on selmer groups and tate-shafarevich groups for …mamsxiong/fx2.pdf · lemma2(proposition4,[16])....

ON SELMER GROUPS AND TATE-SHAFAREVICH GROUPS FORELLIPTIC CURVES y2 = x3 − n3

KEQIN FENG AND MAOSHENG XIONG?

Abstract. We study the distribution of the size of Selmer groups and Tate-

Shafarevich groups arising from a 2-isogeny and its dual 2-isogeny for elliptic

curves En : y2 = x3 − n3. We show that the 2-ranks of these groups all follow the

same distribution. The result also implies that the mean value of the 2-rank of the

corresponding Tate-Shafarevich groups for square-free positive integers n ≤ X is√12 log log X as X → ∞. This is quite different from quadratic twists of elliptic

curves with full 2-torsion points over Q ([29]), where one Tate-Shafarevich group

is almost always trivial while the other is much larger.

1. Introduction

In [29] one of the authors has studied asymptotic behavior of the size of Selmer

groups arising from a 2-isogeny and its dual 2-isogeny for quadratic twists of the

elliptic curve Ea,b : y2 = x(x+a)(x+ b), where a, b are any integers satisfying ab(a−

b) 6= 0 and that ab is not a square. This result, combined with a result obtained by

Yu [32], implies explicit information of the corresponding Tate-Shafarevich groups.

Interested readers may refer to [29] for details.

The curve Ea,b can be characterized as elliptic curves with full 2-torsion points

over Q. It would be interesting to do a similar analysis for other family of elliptic

curves, for example for elliptic curves with one non-trivial 2-torsion point over Q.

It turns out, however, to be quite difficult in general. In this paper we focus on

2000 Mathematics Subject Classification. 11G05, 14H52, 11L40, 11N45.

Key words and phrases. Elliptic curves, Selmer group, Tate-Shafarevich group,distribution.

The author is supported by RGC grant number 606211 and DAG11SC02 from Hong Kong.

1

2 FENG AND XIONG

the most interesting case, the elliptic curve E : y2 = x3 − 1 and consider, for a

square-free integer n, the quadratic twist En given by the equation

En : y2 = x3 − n3.(1)

En hase complex multiplication√−3, it various properties have been studied exten-

sively in the literature.

Corresponding to the 2-torsion point (n, 0) there is a 2-isogeny φ : En −→ E ′n,

where E ′n is another elliptic curve

E ′n : Y 2 = X3 − 6nX2 − 3n2X,

and the 2-isogeny φ is given by ([25, pp. 74])

φ(x, y) =

(y2

(x− n)2,y(3n2 − (x− n)2)

(x− n)2

).

Via Galois cohomology we obtain the short exact sequence

0 // E′n(Q)φ(En(Q))

// Sel(φ)(En/Q) // X(En/Q)[φ] // 0,

where Sel(φ)(En/Q) is the φ−Selmer group and X(En/Q)[φ] is the φ-torsion of the

Tate-Shafarevich group. Denote by φ the dual isogeny of φ. We also have the short

exact sequence

0 // En(Q)

φ(E′n(Q))// Sel(φ)(E ′n/Q) // X(E ′n/Q)[φ] // 0.

For X > 0 and integer h, define

(2) S(X, h) = {1 ≤ n ≤ X : n ≡ h (mod 12) and n is square-free}.

We prove the following result.

Theorem 1. Let En be the elliptic curve given in (1). Put

#Sel(φ)(En/Q) = 2s(n)+1, #Sel(φ)(E ′n/Q) = 2s(n)+1,

#X(En/Q)[φ] = 2t(n), #X(E ′n/Q)[φ] = 2t(n),

ON ELLIPTIC CURVES y2 = x3 − n3 3

where the quantities s(n), s(n), t(n), t(n) are called the “2-rank” of those groups.

For h = 1 or 5 and X → ∞, as n varies in S(X, h), the value γ(n) = s(n) (or

s(n), t(n), t(n) respectively) follows the same distribution. More precisely, for any

a > 0, we have

limX→∞

1

#S(X, h)#

n ∈ S(X, h) :γ(n)√

12

log log n≤ a

=1√2π

∫ a

−∞e−

t2

2 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 explain in

the beginning of Section 4. Secondly, Theorem 1 shows that the “normalized” 2-

rank of Sel(φ)(En/Q) and Sel(φ)(E ′n/Q), i.e. the quantity γ(n)/√

12

log log n behaves

asymptotically like the random variable max{0, Y }, where Y is a standard Gaussian

distribution. This distribution is different from that obtained in [29], where one

Selmer group is much smaller, and the other is much larger. It is also different

from results obtained in [18, 19, 32], where on average the 2-Selmer groups have

intermediate size. Thirdly, for the sake of simplicity we only consider square-free

integers modulo 12, there is no essential difficulty for general moduli under some

congruence constraints as in [29]. Finally, the distribution of Selmer groups is with

respect to the set S(X, h) as X →∞. It would be interesting to study the statistics

with the number of prime factors of n fixed, as was done by Swinnerton-Dyer for

2-Selmer groups in [28].

Theorem 2. Under the assumptions of Theorem 1, define

#X(En/Q)[φ] = 2t(n), #X(E ′n/Q)[φ] = 2t(n).

Let γ(n) = t(n) (or t(n)). Then for any integer k > 0, we have

1

#S(X, h)

∑n∈S(X,h)

γ(n)k ∼ λk ·(

log logX

2

)k/2,(3)

4 FENG AND XIONG

and

1

#S(X, h)

∑n∈S(X,h)

(t(n) + t(n)

)k ∼ 2λk ·(

log logX

2

)k/2,(4)

as X →∞, where

λk =

2−1− k2 k!/(k/2)! : k is even,

2k−12

(k−12

)!/√

2π : k is odd.(5)

Equation (3) can be derived from Theorem 1 by computing the k-th moment of the

random variable max{0, X} (see the remark after Theorem 1). Taking k = 1, we see

that the mean value of t(n) and t(n) is√

12

log logX. This behavior is again different

from [29], where one Tate-Shafarevich group is almost always trivial, and the other

has a much larger mean value 12

log logX. Since X(En/Q)[φ] ⊂ X(En/Q)[2] and

X(E ′n/Q)[φ] ⊂X(E ′n/Q)[2], equation (3) shows that the 2-part of X(En/Q) and

X(E ′n/Q) can be arbitrarily large. Finally, equation (4) implies a strong correlation

between t(n) and t(n), namely, t(n) and t(n) can not be large at the same time, and

this phenomenon seems independent of Theorem 1.

We also remark that while Selmer groups are relatively easy to handle, the Tate-

Shafarevich groups are more mysterious. They appear in the Birch and Swinnerton-

Dyer conjecture, and measure the degree of deviation from the Hasse principle. Even

the finiteness of such groups are not known in general. Various families of elliptic

curves with large Tate-Shafarevich groups were identified by a number of authors

(see [2],[3],[4],[5],[20],[21],[22],[23]). Moments ([8]), heuristic results ([7]), and upper

bounds ([14],[15]) on the order of Tate-Shafarevitch Groups were also considered.

There are three main ingredients in the proofs of the above results. First, to com-

pute the Selmer groups, we use a graphical method, which plays an essential role

in identifying the main contribution and reducing the complexity of the problem.

Second, we employ Heath-Brown’s method based on character sums to obtain as-

ymptotic formulas on the size of Selmer groups. Third, combining our results with


results obtained by Chang [6] and Stoll [27], we obtain information on the corre-

sponding Tate-Shafarevich groups. These ingredients are similar to those in [29],

however, the techniques involved here are more complicated. The main reason is

that the graphs constructed for the purpose of computing Selmer groups are more

complicated, which result in a character sum with 18 variables. By comparison, the

character sum appearing in [29] has only 6 variables. Moreover, for each Selmer

group, one needs to deal with 4 such character sums, comparing with a single one in

[29]. This increases the complexity of the problem. As it turns out, all the 4 char-

acter sums need to be combined together at the end so that extra savings can be

exploited by combinatorial argument. Finally, the result is more difficult to analyze

than that in [29]. Here we rely on the powerful machinery developed by Granville

and Soundararajan [16] on the Erdős-Kac theorem to obtain the distribution.

Acknowledgement The authors want to express their gratitude to the referee

for careful reading and many useful suggestions.

2. Preliminaries

2.1. 2-descent and Selmer groups. The 2-descent method is explained in the

last chapter of Silverman’s book ([25]). For the particular curve En in this paper, it

can be specified as follows (see also [3]).

For a square-free positive integer n, define a finite set S of prime divisors of the

rational number field Q by

S = {∞}⋃{p : p|6n}.

Let M be the multiplicative subgroup of Q∗/(Q∗)2 generated by −1 and the prime

divisors of 6n. For each d ∈ M we have the homogeneous space Cd (respectively

C ′d) defined by

Cd : dw2 = d2t4 − 6ndt2z2 − 3n2z4 ,

6 FENG AND XIONG

C ′d : dw2 = d2t4 + 3dnt2z2 + 3n2z4 .

The Selmer group Sel(φ)(En/Q) (respectively Sel(φ)(E ′n/Q)) measures the possibility

of Cd (or C ′d) having non-trivial solutions in the local field Qv for all v ∈ S. Namely,

Sel(φ)(En/Q) ∼= {d ∈M : Cd(Qv) 6= ∅ for all v ∈ S} ,

Sel(φ)(E ′n/Q) ∼= {d ∈M : C ′d(Qv) 6= ∅ for all v ∈ S} ,

where Cd(Qv) 6= ∅ ( or C ′d(Qv) 6= ∅) means that the homogeneous space Cd (or C ′d)

has non-trivial solutions (w, t, z) 6= (0, 0, 0) in Qv. We have

{1,−3} ⊆ Sel(φ)(En/Q), {1, 3} ⊆ Sel(φ)(E ′n/Q),

since each of the homogeneous spaces C1, C−3, C′1, C

′3 has non-trivial solution in Q.

