Searching for simultaneous arithmetic progressions on elliptic curves

arXiv:math/0703624v1 [math.NT] 21 Mar 2007 Searching for simultaneous arithmetic progressions on elliptic curves Irene Garc´ıa–Selfa∗Jos´e M. Tornero∗ November, 2.004 Abstract We look for elliptic curves featuring rational points whose coordinates form two arithmetic progressions, one for each coordinate. A constructive method for creating such curves is shown, for lengths up to 5. 2000 Mathematics Subject Classification: 11G05, 11B25. Keywords: Elliptic curves, Arithmetic progressions. 1 Introduction Let us consider an elliptic curve Edefined over Qby a general Weierstrass equation Y2+a1XY +a3Y=X3+a2X2+a4X+a6, ai∈Q. Definition.– We will say that the points P0, ..., Pn∈Eare in (or form an) x–arithmetic progression if their x–coordinates are. The symmetric concept of y–arithmetic progression is defined analogously. We will note by Sx(E) and Sy(E) the maximal number of points in x–arithmetic progression and y–arithmetic progression respectively that can be found in E. ∗Both authors supported by FQM 218 and BFM 2001–3207 and FEDER. 1 Remark.– What we ignore about arithmetic progressions on elliptic curves is far more than what we know. Apparently the first one in considering the problem was S.P. Mohanty ([5]) who focused on the Mordell equation Y2=X3+kand looked for integral points forming arithmetic progressions of difference 1. He proved that for all these curves and for just these progressions Sx(E)≤2 and Sy(E)≤4. Later on, Lee and V´elez ([4]) fixed their attention in the same family, but they took into consideration all possible progressions. They found infinite families of such curves verifying Sx(E)≥4 and (not simultaneously) Sy(E)≥6. Bremner, Silverman and Tzanakis ([2]) took another quite popular family of curves, Y2=X(X2−n2), and proved that for all these curves and considering only integral points Sx(E)≤5. They went further and proved very interesting results on this line concerning free subgroups of rank one in arbitrary elliptic curves. Their proofs were quite lengthy, involving very delicate computations of local heights. After this (although it was published earlier) Bremner ([1]) carried out very clever computations in order to show that there are infinitely many curves verifying Sx(E)≥8. Campbell ([3]) followed his lines to produce curves with eight points in x–arithmetic progression and also a genus 1 curve with 12 such points (unfortunately the curve was not in Weierstrass form!). Maybe the more intriguing parts of Bremner’s results are, on one side, the numerical evidence of the fact that the length of x–arithmetic progressions on elliptic curves may well not be bounded (although Bremner himself did not risk to state such a conjecture) and, on the other hand, the apparent connection between long arithmetic progressions and high ranks of the Mordell-Weil group. From these last papers it becomes clear that the main problem, when one deals with arithmetic progressions on elliptic curves, is that the number of parameters involved (if one wants to work with full generality) becomes unmanageable. This is why the only precise results known are bounded to one–parametric families. This paper is devoted to finding simultaneous arithmetic progressions on elliptic curves. The precise definition goes as follows: Definition.– We will say that the points P0, ..., Pn∈Eare in (or form a) simultaneous arithmetic progression if their x–coordinates and its y–coordinates are (maybe not in the same order). 2 We will note by Sx,y(E) the maximal number of points in simultaneous arithmetic progressions that can be found in E. Remark.– When one looks for such progressions with more than three points, things start to become difficult, as the ordering of the points in both progressions can not coincide (see below for a precise explanation of this). Our initial aim was following [1] and imposing the conditions with full generality in order to narrow the search with respect to the x– arithmetic progression problem. Unfortunately it eventually became too complicated to deal with as well. So, we took a different point of view from Bremner’s ([1]): we have worked with (almost) arbitrary elliptic curves, but we have looked for a restricted class of arithmetic progressions. With this starting point, we have been able to prove the following results, which were unknown so far: Theorem 1.– There are elliptic curves over Qwith Sy(E)≥7. Theorem 2.– There are elliptic curves over Qwith Sx,y(E)≥5. 