛ ɛ ɛ 8, ɛ = ɛ 7 ɛ 8, ɛ = ɛ ɛ ɛ 3 ɛ ɛ 3 ɛ 4 ɛ 5 ɛ 6, ɛ 3 = ɛ 4 ɛ 5 ɛ 6 ɛ 7 ɛ 8 ɛ 9 ɛ 0 ɛ, and the function ω counts the number of distinct rime factors. We use similar techniques to estimate ŝn = F n, that is, we sum over the 8 variables ɛ i, subject to the conditions that each ɛ i is square-free, ositive, that they are relatively rime in airs, and that their roduct n satisfies n X, n h mod, where h = or 5, and we divide the range of each variable ɛ i into intervals [A i, A i, where A i runs over owers of. Following the stes in Section 4 6 carefully we can obtain Define ŝn = #SX, h + It is known from 9 that g n = ω 7 n ω n, 4 ω 7n ω n + O Xlog X /0. ŝn λ n g n. λ n = max{0, g n}.

47 ON ELLIPTIC CURVES y = x 3 n 3 47 Following the argument in Section 7 in the same way, we find that for any fixed ositive integer k, #SX, h and hence asymtotically as X, k/ ŝn k = λ k + o log log X, ŝn max {0, Y }, log log X where Y is a standard Gaussian distribution. This comletes the roof of Theorem for ŝn. As for ˆtn, define r n = ŝn ˆtn. From the commutative diagrams in Section 8, we have 0 r n rn = ranke n Q. Following the argument of Section 8 we obtain that for any ositive integer k, k/ ˆtn k = λ k + o log log X. #SX, h This comletes the roof of Theorem for ˆtn. 9.. Proof of Theorem. It remains to rove the asymtotic formula 4 of Theorem. We summarize our results as follows. Let λn = max {0, ω n ω 7 n}, λ n = max {0, ω 7 n ω n}. sn = λn + δn = r n + tn, ŝn = λ n + δ n = r n + ˆtn, where for each n, δn, r n, δ n, r n 0.

48 48 FENG AND XIONG We have for any fixed ositive integers a, b, c, d, by Cauchy-Schwartz inequality, 6 #SX, h δn a δ n b r n c r n d. For any fixed ositive integer k, we also have #SX, h k/ λn k = λ k + o log log X, Since #SX, h k/ λ n k = λ k + o log log X. λn λ n = 0, n, we have 7 #SX, h k/ λn + λ n k = λ k + o log log X. Now 4 can be roved easily: since tn + ˆtn = λn + λ n + δn + δ n r n r n, for any fixed ositive integer k, exanding the k-th ower, we obtain where E = #SX, h #SX, h tn + ˆtn k = #SX, h k i=0 We deduce from 6 and 7 that λn + λ n k + E, k λn + λ n i δn + δ n r n r n k i. i E log log X k.

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