The average number of integral points on the congruent number curves

We show that the total number of non-torsion integral points on the elliptic curves $\mathcal{E}_D:y^2=x^3-D^2x$, where $D$ ranges over positive squarefree integers less than $N$, is $O( N(\log N)^{-1/4+\epsilon})$. The proof involves a discriminant-lowering procedure on integral binary quartic forms and an application of Heath-Brown's method on estimating the average size of the $2$-Selmer group of the curves in this family.


Introduction
Given an elliptic curve over Q with short Weierstrass model (1) E : we study the quadratic twists of E, with the model (2) where D denotes a positive squarefree integer.Consider the set of integral points E D (Z) := (x, y) ∈ Z 2 : y 2 = x 3 + AD 2 x + BD 3 .
It follows from a result of Mordell [17] that #E D (Z) is always finite.We are interested in the distribution of the number of integral points #E D (Z) in quadratic twist families, when E D are ordered according to the size of D. If E(Q) contains a 2-torsion point, this point must have the form (a, 0) for some integer a under the model (1), then (aD, 0) ∈ E D (Z) for all squarefree integers D. Therefore we call an integral point non-trivial if it is not a 2-torsion point of E D (Q).Define the set of nontrivial integral points on E D to be Granville [9] conjectured that almost all curves within a quadratic twist family have no non-trivial integral point.We state the conjecture adapted to our model (2).Conjecture 1.1 (Granville [9]).Fix A, B ∈ Z such that 4A 3 + 27B 2 = 0. Let E D : y 2 = x 3 + AD 2 x + BD 3 , D ∈ D. Then , where C A,B is a constant that depends only on A, B.
We note that Granville's original conjecture considers a different model Dy 2 = f (x), where f ∈ Z[x] and deg f = 3.When f (x) = x 3 + Ax + B, any point (x, y) ∈ Z 2 satisfying Dy 2 = f (x) corresponds to a point (Dx, Dy) ∈ E D (Z), so there are fewer integral points using the model Dy 2 = f (x) when compared to our model (2).The exponent 1 2 stated in Conjecture 1.1 replaces 1  3 in the original conjecture because of this discrepancy.The exponent 1  2 is suggested by some heuristics we gave in [5, p. 6677-6678] for the family y 2 = x 3 − D 2 x.
In this direction, Matschke and Mudigonda [16] handled the case when f (x) is reducible, assuming the abc conjecture.
Our goal here is to gain progress towards Conjecture 1.1 on a specific quadratic twist family.We restrict our attention to the congruent number curve E : y 2 = x 3 − x, and study its twists It is well known that the torsion subgroup of E D (Q) is {O, (0, 0), (±D, 0)} ∼ = Z/2Z × Z/2Z (see for example [13, Chapter I, Proposition 17]), where O denotes the point at infinity.
For this family E D , we can deduce from existing results that all moments of #E D (Z) are finite.The 2-Selmer groups of E D , which we denote by Sel 2 (E D ), is a finite group with order 2 that is defined via local conditions and admits an injection E D (Q)/2E D (Q) ֒→ Sel 2 (E D ) (see for example [21,Chapter X]).In particular, the 2-Selmer rank provides an upper bound to the rank rank(E D (Q)) of the Mordell-Weil group of E D (Q).It is usually easier to compute the 2-Selmer groups of elliptic curves with a torsion subgroup Z/2Z×Z/2Z over Q, since then most of the work can be done over Q. Heath-Brown [11, Theorem 1] computed all the moments of the size of the 2-Selmer groups of E D .For any fixed positive integer k, he showed that (3) lim where c k are explicit constants that can be bounded by 3 k(k+1) .Since the 2-Selmer rank provides an upper bound to the the rank of E D , (3) implies that (4) lim sup Lang [14, page 140] conjectured that the number of integral points on a quasi-minimal Weierstrass equation of an elliptic curve E should be bounded only in terms of rank E(Q).For the family E D , if follows from known results in this direction [20, Theorem A], [12,Theorem 0.7], that there exists some absolute constant C 1 , such that .
In [5], we showed that C 1 in ( 5) can be taken as 3.8.Combining the upper bound ( 5) and ( 4), we can bound the k-th moment (6) lim sup , where C 2 is an absolute constant.We will show that in fact the moments of #E D (Z) should each tend to 0. The following is our main result.