2 A construction scheme As it is well–known ([7]), any change of variables preserving the Weierstrass form of Emust be of the form X′=u2X+r, Y ′=u3Y+sX +t, so the existence (and the length) of x–arithmetic progressions is not affected by changes of variables. This also implies that, up to such a change, we can consider all the points in a certain x–arithmetic progression to be integral (the ycase being symmetric). Therefore as an immediate corollary of Siegel’s theorem (see [6] for the original proof or [7] for a modern one) Ecannot contain infinite x–arithmetic progressions. Hence we are left to study if there is a universal bound, independent of the chosen curve, for the length of arithmetic progressions. So, assume we have a curve with an x–arithmetic progression on it, say P0, ..., Pn, and assume [2]P06=O. Then we can take P0to (0,0) and rotate the axes in order to take the tangent at (0,0) to the line X= 0 (warning: this may dismantle y–arithmetic progressions in the 3 original model). Then our curve must look like E:Y2+aXY +bY =X3+cX2. Now, if we make the change X7−→ −c b3 X, Y 7−→ −c b2 Y, our curve will be defined by E(a, b) : Y2+aXY +bY =X3−bX2. This equation (also called Tate normal form) features two another obvious points in E(a, b) other than (0,0): (b, 0) and (0,−b). Hence, in what follows we will make a new (strong indeed) assumption and suppose P1= (b, 0). This implies that the difference in our x–arithmetic progression must be precisely b. In fact, we will actually take P0= (0,−b) in order to avoid the repetition of 0 in the y–progression. We must look for conditions which assure us that points (kb, yk) appear in E(a, b). Furthermore, we want these points to form as well a y–arithmetic progression. Although the subindex of the points will represent the increasing order in the first coordinate it is obvious that, as for the second coordinate is concerned, the subindex of a point might have nothing to do with its position in the y–arithmetic progression. In order to do that mind that if Pk= (kb, yk)∈E(a, b) then it must hold yk=−b(ak + 1) ±bp(ak + 1)2+ 4k2b(k−1) 2, so proving the existence of Pkis equivalent to finding a rational solution for the diophantine equation Z2 k= (ak + 1)2+ 4k2(k−1)b. We make now the change of variables αk=ak + 1 + Zk, βk=ak + 1 −Zk, and so our previous equation becomes αkβk+ 4k2(k−1)b= 0. 4 Now, when we gather together the equations for P2, ..., Pnwe must take into account that, for all k,αk+βk= 2ak + 2. The diophantine system which is equivalent to the existence of our x–arithmetic progression is, therefore α2β2+16b= 0 α3β3+72b= 0 .... . .. . .. . . αnβn+4n2(n−1)b= 0 3α2+ 3β2−2α3−2β3= 2 . . ..... . .. . . nα2+nβ2−2αn−2βn= 2(n−2) This can be seen as the intersection of (n−1) hyperquadrics and (n−2) hyperplanes in the affine (2n−1)–dimensional space over the rationals. Such a system is clearly unmanageable for, say n= 10 (not to say less). Observe that, as b= 0 leads to no progression at all, we must in fact ask for all αiand βito be non–zero. What looks particularly useful with this formulation of the problem is that, in this context, we can write yk=b 2(−(ak + 1) ±Zk), and hence the two solutions for ykare precisely, −bαk/2 and −bβk/2. We can choose freely one of them, as the above equations are symmetric in {αi, βi}. This is quite useful in order to force the existence of a simultaneous y–arithmetic progression. 3 Numerical results Unlike the case of x–arithmetic progressions (for instance, in [1]), where the conditions for the existence of a progression of length, say, kare also to be filled for the existence of a longer progression, in our case, if a set P0, ..., Pndisplays a simultaneous arithmetic progression, that does not mean, in principle, that P0, ..., Pkalso does (although we have not been able so far to find an example of this). All the calculations in this section have been carried out with MapleV and PARI/GP. 5 Lenght 3.There are infinitely many curves with a simultaneous arithmetic progression of length 3. In fact, we may even ask both coordinates to share the same order in the progression. This clearly implies the three points must be collinear, and it also explains why one cannot hope to have such examples with longer lengths. For instance, all curves of the family E(b) : Y2+ (2b−1)XY +bY =X3−bX2 have such a progression: {(0,−b),(b, 0),(2b, b)}. Length 4.The system needed for the existence of an x–progression of length 4 is α2β2+16b= 0 α3β3+72b= 0 3(α2+β2)−2(α3+β3) = 2 with y2=−β2b/2, y3=−β3b/2. Now, we can choose two values for β2and β3which will guarantee the existence of the y–arithmetic progression. These values, when substituted in the system will lead to a system of three linear equations in α2, α3, b whose matrix is    β20 16 0 0β372 0 3−2 0 2−3β2+ 2β3   . Hence we will have a unique solution if β26=β3/3, a family of solutions if β2=−2/3, β3=−2 and no solutions at all otherwise. The pair (β2=−2/3, β3=−2) does not guarantee a length 4 y– progression but it will appear later, in the length 5 study. If we want 0 and −bto be in the y–progression, there are only six possibilities for this sequence and they are precisely (2b, b, 0,−b),(b, 0,−b, −2b),(0,−b, −2b, −3b),b 2,0,−b 2,−b, 0,−b 2,−b, −3b 2,0,−b 3,−2b 3,−b. Each of them allows two