For the rank of the elliptic curve En we obtain the formula (see pp 286, [3])

rank(En(Q)) = dimF2Sel(φ)(En/Q)− dimF2X(En/Q)[φ]

+ dimF2Sel(φ)(E ′n/Q)− dimF2X(E ′n/Q)[φ]− 2.

Thus we can calculate the rank from the dimensions of the Selmer groups and the

Tate-Shafarevich groups.

2.2. Even partitions. We use standard terminology in graph theory ([17]). Let

G = (V,A) be a simple directed graph where V = V (G) = {v1, · · · , vm} is the set

of vertices, A = A(G) is the set of arcs. The adjacency matrix of G is defined by

M(G) = (aij)16i,j6m ,

where

aij =

1, if (vi, vj) ∈ A (1 6 i 6= j 6 m),

0, otherwise .

For the vertex vi, let di =∑m

j=1 aij . The Laplace matrix of the graph G is defined

by

L(G) = diag(d1, · · · , dm)−M(G) .


Definition 1. Let G = (V,A) be a directed graph. A partition of vertices V1∪V2 = V

is called even if for any v1 ∈ V1, #{v1 → V2}, the total number of arcs from v1 to

vertices in V2 is even, and for any v2 ∈ V2, #{v2 → V1} is also even.

Lemma 1. Let G = (V,A) be a directed graph, V = {v1, . . . , vs+t} (s, t ≥ 0) . Then

the number of even partition {V1, V2} of V such that {vs+1, . . . , vs+t} ⊂ V2 is equal to

the number of vectors (x1, . . . , xs) ∈ Fs2 such that L(G) · (x1, . . . , xs, 0, . . . , 0)T = 0.

Lemma 1 in various forms has been used by Xue and the authors in the study of

new families of non-congruent numbers ([11],[12],[13]). It was also used by Faulkner

and James in [9]. The proof of Lemma 1 is elementary. For example, it can be

derived easily by adapting the proof of Lemma 2.2 in [11].

2.3. A theorem of Granville and Soundararajan. We use a result of Granville

and Soundararajan [16], which provides a most transparent account of the Erdös-Kac

theorem. Recall that g is an additive function if g(1) = 0, and g(mn) = g(m)+g(n)

whenever gcd(m,n) = 1. If in addition g(pk) = g(p) for all k ≥ 1 where p is a prime,

we say the function g is “strongly additive”.

Let A = {a1, . . . , ax} be a (multi)-set of x (not necessarily distinct) natural num-

bers. Let Ad = #{n ≤ x : d|an}. We suppose that there is a real valued, non-

negative multiplicative function h(d) such that for square-free d we may write

Ad =h(d)

dx+ rd.

Lemma 2 (Proposition 4, [16]). Let A be a (multi)-set of x integers, and let h(d)

and rd be as above. Let P be a set of primes, and let g be a real-valued, strongly

additive function with |g(p)| ≤M for all p ∈ P. Let

µP(g) =∑p∈P

g(p)h(p)

p, σ2

P(g) =∑p∈P

g(p)2 h(p)

p

(1− h(p)

p

).

8 FENG AND XIONG

Then, uniformly for all even natural numbers k ≤ (σP(g)/M)2/3,

∑a∈A

∑p|ap∈P

g(p)− µP(g)

k

= CkxσP(g)k + E1,

where

E1 � CkxσP(g)k−2k3M2 +Mk

(∑p∈P

h(p)

p

)k ∑d∈Dk(P)

|rd|,

and for all odd natural numbers k ≤ (σP(g)/M)2/3,

∑a∈A

∑p|ap∈P

g(p)− µP(g)

k

≤ E2,

where

E2 � CkxσP(g)k−1k3/2M +Mk

(∑p∈P

h(p)

p

)k ∑d∈Dk(P)

|rd|.

Here for any natural number k,

Ck =Γ(k + 1)

2k/2Γ(k/2 + 1),

and Dk(P) denotes the set of square-free integers which are the product of at most

k primes all from the set P.

3. Explicit formulas of the Selmer groups

3.1. Solvability conditions. The problem of finding the size of Selmer groups

Sel(φ)(En/Q) (and Sel(φ)(E ′n/Q)) is equivalent to the problem of determining how

many homogeneous spaces Cd (respectively C ′d) have non-trivial solutions over cer-

tain local fields. We collect solvability conditions for Cd and C ′d in the following two

lemmas.


Lemma 3. Let n be a square-free integer with gcd(n, 6) = 1, and M ⊆ Q∗/Q∗2, the

multiplicative subgroup generated by −1 and the prime divisors of 6n. Let p denote

an odd prime. For any d ∈M , we have

(1). 2|d =⇒ Cd(Q2) = ∅

(2). For p|d we have

Cd(Qp) = ∅ ⇐⇒

(3p) = −1; or

(3p) = (−1

p) = −(−3

p)4(

n/dp

) = 1.

(3). If 3 - d, then Cd(Q3) = ∅ ⇐⇒ d ≡ 2 (mod 3).

(4). If p|nd, then Cd(Qp) = ∅ ⇐⇒ p ≡ 1 (mod 3) and (d

p) = −1.

Proof. (1). For a prime p, we denote by vp the standard p-adic exponential val-

uation. Let (w, t, z) 6= (0, 0, 0) be a solution of Cd in Q2. From the equation of

Cd and 2‖d we know that at least two numbers in {1 + 2v2(w), 4v2(t), 2v2(t) +

2v2(z), 4v2(z) − 2} reach the minimal value, which is denoted by min. It must be

2v2(t) + 2v2(z) = min since the other three numbers are distinct modulo 4. Then

we must have

2(2v2(t) + 2v2(z)) ≤ (4v2(t)) + (4v2(z)− 2),

which is a contradiction. Therefore in this case Cd(Q2) = ∅.

(2). “⇐=”: Let (w, t, z) 6= (0, 0, 0) be a solution of Cd in Qp. We may assume

that (w, t, z) ∈ Z3p and 0 = min{1 + 2vp(w), 4vp(t), 2vp(t) + 2vp(z), 4vp(z)}. It is

easy to see that 4vp(t) = 2vp(t) + 2vp(z) = 4vp(z) < 1 + 2vp(w) so that 0 = vp(t) =

vp(z) ≤ vp(w). Thus

0 ≡ t4 − 6n

dt2z2 − 3(

n

d)2z4 =

(t2 − 3

n

dz2)2

− 12(ndz2)2

(mod p).

Therefore(

3p

)= −1 =⇒ Cd(Qp) = ∅. Suppose that

(3p

)=(−1p

)= 1, then

√3 ∈ Qp and

(t2 − (3 + 2

√3)n

dz2) (t2 − (3− 2

√3)n

dz2)≡ 0 (mod p), which means

that

10 FENG AND XIONG

(A). t2 ≡(3 + 2

√3)ndz2 (mod p),

so that

1 =

((3 + 2

√3)n

d

p

)=

((1 +

√3)22√

3

p

)(n/d

p

)=

(−3

p

)4

(n/d

p

);

or

(B). t2 ≡(3− 2

√3)ndz2 (mod p),

so that

1 =

((3− 2

√3)n

d

p

)=

(−3

p

)4

(n/d

p

).

We have used the relation((3 + 2

√3)(3− 2

√3)

p

)=

(−3

p

)= 1 .

Therefore Cd(Qp) = ∅ if(

3p

)=(−1p

)= −

(−3p

)4

(n/dp

)= 1.

“=⇒”: Taking w = 0 and z = 1, the equation of Cd is reduced to t4−6ndt2−3(n

d)2 =

0. If(

3p

)= 1, then

√3 ∈ Qp and the equation is t2 =

(3± 2

√3)nd. If

(3p

)= 1 and(

−1p

)= −1, then

((3+2

√3)n/dp

)((3−2

√3)n/dp

)=(−3p

)= −1 so that

((3+2

√3)D/dp

)= 1

or(

(3−2√

3)n/dp

)= 1 which implies by Hensel’s lemma that there exists t = α ∈ Qp

such that α2 =(3 + 2

√3)ndor α2 =

(3− 2

√3)nd. Therefore (w, z, t) = (0, 1, α) is

a solution of Cd in Qp. If(

3p

)=(−1p

)= 1 =

(−3p

)4

(n/dp

), then

((3±2

√3)n/dp

)= 1

and we also have Cd(Qp) 6= ∅ by Hensel’s lemma.

(3). “=⇒”: Suppose that d ≡ 1 (mod 3). Choose z = 0, t = 1. Since(d3

)= 1, by

Hensel’s lemma, the equation dw2 = 1 has a solution in Q3. Hence Cd(Q3) 6= ∅.

“⇐=”: Let (w, t, z) 6= (0, 0, 0) be a solution of Cd in Q3. Then at least two terms

in {2v3(w), 4v3(t), 1 + 2v3(t) + 2v3(z), 4v3(z) + 1} reach the minimal value, which

is denoted by min. The only possibility is that 2v3(w) = 4v3(t) = min. We may

assume that v3(w) = v3(t) = 0, hence v3(z) ≥ 0. The equation Cd modulo 3 gives


us dw2 ≡ t4 (mod 3) and hence(d3

)= 1, which contradicts d ≡ 2 (mod 3). This

completes the proof.

(4). “=⇒”: If p ≡ 2 (mod 3), then(−3p

)= −1. Choose z = 0, t = 1 we have

dw2 = 1, which is solvable in Qp if(dp

)= 1; choose z = 1, t = 0 we have dw2 = −3n

2

d2,

which is solvable in Qp if(−3dp

)= 1. Since

(−3p

)= −1, it is easy to see that one of

the two equations is always solvable. Hence Cd(Qp) 6= ∅.