Theorem 1.3.For any ǫ > 0, we have This shows that the average size of #E * D (Z) tends to 0 as N tends to infinity, since # D N ∼ 6 π 2 N. Theorem 1.3 implies that (7) #{D ∈ D N : +ǫ .An application of Hölder's inequality using ( 6) and (7), gives Corollary 1.4.For any ǫ > 0 and k > 0, we have We now give an outline of the proof of Theorem 1.3.In Section 2, for each integral point (x, y) ∈ E D (Z), we use Mordell's correspondence [18,Chapter 25] to construct a corresponding integral binary quartic form f that represents 1 and has discriminant related to the discriminant of E. Then in Section 3, we show that by picking an auxiliary prime p | D/ gcd(x, D), we can transform f into an integral binary quartic form F that represents p and has discriminant lowered by a factor of p 6 .In Section 4, we show that gcd(x, D) can be controlled by the image of (x, y) in the 2-Selmer group of E D under the map Then in Section 5 we extract some information about the distribution of 2-Selmer elements from Heath-Brown's work [10,11] to show that for almost all D, we are able to pick a prime p of a suitable size.In particular, this p is not too small, so that there are o(N) many discriminants for the discriminant-lowered quartic F to take.At the same time, this p is not too large, so that each GL 2 (Z)-equivalence class of F can only be the image of finitely many integral points by applying bounds on the number of solutions to Thue inequalities.In Section 6, we use Hölder's theorem and (6) to bound the contribution from the the exceptional curves to the number of integral points.In Section 7, we proceed to count the set of those quartics F that were discriminant-lowered by some suitable p.We make use of the fact that every integral binary quartic form is SL 2 (Z)-equivalent to at least one reduced form with bounded seminvariants [6].Applying the syzygy satisfied by the seminvariants returns a set of integral points on twists of E with bounded height.Then Theorem 1.3 follows from an application of an upper bound by Le Boudec [15].

Integer-matrix binary quartic forms
We say that a binary quartic form is integer-matrix if it has the form Given any integral binary quartic form f and (x 0 , y 0 ) ∈ Z 2 , define the action of on the pair (f, (x 0 , y 0 )) by This action preserves the value of f (x 0 , y 0 ).We recall some facts about the seminvariants of quartic forms [6, Section 4.1.1].For our convenience, we choose to scale the seminvariants differently than in [6] , since we will only be dealing with integer-matrix binary forms.The invariants of f are 2  2 , and The discriminant of f is 3 ) The seminvariants attached to the form are I, J, a = a(f ) = a 0 , Comparing to the formulas in [6, Section 4.1.1],here we have taken out a factor of −48 from their H, a factor of 32 from their R, a factor of 12 from their I, a factor 432 from their J, and a factor of 256 • 27 from their ∆.The seminvariants are related by the syzygy (8) Notice that (H, 1 2 R) defines an integral point on a twist of the elliptic curve y 2 = x 3 − I 4 − J 4 .2.1.Mordell's correspondence.For integers A, B ∈ Z such that 4A 3 + 27B 2 = 0, define an elliptic curve over Q with affine integral Weierstrass model The discriminant of E A,B is given by For integers c, d, z ∈ Z, define an integer-matrix binary quartic form The following correspondence is given by Mordell [18, Chapter 25] (or see [3,Section 2.3] for a modern interpretation).
Theorem 2.1 (Mordell).Fix an integer L = 0.There is a bijection The inverse map comes from the syzygy (8) satisfied by the seminvariants, but we will only make use of the the direction from A to B in Theorem 2.1.

Lowering the discriminant
We now fix an elliptic curve E : y 2 = x 3 + Ax + B, A, B ∈ Z and consider its quadratic twists E D : Denote the space of integer-matrix binary quartic forms by V .Define (10) Ψ : given by where ).We will show that Ψ is well-defined in Lemma 3.1 and injective in Lemma 3.2 In work of Bombieri and Schmidt [4], to bound the number of solutions to a Thue equation F 1 (X, Y ) = h, they transformed the integral binary form F 1 to a different form F 2 , whose discriminant is raised by a factor of h 6 , and so that there is a solution to F 2 (X, Y ) = 1.Some applications of this idea can be found in [2,1].Here we attempt to carry out the reverse process on the integral quartic forms f P to lower their discriminants.Lemma 3.1.Let P = (c, d) ∈ E D (Z) and take f P as defined in (9).Fix a positive squarefree integer M dividing D that is coprime to 6c.Then c is a square modulo M, and for any integer k such that k 2 = c mod M, we have that is an integer-matrix binary quartic form.Moreover, we have ( 1) , therefore c is a square modulo M. Taking any integer k such that k 2 ≡ c mod M, by Hensel's lemma we can find a lift of k such that k ≡ K mod M and ( 11) It suffices to show that F is an integer-matrix binary quartic form with this choice of K, since otherwise k = K + uM for some integer u, and we can consider By ( 11) and ( 12), we see that the coefficients of are all divisible by M 3 .Therefore F is integer-matrix from the coefficients.The remaining properties are then a straightforward check from the definition of F . and Then since we have The X 3 Y -coefficients of f P and f Q are both 0, it must be that uM Heath-Brown [10,11] computed the moments of the size of the 2-Selmer group modulo 2-torsion in this family is 3.We will extract some information about the 2-Selmer elements in this family from the argument in [10,11], in order to show that we can usually pick a suitable M to apply Lemma 3.1.

