Squares in a certain sequence related to L-functions of elliptic curves

Finite Fields and Their Applications 21 (2013) 1–10

Contents lists available at SciVerse ScienceDirect

Finite Fields and Their Applications www.elsevier.com/locate/ffa

Squares in a certain sequence related to L-functions of elliptic curves Florian Luca a,∗ , Aynur Yalçiner b a

Fundacion Marcos Moshinsky, Instituto de Ciencias Nucleares UNAM, Circuito Exterior, C.U., Apdo. Postal 70-543, Mexico D.F. 04510, Mexico b Department of Mathematics, Faculty of Science, Selçuk University, Campus 42075 Konya, Turkey

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 7 November 2012 Revised 18 January 2013 Accepted 19 January 2013 Available online 31 January 2013 Communicated by Arne Winterhof



−s be the L-series corresponding to an Let L (s, E ) = n1 an n elliptic curve E defined over Q and satisfying certain technical conditions. We prove that the set of positive integers n such that n2 − an2 + 1 =  has asymptotic density 0. © 2013 Elsevier Inc. All rights reserved.

MSC: 11G40 11B39 11N36 Keywords: L-functions of elliptic curves Linear recurrence sequences

1. Introduction Let E be an elliptic curve over the field of rational numbers Q given by the minimal global Weierstraß equation:

E : y 2 + A 1 xy + A 3 y = x3 + A 2 x2 + A 4 x + A 6 and let  be its discriminant. For each prime p we put

a p = p + 1 − #E (F p ),

*

Corresponding author. E-mail addresses: fl[email protected] (F. Luca), [email protected] (A. Yalçiner).

1071-5797/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.ffa.2013.01.005

(1)

2

F. Luca, A. Yalçiner / Finite Fields and Their Applications 21 (2013) 1–10

where E (F p ) is the reduction of E modulo p. If p | , then E (F p ) has a singularity and we put

 ap =

√

If p  , we have |a p | < 2

0 1

for the case of a cusp, for the case of a split node, −1 for the case of a non–split node.

p. The L-function associated to E is given by

L (s, E ) =

 p |



1 1 − a p p −s

p 

1 1 − a p p −s

+ p 1−2s

.

The infinite  product above is convergent for Re(s) > 3/2 and therefore we can expand it into a series L (s, E ) = n1 an n−s . Here we study the set

  N E = n: n2 − an2 + 1 =  . We address only the instance of non-CM curves E, which are those curves whose endomorphism ring over the field of complex numbers consists of the ring of integers; that is, their only endomorphisms are the maps n E which send P to n P for all P ∈ E (C), where n is some arbitrary integer. We also assume that the Sato–Tate conjecture holds for E. This has recently been proved to hold for a large class of elliptic curves such as the elliptic curves with non-integral j-invariant (see [1]). We have the following result. For a positive real number x we put





#N E (x) = # N E ∩ [1, x] . Theorem 1. Let E be a non-CM curve for which the Sato–Tate conjecture holds. The estimate

#N E (x) = O

x



(log x)0.00001

holds for all x  2. The implied constant depends on E. In particular, N E is of asymptotic density 0. Our motivation in studying N E started upon noting that if p |  and a p = ±1, and   1, then a p  = (a p ) = (±1) , which implies that n = p  ∈ N E . Moreover if all prime factors p of n divide  and have a p = ±1, then n ∈ N E . However, the set of such positive integers n is very thin since the number of such integers n  x is O ((log x)c ) for some constant c  ω(). We use the usual notations. The letters p and q with or without subscripts stand for prime numbers. We use ω(n), Ω(n) and τ (n) for the number of prime divisors of n with and without repetitions, and for the total number of divisors of n, respectively. For a subset P of primes we use ωP (n) for the number of distinct prime factors of n which belong to P . We write P (n) for the largest prime factor of n. We use α , β and γ with or without subscripts for positive absolute constants. Finally, we use the Landau notation O and o as well as the Vinogradov notations  and  with their regular meanings. The constants implied by them might depend on E. We let x0 be a large positive real number not necessarily the same at each occurrence.

F. Luca, A. Yalçiner / Finite Fields and Their Applications 21 (2013) 1–10

3

2. The proof 2.1. Weierstraß equations With the standard birational transformation (see Chapter III, Section 1 in [4]), replacing y in (1) by ( y − A 1 x − A 3 )/2 gives an equation of the form

y 2 = 4x3 + B 2 x2 + 2B 4 x + B 6 , where

B 2 = A 21 + 4 A 2 ;

B 4 = 2 A4 + A1 A3;

B 6 = A 23 + 4 A 6 .