“⇐=”: Let (w, t, z) 6= (0, 0, 0) be a solution of Cd in Qp. Then at least two terms

in {2vp(w), 4vp(t), 1 + 2vp(t) + 2vp(z), 4vp(z) + 2} reach the minimal value, which is

denoted by min. There are two possibilities. First, if 2vp(w) = 4vp(t) = min, we

may assume that vp(w) = vp(t) = 0, hence vp(z) ≥ 0. The equation Cd modulo p

gives us dw2 ≡ t4 (mod p) and hence(dp

)= 1, which contradicts the assumption

that(dp

)= −1. Second, if 2vp(w) = 4vp(z) + 2 = min, we may assume that

vp(z) = 0, vp(w) = 1, hence vp(t) ≥ 1. Let w = pw′, t = pt′, then vp(w′) = 0 ≤

vp(t′). Dividing both sides of the equation Cd by p2 and then modulo p we find

that dw′2 ≡ −3(ndp

)2

z4 (mod p). Hence(−3dp

)= 1. Since p ≡ 1 (mod 3), we

have(−3p

)= 1, which implies that

(dp

)= 1, which contradicts the assumption that(

dp

)= −1. This concludes the proof. �

Lemma 4. Under the same assumption of lemma 3, we have

(1). C ′d(Q2) = ∅ ⇐⇒ 2|d

(2). If p|d then

C ′d(Qp) = ∅ ⇐⇒

(−3p

) = −1; or

(−3p

) = (−1p

) = 1 = −(−3p

)4(D/dp

)

(3). If 3 - d, then C ′d(Q3) = ∅ ⇐⇒ d ≡ 2 (mod 3).

(4). If p|nd, then C ′d(Qp) = ∅ ⇐⇒ (3

p) = 1 = −(d

p).

(5). d < 0 ⇐⇒ C ′d(Q∞) = ∅.

12 FENG AND XIONG

Proof. The proof is very similar to that of Lemma 3, so we omit the details. �

3.2. The size of the Selmer groups. For any specific value n we may determine if

Cd(Qv) 6= ∅ and C ′d(Qv) 6= ∅ for v ∈ S, d ∈M by Lemma 3 and Lemma 4, so that we

can determine the Selmer groups Sel(φ)(En/Q) and Sel(φ)(E ′n/Q). However when the

number of prime factors of n gets larger, the computation becomes more complicated

and it seems difficult to express the size of Sel(φ)(En/Q) and Sel(φ)(E ′n/Q) explicitly

with certain knowledge of n. We find a reasonable way to calculate the size of

the Selmer groups in terms of some specific graphs associated to n. This graph-

theoretical method has been used to find rank zero elliptic curves y2 = x3 − n2x

([11],[12],[13]) (so we find new non-congruent numbers n). This graphical method

also works nicely for the curve En, and the final result can be stated in a very simple

way.

For a square-free positive integer n with gcd(n, 6) = 1, we factor it as a product

of primes

n = P1 · · ·PtQ1 · · ·Qsp1 · · · pt′ · · · q1 · · · qs′ ,

where

Pi ≡ 1, Qj ≡ 5, pλ ≡ 7, qµ ≡ 11 (mod 12),

t, s, t′, s′ ≥ 0, t+ s+ t′ + s′ ≥ 1.

Put

P = {P1, . . . , Pt}, Q = {Q1, . . . , Qs}, p = {p1, . . . , pt′}, q = {q1, . . . , qs′}.

Then

S = {∞, 2, 3} ∪P ∪Q ∪ p ∪ q,

and

M = 〈−1, 2, 3, P1, . . . , Pt, Q1, . . . , Qs, p1, . . . , pt′ , q1, . . . , qs′〉 ⊆ Q∗/(Q∗)2.


We construct a graph G′(n) = (V,E) as follows:

V = P ∪Q ∪ p ∪ q ∪ {3},

E = {−→ππ′ | (π

′

π) = −1, π ∈ P, π′ ∈ P ∪Q ∪ p ∪ q}

∪ {−→π3 | (−3

π)4 = −1, π ∈ P}

∪ {−→ππ′ | (π

′

π) = −1, π ∈ q, π′ ∈ P ∪ p}.

(1) and (5) of Lemma 4 imply that if 2|d or d < 0, then d /∈ Sel(φ)(E ′n/Q). (2)

of Lemma 4 implies that if p|d and (−3p

) = −1 (which means that p ∈ Q ∪ q),

then d 6∈ Sel(φ)(E ′n/Q). Hence Sel(φ)(E ′n/Q) ⊆ 〈3, P1, . . . , Pt, p1, . . . , pt′〉. Since 3 ∈

Sel(φ)(E ′n/Q), we may restrict our attention to the subgroupM ′ = 〈P1, . . . , Pt, p1, . . . , pt′〉.

Each d ∈M ′ is associated to a partition V = V1(d) ∪ V2(d) given by

V1(d) = {π ∈ P ∪ p : π|d} , V2(d) = V − V1(d),

and this partition satisfies the property that V2(d) ⊇ Q ∪ q ∪ {3}. There is a one-

to-one correspondence between the elements of M ′ and the partitions V = V1 ∪ V2

of G′(n) with V2 ⊇ Q ∪ q ∪ {3}, and it is easy to see from Lemma 4 that for any

d ∈ M ′, d ∈ Sel(φ)(E ′n/Q) if and only if the partition V = V1(d) ∪ V2(d) is an even

partition with V2(d) ⊇ Q ∪ q ∪ {3}. To compute this cardinality 2s(n), we order the

vertices of V as

V = {P1, . . . , Pt, p1, . . . , pt′ , q1, . . . , qs′ , Q1, . . . , Qs, 3} = {v1, . . . , vm},(6)

where m = t + t′ + s + s′ + 1 ≥ 2. Let L(G′(n)) be the Laplace matrix of

the graph G′(n). By Lemma 1, the quantity 2s(n) equals the number of vectors

(x1, . . . , xt, y1, . . . , yt′ , 0, . . . , 0) ∈ Fm2 such that

L(G′(n)) · (x1, . . . , xt, y1, . . . , yt′ , 0, . . . , 0)T = 0.(7)

14 FENG AND XIONG

The matrix L(G′(n)) can be written explicitly in the form

L(G′(n)) =

APP APp ∗ ∗ ∗

0 0 0 0 0

AqP Aqp 0 0 0

0 0 0 0 0

0 0 0 0 0

,

where APP is a symmetric t× t matrix, APp is a t× t′ matrix, AqP is a s′× t matrix,

and Aqp is a s′ × t′ matrix. Define

Γ(n) =

APP APp

AqP Aqp

.(8)

and a = rankF2Γ(n). Then the number of solutions of the equation (7) is 2t+t′−a.

Hence

s(n) = t+ t′ − a.

For any integer b with gcd(b, 12) = 1, let

ωb(n) =∑p|n

p≡b (mod 12)

1.

It is worth noting that a ≤ ω1(n) + min{ω7(n), ω11(n)}, hence

s(n) ≥ max {0, ω7(n)− ω11(n)} .(9)


To compute #Sel(φ)(En/Q), we use a similar method. We construct a graph

G(n) = (V,E) as follows:

V = P ∪Q ∪ p ∪ q ∪ {3}

E = {−→ππ′ | (π

′

π) = −1, π ∈ P, π′ ∈ P ∪Q ∪ p ∪ q}

∪ {−→π3 | (−3

π)4 = −1, π ∈ P}

∪ {−→ππ′ | (π

′

π) = −1, π ∈ p, π′ ∈ P}

∪ {−→ππ′ | ( π

π′) = −1, π ∈ p, π′ ∈ q}.

Since −3 ∈ Sel(φ)(En/Q), by Lemma 3, it is enough to consider the subgroup

T = {d ∈ 〈−1, P1, . . . , Pt, q1, . . . , qs′〉 : d ≡ 1 (mod 3)}.

Each d ∈ T is associated to a partition V = V1(d) ∪ V2(d) where

V1(d) = {p ∈ P ∪ q : p|d} , V2(d) = V − V1(d),

and this partition satisfies the property that V2(d) ⊇ Q ∪ p ∪ {3}. There is a one-

to-one correspondence between the elements of T and the partitions V = V1 ∪ V2

with Q ∪ p ∪ {3} ⊆ V2, and it is also easy to see from Lemma 3 that for any

d ∈ T , d ∈ Sel(φ)(En/Q) if and only if V = V1(d) ∪ V2(d) is an even partition

in the graph G(n) with Q ∪ p ∪ {3} ⊆ V2. To compute this cardinality 2s(n),

similarly we order the vertices as in (6). Let L(G(n)) be the Laplace matrix of

the graph G(n). By Lemma 1, the quantity 2s(n) equals the number of vectors

(x1, . . . , xt, 0, . . . , 0, y1, . . . , ys′ , 0, . . . , 0) ∈ Fm2 such that

L(G(n)) · (x1, . . . , xt, 0, . . . , 0, y1, . . . , ys′ , 0, . . . , 0)T = 0.(10)

16 FENG AND XIONG

Comparing with L(G′(n)), the matrix L(G(n)) can be written explicitly as

L(G(n)) =

APP ∗ APq ∗ ∗

ApP ∗ Apq 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

,

where APP is symmetric, ApP = ATPp, APq = ATqP, and Apq = ATqp. Using the same

matrix Γ(n) in (8), and a = rankF2Γ(n), we find that the number of solutions of the

equation (10) is 2t+s′−a. Hence

s(n) = t+ s′ − a.(11)

It is also worth noting that

s(n) ≥ max {0, ω11(n)− ω7(n)} .(12)

The matrices APP, APp, AqP and Aqp depend on the factorization of the integer

n and can be written down explicitly. We choose not to do that for the sake of

simplicity. As applications we present three interesting consequences, which can be

derived immediately from above.

Corollary 1. (i). Sel(φ)(En/Q) = {1,−3} and Sel(φ)(E ′n/Q) = {1, 3} (hence

rank(En(Q)) = 0) if and only if s′ = t′ and Γ(n) is nonsingular over F2.

(ii). If n is a product of distinct primes which are all congruent to 5 modulo 12,

then Sel(φ)(En/Q) = {1,−3} and Sel(φ)(E ′n/Q) = {1, 3}, hence rank(En(Q)) = 0.

