Arithmetic Progressions on y^{2} = x^{3} + k

Pallab Kanti Dey and Bibekananda Maji Harish-Chandra Research Institute Allahabad - 211019 India

                   pallabkantidey@gmail.com
                     bibekmaji@hri.res.in

Abstract

Many authors have studied the problem of finding sequences of rational points on elliptic curves such that either the abscissae or the ordinates of these points are in arithmetic progression. In this paper we obtain upper bounds for the lengths of sequences of rational points on curves of the type y^{2} = x^{3} + k, k ∈ ℚ \ {0}, such that the ordinates of the points are in arithmetic progression, and also when both the abscissae and the ordinates of the points are separately the terms of two arithmetic progressions.

1 Introduction

Let there exist n rational points P_{i} = (x_{i}, y_{i}), i = 1, 2, …, n on an elliptic curve E over ℚ such that either the n numbers x_{i}, i = 1, 2, …, n, or the n numbers y_{i}, i = 1, 2, …, n, are the terms of an arithmetic progression. We say that there exists an x-arithmetic progression, or a y-arithmetic progression, of length n on the elliptic curve E depending on whether the abscissae or the ordinates of the points P_{i} are in arithmetic progression. Further, we say there exists a simultaneous arithmetic progression of length n on the curve E if the n numbers x_{i}, i = 1, 2, …, n are in arithmetic progression and simultaneously, the n numbers y_{i}, i = 1, 2, …, n are also, in some order, the terms of an arithmetic progression.

The problem of finding x-arithmetic progressions on various models of elliptic curves has been studied by several mathematicians [3, 4, 5, 6, 10, 13, 13, 15]. The existence of simultaneous arithmetic progressions on the cubic model of elliptic curves has also been studied [7].

This paper is concerned with the existence of y-arithmetic progressions as well as simultaneous arithmetic progressions on Mordell curves which are elliptic curves defined by equations of the type,

y^{2} = x^{3} + k,  (1)

where k is a rational number. Mohanty [11] found an y-arithmetic progression of four consecutive integers on the curve (1) when k = 1025 and conjectured [12] that the length of an x-arithmetic progression on a Mordell curve cannot exceed 4. Later, Lee and Vélez [9] found infinitely many families of Mordell curves with x-arithmetic progressions of length 4 and y-arithmetic progressions of length 6.

In this paper we show that the maximum length of a y-arithmetic progression on a Mordell curve is 6, and we explicitly find infinitely many Mordell curves on which there exist y-arithmetic progressions of length 6. Further, we show that the maximum length of a simultaneous arithmetic progression on a Mordell curve is 3, and we explicitly find infinitely many Mordell curves on which there exist simultaneous arithmetic progressions of length 3.

2 Notation and preliminaries

Now we will introduce some basic notation. We denote by S_{x}(E) (respectively, S_{y}(E)) as the maximal number length of x-arithmetic progression (respectively, y-arithmetic progression) on the elliptic curve E. Also, let S_{x,y}(E) denote the maximal length of simultaneous arithmetic progression on the elliptic curve E

To prove one of our results we need the following theorem of Bremner [2].

Theorem 1. The only general rational solution of the surface

x^{3} + y^{3} + cz^{3} = c (c ≠ 1)  (2)

are given, up to symmetry, by

(i) (λ, −λ, 1) and (ii) (\frac{9}{c}λ^{4} − 3λ, −\frac{9}{c}λ^{4}, \frac{9}{c}λ^{3} − 1)

and the additional case when c = 2

\begin{matrix}(iii) (x, y, z) = & (−4λ^{2} + 6λ − 1, −4λ^{2} + 2λ + 1, 4λ^{2} − 4λ + 1) \\ (iv) (x, y, z) = & (\frac{2}{27}(4λ^{4} − 4λ^{3} − 6λ^{2} + 17λ − 2), \\ \frac{4}{27}(2λ^{4} − 8λ^{3} + 6λ^{2} + 4λ − 13), \\ \frac{−1}{27}(8λ^{4} − 20λ^{3} + 24λ^{2} + 16λ − 37)) \end{matrix}  (3)

with λ ∈ ℚ.