The 2-Selmer group of y
In the following sections we will specialise in the case when A = −1 and B = 0, that is, the quadratic twists family Heath-Brown [10,11] computed the moments of the size of the 2-Selmer group modulo 2-torsion in this family is 3.We will extract some information about the 2-Selmer elements in this family from the argument in [10,11], in order to show that we can usually pick a suitable M to apply Lemma 3.1.
The 2-Selmer group of E D is defined to be Since E D has full 2-torsion, there is an isomorphism , and it is possible to obtain explicit equations for the homogeneous spaces (See for example [21, Chapter X, Proposition 1.4]).
Explicit equations for homogeneous spaces for the curves E D were found as part of Heath-Brown's argument [10, Section 2].As we will see, each 2-Selmer element of E D (Q) corresponds to a system of two binary quadratic forms that is everywhere locally solvable.We will follow [10, Section 2] to recover the equations.(1) the system is everywhere locally solvable, and (2) Consider the injective homomorphism (14) θ : given by ( Lemma 4.2.The set W D is in bijection to where ψ here denotes the natural map In particular, if the image of (c, d) ∈ E D (Z) under the map Proof.We study im θ following [10, Section 2].Suppose (x, y) ∈ E D (Q), and write x = r/s and y = t/u, where r, s, t, u are integers and gcd(r, s) = 1 and gcd(t, u)=1.Putting this into y 2 = x 3 − D 2 x, we have r(r + sD)(r − sD)u 2 = t 2 s 3 .
Then since gcd(t, u) = 1 and gcd(r, s) = 1, we must have s 3 = u 2 , so s = W 2 for some integer W . Now write gcd(r, D) = D 0 , and we see that D 3 0 | t 2 , hence D 2 0 | t since D 0 is squarefree.Then writing D 4 = D/D 0 , we have gcd(r ′ , sD 4 ) = 1 by construction, and we have The factors on the left are pairwise coprime except possibly a common factor of 2, which only occurs when r ′ and sD 4 are both odd, in this case r ′ , (r ′ + sD 4 )/2, (r ′ − sD 4 )/2 are pairwise coprime.Now we can write r would need to be odd, and D 2 , D 3 would need to be even and squarefree.This produces a solution to the system An element in Sel 2 (E D ) corresponds to a system of this form that is everywhere locally solvable.To fix the signs of D 1 , D 2 , D 3 , D 4 and their valuations at 2, we will use the torsion points of E D .

Generic 2-Selmer elements
We want to show that those (D 1 , D 2 , D 3 , D 4 ) that appears usually satisfies some nice properties that will eventually allow us to pick M for Lemma 3.1.Take where κ > 0 is an absolute constant as defined in [10, Lemma 7] or [11,Section 3].This N ‡ is C 4 in the notation of [10].
Henceforth 0 < ǫ < 1 2 will be a fixed constant.Let S be the interval S := exp((log N) 2ǫ ), exp((log N) 1−2ǫ ) .We will prove the following.(1 and some η ∈ {0, 1, 2, 3, 4} , and (2) one of the conditions (W1) and (W2) listed below.(W1) For some i = 1, 2, 3, 4, we have D i < N ‡ , and (W2) We have D i ≥ N ‡ for all i = 1, 2, 3, 4, and there exists an i such that D i has no prime factor in S. In the above notation, η = 0 implies that We first consider the case when D is odd.By [10, Lemma 3], considering the local solvability of (13) at each prime, we can package the condition (D 1 , D 2 , D 3 , D 4 ) ∈ W D as a sum of product of Jacobi symbols.Write D = (D ij ) as a 16-tuple of positive odd integers, where the indices are in the range For even D, since we are only aiming for an upper bound, we can ignore any local conditions at 2 when considering the solvability of (13).Following the proof of [10, Lemma 3], the only difference being essentially pulling the factor of 2 in D into the 2 • term, set where Then For each η = 0, 1, 2, 3, 4, we will want to estimate the sum where the sum is taken over all positive odd integers Following [10, Section 3] and [11, Section 3], dissect the sum according to the size of each D ij in the factorisation.For each (i, j), take A ij to run over powers of 2. For A = (A ij ), define the restricted sum where the sum is taken over all 16-tuples of odd positive integers We can summarise the error term estimates in [10,11] as follows.