(iii) For the following cases rankEn(Q) = 0 (in what follows p, q are distinct odd

primes):

(1) n = p,

p ≡ 5 (mod 12); or

p ≡ 1 (mod 12),(−3p

)4

= −1;


(2) n = pq,

(p, q) ≡ (5, 5) (mod 12);

(p, q) ≡ (1, 1) (mod 12),(−3p

)4

=(−3q

)4

= −1;

(p, q) ≡ (1, 1) (mod 12),(qp

)=(−3p

)4

(−3q

)4

= −1;

(p, q) ≡ (1, 5) (mod 12),(−3p

)4

= −(qp

);

(p, q) ≡ (7, 11) (mod 12),(qp

)= 1;

We point out that (iii) of Corollary 1 was first obtained by the first author in [10].

4. Averaging #Sel(φ)(En/Q)

Let n be a square-free positive integer such that gcd(n, 6) = 1. Our task now is to

study how the 2-rank s(n) (and s(n)) is distributed as n varies in the set S(X, h).

We can give a heuristic argument first.

The matrix Γ(n) in (8) depends on the factorization of n, and can be considered

as a random matrix when n varies in S(X, h). Its F2-rank tends to be as large as

possible, if there are no obvious obstructions, with high probability. Considering the

width and height of Γ(n), its largest rank is obviously min{ω1(n) + ω11(n), ω1(n) +

ω7(n)}. By this argument and (11), the quantity s(n) should be distributed as

max{0, ω11(n) − ω7(n)}, which behaves like a Gaussian random variable by the

famous Erdős-Kac theorem. The argument for s(n) is the same. We will prove

below that it is indeed the case.

From Lemma 3, we have

2s(n) = F1(n) + F2(n),

18 FENG AND XIONG

where

F1(n) =∑n=dd′

1

2

(1 +

(d

3

))∏p|d′

{1

4

(1 +

(p3

))(1 +

(d

p

))+

1

2

(1−

(p3

))}×

∏p|d

{1

4

(1 +

(3

p

))(1−

(−1

p

))+

1

8

(1 +

(3

p

))(1 +

(−1

p

))(1 +

(−3

p

)4

(d′

p

))},

and

F2(n) =∑n=dd′

1

2

(1 +

(−d3

))∏p|d′

{1

4

(1 +

(p3

))(1 +

(−dp

))+

1

2

(1−

(p3

))}×

∏p|d

{1

4

(1 +

(3

p

))(1−

(−1

p

))+

1

8

(1 +

(3

p

))(1 +

(−1

p

))(1 +

(−3

p

)4

(−d′

p

))}.

Here F1(n) accounts for d ∈ Sel(φ)(En/Q) with d > 0 and F2(n) accounts for

d ∈ Sel(φ)(En/Q) with d < 0. We deal with F1(n) first. Expanding the product, we

obtain

F1(n) =1

2(F1,1(n) + F1,2(n)) ,

where

F1,1(n) =∑

n=ε1ε2···ε18

g1,1(ε),

F1,2(n) =∑

n=ε1ε2···ε18

g1,2(ε),

g1,1(ε) = 2−ω(ε(1))−2ω(ε(2))−3ω(ε(3))(−1)ω(ε2ε3ε17) ×(−1

ε2ε3ε4ε6ε7ε9

)(3

ε1ε2ε4ε5ε7ε8

)(ε13ε14ε17

3

)( −3

ε5ε6ε7ε10

)4

×(ε14ε16ε17ε18

ε5ε6ε7ε10

)(ε1ε2ε3ε4ε8ε9ε11ε12

ε13ε15

)(ε13ε15

ε5ε6ε7ε10

)(ε5ε6ε7ε10

ε13ε15

),


and

g1,2(ε) = 2−ω(ε(1))−2ω(ε(2))−3ω(ε(3))(−1)ω(ε2ε3ε17)(ε1 · · · ε12

3

)×(

−1

ε2ε3ε4ε6ε7ε9

)(3

ε1ε2ε4ε5ε7ε8

)(ε13ε14ε17

3

)( −3

ε5ε6ε7ε10

)4

×(ε14ε16ε17ε18

ε5ε6ε7ε10


ε13ε15

)(ε13ε15

ε5ε6ε7ε10

)(ε5ε6ε7ε10

ε13ε15

).

Here the sums in F1,2(n) and F1,2(n) are over all positive integers ε1, ε2, . . . , ε18 such

that n = ε1ε2 · · · ε18, and we put

ε(1) = ε17ε18, ε(2) = ε1ε2ε3ε11ε13ε14ε15ε16, ε(3) = ε4ε5ε6ε7ε8ε9ε10ε12,

and the function ω(n) counts the number of distinct prime factors of n.

Our goal is to estimate∑n∈S(X,h)

2s(n) =∑

n∈S(X,h)

F1(n) +∑

n∈S(X,h)

F2(n) .

We shall first estimate ∑n∈S(X,h)

F1,1(n) .

We sum over the 18 variables εi, subject to the conditions that each εi is square-free,

positive, that they are relatively prime in pairs, and that their product n satisfies

1 ≤ n ≤ X, n ≡ h (mod 12),

where h = 1 or 5. We divide the range of each variable εi into intervals [Ai, 2Ai),

where Ai runs over powers of 2. For such Ai’s, we define A = (A1, . . . , A18) and

S(A) =∑εi∼Ai

1≤i≤18

g1,1(ε),

where εi ∼ Ai means Ai ≤ εi < 2Ai, and the sum is under the restriction that

ε1 · · · ε18 ≤ X, εi’s are relatively prime in pairs and ε1 · · · ε18 ≡ h (mod 12). We

20 FENG AND XIONG

collect all such admissible A’s together and we call it the set L . It is clear that

#L = O(log18X). We may assume that for any A ∈ L ,

1 ≤18∏i=1

Ai � X.

Now we estimate the sum ∑A∈L

S(A).

Following Heath-Brown [18, 19], we shall describe the variables εi and εj as being

“linked” if exactly one of the Jacobi symbols(εiεj

),

(εjεi

)occurs in the expression for g1,1(ε). It is easy to see that the pairs (εi, εj) with i ∈

{14, 16, 17, 18}, j ∈ {5, 6, 7, 10} are linked. The pairs (εi, εj) with i ∈ {13, 15}, j ∈

{1, 2, 3, 4, 8, 9, 11, 12} are also linked.

4.1. Case one. Given A ∈ L , suppose that Ai, Aj ≥ (logX)640 and that εi, εj are

linked variables. We may write g1,1(ε) in the form

g1,1(ε) =

(εiεj

)a(εi)b(εj),

where the function a(εi) depends all the variables εk except εj, and similarly the

function b(εj) depends on all the variables εk except εi. Moreover we have

|a(εi)|, |b(εj)| ≤ 1.

We can now write

|S(A)| ≤∑

εk,1≤k≤18k 6=i,j

∣∣∣∣∣∣∑ε1,εj

(εiεj

)a(εi)b(εj)

∣∣∣∣∣∣ .We need the following result.


Lemma 5 (Lemma 4 in [18] and Lemma 4.1 in [30]). Suppose ε > 0 is any fixed

number, X,M and N are sufficiently large real numbers, and {am}, {bn} are two

complex sequences, supported on odd integers, satisfying |am|, |bn| ≤ 1. Fix positive

integers h, q satisfying gcd(h, q) = 1 and q ≤ {min(M,N)}ε/3. Let

S :=∑m,n

ambn

(mn

),

where the summation is subject to

M ≤ m < 2M,N ≤ n < 2N,mn ≤ X and mn ≡ h (mod q).

Then

S �MN15/16+ε +M15/16+εN ,

where the constant involved in the � symbol depends on ε only.

As a consequence of this lemma one finds that

S(A)�18∏k=1

Ak {min(Ai, Aj)}−1/32 � X(logX)−20.

Therefore

Lemma 6. We have

S(A)� X(logX)−20

whenever there is a pair of linked variables (εi, εj) with Ai, Aj ≥ (logX)640.

4.2. Case two. We now examine the case that Ai ≥ (logX)640, Aj < (logX)640

while (εi, εj) is a pair of linked variables. Using the quadratic reciprocity law we put

g1,1(ε) in the form

g1,1(ε) = r−ω(εi)

(εiεj

)χ(εi)c ,

22 FENG AND XIONG

where χ is a character modulo 8, which may depend on the variables εk’s other than

εi, and the factor c is independent of εi and satisfies |c| ≤ 1, and r = ±2,±4 or ±8,

depending on the choice of i. It follows that

(13) |S(A)| ≤∑εk,k 6=i

1≤k≤18

∣∣∣∣∣∑εi

r−ω(εi)

(εiεj

)χ(εi)

∣∣∣∣∣ ,

where the inner sum is restricted by the conditions that εi must be square-free and

coprime to all the other variables εk, k 6= i. Next, we employ the following result,

which generalizes Lemma 4 in [18].

Lemma 7. (Lemma 4.2, [30]) Suppose s is a fixed rational number. Let N be

sufficiently large. Then for arbitrary positive integers q, r and any nonprincipal

character χ (mod q), we have

∑n≤X,gcd(n,r)=1

µ2(n)sω(n)χ(n)� Xτ(r) exp(−η√

logX)

with a positive constant η = ηs,N , uniformly for q ≤ logN X. Here τ is the usual

divisor function and µ is the Möbius function.

To use this result we remove the condition εi ≡ h′ (mod 12) from the inner sum

on the right side of (13) and insert instead a factor

1

4

∑ψ (mod 12)

ψ(εi)ψ(h′).


One has

S(A) �∑εk,k 6=i

1≤k≤18

Ai exp(−η√

logAi)∏k 6=i

τ(εk)

= Ai exp(−η√

logAi)∏k 6=i

∑εk

τ(εk)

� Ai exp(−η√

logAi)∏k 6=i

Ak logX

� X(logX)17 exp(−η√

logAi) ,

provided that εj 6= 1 and 8 · 12 · εj � logN Ai for some N > 0. We summarize the

above results as follows.

Lemma 8. For any constant κ with 0 < κ < 1 one has

S(A)� X(logX)−19

whenever there are linked variables εi, εj for which

Ai ≥ exp{(logX)κ}

and εj > 1.

4.3. Case Three. Put

I1 = {17, 18}, I2 = {1, 2, 3, 11, 13, 14, 15, 16},

and

I3 = {4, 5, 6, 7, 8, 9, 10, 12}.