3 Main results

We obtained an upper bound for S_{y}(E) and S_{x,y}(E) where E : y^{2} = x^{3} + k with k ∈ ℚ^{∗}. Indeed, we prove the following results.

Theorem 2. Let E : y^{2} = x^{3} + k be an elliptic curve for some k ∈ ℚ. Then S_{y}(E) ≤ 6. Moreover, there exist infinitely many such elliptic curves E : y^{2} = x^{3} + k with S_{y}(E) = 6.

Proof. At first we will construct an infinite family of elliptic curves of the form E : y^{2} = x^{3} + k with six rational points such that the y-coordinates of these six rational points are in arithmetic progression. After that, we will show that there exist at most six rational points whose y-coordinates are in arithmetic progression i.e., S_{y}(E) ≤ 6.

Let r and d be two arbitrary non-zero rational numbers and k = r^{2} − d^{3}. Consequently (d, ±r) lie on E. Now we want (x_{1}, ±3r) and (x_{2}, ±5r) (with x_{1}, x_{2} ∈ ℚ) to be on E and that would give,

8r^{2} + d^{3} = x_{1}^{3}, 24r^{2} + d^{3} = x_{2}^{3}.

Hence,

16r^{2} = x_{2}^{3} − x_{1}^{3}  (4)

and

2d^{3} = 3x_{1}^{3} − x_{2}^{3}.  (5)

Considering X_{1} = \frac{x_{1}}{d} and X_{2} = \frac{x_{2}}{d} in (5), we get

3X_{1}^{3} − X_{2}^{3} = 2.  (6)

Now (X_{1}, X_{2}) = (\frac{1}{4}, \frac{−5}{4}) is a point on (6) and correspondingly one gets x_{1} = \frac{d}{4} and x_{2} = \frac{−5d}{4}. Now putting the values of x_{1} and x_{2} in (4) we get

2^{10}r^{2} = −126d^{3}.  (7)

At this point if we choose d = −14q^{2} for any q ∈ ℚ^{∗}, then r = ±\frac{147}{8}q^{3}. Hence

k = r^{2} − d^{3} = \frac{197225}{64}q^{6}.  (8)

Therefore corresponding to every rational number q we can find d, r and k; i.e., we can find infinitely many elliptic curves of the form y^{2} = x^{3} + k with at least six rational points whose y-coordinates are in arithmetic progression.

Next we will show that there do not exist more than six rational points whose ycoordinates are in arithmetic progression.

Let d be any non-zero rational number. Let us assume that there exist seven rational points on E, whose y-coordinates are in arithmetic progression. Without loss of generality, we may take the points to be

P_{1} = (x_{1}, y_{1}), P_{2} = (x_{2}, y_{1} + d), P_{3} = (x_{3}, y_{1} − d), P_{4} = (x_{4}, y_{1} + 2d),

P_{5} = (x_{5}, y_{1} − 2d), P_{6} = (x_{6}, y_{1} + 3d), P_{7} = (x_{7}, y_{1} − 3d).

One has

y_{1}^{2} = x_{1}^{3} + k,  (9)
(y_{1} + d)^{2} = x_{2}^{3} + k,  (10)
(y_{1} − d)^{2} = x_{3}^{3} + k,  (11)
(y_{1} + 2d)^{2} = x_{4}^{3} + k,  (12)
(y_{1} − 2d)^{2} = x_{5}^{3} + k,  (13)
(y_{1} + 3d)^{2} = x_{6}^{3} + k,  (14)
(y_{1} − 3d)^{2} = x_{7}^{3} + k.  (15)

Now calculating [(10) − (11)] , [(12) − (13)] , and [(14) − (15)] one has

4y_{1}d = x_{2}^{3} − x_{3}^{3},  (16)

8y_{1}d = x_{4}^{3} − x_{5}^{3},  (17)
12y_{1}d = x_{6}^{3} − x_{7}^{3}.  (18)

Case 1: In this case we are considering y_{1} ≠ 0. From (17) it is clear that one of x_{4} and x_{5} has to be non-zero as d ≠ 0 and y_{1} ≠ 0. So, without loss of generality we can assume that x_{5} is non-zero.