Further, defining the quantities

C 4 = B 22 − 24B 4 ; C 6 = − B 32 + 36B 2 B 4 − 216B 6 , and then replacing (x, y ) by ((x − 3B 2 )/36, y /108) yields the simpler Weierstraß equation

E : y 2 = x3 − 27C 4 x − 54C 6 . We put A = −27C 4 and B = −54C 6 . From now on, we assume that p > 3 is a prime so the above transformations are well-defined modulo p and we work with the equation

E : y 2 = x3 + Ax + B . 2.2. The numbers a p  for p  6 and   1

√

Let p > 3 be such that p  . Let λ p = pe i θ p with θ p ∈ [0, π ) be the root of λ2 − a p λ + p = 0 √ √ with positive imaginary part. Then the other root is λ p = pe −i θ p , so a p = 2 p cos θ p . Furthermore,

a p =

1 − λp λ+ p

+1

λp − λp

for all   1.

In particular, 3

a p2 =

λ3p − λ p λp − λp

2

= λ2p + λ p λ p + λ p = (λ p + λ p )2 − λ p λ p = a2p − p .

2.3. Removing n with a large square full part Recall that s is a square full number if p 2 | s whenever p | s. Let be determined later and put y = (log x)α0 . For each n we write

t (n) =

 p n p 6

p

and

s(n) = n/t (n).

α0 ∈ (0, 1) be some constant to

4

F. Luca, A. Yalçiner / Finite Fields and Their Applications 21 (2013) 1–10

Then s(n) = ab, where a is square free and a | 6 and b is squarefull. We put

  N1 (x) = n  x: s(n) > y .

(2)

If n ∈ N1 (x), then n is divisible by a number of the form ab, where a | 6 is square free and b > y /a  y /(6||) is square full. For fixed a and b, the number of such n  x is x/ab  x/ab  x/b. Making a and b vary, we get that

#N1 (x) 

 b> y /(6||) b squarefull a|6

x b



   xτ 6||

1

b> y /(6||) b squarefull

b



x y 1/2

=

x

(log x)α0 /2

,

(3)

where in the above calculation we use the Abel summation formula together with the fact that the counting function of the number of square full numbers s  t is O (t 1/2 ). 2.4. Removing n with many prime factors Here, we take K = 10 log log x and put

  N2 (x) = n  x: Ω(n)  K .

(4)

By Lemma 12 in [3], we have

#N2 (x) 

K x log x 2K



x log x

(5)

.

Furthermore, from now on we shall also remove the integers n  x/ log x since the number of them is O (x/ log x), as indicated in the upper bound for #N2 (x). So, we shall assume that n > x/ log x. 2.5. Rewriting the condition that n ∈ N E We let n ∈ N E (x) and write n2 − an2 + 1 = m2 for some m  0. Put m = n − k for some integer k. If k  0, then k  n because we are assuming that m  0. If k < 0, then, since |an2 |  τ (n2 )n  x1+o(1) as x → ∞, it follows that



2

n + |k|

  = m2 = n2 − an2 + 1  n2 + τ n2 n + 1 < (2n)2 ,

for x > x0 ,

therefore |k| < n. Thus, in both cases |k| < n if x is sufficiently large. Further,

n2 − an2 + 1 = (n − k)2 = n2 − 2nk + k2 ,





so an2 = 2nk − k2 − 1 .

(6)

From now on, we work with n ∈ N E (x) such that n ∈ / N1 (x) ∪ N2 (x). Since n ∈ / N1 (x), we may write

n = up 1 · · · p  ,

u  y, p1 < · · · < p





gcd 6||u , p 1 · · · p  = 1.

and

Here, u = s(n). Then













2kn − k2 − 1 = an2 = au 2 a p 2 · · · a p 2 = au 2 a2p 1 − p 1 · · · a2p  − p  . 1



(7)

F. Luca, A. Yalçiner / Finite Fields and Their Applications 21 (2013) 1–10

5

2.6. Removing n ∈ N E with k = 0 Assume that k = 0 and let N3 (x) be the subset of such n ∈ N E (x). We then get that an2 = 1, therefore a2p i − p i ∈ {±1} for all i = 1, . . . , . In particular, p i − 1 =  for all i = 1, . . . , . Put Q for the / N2 (x), we have set of primes of the form  + 1. Let P = P (n), and observe that since n ∈

P >n

1/Ω(n)

>

1/10 log log x

x

> (log x)2 + 1 for x > x0 .

log x

Since P = a2 + 1 for some positive integer a, we get that a > log x. Fix such a prime P = a2 + 1 and look at the set of n  x which are multiples of P . The number of them is x/ P  x/ P < x/a2 . Summing up over all primes P > (log x)2 + 1 of the form P = a2 + 1, we get that

 x

#N3 (x) 

a>log x

a2



x log x

(8)

.