For κ > 0 to be a sufficiently small real number, let

(14) C = exp {(logX)κ} ,

24 FENG AND XIONG

For any positive integers λ1, λ2, λ3, we define

L (λ1, λ2, λ3) =

A ∈ L :

#{i ∈ I1 : Ai > C} = λ1

#{i ∈ I2 : Ai > C} = λ2

#{i ∈ I3 : Ai > C} = λ3

.

For each j = 1, 2, 3, write

mj =∏εi<2Ci∈Ij

εi, nj =∏εi≥2Ci∈Ij

εi.

For A ∈ L (λ1, λ2, λ3), fix mj and nj, the number of εi such that∏

i∈Ij ,εi<2C εi = mj

is at most (#Ij)ω(mj), and the number of εi such that

∏i∈Ij ,εi>2C εi = nj is at most(

#Ij

λj

)λω(nj)j . Hence∣∣∣∣∣∣∑

A∈L (λ1,λ2,λ3)

S(A)

∣∣∣∣∣∣ �∑

m1,m2,m3≤(2C)8

2−ω(m1)−2ω(m2)−3ω(m3)2ω(m1)+3ω(m2)+3ω(m3) ×

∑n1n2n3≤X/(m1m2m3)

2−ω(n1)−2ω(n2)−3ω(n3)λω(n1)1 λ

ω(n2)2 λ

ω(n3)3 .

For j = 1, 2, 3, define

fj(n) = αω(n)j µ2(n),

where µ is the Möbius function, and

α1 =λ1

2, α2 =

λ2

4, α3 =

λ3

8.

Then ∑n≤X

fj(n)� X

logX

∑n≤X

αω(n)j µ2(n)

n≤ X

logX

∏p≤X

(1 +

αjp

).

By Mertens’ estimate, this gives us

∑n≤X

fj(n)� X

logXexp

(∑p≤X

αjp

)� X(logX)αj−1.(15)


We also have∑n1n2≤X

f1(n1)f2(n2) ≤∑

n1≤√X

f1(n1)∑

n2≤X/n1

f2(n2) +∑

n2≤√X

f2(n1)∑

n1≤X/n2

f1(n1).

Applying (15) we obtain that∑n1n2≤X

f1(n1)f2(n2)� X(logX)α1+α2−1.

Similarly ∑n1n2n3≤X

f1(n1)f2(n2)f3(n3)� X(logX)α1+α2+α3−1.

Now we can compute∣∣∣∣∣∣∑

A∈L (λ1,λ2,λ3)

S(A)

∣∣∣∣∣∣ �∑

m1,m2,m3≤(2C)8

2ω(m2)∑

n1n2n3≤X/(m1m2m3)

f1(n1)f2(n2)f3(n3)

�∑

m1,m2,m3≤(2C)8

2ω(m2) X

m1m2m3

(logX)α1+α2+α3−1.

Using the bound ∑n≤X

γω(n)

n≤∏p≤X

(1 +

γ

p

)� (logX)γ,

which is valid for any fixed γ > 0, we find that∣∣∣∣∣∣∑

A∈L (λ1,λ2,λ3)

S(A)

∣∣∣∣∣∣ � X(logX)α1+α2+α3+4κ−1,

where

α1 =λ1

2, α2 =

λ2

4, α3 =

λ3

8.

If λ1

2+ λ2

4+ λ3

8< 1, we may choose κ > 0 sufficiently small so that

λ1

2+λ2

4+λ3

8+ 4κ− 1 < − 1

10.

Then in this case the total error is bounded by O(X(logX)−1/10

).

26 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

λ1

2+λ2

4+λ3

8≥ 1.

We list all possibilities as follows.

(1) λ1 = 2;

(2) λ1 = 0;

(2.1) A13 > C or A15 > C;

(2.1.1) A14 > C or A16 > C;

(2.1.2) A14 ≤ C and A16 ≤ C;

(2.2) A13 ≤ C and A15 ≤ C;

(2.2.1) A14 > C or A16 > C;

(2.2.2) A14 ≤ C and A16 ≤ C;

(3) λ1 = 1;

(3.1) A18 > C and A17 ≤ C;

(3.1.1) A13 > C or A15 > C;

(3.1.2) A13 ≤ C and A15 ≤ C;

(3.2) A17 > C and A18 ≤ C;

(3.2.1) A13 > C or A15 > C;

(3.2.2) A13 ≤ C and A15 ≤ C;

We deal with each case individually.

5.1. Case (1). If λ1 = 2, so that A17 > C and A18 > C. This implies that

A5 = A6 = A7 = A10 = 1. For such A, we can obtain

|S(A)| ≤∑εi∼Aii 6=17,18

∣∣∣∣∣ ∑ε17,ε18

(−2)−ω(ε17)2−ω(ε18)(ε17

3

)∣∣∣∣∣ .


The next lemma can be used to get a desired upper bound.

Lemma 9. Let s1, s2 be non-zero rational numbers. Let C be a positive integer, and

let A > 0 be any fixed number. For X > 1, let T ≤ exp(√

logX) and M,N ≥ T

be given. There exists some constant η > 0 and B > 0 such that, for any positive

integer r, any integer h coprime to C, and any distinct characters χ1, χ2 (mod q),

where Cq � (logX)A, we have

∑m,n

µ2(m)µ2(n)s−ω(m)1 s

−ω(n)2 χ1(m)χ2(n)� Xτ(r) exp

(−η√

log T)

(logX)B,

where the sum is over coprime variables satisfying the conditions

M < m ≤ 2M, N < n ≤ 2N, mn ≤ X, mn ≡ h (mod C), gcd(mn, r) = 1,

and the constant involved in the �-symbol depends on s1, s2 and C only.

We remark that Lemma 9 slightly generalizes Lemma 2.4 of [31], which proves

the result when s1 = s2 is a positive integer. Equipped with Lemma 7, however, the

proof of Lemma 2.4 of [31] can be adapted easily to obtain Lemma 9. The proof

of Lemma 9 also follows the same line as that of Lemma 10 in [18]. We omit the

details here.

By Lemma 9, we find that

|S(A)| � X(logX)−19.

The total contribution from this case is bounded by O (X(logX)−1).

5.2. Case (2.1.1). This is the case that λ1 = 0, “A13 > C or A15 > C”, and

“A14 > C or A16 > C”. The conditions imply that A17, A18 ≤ C and A1 = A2 =

A3 = A4 = A5 = A6 = A7 = A8 = A9 = A10 = A11 = A12 = 1. Hence λ1 = λ3 = 0.

28 FENG AND XIONG

Since λ1

2+ λ2

4+ λ3

8≥ 1, we must have λ2 = 4, which implies that A13, A15 > C.

However, in this case we have

|S(A)| ≤∑εi∼Aii 6=13,1513≤i≤18

∣∣∣∣∣ ∑ε13,ε15

2−2ω(ε13)2−2ω(ε15)(ε13

3

)∣∣∣∣∣ .By Lemma 9 again, |S(A)| is bounded by O (X(logX)−19). So the total contribution

from this case is O (X(logX)−1).

5.2.1. Case 2.1.2. This is the case that λ1 = 0, “A13 > C or A15 > C”, and “A14 ≤ C

and A16 ≤ C”. The conditions imply that A17, A18 ≤ C and A1 = A2 = A3 = A4 =

A8 = A9 = A11 = A12 = 1. Hence λ1 = 0, λ2 ≤ 2 and λ3 ≤ 4.

Since λ1

2+ λ2

4+ λ3

8≥ 1, we must have λ2 = 2 and λ3 = 4, which implies that

A13, A15 > C. By Lemma 9, similar to the argument of Case 2.1.1, we find that

|S(A)| is bounded by O (X(logX)−19). So the total contribution from this case is

O (X(logX)−1).

5.2.2. Case 2.2.1. This is the case that λ1 = 0, “A13 ≤ C and A15 ≤ C”, and

“A14 > C or A16 > C”. The conditions imply that A17, A18 ≤ C and A5 = A6 =

A7 = A10 = 1.

If A14 > C and A16 > C, by Lemma 9, similar to the argument of Case 2.1.1,

|S(A)| is bounded by O (X(logX)−19). So we can assume that exactly one element

of {A14, A16} is larger than C.

If at least two elements of {A1, A2, A3, A11} are larger than C, then A13 = A15 = 1,

and by Lemma 9, similar to the argument of Case 2.1.1, |S(A)| is bounded by

O (X(logX)−19). So we can assume that at most one element of {A1, A2, A3, A11}

is larger than C.





is larger than C.

We are left with λ1 = 0, λ1 ≤ 2, λ3 ≤ 1, and

λ1

2+λ2

4+λ3

8<

3

4.

This has been dealt with in Case three.

5.2.3. Case 2.2.2. This is the case that λ1 = 0, “A13 ≤ C and A15 ≤ C”, and

“A14 ≤ C and A16 ≤ C”. The conditions imply that A17, A18 ≤ C.



O (X(logX)−19). So we can assume that at most one element of {A1, A2, A3, A11} is

larger than C. Hence λ2 ≤ 1. Similarly, if at least two elements of {A4, A8, A9, A12}

are larger than C, then A13 = A15 = 1, and by Lemma 9, |S(A)| is bounded by


is larger than C. Hence λ3 ≤ 5.

We are left with λ1 = 0, λ1 ≤ 1, λ3 ≤ 5, and

λ1

2+λ2

4+λ3

8<

7

8.

This has been dealt with in Case three.

5.2.4. Case 3.1.1. This is the case that A18 > C,A17 ≤ C, “A13 > C or A15 > C”.

The conditions imply that A1 = A2 = A3 = A4 = A5 = A6 = A7 = A8 = A9 =

A10 = A11 = A12 = 1. Hence λ1 = 1, λ3 = 0.

If A13 > C, since A18 > C, by Lemma 9, similar to the argument of Case 2.1.1,

|S(A)| is bounded by O (X(logX)−19). So we can assume that A13 ≤ C, and hence

A15 > C.

IfA14 > C, sinceA18 > C, by Lemma 9 again, |S(A)| is bounded byO (X(logX)−19).

So we can assume that A14 ≤ C.