where the sum is over all A other than those that satisfy for all indices u, where U is one of Moreover, the estimate still holds if we further impose that j D ij ≤ N i for each i = 1, 2, 3, 4.
Strictly speaking [10] only applies to the case for odd D, namely η = 0.The modification made in Lemma 5.2 for even D was in the sets in (17), since the possible U depends on the variables in β η .Following the case analysis in the proof of [10,Lemma 11] with this change, the sets in ( 17) can be found for each of η = 1, 2, 3, 4.
We are ready to bound the contribution from (W1).
Lemma 5.3.For each η ∈ {0, 1, 2, 3, 4}, we have Proof.We want to bound ( 18) Dissect the sum using A, then apply Lemma 5.2 to (18).The only possibility that this S η (A) is not covered by Lemma 5.2 is if A satisfies ( 16) with {40, 41, 42, 43} when η = 0, {30, 31, 32, 34} when η = 2, {40, 41, 42, 43} when η = 4.This implies that which are from the excluded torsion points.Therefore the required estimate follows from that in Lemma 5.2.5.2.Prime divisor of a large D i .We now bound the contribution from (W2).Assume that it is D 4 ≥ N ‡ that is not divisible by any prime in S. The cases with D 4 replaced by D 1 , D 2 , D 3 the same after relabelling.We want to bound (19) For each D that appears in the sum of g η (D), we can find a set of indices U = {1i, 2j, 3k, 4l}, where i, j, k, l are not necessarily distinct, such that D 1i , D 2j , D 3k , D 4l > (N ‡ ) 1 4 .Lemma 5.2 do not immediately apply because of the restrictions on the primes dividing D 4 .However it will be sufficient to follow [10, Section 3].Definition 5.4.We call two indices (i, j) and (k, l) linked if i = k and precisely one of the conditions l = 0, i, j = 0, k holds.
We will show that the indices in U are pairwise unlinked with the number of exceptions contributing O(N(log N) − 1 4 +ǫ ) to (19).When there are linked indices u and v, the Jacobi symbol between D u and D v appears in g η non-trivially.When both D u and D v are large, then the large sieve ([10, Lemma 4]) can be applied to obtain cancellation.Imposing further restrictions to each of D u , D v , (D w ) w =u,v do not affect the estimates in [10, Section 3].We state the modified statement below.Lemma 5.5 ([10, Lemma 5]).Suppose u and v are linked indices.Take C 1 to be any collection of D u ≥ (log N) 544 .Take C 2 to be any collection of D v ≥ (log N) 544 .Take C 3 to be any collection of B = (D w ) w =u,v .Then Suppose u and v are linked.When one of D u is large, and D v = 1 is small, Siegel-Walfisz for character sums ([10, Lemma 6]) is used instead.For this we need that the sum over D u is over an interval, but we can still impose restrictions on D v and (D w ) w =u,v without affecting the estimates.Lemma 5.6 ([10, Lemma 7]).Suppose u and v are linked indices.Take C 2 to be any collection of 1 < D v < (log N) 544 .Take C 3 to be any collection of B = (D w ) w =u,v .Define where the sum is over all A u ≥ (N ‡ ) 1 4 that are powers 2.
Combining Lemma 5.5 and Lemma 5.6 we have the following.
Lemma 5.7.Suppose u and v are linked indices.Take C 2 to be any collection of D v > 1.Take C 3 to be any collection of B = (D w ) w =u,v .Define where the sum is over all A u ≥ (N ‡ ) 1 4 that are powers 2.