From now on, we assume that n ∈ / N3 (x), and therefore that k = 0. 2.7. Special primes I: Sato–Tate

/ N1 (x) ∪ N2 (x) ∪ N3 (x). Didiving both sides From now on, we work with n ∈ N E (x) such that n ∈ of Eq. (7) by n, we get 2k −

(k2 − 1) n



=

 au 2 

u

a pi

√

i =1

2

pi

−1 .

Since 1  |k| < n for x > x0 and n > x/ log x, we get that (k2 − 1)/n < |k|, so the absolute value of the √ left-hand side above exceeds |k| for large x. Since |a p i | < 2 p i for i = 1, . . . , , it follows that



2 k2 − 1 au 2  a p i |k| < 2k − − 1 = √p n u 

i =1

 2

i

 τ u 3  22Ω(u ) 4 = 22Ω(u )+2 = 22Ω(n) < 22K = (log x)20 log 2 < (log x)15 .

(9)

In what follows, we will improve the above bound for |k|. For this, we take a closer look at the ratio

ap

√ = 2 cos θ p , p

where θ p ∈ [0, π ).

Let

  P1 = p: θ p ∈ [0, π /4) ∪ (3π /4, π ) . √

Observe that if p ∈ / P1 , then |cos θ p |  1/ 2, therefore



ap

√

p

2 = 4 cos2 θ p  2.

6

F. Luca, A. Yalçiner / Finite Fields and Their Applications 21 (2013) 1–10

Thus,



ap 2 √  1, − 1 p

whenever p ∈ / P1 .

(10)

To estimate the density of P1 , we recall that by the Sato–Tate conjecture which we assume, we have that if we put P1 (t ) = P1 ∩ [1, t ], then

#P1 (t )

π (t )

=



π /4 π 2 2 + o(1) sin θ dθ + sin θ dθ

2

π

3π /4

0

=

π /4 + o(1) sin2 θ dθ

4

π

0

as t → ∞. Computing the integral, we get

π /4

π /4

sin2 θ dθ = 0

2





1 − cos(2θ) dθ =

θ 2

−



θ =π /4 sin(2θ)

4

θ =0

0

= Thus, putting

1

π 8

−

1 4

=

π −2 8

.

ρ1 for the density of P1 , we get that #P1 (t ) =





ρ1 + o(1) π (t ), where ρ1 =

4

π

×

π −2 8

=

π −2 (t → ∞). 2π

2.8. Special primes II: Chebotarev Let P2 be the set of primes p   such that p ≡ 1 (mod 3) and a p ≡ 2 (mod 3). In this section, we give a lower bound on the density ρ2 of P2 as a subset of all the primes. Observe that if p ≡ 1 (mod 3), then

#E (F p ) = p + 1 − a p is a multiple of 3 if and only if a p ≡ 2 (mod 3). So, an equivalent condition for p ≡ 1 (mod 3) to belong to P2 is that E (F p ) has an element of order 3. Recall that E (F p ) has an element of order 3 if and only if there is a point P = (x0 , y 0 ) on E (F p ) such that ψ3 (x0 , y 0 ) = 0, where for an integer s  2, ψs (x, y ) ∈ Z[ A , B , x, y ] stands for the sth-division polynomial (see Exercise 3.7 on page 105 in [4]). For us,

ψ3 (x, y ) = 3x4 + 6 Ax2 + 12Bx − A 2 ∈ Z[ A , B , x]. With fixed A and B, we can regard f (x) = ψ3 (x, y ) as a polynomial in the variable x. Its discriminant is

 2  f = −28 33 4 A 3 + 27B 2 ,

F. Luca, A. Yalçiner / Finite Fields and Their Applications 21 (2013) 1–10

7

therefore the splitting field L or f (x) over Q contains K = Q(e 2π i /3 ). Since p ≡ 1 (mod 3), it follows that the cubic roots of unity belong to F p . Let x1 , x2 , x3 , x4 be the roots of f (x) in Q. Formally, these four numbers depend on A and Bbut we shall omit this dependence in what follows. Then P2 is the set of all primes such that one of

