xε and AB > x log10 x). For the first term on the right-hand side of (6) we will obtain an exact formula with main term μ2 . We proceed as follows. As in our deduction of formula (3), by partitioning the elliptic curves in C according to their isomorphism classes over Fp and Fp , we obtain 1 |C|
E∈C
1=
p,p ≤x p=p |E(Fp )|,|E(Fp )| prime
1 1 |C| (p − 1)(p − 1) p,p ≤x p=p
×
wp,s,twp ,s ,t Aut E(s, t)
s,t∈Fp s ,t ∈Fp
· Aut E(s , t ) · S(p, p , s, t, s , t ), where wp,s,t, wp ,s ,t are 1 or 0 according to whether |E(s, t)|, |E(s , t )|, respectively, are prime or not, and S(p, p , s, t, s , t ) is the number of integers |a| ≤ A, |b| ≤ B such that there exist 1 ≤ u < p, 1 ≤ u < p satisfying a ≡ su4 (mod p), a ≡ s u 4 (mod p ),
b ≡ tu6 (mod p), b ≡ t u 6 (mod p ).
We now use the following generalization of Lemma 6, which will be proved at the end of the section:
1191
AVERAGE TWIN PRIME CONJECTURE FOR ELLIPTIC CURVES
L EMMA 7. For fixed primes p, p and fixed s, t ∈ F∗p , s , t ∈ F∗p , let wp,s,t, wp ,s ,t ∈ C be such that |wp,s,t |, |wp ,s ,t | ≤ 1. Let A, B > 0. Then, for any positive integer k, we have that, as x → ∞, 1 pp
p,p ≤x p=p
1≤s,t
x2 log14 x, where ε > 0 is fixed and arbitrarily small. Now we apply Theorem 5 to deduce x 2 x2 1 twin (10) , πE (x) − C(E) 2 ε 6 |C| log x log x E∈C 7
A = 0. provided that A, B > xε , AB > x2 log14 x, and limA,B→∞ log B·log B Finally, let f (x) be any real function such that f (x) = o(log x). Then (10) implies that " ! twin x x # E ∈ C : πE (x) − C(E) 2 log x f (x) log2 x x 2 f (x)2 log4 x twin πE (x) − C(E) log2 x x2 E∈C
ε
f (x)2 |C|. log2 x
This completes the proof of Corollary 2.
3.3. Character sums. The rest of this section consists of proving Lemmas 6 and 7. Let us note that we make no effort to write the error term in terms of the weights wp,s,t, as we will use the trivial bound |wp,s,t| ≤ 1 for our application. (The present method using the trivial bound on the weights also works in the case when the weight function wp,s,t is supported on a sparse set. However, in that case the result is less satisfactory: the method recovers the results of Baier [Bai] and Fouvry and Murty [FoMu] about the average Lang-Trotter conjecture on Frobenius traces, but does not seem to improve upon them, even after a careful analysis of the contribution of the weights to the error term.) Our proofs will rely on two important analytic results, which we state below. T HEOREM 8. (The large sieve inequality) Let M, N, Q be positive integers and (an )n a sequence of complex numbers. For a fixed q ≤ Q, we denote by χ Dirichlet
AVERAGE TWIN PRIME CONJECTURE FOR ELLIPTIC CURVES
1193
characters modulo q. Then q ϕ(q)
q≤Q
χ(mod q) M 0, provided that 4U V 2 ≤ x log−N x.
(46)
Using the estimates (44) and (45) in (43), we finally obtain that 2 2 a 1 2 r − af E X, Y ; r − 1, nf , nf n 4 (47)
|r|≤R, r=1 f ≤V n≤U
a (mod 4n)
1 Rx Rx log 2 U 1 M M −3 f log 2 x f ≤V log 2 x
for any M > 0. This estimates the error term (42), and thus the error term of Proposition 13 (after renaming M ).