Now, eliminating y_{1} from the equations (16) and (17), we have

\begin{matrix}2(x_{2}^{3} − x_{3}^{3}) = x_{4}^{3} − x_{5}^{3} \\ ⇒ & 8(x_{2}^{3} − x_{3}^{3}) = 4(x_{4}^{3} − x_{5}^{3}) \\ ⇒ & − 8x_{2}^{3} + 8x_{3}^{3} + 4x_{4}^{3} = 4x_{5}^{3} \\ ⇒ & (\frac{−2x_{2}}{x_{5}} )^{3} + (\frac{2x_{3}}{x_{5}} )^{3} + 4 (\frac{x_{4}}{x_{5}} )^{3} = 4 \\ ⇒ & X^{3} + Y^{3} + 4Z^{3} = 4 \end{matrix}  (19)

where X = \frac{−2x_{2}}{x_{5}} , Y = \frac{2x_{3}}{x_{5}} , and Z = \frac{x_{4}}{x_{5}} .

Again, eliminating y_{1} from the equations (18) and (17), we get

\begin{matrix}2(x_{7}^{3} − x_{6}^{3}) = 3(x_{5}^{3} − x_{4}^{3}) \\ ⇒ & 8x_{7}^{3} − 8x_{6}^{3} + 12x_{4}^{3} = 12x_{5}^{3} \\ ⇒ & (\frac{2x_{7}}{x_{5}} )^{3} + (\frac{−2x_{6}}{x_{5}} )^{3} + 12 (\frac{x_{4}}{x_{5}} )^{3} = 12 \\ ⇒ & U^{3} + V^{3} + 12W^{3} = 12 \end{matrix}  (20)

where U = \frac{2x_{7}}{x_{5}} , V = \frac{−2x_{6}}{x_{5}} , and W = \frac{x_{4}}{x_{5}} .

To find rational solution of (19) and (20) we use Bremner’s result. Using Theorem (1) we get all possible rational solutions of (19) and (20). The rational solutions of (19) are

(i) (λ, −λ, 1) and (ii) (\frac{9}{4}λ^{4} − 3λ, −\frac{9}{4}λ^{4}, \frac{9}{4}λ^{3} − 1)

with λ ∈ ℚ.

The rational solutions of (20) are

(i) (μ, −μ, 1) and (ii) (\frac{3}{4}μ^{4} − 3μ, −\frac{3}{4}μ^{4}, \frac{3}{4}μ^{3} − 1)

with μ ∈ ℚ.

One can see from (19) and (20) that,

Z = W = \frac{x_{4}}{x_{5}} .

Clearly,

\frac{9}{4}λ^{3} − 1 = 1 and \frac{3}{4}μ^{3} − 1 = 1

cannot hold for any λ, μ ∈ ℚ.

Now if possible let us assume that for some λ and μ,

\frac{9}{4}λ^{3} − 1 = \frac{3}{4}μ^{3} − 1.

That gives 3λ^{3} = μ^{3}, which is again not possible as λ, μ ∈ ℚ.

Thus we get Z = W = 1 and that implies x_{4} = x_{5}. Now from (12) and (13) we can write,

(y_{1} + 2d) = ±(y_{1} − 2d).

This forces y_{1} = 0 as d ≠ 0, which is a contradiction.