x3i + Axi + B for i = 1, . . . , 4 is defined in F p . Chebotarev’s density

theorem now shows that ρ2 exists. To get a lower bound on it, we consider various cases. Assume first that f (x) has a root in K. Say x1 ∈ K. Then P2 contains the set of primes p such that p ≡ 1 (mod 3) and x31 + Ax1 + B is a quadratic residue modulo p. Thus, ρ2  1/4. Assume next that f (x) is a product of two irreducible factors of degree 2 over K. Say x1 and x2 are conjugated in a quadratic field K1 and x3 and x4 are conjugated in a quadratic field K2 . If K1 = K2 , then, as in the previous case, we get easily that ρ2  1/4. If K1 = K2 , then the Galois group of L over K, as a subset of S 4 , is the four element group G = {id, (12), (34), (12)(34)}. Of these 4 permutations, 3 of them have a fixed point. This shows that for a proportion 3/4 of all primes p ≡ 1 (mod 3) have the property that one of the xi ’s is defined modulo p for some i = 1, . . . , 4. Accounting also for the condition that x3i + Axi + B must be a square modulo p immediately implies that ρ2  3/16. Assume finally that f (x) is irreducible in K[x]. The Galois group of L over K is a transitive subgroup of A 4 . There are only two such, namely the subgroup G above and the entire A 4 . Let us show that the case Gal(L/K) = G is impossible. If this were so, √ then there exist√two elements a, b in K which√are√not squares in K such that if we put K1 = K( a ) and K2 = K( b ), then K1 = K2 and L = K( a, b ). Write

√

√

x1 = c + d a + e b + f

√

with c , d, e , f ∈ K.

ab

Then, since the minimal polynomial of x1 over K is the same as the minimal polynomial of x1 over Q, we get that the conjugates over K of x1 are exactly

√

√

x i = c + η1 d a + η 2 e b + η 1 η 2 f

√

for all η1 , η2 ∈ {±1}.

ab

Since TrL/K (x1 ) = x1 + x2 + x3 + x4 = 4c = 0, it follows that c = 0. But then

−4B = x1 x2 x3 + x1 x2 x4 + x1 x3 x4 + x2 x3 x4

1 1 1 1 = x1 x2 x3 x4 + + + =

−A

2

x1

x2

x3

1

√

d a+e b+ f

√

+

1

√ −d a + e b − f ab

1 1 + √ √ √ + √ √ √ d a − e b − f ab −d a − e b + f ab √ √



− A2 −2e b 2e b = + = 0, √ √ 3 e 2 b − a(d + f b)2 e 2 b − a(d + f b)2 3

√

x4

ab

√

√

giving B = 0. But if B = 0, then



E : y 2 = x x2 + A



has CM by Z[i ], a contradiction. So, it is not possible that Gal(L/K) = G, therefore Gal(L/K) = A 4 . Since A 4 has 9 permutations with at least one fixed point, it follows that a proportion of 9/12 = 3/4 primes p ≡ 1 (mod 3) have the property that one of xi is defined modulo p for some i = 1, . . . , 4,

8

F. Luca, A. Yalçiner / Finite Fields and Their Applications 21 (2013) 1–10

which by a previous argument implies that ρ2  3/16. From now on, we shrink if necessary P2 and work with a subset of it having the property that

#P2 (t ) =





ρ2 + o(1) π (t ), where ρ2 =

3 16

(t → ∞).

3. Removing n with too many primes in P1 or too few primes in P2 Let

ρ ∈ (0, 1) be fixed and assume that P is a subset of primes such that 1 p t

p

  = ρ + o(1) log log t as t → ∞.

Then “most” positive integers n have ωP (n)/ log log n ∈ [ρ − ε , ρ + ε ] for any statement quantitative, assume that α ∈ (0, ρ ) and let

ε > 0. To make this

  M−,α ,P (x) = n  x: ωP (n) < (ρ − α ) log log x .

(11)

Then

#M−,α ,P (x) 

x

(log x)β+o(1)

(x → ∞),

(12)

where

β = ρ − (ρ − α ) log

eρ

ρ −α

.

(13)

Similarly, putting

  M+,α ,P (x) = n  x: ωP (n) > (ρ + α ) log log x ,

(14)

then

#M+,α ,P (x) 

x

(log x)γ +o(1)

(x → ∞),

(15)

where



γ = ρ − (ρ + α ) log

eρ

ρ +α

.

All this can be found in Chapter 0 of [2]. We now choose P = P1 and P2 , respectively, let be two small numbers in (0, ρ1 ) and (0, ρ2 ), respectively, to be determined later, and put

N4 (x) = M+,α1 ,P1 (x) and N5 (x) = M−,α2 ,P2 (x).