30 FENG AND XIONG

If A16 ≤ C, then λ2 = 1, and

λ1

2+λ2

4+λ3

8≤ 3

4,

which has been dealt with in Case three.

We are left with the case that A1 = A2 = A3 = A4 = A5 = A6 = A7 = A8 = A9 =

A10 = A11 = A12 = 1, A13, A14, A17 ≤ C and A15, A16, A18 > C. Collecting this case

together we define

L1 =

A ∈ L :

A1 = A2 = . . . = A12 = 1,

A13, A14, A17 ≤ C,

A15, A16, A18 > C.

.

For any subset B ⊂ L , let

S(B) =∑A∈B

S(A).

We obtain for L1

S(L1) =∑A∈L1

2−ω(ε17ε18)−2ω(ε13ε14ε15ε16)(−1)ω(ε17)(ε13ε14ε17

3

).(16)

5.2.5. Case 3.1.2. This is the case that A18 > C,A17 ≤ C, “A13 ≤ C and A15 ≤ C”.

The conditions imply that A5 = A6 = A7 = A10 = 1.

Fist, since A18 > C, by Lemma 9, using the same argument as in Case 3.1.1, we

may assume that A14 ≤ C.

If max{A1, A2, A3, A4, A8, A9, A11, A12} ≤ C, then λ1 = 1, λ2 ≤ 1 and λ3 = 0, and

λ1

2+λ2

4+λ3

8≤ 3

4,


Now assume that max{A1, A2, A3, A4, A8, A9, A11, A12} > C. Then A13 = A15 =

1. Since A18 > C, by Lemma 9 again, using the same argument as in Case 3.1.1,

we may assume that A1, A2, A3, A4, A8, A9 ≤ C.


If either A11 ≤ C or A16 ≤ C, then λ1 = 1, λ2 ≤ 1 and λ3 ≤ 1, and

λ1

2+λ2

4+λ3

8≤ 7

8,

which has been dealt with in Case three. So we may assume that A11, A16 > C.

Collecting this case which is left together we define

L2 =

A ∈ L :

A1, A2, A3, A4, A8, A9, A14, A17 ≤ C,

A5 = A6 = A7 = A10 = 1,

A11, A16, A18 > C.

.

We obtain

S(L2) =∑

A∈L22−2ω(ε1ε2ε3ε11ε14ε16)−3ω(ε4ε8ε9ε12)−ω(ε17ε18)×

(−1)−ω(ε2ε3ε17)(

−1ε2ε3ε4ε9

)(3

ε1ε2ε4ε8

) (ε14ε17

3

).

(17)

5.2.6. Case 3.2.1. This is the case that A17 > C,A18 ≤ C, “A13 > C or A15 > C”.

The conditions imply that A1 = A2 = A3 = A4 = A5 = A6 = A7 = A8 = A9 =

A10 = A11 = A12 = 1. Hence λ1 = 1, λ3 = 0.

If A15 > C or A16 > C, since A17 > C, by Lemma 9, similar to the argument of

Case 2.1.1, |S(A)| is bounded by O (X(logX)−19). So we can assume that A15 ≤ C

and A16 ≤ C. Hence A13 > C.

If A14 ≤ C, then λ2 = 1, and

λ1

2+λ2

4+λ3

8=

3

4,

which has been dealt with in Case three. So A14 > C.

We are left with the case that A1 = A2 = A3 = A4 = A5 = A6 = A7 = A8 = A9 =

A10 = A11 = A12 = 1, A13, A14, A17 > C and A15, A16, A18 ≤ C. Collecting this case

together we define

L3 =

A ∈ L :

A1 = A2 = . . . = A12 = 1,

A13, A14, A17 > C,

A15, A16, A18 ≤ C.

.

32 FENG AND XIONG

We obtain

S(L3) =∑A∈L3

2−ω(ε17ε18)−2ω(ε13ε14ε15ε16)(−1)ω(ε17)(ε13ε14ε17

3

).(18)

5.2.7. Case 3.2.2. This is the case that A17 > C,A18 ≤ C, “A13 ≤ C and A15 ≤ C”.

The conditions imply that A5 = A6 = A7 = A10 = 1.

Fist, since A17 > C, by Lemma 9, using the same argument as in Case 3.1.1, we

may assume that A16 ≤ C.

If max{A1, A2, A3, A4, A8, A9, A11, A12} ≤ C, then λ1 = 1, λ2 ≤ 1 and λ3 = 0, and

λ1

2+λ2

4+λ3

8≤ 3

4,


Now assume that max{A1, A2, A3, A4, A8, A9, A11, A12} > C. Then A13 = A15 =

1. Since A17 > C, by Lemma 9 again, using the same argument as in Case 3.1.1,

we may assume that A1, A3, A8, A9, A11, A12 ≤ C.

If either A2 ≤ C or A14 ≤ C, then λ1 = 1, λ2 ≤ 1 and λ3 ≤ 1, and

λ1

2+λ2

4+λ3

8≤ 7

8,

which has been dealt with in Case three. So we may assume that A2, A14 > C.

Collecting this case which is left together we define

L4 =

A ∈ L :

A1, A3, A8, A9, A11, A12, A16, A18 ≤ C,

A5 = A6 = A7 = A10 = 1,

A2, A14, A17 > C.

.

We obtain

S(L4) =∑

A∈L42−2ω(ε1ε2ε3ε11ε14ε16)−3ω(ε4ε8ε9ε12)−ω(ε17ε18)×

(−1)−ω(ε2ε3ε17)(

−1ε2ε3ε4ε9

)(3

ε1ε2ε4ε8

) (ε14ε17

3

).

(19)


6. Conclusions

Combining all results in Sections 4 and 5, we conclude that∑n∈S(X,h)

F1,1(n) = S(L1) + S(L2) + S(L3) + S(L4) +O(X(logX)−1/10

).

We can also conclude by similar analysis that∑n∈S(X,h)

F1,2(n) = S(L1) + S(L ′2) + S(L3) + S(L ′

4) +O(X(logX)−1/10

),

where

L ′2 =

A ∈ L :

A1, A3, A8, A9, A11, A12, A14, A17 ≤ C,

A5 = A6 = A7 = A10 = 1,

A2, A16, A18 > C.

,

and

L ′4 =

A ∈ L :

A1, A2, A3, A4, A8, A9, A16, A18 ≤ C,

A5 = A6 = A7 = A10 = 1,

A11, A14, A17 > C.

.

As for F2(n), we can also expand the product to obtain

F2(n) =1

2(F2,1(n) + F2,2(n)) ,

where

F2,1(n) =∑

n=ε1ε2···ε18

g2,1(ε),

F2,2(n) =∑

n=ε1ε2···ε18

g2,2(ε),

g2,1(ε) = 2−ω(ε(1))−2ω(ε(2))−3ω(ε(3))(−1)ω(ε2ε3ε17) ×(−1

ε2ε3ε4ε5ε9ε10ε13ε15

)(3

ε1ε2ε4ε5ε7ε8

)(ε13ε14ε17

3

)( −3

ε5ε6ε7ε10

)4

×(ε14ε16ε17ε18

ε5ε6ε7ε10


ε13ε15

)(ε13ε15

ε5ε6ε7ε10

)(ε5ε6ε7ε10

ε13ε15

),

34 FENG AND XIONG

and

g2,2(ε) = 2−ω(ε(1))−2ω(ε(2))−3ω(ε(3))(−1)ω(ε2ε3ε17)(ε1 · · · ε12

3

)×(

−1

3ε2ε3ε4ε5ε9ε10ε13ε15

)(3

ε1ε2ε4ε5ε7ε8

)(ε13ε14ε17

3

)( −3

ε5ε6ε7ε10

)4

×(ε14ε16ε17ε18

ε5ε6ε7ε10


ε13ε15

)(ε13ε15

ε5ε6ε7ε10

)(ε5ε6ε7ε10

ε13ε15

).

Here the sums in F2,1(n) and F2,2(n) are over all positive integers ε1, ε2, . . . , ε18 such

that n = ε1ε2 · · · ε18,


and the function ω counts the number of distinct prime factors.

The functions g2,1(ε) and g2,2(ε) are very similar to g1,1(ε) and g1,2(ε) and can be

treated in the same way. By careful analysis and a little bit of extra work we also

obtain that ∑n∈S(X,h)

F2,1(n) = S(L2) + S(L4) +O(X(logX)−1/10

),

and ∑n∈S(X,h)

F2,2(n) = −S(L ′2)− S(L ′

4) +O(X(logX)−1/10

).

Recall that

∑n∈S(X,h)

2s(n) =1

2

∑n∈S(X,h)

F1,1(n) + F1,2(n) + F2,1(n) + F2,2(n)

,

we obtain∑n∈S(X,h)

2s(n) = S(L1) + S(L2) + S(L3) + S(L4) +O(X(logX)−1/10

).

Our goal now is to identify the main terms.


6.1. Let

H(X) = S(L1) + S(L3).

Since

L1∪L3 =

A ∈ L : A1 = . . . = A12 = 1,A13, A14, A17 ≤ C,A15, A16, A18 > C; or

A13, A14, A17 > C,A15, A16, A18 ≤ C.

.

For A ∈ L1 ∪L3, the summand can be simplified as

2−ω(ε17ε18)−2ω(ε13ε14ε15ε16)(−1)ω(ε17)(ε13ε14ε17

3

).

Hence

H(X) =∑

A∈L1∪L3

2−ω(ε17ε18)−2ω(ε13ε14ε15ε16)(−1)ω(ε17)(ε13ε14ε17

3

).

Removing the constraints on A13, . . . , A18, we consider the set

H1 = {A ∈ L : A1 = . . . = A12 = 1} .

It is easy to see, following the argument in the previous two sections that

S(H1) = H(X) +O(X(logX)−1/10

).

We can rewrite it as

S(H1) =∑

n∈S(X,h)

∑ε13ε14ε15ε16ε17ε18=n

2−ω(ε17ε18)−2ω(ε13ε14ε15ε16)(−1)ω(ε17)(ε13ε14ε17

3

).