Case 2: In this case we consider y_{1} = 0. Thus our points are

P_{1} = (x_{1}, 0), P_{2} = (x_{2}, d), P_{3} = (x_{3}, −d), P_{4} = (x_{4}, 2d),

P_{5} = (x_{5}, −2d), P_{6} = (x_{6}, 3d), P_{7} = (x_{7}, −3d).

One can easily check that, k = (−x_{1})^{3}, x_{2} = x_{3}, and x_{6} = x_{7}. Thus,

d^{2} = x_{2}^{3} − x_{1}^{3}.  (21)

4d^{2} = x_{4}^{3} − x_{1}^{3}.  (22)
9d^{2} = x_{6}^{3} − x_{1}^{3}.  (23)

From (21) it is clear that one of x_{1} and x_{2} has to be non-zero as d ≠ 0. So, without loss of generality we can assume that x_{1} is non-zero.

Eliminating d from the equations (21), (22) and (23), we have

\begin{matrix}x_{6}^{3} − x_{4}^{3} = 5(x_{2}^{3} − x_{1}^{3}) \\ ⇒ & (\frac{x_{4}}{x_{1}} )^{3} + (\frac{−x_{6}}{x_{1}} )^{3} + 5 (\frac{x_{2}}{x_{1}} )^{3} = 5 \\ ⇒ & L^{3} + M^{3} + 5N^{3} = 5 \end{matrix}  (24)

where L = \frac{x_{4}}{x_{1}} , M = \frac{−x_{6}}{x_{1}} , and N = \frac{x_{2}}{x_{1}} . Again using Bremner’s result, the solution set of the equation (24) is

(i) (ν, −ν, 1) and (ii) (\frac{9}{5}ν^{4} − 3ν, −\frac{9}{5}ν^{4}, \frac{9}{5}ν^{3} − 1) ,

where ν ∈ ℚ.

Note that N ≠ 1, as N = 1 implies d = 0, which is not possible. Let x_{1} be any non-zero rational number, say q. Then we have

x_{2} = (\frac{9}{5}ν^{3} − 1) q, x_{4} = (\frac{9}{5}ν^{4} − 3ν) q, x_{6} = (\frac{9}{5}ν^{4}) q.

Substituting in the values of x_{2} and x_{6} into (21) and (23) respectively, we have

d^{2} = [(\frac{9}{5}ν^{3} − 1)^{3} − 1] q^{3},

9d^{2} = [(\frac{9}{5}ν^{4})^{3} − 1] q^{3}.

Therefore we get

9 [(\frac{9}{5}ν^{3} − 1)^{3} − 1] = [(\frac{9}{5}ν^{4})^{3} − 1] ,

which leads us to

9(9t − 5)^{3} = 9^{3}t^{4} + 1000,

where ν^{3} = t.

We will prove that this equation does not have any rational solution. Putting s = 3t we have the following equation:

9s^{4} − 243s^{3} + 1215s^{2} − 2025s + 2125 = 0.  (25)

Using the rational root theorem, we can see that (25) does not have any rational solution. Therefore our system of equations [(9) to (15)] do not have any rational solution. Hence there do not exist seven rational points on y^{2} = x^{3}+k with y-coordinates in arithmetic progression, i.e., S_{y}(E) ≤ 6.

Now we will state the result related to simultaneous arithmetic progression.

Theorem 3. Let E : y^{2} = x^{3} + k be an elliptic curve for some k ∈ ℚ. Then S_{x,y}(E) ≤ 3. Moreover, there exist infinitely many such elliptic curves E : y^{2} = x^{3} + k with S_{x,y}(E) = 3.