1218
A. BALOG, A.-C. COJOCARU, AND C. DAVID
7.2. Computation of the constant in Proposition 13. We now treat the main term (41) in (40), which is essentially a computation of the constant C in Theorem 1. We first analyze the sum over n and f of (41) when r is a fixed integer. We remark that the sum is zero if r is even, thus we can assume that r is odd. We also take r = 1. Our goal in this section is to prove: P ROPOSITION 15. Let r = 1 be an odd integer. Then 1 nf
f ≤V n≤U
a (mod 4n)
2 2 1 1 a 2 r − af , S r − 1, nf , = Cr + O +√ n 4 V2 U
where Cr is the positive constant ∞ ∞ 1 Cr := nf f =1 n=1
a (mod 4n)
2 2 a 2 r − af S r − 1, nf , n 4
+1 1 4 2 (2 − 2 − 2) · · . 1+ 2 1+ 2 = 3 ( − 1)3 ( + 1) − 2 − 2 − 2 − 2 =2
|(r−1) =2
|r(r−2) =2
Proof of Proposition 15. Using the definition of S(·, ·, ·), we rewrite the lefthand side of the desired equation in Proposition 15 as ⎛ (48)
2⎝
( − 2) =2
( − 1)2
⎞ ⎠
⎞
⎛ f ≤V n≤U f odd
⎜ 1
⎜ ⎜ nf φ nf 2 ⎝
|nf 2 (r−1) =2
− 1⎟ ⎟ r ⎟ c (n), − 2⎠ f
where
crf (n) :=
a (mod 4n)
a
n
and indicates that the sum is taken over the invertible residues a modulo 4n such that
r 2 − af 2 2 , nf = 1 4
2 r − af 2 , 4nf 2 = 4
and ⇐⇒ and
r 2 − af 2 2 − (r − 1), nf = 1 4
r 2 − af 2 − 4(r − 1), 4nf 2 = 4.
AVERAGE TWIN PRIME CONJECTURE FOR ELLIPTIC CURVES
1219
As r, f are odd and (r, f ) divides (r 2 − af 2 , 4nf 2 ), we must have that (r, f ) = 1; in this case,
2
r − af 2 , 4nf 2 = 4 ⇐⇒ r 2 − af 2 , 4n = 4. Similarly, as (r − 2, f ) divides (r 2 − af 2 − 4(r − 1), nf 2 ) = ((r − 2)2 − af 2 , nf 2 ), we must have (r − 2, f ) = 1; in this case,
(r − 2)2 − af 2 , 4nf 2 = 4 ⇐⇒ (r − 2)2 − af 2 , 4n = 4. Thus
crf (n) =
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨
a if (r, f ) = (r − 2, f ) = 1, n
a (mod 4n)∗
(r −af ,4n)=4 ⎪ ⎪ 2 ⎪ ((r−2) −af 2 ,4n)=4 ⎪ ⎪ ⎩ 0 2
2
otherwise,
where a (mod 4n)∗ denotes invertible residue classes a modulo 4n. To continue the proof, we need some properties of the function crf (n): L EMMA 16. Let r = 1 be an odd integer and let f be a positive odd integer such that (r, f ) = (r − 2, f ) = 1. Let crf (n) be as defined above. The following statements hold: (1) if n is odd, then a , crf (n) = n ∗ a (mod n) (r 2 −af 2 ,n)=1
((r−2)2 −af 2 ,n)=1
where a (mod n)∗ denotes invertible residue classes a modulo n; (2) crf (n) is a multiplicative function of n; (3) if is an odd prime and (, f ) = 1, then ⎧ ⎪ − 2 if α is even and | r(r − 2)(r − 1), ⎪
α ⎪ ⎪ r ⎨ − 3 if α is even and r(r − 2)(r − 1), cf = ⎪ α−1 −1 if α is odd and | r(r − 2)(r − 1), ⎪ ⎪ ⎪ ⎩−2 if α is odd and r(r − 2)(r − 1); (4) if is an odd prime and | f (which implies that (, r) = (, r − 2) = 1 by the hypotheses on f ), then # crf (α ) 0 if α is odd, = α−1 − 1 if α is even;
1220 (5)
A. BALOG, A.-C. COJOCARU, AND C. DAVID crf (2α ) 2α−1
= (−1)α .
Proof. (1) If n is odd, then crf (n) =
a n
a (mod 4n)∗ , a≡1 (mod 4) (r 2 −af 2 ,n)=1
a
=
a (mod n)∗ (r 2 −af 2 ,n)=1
((r−2)2 −af 2 ,n)=1
n
,
((r−2)2 −af 2 ,n)=1
where follows from the Chinese Remainder theorem and the fact last
a2equality
a1the that n = n when a1 ≡ a2 (mod n) for n odd. (2) Let n1 , n2 be two co-prime positive integers with n1 odd, and let n = n1 n2 . Then, using the Chinese Remainder theorem, we obtain
r a1 a2 r × cf n1 cf n2 = n1 n2 ∗ ∗ a1 (mod n1 ) (r 2 −a1 f 2 ,n1 )=1
a2 (mod 4n2 ) (r 2 −a2 f 2 ,4n2 )=4
((r−2)2 −a1 f 2 ,n1 )=1
=
a (mod 4n1 n2 )∗ (r 2 −af 2 ,4n1 n2 )=4
((r−2)2 −a2 f 2 ,4n2 )=4
a n1
a n2
= crf n1 n2 .