(16)

α1 and α2

(17)

F. Luca, A. Yalçiner / Finite Fields and Their Applications 21 (2013) 1–10

9

Then the above arguments show that

#N4 (x)  #N5 (x) 



x

; (log x)γ1 x

(log x)β2

;

with γ1 = ρ1 − (ρ1 + α1 ) log

with β2 = ρ2 − (ρ2 − α2 ) log

e ρ1

ρ1 + α1 e ρ2

ρ2 − α2

;

.

(18)

4. The final argument

5

From now on, we assume that n ∈ N6 (x) = N E (x)\( i =1 Ni (x)). We go back to formula (7) and / N4 (x). We have first rework the upper bound for |k| given by (9) using the fact that n ∈



2 k2 − 1 au 2  a p = √ . |k| < 2k − − 1 2 n p u p |n

We have

au 2  2 2 2+o(1) = (log x)2α0 +o(1) (x → ∞) u2  τ u u  y

(19)

because n ∈ / N1 (x). Further, by (10), (19) and the fact that n ∈ / N4 (x), we get that

|k| < (log x)2α0 +o(1)

 a p 2 √ − 1 p

p |n p ∈P1

 (log x)2α0 +o(1) 3ωP1 (n)  (log x)2α0 +(ρ1 +α1 ) log 3+o(1)

(20)

as x → ∞. Put L for the above upper bound on |k|. Now we note that if p ∈ P2 , then a2p − p ≡ 0

(mod 3). In particular, the left-hand side of (7) is a multiple of 3ωP2 (t (n)) . Put M = (ρ2 − α2 ) log log x , and let T = T (x) be the largest positive integer T such that q1 · · · q T  log x, where q j is the jth prime. By the Prime Number Theorem, T = (1 + o(1))π (log log x) = o(log log x) as x → ∞. Then since / N5 (x), we have that ω(t (n)) = ωP2 (n) − u  y < log x, it follows that ωP2 (u )  T , therefore, since n ∈ ωP2 (u )  M − T , so 3M −T | an2 . In particular, 3M −T | 2kn − (k2 − 1). We recall that k = 0 because n ∈/ N3 (x). Fix |k| ∈ [1, L ]. Since k and k2 − 1 are coprime, it follows that k is coprime to 3, so the above congruence puts n into a certain residue class (namely (k2 − 1)(2k)−1 ) modulo 3 M − T . The number of such n  x is therefore at most x/3 M − T + 1  2x/3 M − T . Summing this bound up over all possible values of k with |k|  L, we get that #N6 (x) 

4Lx

=

x

as x → ∞.

(21)

Comparing the bounds (3), (5), (8), (18) and (21), it follows that we need to choose that

α0 , α1 , α2 such

3M −T

(log x)((ρ2 −α2 )−(ρ1 +α1 )) log 3−2α0 +o(1)

10

F. Luca, A. Yalçiner / Finite Fields and Their Applications 21 (2013) 1–10



α0 /2 = γ1 = ρ1 − (ρ1 + α1 ) log

α0 /2 = β2 = ρ2 − (ρ2 − α2 ) log

e ρ1

ρ1 + α1 e ρ2

ρ2 − α2 α0 /2 = (ρ2 − ρ1 − α1 − α2 ) log 3 − 2α0 .

;

;

Solving the above system we get α0 /2 = 0.0000108058 . . . (with corresponding and α2 = 0.00286982 . . .) implying the desired conclusion.

α1 = 0.0028396 . . .

Acknowledgments We thank the referee for comments which improved the quality of this paper. F.L. worked on this paper in 2012, during a visit to Turkey supported by Tubitak. He thanks Tubitak for financial support. Work of A.Y. was supported by Tubitak and the Scientific Research Office (BAP) of Selçuk University. References [1] L. Clozel, M. Harris, R. Taylor, Automorphy for some l-adic lifts of automorphic mod l Galois representations, with Appendix A, summarizing unpublished work of Russ Mann, and Appendix B by Marie-France Vignéras, Publ. Math. Inst. Hautes Études Sci. 108 (2008) 1–181. [2] R.R. Hall, G. Tenenbaum, Divisors, Cambridge Univ. Press, 1988. [3] F. Luca, C. Pomerance, Irreducible radical extensions and Euler-function chains, in: Landman, Nathanson, Nešetril, Nowakowski, Pomerance (Eds.), Combinatorial Number Theory, Proceedings of the “INTEGERS” Conference in Honor of R. Graham’s 70th Birthday, de Gruyter, 2007, pp. 351–362. [4] J.H. Silverman, The Arithmetic of Elliptic Curves, Springer-Verlag, Berlin, 1995.