We can now return to the set U = {1i, 2j, 3k, 4l}.Since by construction D u > (N ‡ ) 1 4 for all u ∈ U, we can assume that the indices 1i, 2j, 3k, 4l are pairwise unlinked by Lemma 5.5.Now suppose v / ∈ U.If v is not linked any one of 1i, 2j, 3k, then {1i, 2j, 3k, v} is a set of unlinked indices.Comparing against the 24 possible sets of unlinked indices in [10, Lemma 9], if {1i, 2j, 3k, 4l} and {1i, 2j, 3k, v} are both sets of unlinked indices, they must be the same set.Therefore v must be linked to one of {1i, 2j, 3k}, and this allows us to apply Lemma 5.7, so we are left with the terms in the sum (19) with D v = 1 for all v / ∈ U. Noting that there are only a maximum of 24 possible sets of unlinked indices U, and putting in To bound the main term here we make use of the following result.
Lemma 5.8 ([19, Theorem 1]).Fix 0 < ǫ < 1 and some positive constant C. Let f be a multiplicative function such that f (p ℓ ) ≤ C for all prime p and ℓ ≥ 1.Then By Lemma 5.8 and Mertens' theorem, This gives the following estimate.

Counting generic points
By Lemma 6.1, we may exclude any D ∈ G N .Any integral point in E D (Z) maps to 2-Selmer element, and hence to W D under the map Recall that E * D (Z) = E D (Z) \ {(0, 0), (±D, 0)}.For the non-trivial integral points that has image of the type (S2), we have the following bound from [5,Theorem 1.4] and the discussion after [5,Theorem 10.1].This gives a solution (x, y) = (1, 0) • γ −1 to the Thue inequality where h := exp((log N) 1−2ǫ ) is taken so that h is greater than any M in (20).In particular this solution is primitive (x and y coprime), since γ −1 ∈ GL 2 (Z) has determinant ±1 and entries in Z.The solutions to the Thue inequality constructed from different elements (F, (1, 0)) ∈ im Ψ are distinct as long as (F, (1, 0)) are from different GL 2 (Z)-equivalence classes.Indeed, suppose , then if the solutions produced on ( 22) are same, we also have (1, 0) A result by Evertse [8, Theorem 6.3] implies that when 2 8 ∆(F 0 ) ≥ (13h) 10 , the number of solutions to ( 22) is bounded by some absolute constant.Since ∆(F 0 ) = (2 D) 6 ≫ exp(12(log N) 1−2ǫ ) from ( 21), and h 10 = exp(10(log N) 1−2ǫ ), we conclude that the number of possible classes (F, (1, 0)) associated to each class of F 0 is absolutely bounded.7.2.Integral points with bounded height.Every integral binary quartic form is SL 2 (Z)-equivalent to at least one reduced form [6].The seminvariant a, H of the reduced form are bounded in terms of I and J.We restate a theorem in [6] in terms of our rescaled seminvariants.The scale factors of the seminvariants can be found in Section 2. The syzygy in (8) for F 0 now takes the form Notice that this gives an integral point (H, 1 2 R) ∈ E |a D| (Z) when a = 0. Below we show that the possibility that a = 0 does not happen because we restricted our counting to Z N .Lemma 7.4.Suppose F ∈ im Φ.Then any form in the SL 2 (Z)equivalence class of F has non-zero leading coefficient.
Proof.Assume for contradiction that Φ(P ) = F for some P = (c, d) ∈ Z N and F is equivalent to some quartic form with leading coefficient 0. Then there is a non-trivial integral solution to F (X, Y ) = 0. From Φ(P ) = F , we know that F (X, Y ) = 1 M 3 f P (MX + kY, Y ) for some M, k ∈ Z, so f P (X, Y ) = 0 also has a non-trivial solution, say (x 0 , y 0 ).Then from the expression of f P in (9), f P (x 0 , y 0 ) = x 4 0 − 6cx 2 0 y 2 0 + 8dx 0 y 3 0 + (4D 2 − 3c 2 )y 4 0 = 0. We see that y 0 = 0 since the solution is non-trivial.The roots of f P (X, 1) are For x 0 /y 0 to be rational, it must be that θ(P ) = (1, 1, 1), where θ is the 2-descent homomorphism defined in (14).This implies that P ∈ 2E D (Q), but such points are not in Z N .
The SL 2 (Z)-equivalence class of F 0 is determined by (a, D, H, R) and R is fixed by (a, D, H) up to ± sign, so it suffices to count the number of (a, D, H) that arise from this reduction, with the bounds (23).7.3.Torsion points.We have reduced the problem to counting integral points with bounded height on E |a D| , but since there are 2-torsion points on every curve, we need to deal with this possibility separately.