((r−2)2 −af 2 ,4n1 n2 )=4
(3) We have that
crf α = α−1
a(mod )∗
a α
(r 2 −af 2 ,)=1 ((r−2)2 −af 2 ,)=1
⎛ (49)
⎜ ⎜ ⎜ a α ⎜ = α−1 ⎜ − ⎜ ∗ ⎜a (mod ) ⎝
⎞ a (mod )∗ −2 ¯ a≡f r 2 (mod ) or a≡f¯−2 (r−2)2 (mod )
⎟ ⎟ a α ⎟ ⎟ ⎟, ⎟ ⎟ ⎠
where f¯ denotes the inverse of f modulo . We then need to count the number of invertible residues a modulo which are eliminated by the two congruence conditions of the second sum. If r ≡ 0, 1, 2 (mod 4), there is exactly one such residue (notice that r 2 ≡ (r − 2)2 (mod ) ⇔ r ≡ 1 (mod )). In all three cases, this residue is an invertible square modulo , and the second sum on the right-hand side of (49) has value +1. The result follows immediately when α is even, and follows from the orthogonality relations when α is odd. If r ≡ 0, 1, 2 (mod ), there are two invertible residues which are eliminated by the two congruence conditions on a, and
AVERAGE TWIN PRIME CONJECTURE FOR ELLIPTIC CURVES
1221
the second sum on the right-hand side of (49) has value +2. The result follows as above. (4) We have that a α a α
= α−1 crf α = α−1 ∗ ∗ a (mod ) (r 2 −af 2 ,)=1 ((r−2)2 −af 2 ,)=1
a (mod )
since (r 2 − af 2 , ) = ((r − 2)2 − af 2 , ) = 1 for all a when | f and (r, f ) = (r − 2, f ) = 1. The result follows immediately when α is even, and using the orthogonality relations when
aα is odd. (5) Let α ≥ 1. Since 2 is a character modulo 8, we write crf
α
2
a α
=2
α−1
2
a (mod 8)∗
=2
α−1
α 5 = 2α−1 (−1)α . 2
(r 2 −af 2 ,2α+2 )=4 ((r−2)2 −af 2 ,2α+2 )=4
Using parts (3) and (4) of Lemma 16, we write ⎞ ⎛ ⎜ 1
⎜ ⎜ 2 ⎝ nf φ nf n≤U
f ≤V f odd
|nf 2 (r−1) =2
∞
=
f =1 (2,f )=(r,f )=(r−2,f )=1
− 1⎟ ⎟ r ⎟ c (n) − 2⎠ f ⎞
⎛
⎜ 1
⎜ ⎜ nf φ nf 2 ⎝ n=1
∞
|nf 2 (r−1) =2
1 1 . +√ =: Dr + O V2 U
1 1 − 1⎟ ⎟ r +√ ⎟ c (n) + O − 2⎠ f V2 U
Note that Cr = 2
( − 2) =2
( − 1)2
Dr .