possible choices for β2and β3, except the penultimate case, in which (β2= 1, β3= 3) is forbidden. The results 6 found are shown in the following table, except the cases (β2=−2, β3= −4), which leads to a degenerate case b= 0, and (β2= 4, β3= 6), which gives the same curve as (β2= 4, β3=−2). (β1, β2) (a, b)SxSy (−4,−2) (−5/3,−1/6) ≥5≥5 (−2,4) (−7/15,4/15) ≥5≥4 (−1,1) (−29/48,7/192) ≥4≥4 (2/3,4/2) (−7/9,2/27) ≥4≥5 (1,−1) (−5/16,1/64) ≥6≥7 (4/3,2/3) (−7/45,−1/270) ≥4≥4 (3,1) (29/96,−5/128) ≥4≥4 (4,−2) (1/3,1/6) ≥4≥5 (6,4) (25/21,−2/7) ≥6≥4 This search gave us the first interesting example, announced in Theorem 1: the existence of an elliptic curve with a y–arithmetic progression of length 7, a fact not reported until now, as far as we know. We will look closer at this example below. Length 5.The system needed for the x–progression is α2β2+16b= 0 α3β3+72b= 0 α4β4+192b= 0 3(α2+β2)−2(α3+β3) = 2 4(α2+β2)−2(α4+β4) = 4 where yj=−βjb/2, for j= 2,3,4. Observe now that a blind choice of (β2, β3, β4) as in the previous case will lead, in general, to an incompatible system of equations. In fact, the rank of the coefficients matrix is 4, except in the case β2=β3/3 = β4/6 but, in this case the system is incompatible. However, these relations will prove useful later on. There are ten possible progressions of length 5 containing both 0 and −b, each of them permitting six different choices for the triple (β2, β3, β4), which are the permutations of a single choice. From this 60 cases only two led to a compatible system. This two cases, shown below, prove therefore Theorem 2. β2=−4, β3=−2, β4=−6 7 This triple gives the curve E(−5/3,−1/6) with the following points lying on it: 0,1 6,−1 6,0,−2 6,−2 6,−3 6,−1 6,−4 6,−3 6, which form a simultaneous arithmetic progression of length 5. Note that, in this case, α2=−2/3, hence it can be considered as a subcase of the infinite family of x–arithmetic progressions found with lenght 4. β2= 1, β3=−1, β4=−2 The resulting curve is E(−5/16,1/64) with the following points: 0,−2 128,1 64,0,2 64,−1 128,3 64,1 128,4 64,2 128, which form a simultaneous arithmetic progression (note that this curve also appeared in the previous case). Furthermore, other points lying on the curve are 1 8,−4 128−1 32 ,−3 128,5 64,−1 64 , hence as noted above Sx≥6, Sy≥7, although there are no simultaneous progressions in the curve of length 6. Remark.– Please note that, as we mentioned at the beginning of the section, both examples of simultaneous arithmetic progressions of length 5 also have arithmetic progressions of length 4. This is also the case with the nine examples of length 4 found; however, these data seem to us not enough for conjecturing that this holds in general. Back to our search, another reasonable way of constructing simultaneous arithmetic progressions of length 5 seems to be using 0 and −b(a+ 1) as terms of our progression and again we have 60 possibilities, all of them useless. Most cases can be discarded using one of the following arguments, explained with two examples. β2= 4(a+ 1), β3= 6(a+ 1), β4= 8(a+ 1) If this choice made sense, we would have the y–progression {0,−b(a+ 1),−2b(a+ 1),−3b(a+ 1),−4b(a+ 1)}. 8 As we said above, there are 5 more possibilities in order to assure this sequence, permuting the values of β2,β3and β4, but this one suits us well as illustration. Our system matrix is now        −4(a+ 1) 0 0 16 0 0−6(a+ 1) 0 72 0 0 0 −8(a+ 1) 192 0 3−2 0 0 2 4 0 −2 0 4        with determinant −3072(a+ 1)2. Hence we must take a=−1 if we want the system to have solution, but this case is degenerate. β2= 6(a+ 1), β3= 4(a+ 1), β4= 8(a+ 1) This case is also impossible, but for different reasons: it has a matrix with determinant 7168(a+ 1)2(4a+ 5). The choice a=−5/4 leads to a compatible system. But remember that it also must hold a=αi+βi−2 2i, i = 2,3,4, which is not true in this case. In fact, this extra condition annihilates the advantage of making a parameter choice, because the value of the parameter cannot be truly arbitrary. 4 Final remarks The above arguments, when applied to a length nsimultaneous arithmetic progression lead to n(n−1)/2 progressions with (n−2)! possible value choices for each one. That is, n!/2 systems have to be checked. We have done the 360 calculations for n= 6, obtaining no simultaneous progressions. Of course, that does not mean that the case n= 7 will also be unsuccessful, as we noted before, but most probably this kind of search will not produce further results. As a final comment, we would like to stress possible ways for expanding the results in this paper: (a) The first obvious thing to do it is enlarging the scope of the progressions considered. Full generality, although desirable, may be too messy for dealing with, at least with the current state of the art. A simpler way could be, for instance, considering x– progressions in which bappears, although not necessarily as the 9