Since the function in the inner sum is multiplicative, we can compute directly that

S(H1) =∑

n∈S(X,h)

∏p|n

1 =∑

n∈S(X,h)

1 = #S(X, h).

Therefore

H(X) = #S(X, h) +O(X(logX)−1/10

).

36 FENG AND XIONG

6.2. On the other hand, let

G(X) = S(L2) + S(L4).

Then

G(X) =∑

A∈L2∪L42−2ω(ε1ε2ε3ε11ε14ε16)−3ω(ε4ε8ε9ε12)−ω(ε17ε18)×

(−1)−ω(ε2ε3ε17)(

−1ε2ε3ε4ε9

)(3

ε1ε2ε4ε8

) (ε14ε17

3

).

Here

L2 =

A ∈ L :

A1, A2, A3, A4, A8, A9, A14, A17 ≤ C,

A5 = A6 = A7 = A10 = A13 = A15 = 1,

A11, A16, A18 > C.

,

and

L4 =

A ∈ L :

A1, A3, A8, A9, A11, A12, A16, A18 ≤ C,

A5 = A6 = A7 = A10 = A13 = A15 = 1,

A2, A14, A17 > C.

.

Removing the constraints on A1, A2, A3, A4, A8, A9, A11, A12, A14, A16, A17, A18, we

consider the set

H2 = {A ∈ L : A5 = A6 = A7 = A10 = A13 = A15 = 1} .

It is easy to see, following the argument in the previous two sections that

S(H2) = G(X) +O(X(logX)−1/10

).

Moreover, we can rewrite it as

S(H2) =∑

n∈S(X,h)

∑ε1ε2ε3ε4ε8ε9ε11ε12ε14ε16ε17ε18=n

2−2ω(ε1ε2ε3ε11ε14ε16) ×

2−3ω(ε4ε8ε9ε12)−ω(ε17ε18)(−1)−ω(ε2ε3ε17)

(−1

ε2ε3ε4ε9

)(3

ε1ε2ε4ε8

)(ε14ε17

3

).

Since the function in the inner sum is multiplicative, we can compute directly that

S(H2) =∑

n∈S(X,h)

2ω11(n)−ω7(n).


Here for any integer a with gcd(a, 12) = 1, we define

ωa(n) =∑p|n

p≡a (mod 12)

1 .

Therefore

G(X) =∑

n∈S(X,h)

2ω11(n)−ω7(n) +O(X(logX)−1/10

).

We finally conclude that∑n∈S(X,h)

2s(n) = #S(X, h) +∑

n∈S(X,h)

2ω11(n)−ω7(n) +O(X(logX)−1/10

).

7. Distribution of s(n)

For each n, let

g(n) = ω11(n)− ω7(n), λ(n) = max{0, g(n)}.

To find the distribution of s(n) for n ∈ S(X, h) as X → ∞, we first write the

asymptotic equation in the form∑n∈S(X,h)

2s(n) =∑

n∈S(X,h)

2g(n) +O (X) .(20)

It is known from (12) that

s(n) ≥ λ(n) ≥ g(n).

We shall study the distribution of the right hand side first.

7.1. Moments of λ(n). We first apply Lemma 2 for the strongly additive function

g(n), and A = S(X, h), P is the set of primes not exceeding X. From the proof of

[29, Lemma 14] we find that

#A =3

4π2X +O

(X exp(−η

√logX)

),

38 FENG AND XIONG

and for a square-free d with (d, 12) = 1, if Ad = #{n ∈ A : d|n}, then

Ad#A

=

∏p|d (1 + p−1)

−1

d

(1 +O

(τ(d) exp(−η

√logX)

)),

where η is a sufficiently small positive constant, τ is the divisor function. Hence in

application of Lemma 2,

h(d) =∏p|d

(1 + p−1

)−1,

and

|rd| �τ(d)

∏p|d (1 + p−1)

−1

dX exp(−η

√logX) .

Since

g(p) =

1 p ≡ 11 (mod 12),

−1 p ≡ 7 (mod 12),

0 p 6≡ 7, 11 (mod 12),

we can choose M = 1. We obtain

µP(g) =∑p≤X

g(p)h(p)

p=

∑p≡11 (mod 12)

p≤X

1

p+ 1−

∑p≡7 (mod 12)

p≤X

1

p+ 1.

This in turn, by Merten’s estimate, gives us

µP(g) = O(1).

On the other hand,

σP(g)2 =∑p≤X

g(p)2h(p)

p

(1− h(p)

p

)

=∑

p≡11 (mod 12)p≤X

1

p+ 1

(1− 1

p+ 1

)+

∑p≡7 (mod 12)

p≤X

1

p+ 1

(1− 1

p+ 1

).

By Mertens’ estimate, we obtain

σP(g)2 =1

2log logX +O(1).


For a fixed positive integer k, it is also easy to see that∑d∈Dk(P)

|rd| � X exp(−η√

logX) ∑d≤Xk

µ2(d)τ(d)

d� X exp

(−η′√

logX).

Then by Lemma 2, for any fixed even integer k, we have

1

#S(X, h)

∑n∈S(X,h)

(g(n)−O(1))k = Ck

(1

2log logX +O(1)

)k/2+O

((log logX)

k2−1),

and for any fixed odd integer k, we have

1

#S(X, h)

∑n∈S(X,h)

(g(n)−O(1))k � Ckk3/2

(1

2log logX +O(1)

)(k−1)/2

.

We can conclude that for any fixed positive integer k,

1

#S(X, h)

∑n∈S(X,h)

g(n)k = Ck

(1

2log logX

)k/2 (δk/2 +O

((log logX)−1/2

)),

where for any α ∈ R, δα = 1 if α ∈ Z, and δα = 0 if α 6∈ Z.

Noticing that, all the moments of g(n)√12

log logX, as n varies in the set S(X, h) with

X → ∞, are the same as the moments of a standard Gaussian distribution, where

the odd moments vanish and the even moments are

1√2π

∫ ∞∞

x2re−x2/2 dx = C2r,

this implies that asymptotically g(n)√12

log logXis a standard Gaussian distribution.

Let X be a standard Gaussian distribution and Y = max{0, X}. We can compute

easily that for any positive integer k, E(Y k) = λk, where λk’s are the constants

defined in (5). Since λ(n) = max{0, g(n)},

λ(n)√12

log logX∼ Y,

we find that for any positive integer k, as X →∞,

1

#S(X, h)

∑n∈S(X,h)

λ(n)k = (λk + o(1))

(1

2log logX

)k/2.(21)

40 FENG AND XIONG

7.2. Moments of s(n). For each n ∈ S(X, h), write

s(n) = λ(n) + δ(n),

then δ(n) ∈ Z and δ(n) ≥ 0. Since λ(n) ≥ g(n), from (20) we obtain

X �∑

n∈S(X,h)

2λ(n)+δ(n) − 2λ(n) ≥∑

n∈S(X,h)

2s(n) − 2g(n).

Since ∑n∈S(X,h)

2λ(n)(2δ(n) − 1

)≥ 1

2

∑n∈S(X,h)δ(n)≥1

2λ(n)+δ(n),

we obtain ∑n∈S(X,h)δ(n)≥1

2λ(n)+δ(n) � X.(22)

For each positive integer r, put

ar = # {n ∈ S(X, h) : λ(n) + δ(n) = r, δ(n) ≥ 1} .

The asymptotic formula (22) can be written as∑r∈N

2rar � X.

Hence

ar �X

2r.

Then for any fixed positive integer k,∑n∈S(X,h)δ(n)≥1

(λ(n) + δ(n))k =∞∑r=1

rkar �∞∑r=1

rkX

2r� X.

For such a fixed positive integer k, we expand

1

#S(X, h)

∑n∈S(X,h)

s(n)k =1

#S(X, h)

∑n∈S(X,h)

(λ(n) + δ(n))k =1

#S(X, h)

∑n∈S(X,h)

λ(n)k+E,


where

E =1

#S(X, h)

∑n∈S(X,h)

k−1∑r=0

(k

r

)λ(n)rδ(n)k−r.

The right hand side is

1

#S(X, h)


k−1∑r=0

(k

r

)λ(n)rδ(n)k−r ≤ 1

#S(X, h)


k∑r=0

(k

r

)λ(n)rδ(n)k−r.

Hence we can bound E by

0 ≤ E ≤ 1

#S(X, h)


(λ(n) + δ(n))k � 1.

We deduce from (21) that for any fixed positive integer k,

1

#S(X, h)

∑n∈S(X,h)

s(n)k = (λk + o(1))

(1

2log logX

)k/2.

Noticing that, all the moments of s(n)√12

log logX, as n varies in the set S(X, h) with

X →∞, are the same as the moments of the random variable Y , this implies that

asymptoticallys(n)√

12

log logX∼ Y.

This completes the proof of Theorem 1 for s(n). �

8. Distribution of t(n)

8.1. Stoll’s formula. Chang [6] has used Stoll’s formula [26] to obtain upper

bound for average rank of quadratic twists of the elliptic curve y2 = x3 − A for

any integer A under some congruence conditions. For the special elliptic curve

En : y2 = x3 − n3, we use [27], which improves upon [26] by Stoll himself, in order

to treat as many En as possible. We follow Chang’s ideas and focus on En with n

positive, square-free and n ≡ 1, 5 (mod 12). Interested readers may refer to [6] and

[27] for general setting and details.

42 FENG AND XIONG

Let ζ ∈ Q be a primitive third root of unity and K = Q(ζ). The elliptic curve

En : y2 = x3 − n3 can be considered as an elliptic curve over K. We denote simply

by ζ the endomorphism on E given by (x, y) 7→ (ζx, y) which is defined over K.

Denote by λ this endomorphism 1 − ζ. Via Galois cohomology, we have the short

exact sequence

0 // En(K)λ(En(K))

// Sel(λ)(En/K) // X(En/K)[λ] // 0 .

If n is square-free, then

rank(En(Q)) ≤ rankF3

En(K)

λ(En(K))≤ rankF3Sel

(λ)(En/K).