The rest of this section consists of writing Dr as an Euler product. We first write the sum over n as a product. In order to have multiplicative functions of n, we use the formulas
2 φ(n)φ f 2 n, f 2
, φ nf = φ n, f 2
1222
A. BALOG, A.-C. COJOCARU, AND C. DAVID
⎛ |nf 2 (r−1) =2
⎞⎛
⎞⎛
⎟⎜ −1 ⎜ ⎜ − 1 ⎟ ⎜ =⎜ ⎟⎜ −2 ⎝ − 2⎠⎝ 2
|f (r−1) =2
|n =2
⎜ − 1⎟ ⎟⎜ ⎟⎜ − 2⎠⎝
⎞ |(n,f 2 (r−1)) =2
− 2⎟ ⎟ ⎟. − 1⎠
Now we rewrite Dr as ⎞
⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝
⎟
⎟ r cf (n) φ f 2 , n 1 − 1⎟ ⎟
− 2⎟ nφ(n) f 2 , n fφ f2 ⎟ 2 n≥1 |f (r−1) ⎠
f ≥1 f odd (r,f )=(r−2,f )=1
=2
⎞⎛
⎛
⎞
⎜ − 1 ⎟ ⎜ ⎟⎜ ⎜ ×⎜ ⎟⎜ ⎝ − 2⎠⎝ |n =2
|(n,f 2 (r−1)) =2
− 2⎟ ⎟ ⎟. − 1⎠
The sum over n is the sum of a multiplicative function of n whose factors at prime powers α depend on the divisibilities of f, r, r − 1 and r − 2 by . Using Lemma 16, we write the n-sum as ⎞⎛ ⎞ ⎛ ⎛
⎞ α≥0 ar α ⎝
⎠, ar α ⎠ ⎝ br α ⎠ = br α ⎝ b α r α≥0 α≥0 α≥0 α≥0 |f
f
|f
where ar (1) = br (1) = 1 and for = 2 and α ≥ 1,
ar (α ) =
br (α ) =
# 0
if α odd,
( − 1)/(α+1 ) if α even; ⎧ ⎪ −2/α ( − 2) if α odd and r(r − 1)(r − 2), ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ( − 3)/α ( − 2) if α even and r(r − 1)(r − 2), ⎪ ⎪ ⎪ ⎪ ⎨−1/α ( − 2) if α odd and | r(r − 2), ⎪ if α even and | r(r − 2), 1/α ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ if α odd and | r − 1, −1/α ( − 1) ⎪ ⎪ ⎪ ⎪ ⎩ ( − 2)/α ( − 1) if α even and | r − 1.
AVERAGE TWIN PRIME CONJECTURE FOR ELLIPTIC CURVES
1223
Replacing in Dr , this gives ⎛
Dr = br α
f ≥1 f odd (r,f )=(r−2,f )=1
α≥0
=
1 ⎜
⎜ ⎜ fφ f2 ⎝ 2
⎛
|f
⎞
⎟ ⎜ ⎜ − 1⎟ α br ⎟ ⎜ − 2⎠ ⎝ α≥0
|(r−1) =2
⎞⎛
⎛
×
|f (r−1) =2
⎞ ⎛
α ⎞ a −1⎟ r ⎟⎝ α≥0 α ⎠ ⎟ −2⎠ α≥0 br
⎜ 1 ⎜ − 1⎟ ⎟⎜
⎜ ⎟ ⎜ ⎜ f φ f 2 ⎝ 2 − 2⎠⎝ 2
f ≥1 f odd (r,f )=(r−2,f )=1
|f =2
|(f ,r−1) =2
⎞ ⎛
α ⎞ ⎟ ar − 2⎟⎝ α≥0 α ⎠ . ⎟ − 1⎠ α≥0 br |f
The sum over f in the last expression is a sum of multiplicative functions of f , which we write as cr α . 2r(r−2) α≥0
Here cr (1) = 1 and for any prime with 2r(r − 2) and any α ≥ 1, we have
β ⎧ 1 ⎪ β≥0 ar ⎪
· ⎪ ⎪ ⎨ 3α−1 ( − 2) br β β≥0 cr (α ) =
⎪ ⎪ ar β 1 ⎪ β≥0 ⎪
· ⎩ 3α−1 β ( − 1) β≥0 br
if r − 1, if | r − 1.
Then (50)
Dr =
⎛
⎞
⎛ ⎜ − 1⎟ ⎟⎝ br α ⎜ ⎝ − 2⎠
α≥0
|(r−1) =2
⎞ cr α ⎠ .