Proof. Let d and d′ be given nonzero rational numbers. Suppose P_{1} = (x_{1}, y_{1}) is a rational point on E : y^{2} = x^{3} + k for some k ∈ ℚ. Therefore, P_{1} satisfies

y_{1}^{2} = x_{1}^{3} + k.  (26)

Suppose P_{2} and P_{3} are other rational points on E such that P_{1}, P_{2} and P_{3} form a simultaneous arithmetic progression with differences d and d′ of length 3. Therefore, we let P_{2} = (x_{1} + d, y_{1} + d′) and P_{3} = (x_{1} − d, y_{1} − d′) and we get

(y_{1} + d′)^{2} = (x_{1} + d)^{3} + k  (27)

and

(y_{1} − d′)^{2} = (x_{1} − d)^{3} + k.  (28)

By subtracting (28) from (27), we get

4y_{1}d′ = 6x_{1}^{2}d + 2d^{3}.  (29)

From (26) and (27), we get

2y_{1}d′ + d′^{2} = 3x_{1}^{2}d + 3x_{1}d^{2} + d^{3}.  (30)

Therefore, from (29) and (30), we get

2d′^{2} = 6x_{1}d^{2} ⟺ x_{1} = \frac{1}{3} \frac{d′^{2}}{d^{2}} .  (31)

Once we get x_{1} as a function of d and d′, we can get y_{1} and k as a function of d and d′. Since this is true for a given non-zero rational numbers d and d′, we conclude that there are infinitely many elliptic curves E of the form y^{2} = x^{3} + k which admits S_{x,y}(E) ≥ 3.

In order to finish the proof of the theorem, now it is enough to prove that S_{x,y}(E) ≤ 3 for all elliptic curves E : y^{2} = x^{3} + k.

Let E be one such curve. If possible, let P_{1}, P_{2}, P_{3} and P_{4} be the rational points on E which form a simultaneous arithmetic progression of length 4 with some differences d and d′ for the x and y coordinates respectively. Now, we need to consider several cases depending on the arrangement of coordinates of P_{i}’s.

Case 1: Let P_{1} = (x_{1} −d, y_{1} −d′), P_{2} = (x_{1}, y_{1}), P_{3} = (x_{1} +d, y_{1} +d′) and P_{4} = (x_{1} +2d, y_{1} + 2d′).

Since P_{1}, P_{2} and P_{3} are in simultaneous arithmetic progression, by the previous discussion,

we conclude that x_{1} = \frac{1}{3} \frac{d′^{2}}{d^{2}} . Since P_{4} is a rational point on E, we get

(y_{1} + 2d′)^{2} = (x_{1} + 2d)^{3} + k.  (32)

From the equations (32) and (26), we get

4y_{1}d′ + 4d′^{2} = 6x_{1}^{2}d + 12x_{1}d^{2} + 8d^{3}.  (33)

Now, by putting x_{1} = d^{′2}/3d^{2} in (33), we arrive at d = 0, which is a contradiction. Thus, we conclude, in this case, S_{x,y}(E) ≤ 3.

Case 2: Let P_{1} = (x_{1}, y_{1}), P_{2} = (x_{1}+d, y_{1}+d′), P_{3} = (x_{1}+2d, y_{1}+3d′), P_{4} = (x_{1}+3d, y_{1}+2d′). Since these are rational points on E, we have

y_{1}^{2} = x_{1}^{3} + k.  (34)

(y_{1} + d′)^{2} = (x_{1} + d)^{3} + k,  (35)
(y_{1} + 3d′)^{2} = (x_{1} + 2d)^{3} + k,  (36)

and

(y_{1} + 2d′)^{2} = (x_{1} + 3d)^{3} + k.  (37)

Eliminating y_{1} and k, from the equations (34), (35) and (36), we get

6d′^{2} = −3x_{1}^{2}d + 3x_{1}d^{2} + 5d^{3}.  (38)

Again, eliminating y_{1} and k, from the equations (34), (36) and (37), we have

6d′^{2} = −15x_{1}^{2}d − 57x_{1}d^{2} − 65d^{3}.  (39)

Now eliminating d′ from the equations (38) and (39), we get

6x_{1}^{2} + 30x_{1}d + 35d^{2} = 0.  (40)

Clearly, the quadratic equation (40) in x_{1} does not have any rational solutions and hence we conclude, in this case, S_{x,y}(E) ≤ 3.