For n ≡ 1, 5 (mod 12) and n > 0, Stoll’s formula (Corollary 2.1 and Theorem 2.1

of [27]) states that

rankF3Sel(λ)(En/K) = 2 dimF3 Cl

(Q(√−n)

)[3],(23)

where Cl(Q(√−n)

)[3] denotes the 3-torsion part of the class group of the number

field Q(√−n), whose discriminant equals −4n.

Let h3(4) denote #Cl(F )[3], where F is the quadratic extension of Q with dis-

criminant 4. For positive integers N and m, denote by N−2 (X,m,N) the set of

fundamental discriminants 4 such that −X < 4 < 0 and 4 ≡ m (mod N). For

h = 1, 5, recall the following special case of a general theorem proved by Nakagawa

and Horie [24]

limX→∞

1

#N−2 (4X,−4h, 48)

∑4∈N−2 (4X,−4h,48)

h3(4) = 2.(24)

The set N−2 (4X,−4h, 48) can be identified with S(X, h) by the mapping −4n 7→ n.

Define

#Sel(λ)(En/K) = 3s′(n).


The equations (23) and (24) can be restated as

limX→∞

1

#S(X, h)

∑n∈S(X,h)

√3s′(n)

= 2.

Let rank(En(Q)) = r(n). Since r(n) ≤ s′(n), from above we obtain∑n∈S(X,h)

√3r(n)� X.(25)

8.2. Moments of t(n). Recall that φ : En → E ′n is a 2-isogeny and φ : E ′n −→ En

is the dual 2-isogeny. Hence φ ◦ φ = [ 2 ], and one has the following commutative

diagrams (see pp 97, [1]):

0

��

0

��

0

��

0 // E′n(Q)φ(En(Q))

//

��

Sel(φ)(En/Q) //

��

X(En/Q)[φ] //

��

0

0 // En(Q)2En(Q))

//

��

Sel(2)(En/Q) //

��

X(En/Q)[2] //

��

0

0 // En(Q)

φ(E′n(Q))//

��

Sel(φ)(E ′n/Q) //

��

X(E ′n/Q)[φ] //

��

0

0 // C //

��

C //

��

0

0 0

Define#Sel(φ)(En/Q) = 2s(n)+1, #Sel(φ)(E ′n/Q) = 2s(n)+1,

#X(En/Q)[φ] = 2t(n), #X(E ′n/Q)[φ] = 2t(n).

From the commutative diagrams, we have

r(n) = s(n)− t(n) + s(n)− t(n),

44 FENG AND XIONG

where r(n) is the rank of En over Q. Let

r1(n) = s(n)− t(n).

Using t(n) = s(n)− r1(n), we expand

1

#S(X, h)

∑n∈S(X,h)

t(n)k =1

#S(X, h)

∑n∈S(X,h)

s(n)k + E,

where

E =1

#S(X, h)

∑n∈S(X,h)

k−1∑i=0

(k

i

)(−1)is(n)ir1(n)k−i.

Since

0 ≤ r1(n) ≤ r(n),

for each fixed positive integer k, we deduce from (25) that

1

#S(X, h)

∑n∈S(X,h)

r1(n)k ≤ 1

#S(X, h)

∑n∈S(X,h)

r(n)k � 1.

Recalling

1

#S(X, h)

∑n∈S(X,h)

s(n)k = (λk + o(1))

(1

2log logX

)k/2,

we find that

|E| � max0≤i≤k−1

1

#S(X, h)

∑n∈S(X,h)

s(n)2i

1/2 1

#S(X, h)

∑n∈S(X,h)

r(n)2k−2i

1/2

� (log logX)k−12 .

We conclude that as X →∞,

1

#S(X, h)

∑n∈S(X,h)

t(n)k = (λk + o(1))

(1

2log logX

)k/2.


Noticing again that, all the moments of t(n)√12

log logX, as n varies in the set S(X, h)

with X →∞, are the same as the moments of the random variable Y described at

the end of Section 7, this implies that asymptotically

t(n)√12

log logX∼ Y.

This completes the proof of Theorem 1 for t(n). �

9. Distribution of s(n) and t(n)

9.1. s(n) and t(n). Let n be a square-free positive integer such that gcd(n, 6) = 1.

From Lemma 4, we have

2s(n) = F (n),

where

F (n) =∑n=dd′

1

2

(1 +

(d

3

))∏p|d′

{1

4

(1 +

(3

p

))(1 +

(d

p

))+

1

2

(1−

(3

p

))}×

∏p|d

{1

4

(1 +

(−3

p

))(1−

(−1

p

))+

1

8

(1 +

(−3

p

))(1 +

(−1

p

))(1 +

(−3

p

)4

(d′

p

))}.

Expanding the product, we obtain

F (n) =1

2(F1(n) + F2(n)) ,

where

F1(n) =∑

n=ε1ε2···ε18

g1(ε), F2(n) =∑

n=ε1ε2···ε18

g2(ε).

Here

g1(ε) = 2−ω(ε(1))−2ω(ε(2))−3ω(ε(3))(−1)ω(ε2ε3ε17) ×(−1

ε1ε3ε5ε6ε8ε9

)(3

ε1ε2ε4ε5ε7ε8ε13ε14ε17

)(−3

ε5ε6ε7ε10

)4

×(ε13ε14ε15ε16ε17ε18

ε5ε6ε7ε10

)(ε1ε2 . . . ε12

ε13ε15

),

46 FENG AND XIONG

and

g1(ε) = 2−ω(ε(1))−2ω(ε(2))−3ω(ε(3))(−1)ω(ε2ε3ε17)(ε1ε2 . . . ε12

3

)×(

−1

ε1ε3ε5ε6ε8ε9

)(3

ε1ε2ε4ε5ε7ε8ε13ε14ε17

)(−3

ε5ε6ε7ε10

)4

×(ε13ε14ε15ε16ε17ε18

ε5ε6ε7ε10

)(ε1ε2 . . . ε12

ε13ε15

).

The sums in F1(n) and F2(n) are over all positive integers ε1, ε2, . . . , ε18 such that

n = ε1ε2 · · · ε18,


and the function ω counts the number of distinct prime factors.

We use similar techniques to estimate∑n∈S(X,h)

2s(n) =∑

n∈S(X,h)

F (n) ,

that is, we sum over the 18 variables εi, subject to the conditions that each εi is

square-free, positive, that they are relatively prime in pairs, and that their product

n satisfies

1 ≤ n ≤ X, n ≡ h (mod 12),

where h = 1 or 5, and we divide the range of each variable εi into intervals [Ai, 2Ai),

where Ai runs over powers of 2. Following the steps in Section 4–6 carefully we can

obtain∑n∈S(X,h)

2s(n) = #S(X, h) +∑

n∈S(X,h)

2ω7(n)−ω11(n) +O(X(logX)−1/10

).

Define

g′(n) = ω7(n)− ω11(n), λ′(n) = max{0, g′(n)}.

It is known from (9) that

s(n) ≥ λ′(n) ≥ g′(n).


Following the argument in Section 7 in the same way, we find that for any fixed

positive integer k,

1

#S(X, h)

∑n∈S(X,h)

s(n)k = (λk + o(1))

(1

2log logX

)k/2,

and hence asymptotically as X →∞,

s(n)√12

log logX∼ max {0, Y } ,

where Y is a standard Gaussian distribution. This completes the proof of Theorem

1 for s(n).

As for t(n), define

r2(n) = s(n)− t(n).

From the commutative diagrams in Section 8, we have

0 ≤ r2(n) ≤ r(n) = rank(En(Q)).

Following the argument of Section 8 we obtain that for any positive integer k,

1

#S(X, h)

∑n∈S(X,h)

t(n)k = (λk + o(1))

(1

2log logX

)k/2.

This completes the proof of Theorem 1 for t(n). �

9.2. Proof of Theorem 2. It remains to prove the asymptotic formula (4) of

Theorem 2. We summarize our results as follows.

Let

λ(n) = max {0, ω11(n)− ω7(n)} , λ′(n) = max {0, ω7(n)− ω11(n)} .

s(n) = λ(n) + δ(n) = r1(n) + t(n), s(n) = λ′(n) + δ′(n) = r2(n) + t(n),

where for each n,

δ(n), r1(n), δ′(n), r2(n) ≥ 0.

48 FENG AND XIONG

We have for any fixed positive integers a, b, c, d, by Cauchy-Schwartz inequality,

1

#S(X, h)

∑n∈S(X,h)

δ(n)aδ′(n)br1(n)cr2(n)d � 1.(26)

For any fixed positive integer k, we also have

1

#S(X, h)

∑n∈S(X,h)

λ(n)k = (λk + o(1))

(1

2log logX

)k/2,

1

#S(X, h)

∑n∈S(X,h)

λ′(n)k = (λk + o(1))

(1

2log logX

)k/2.

Since

λ(n) · λ′(n) = 0,∀n,

we have

1

#S(X, h)

∑n∈S(X,h)

(λ(n) + λ′(n))k

= (2λk + o(1))

(1

2log logX

)k/2.(27)

Now (4) can be proved easily: since

t(n) + t(n) = λ(n) + λ′(n) + δ(n) + δ′(n)− r1(n)− r2(n),

for any fixed positive integer k, expanding the k-th power, we obtain

1

#S(X, h)

∑n∈S(X,h)

(t(n) + t(n)

)k=

1

#S(X, h)

∑n∈S(X,h)

(λ(n) + λ′(n))k

+ E,

where

E =1

#S(X, h)

∑n∈S(X,h)

k−1∑i=0

(k

i

)(λ(n) + λ′(n))

i(δ(n) + δ′(n)− r1(n)− r2(n))

k−i.

We deduce from (26) and (27) that

|E| � (log logX)k−12 .


This implies

1

#S(X, h)

∑n∈S(X,h)

(t(n) + t(n)

)k/2= (2λk + o(1))

(log logX

2

)k.

Now the proof of (4) is complete. �

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Feng Keqing: Department of Mathematical science, Tshinghua University, Bei-

jing, P.R. China

E-mail address: [email protected]

Maosheng Xiong: Department of Mathematics, Hong Kong University of Science

and Technology, Clear Water Bay, Kowloon, P. R. China

E-mail address: [email protected]

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