2r(r−2) α≥0
We now compute the sums appearing in (50), using the formulas for ar () and br () listed above: L EMMA 17. Let be an odd prime and α ≥ 1. Let ar (α ), br (α ) and cr (α ) be as defined above. We have:
1224
A. BALOG, A.-C. COJOCARU, AND C. DAVID
(1) A() :=
α≥0 ar (
α ) = 2 ++1 ; (+1)
3 −22 −2−1 (2) if r(r − 1)(r − 2), then B (1) () := α≥0 br (α ) = (−2)( 2 −1) ; 3 −2 −−1 (3) if | r − 1, then B (2) () := α≥0 br (α ) = (−1) 2 (+1) ; (2 −2−1) (4) if r − 1, but | r(r − 2), then B (3) () := α≥0 br (α ) = (−2)( 2 −1) ; 3 −22 −2 (5) if 2r(r − 2)(r − 1), then C (1) () := α≥0 cr (α ) = 3−2 2 −2−1 ; 2 −−1) (6) if | r − 1, then C (2) () := α≥0 cr (α ) = ( 3 −2 −−1 ; (7) if = 2, then B(2) := α≥0 br (α ) = 23 . Proof. All the computations are straightforward, following from the formula for the sum of the geometric series. We extend the definitions of A(), B (1) (), B (2) (), B (3) (), C (1) (), C (2) () introduced in Lemma 17 to any odd prime (independently of the relation between and r). Then we rewrite Dr as ⎞
⎛
⎛ ⎟ ⎜ − 1 ⎟⎝ ⎜ br α ⎠ ⎜ Dr = B(2) ⎝ ⎟ − 2⎠ ⎝ ⎛
⎞
=2 α≥0
⎛ = B(2) ⎝
=2
|r−1 =2
⎞
⎞ cr α ⎠
2r(r−2) α≥0
⎛
⎞
⎜ ⎟ ⎜ ⎟ B (1) ()C (1) ()⎠ ⎜ B (1) ()−1 C (1) ()−1 B (3) ()⎟ ⎝ ⎠ |r(r−2) =2
⎛
⎞
⎜ − 1 (2) ⎟ ⎟ ⎜ (1) −1 (1) −1 (2) C ()⎟ . B () C () B () ×⎜ −2 ⎠ ⎝ |r−1 =2
Using Lemma 17, we compute B (1) ()C (1) () =
(2 − 2 − 2) 3 − 22 − 2 = , 3 − 22 − + 2 ( − 2)(2 − 1)
B (1) ()−1 C (1) ()−1 B (3) () = B (1) ()−1 C (1) ()−1 B (2) ()C (2) ()
1 2 − 2 − 1 = 1+ 2 , 2 − 2 − 2 − 2 − 2
2 − − 1 +1 −1 = 2 = 1+ 2 . − 2 − 2 − 2 − 2 − 2
AVERAGE TWIN PRIME CONJECTURE FOR ELLIPTIC CURVES
1225
Finally, replacing all the above in (48), we obtain ⎛
2 ⎞ − 2 − 2 4 ( − 2)
⎠ · Cr = ⎝ 3 ( − 1)2 ( − 2) 2 − 1 =2 ·
1+
|r−1 =2
+1 1 · 1 + 2 − 2 − 2 2 − 2 − 2 |r(r−2) =2
+1 4 2 2 − 2 − 2 1+ 2 · = 3 ( − 1)3 ( + 1) − 2 − 2 =2
·
|r(r−2) =2
|(r−1) =2
1 . 1+ 2 − 2 − 2
This completes the proof of Proposition 15.
7.3. The average constant. Using Proposition 15, the main term (41) of (40) is Y |r|≤R, r=1 odd Cr ; thus now we need to average the constant Cr over r. This calculation has similarities with the one done by Gallagher in [Gal] for the average of the standard twin prime constant. As such, we will follow the notation used in [Gal]. The main result of this section is: L EMMA 18. As R → ∞,
Cr = 2CR + O log2 R .
|r|≤R r=1 odd
Proof. Let us write (51)
4 2 2 − 2 − 2 · Cr = 3 ( − 1)3 ( + 1) =2
1 + er () ,
=2 |r(r−1)(r−2)
where ⎧ +1 ⎧ ⎪ ⎪ (1) ⎪ ⎪ e () if | r − 1 2 ⎪ ⎪ ⎨ − − 2 ⎨ (2) 1 er () := e () if | r(r − 2) = ⎪ ⎪ 2 ⎪ ⎪ − 2 − 2 ⎪ ⎩0 ⎪ otherwise ⎩ 0
if | r − 1, if | r(r − 2), otherwise.
1226
A. BALOG, A.-C. COJOCARU, AND C. DAVID
Let us also fix the following notation: for r = 1 odd, we take P(r) := odd prime : | r(r − 1)(r − 2) , F(r) := q positive square-free integer : | q ⇒ ∈ P(r) , D(R) := ∪
|r|≤R F(r). r=1 odd
We want to evaluate (52) S :=
1 + er () =
=2 |r|≤R r=1 odd |r(r−1)(r−2)
where er (1) = 1 and, for q = 1, er (q) =
S=
er (q) =
q∈D(R) |r|≤R r=1 odd
=
er (q),
|r|≤R q∈F (r) r=1 odd
|q er ().