The remaining 22 cases can be proved similarly and we list them in Table 1.

From Table 1, we can see that the discriminant D is not a perfect square for a rational number d in all the cases. Hence, we conclude that x_{1} cannot be rational, which is a contradiction. Therefore, in all the cases, we get S_{x,y}(E) ≤ 3. This completes the proof of Theorem 3.

4 Some open problems

While we obtained upper bounds for the lengths of y−arithmetic progressions and simultaneous arithmetic progressions on Mordell curves, we could not obtain an upper bound for the length of x−arithmetic progressions on Mordell curves. It would be of interest to determine such an upper bound. Similarly it would be of interest to determine upper bounds for the length of x−arithmetic progressions, y−arithmetic progressions and simultaneous arithmetic progressions on various models of elliptic curves.

5 Acknowledgments

We want to thank Ajai Choudhry for suggesting and discussing this problem. We also want to thank Kalyan Chakraborty and R. Thangadurai for going through this paper carefully. Finally, we would like to thank the anonymous referee for his/her valuable comments.

Table 1

Cases  Points  Final Equation  Discriminant(D)
Case 3  (x_{1}, y_{1}), (x_{1} + d, y_{1} + 2d′)  3x_{1}^{2} + 9x_{1}d + 8d^{2} = 0  −15d^{2}

(x_{1} + 2d, y_{1} + d′), (x_{1} + 3d, y_{1} + 3d′)

Case 4 (x_{1}, y_{1}), (x_{1} + d, y_{1} + 2d′) 24x_{1}^{2} + 84x_{1}d + 86d^{2} = 0 −1200d^{2}

(x_{1} + 2d, y_{1} + 3d′), (x_{1} + 3d, y_{1} + d′)

Case 5 (x_{1}, y_{1}), (x_{1} + d, y_{1} + 3d′) 3x_{1}^{2} + 14x_{1}d + 28d^{2} = 0 −140d^{2}

(x_{1} + 2d, y_{1} + d′), (x_{1} + 3d, y_{1} + 2d′)

Case 6 (x_{1}, y_{1}), (x_{1} + d, y_{1} + 3d′) 6x_{1}^{2} + 24x_{1}d + 29d^{2} = 0 −120d^{2}

(x_{1} + 2d, y_{1} + 2d′), (x_{1} + 3d, y_{1} + d′)

Case 7 (x_{1}, y_{1} + d′), (x_{1} + d, y_{1} + 2d′), 3x_{1}^{2} − 3x_{1}d − 8d^{2} = 0 105d^{2}
(x_{1} + 2d, y_{1}), (x_{1} + 3d, y_{1} + 3d′)
Case 8 (x_{1}, y_{1} + d′), (x_{1} + d, y_{1} + 2d′), 6x_{1}^{2} + 12x_{1}d + 11d^{2} = 0 −120d^{2}
(x_{1} + 2d, y_{1} + 3d′), (x_{1} + 3d, y_{1})
Case 9 (x_{1}, y_{1} + d′), (x_{1} + d, y_{1}) 6x_{1}^{2} + 6x_{1}d − d^{2} = 0 60d^{2}

(x_{1} + 2d, y_{1} + 2d′), (x_{1} + 3d, y_{1} + 3d′)

Case 10 (x_{1}, y_{1} + d′), (x_{1} + d, y_{1}) 12x_{1}^{2} + 36x_{1}d + 37d^{2} = 0 −480d^{2}

(x_{1} + 2d, y_{1} + 3d′), (x_{1} + 3d, y_{1} + 2d′)