We write
er (q)
q∈D(R) all possible |r|≤R e=e(q) r=1 odd er (q)=e
# |r| ≤ R : r = 1 odd, er (q) = e
q∈D(R) all possible e=e(q)
=
ev() ()N (q, v),
q∈D(R) v=v(q) |q
where the sum
v=v(q)
is over all maps v : { : | q} → {1, 2} and where
N (q, v) := # |r| ≤ R : r = 1 odd, er () = ev() () ∀ | q . By looking at the conditions imposed on when defining e(1) () and e(2) (), we see that N (q, v) is the number of integers |r| ≤ R with r = 1 odd such that r ≡ 1 (mod 2), r ≡ 1 (mod )
∀ | q with v() = 1,
r ≡ 0 or 2 (mod ) ∀ | q with v() = 2. Therefore, by using the Chinese Remainder Theorem, r as above lies in one of v()−1 distinct residue classes modulo 2q. Consequently, |q 2 ⎛ N (q, v) = ⎝
|q
⎞ 2v()−1 ⎠
R v()−1 2R + 1 + O(1) = 2 + O 2ω(q) , 2q q |q
AVERAGE TWIN PRIME CONJECTURE FOR ELLIPTIC CURVES
1227
where ω(q) denotes the number of distinct prime factors of q. We plug this in the formula for S and obtain ⎛ ⎞ 1 ev() ()2v()−1 + O ⎝ 2ω(q) ev() ()⎠ S =R q q∈D(R)
v=v(q) |q
q∈D(R)
v=v(q) |q
=: Smain + Serror . To estimate Smain , we observe that we have Smain = R
G(q)
q∈D(R)
for some multiplicative function G(q). Therefore
e(1) () + 2e(2) () 1 + G() = R 1+ Smain = R ≤R =2
=R
≤R =2
3 − 22 − + 3 3 − 22 − + 3
=R
+ O(1). 2 − 2 − 2 2 − 2 − 2 ≤R =2 =2
Now let us estimate Serror . As for Smain , we observe that we have ⎛ ⎞ F (q)⎠ Serror = O ⎝ q∈D(R)
for some multiplicative function F (q). Therefore ⎛
⎞
⎜ $ %⎟
1 + 2 e(1) () + e(2) () ⎟ Serror = O ⎜ ⎝ ⎠ ⎛ ⎜ = O⎜ ⎝
≤R =2
≤R =2
⎞
2
2 +−1 ⎟ ⎟ = O (log R)2 . 1+ 2 ⎠ − 2 − 2
We put the two estimates together and obtain S =R
3 − 22 − + 3
+ O (log R)2 . 2 − 2 − 2 =2
1228
A. BALOG, A.-C. COJOCARU, AND C. DAVID
By replacing (52) in (51), this gives
4R 2 2 − 2 − 2 3 − 22 − + 3 + O log2 R · 2 Cr = 3 3 ( − 1) ( + 1) − 2 − 2 =2
|r|≤R r=1 odd
=
4R 4 − 23 − 2 + 3 + O log2 R , 3 3 ( − 1) ( + 1) =2
which completes the proof of Lemma 18.
Replacing Proposition 15 and Lemma 18 in (41), this concludes the proof of Proposition 13. A LFR E´ D R E´ NYI I NSTITUTE OF M ATHEMATICS , H UNGARIAN ACADEMY OF S CIENCES , ´ H-1053 B UDAPEST, R E ALTANODA U . 13-15, H UNGARY E-mail: [email protected] D EPARTMENT OF M ATHEMATICS , S TATISTICS AND C OMPUTER S CIENCE , U NIVERSITY OF I LLINOIS AT C HICAGO , 851 S. M ORGAN S T., 322 SEO, C HICAGO , IL 60607; I NSTITUTE OF M ATHEMATICS “S IMION S TOILOW ” OF THE ROMANIAN ACADEMY, 21 C ALEA G RIVITEI S T., B UCHAREST, 010702, S ECTOR 1, ROMANIA E-mail: [email protected] D EPARTMENT OF M ATHEMATICS AND S TATISTICS , C ONCORDIA U NIVERSITY, 1455 DE M AISONNEUVE W EST, M ONTR E´ AL , QC, H3G 1M8, C ANADA ; I NSTITUTE FOR A DVANCED S TUDY, E INSTEIN D RIVE , P RINCETON , NJ 08540 E-mail: [email protected]
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AVERAGE TWIN PRIME CONJECTURE FOR ELLIPTIC CURVES [DaPa1] [DaPa2] [DaWu] [De] [FoMu] [FrIw] [Gal] [HaLi] [Hu] [IwJU] [JU] [Jo1] [Jo2] [Jo3]
[Ko] [LaTr] [Lav] [MaPo] [MiMu]
[MoVa] [PePi] [Se] [So]
[StWe] [Tch] [Te] [Wa] [Zy1] [Zy2]
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