Case 11 (x_{1}, y_{1} + d′), (x_{1} + d, y_{1} + 3d′) 15x_{1}^{2} + 45x_{1}d + 44d^{2} = 0 −615d^{2}
(x_{1} + 2d, y_{1}), (x_{1} + 3d, y_{1} + 2d′)
Case 12 (x_{1}, y_{1} + d′), (x_{1} + d, y_{1} + 3d′) 12x_{1}^{2} + 30x_{1}d + 25d^{2} = 0 −300d^{2}
(x_{1} + 2d, y_{1} + 2d′), (x_{1} + 3d, y_{1})
Case 13 (x_{1}, y_{1} + 2d′), (x_{1} + d, y_{1}) 12x_{1}^{2} + 30x_{1}d + 25d^{2} = 0 −300d^{2}

(x_{1} + 2d, y_{1} + d′), (x_{1} + 3d, y_{1} + 3d′)

Case 14 (x_{1}, y_{1} + 2d′), (x_{1} + d, y_{1}) 15x_{1}^{2} + 45x_{1}d + 44d^{2} = 0 −615d^{2}

(x_{1} + 2d, y_{1} + 3d′), (x_{1} + 3d, y_{1} + d′)

Case 15 (x_{1}, y_{1} + 2d′), (x_{1} + d, y_{1} + d′) 6x_{1}^{2} + 12x_{1}d + 11d^{2} = 0 −120d^{2}
(x_{1} + 2d, y_{1}), (x_{1} + 3d, y_{1} + 3d′)
Case 16 (x_{1}, y_{1} + 2d′), (x_{1} + d, y_{1} + d′) 3x_{1}^{2} − 3x_{1}d − 8d^{2} = 0 105d^{2}
(x_{1} + 2d, y_{1} + 3d′), (x_{1} + 3d, y_{1})
Case 17 (x_{1}, y_{1} + 2d′), (x_{1} + d, y_{1} + 3d′) 12x_{1}^{2} + 36x_{1}d + 37d^{2} = 0 −480d^{2}
(x_{1} + 2d, y_{1}), (x_{1} + 3d, y_{1} + d′)
Case 18 (x_{1}, y_{1} + 2d′), (x_{1} + d, y_{1} + 3d′) 6x_{1}^{2} + 6x_{1}d − d^{2} = 0 60d^{2}
(x_{1} + 2d, y_{1} + d′), (x_{1} + 3d, y_{1})
Case 19 (x_{1}, y_{1} + 3d′), (x_{1} + d, y_{1}) 6x_{1}^{2} + 24x_{1}d + 29d^{2} = 0 −120d^{2}

(x_{1} + 2d, y_{1} + d′), (x_{1} + 3d, y_{1} + 2d′)

Case 20 (x_{1}, y_{1} + 3d′), (x_{1} + d, y_{1}) 3x_{1}^{2} + 21x_{1}d + 28d^{2} = 0 105d^{2}

(x_{1} + 2d, y_{1} + 2d′), (x_{1} + 3d, y_{1} + d′)

Case 21 (x_{1}, y_{1} + 3d′), (x_{1} + d, y_{1} + d′) 12x_{1}^{2} + 42x_{1}d + 43d^{2} = 0 −300d^{2}
(x_{1} + 2d, y_{1}), (x_{1} + 3d, y_{1} + 2d′)
Case 22 (x_{1}, y_{1} + 3d′), (x_{1} + d, y_{1} + d′) 9x_{1}^{2} + 27x_{1}d + 24d^{2} = 0 −135d^{2}
(x_{1} + 2d, y_{1} + 2d′), (x_{1} + 3d, y_{1})
Case 23 (x_{1}, y_{1} + 3d′), (x_{1} + d, y_{1} + 2d′) 6x_{1}^{2} + 30x_{1}d + 35d^{2} = 0 60d^{2}
(x_{1} + 2d, y_{1}), (x_{1} + 3d, y_{1} + d′)
Case 24 (x_{1}, y_{1} + 3d′), (x_{1} + d, y_{1} + 2d′) d = 0
(x_{1} + 2d, y_{1} + d′), (x_{1} + 3d, y_{1})

References

[1] A. Alvarado, An arithmetic progression on quintic curves, J. Integer Seq. 12 (2009), Article 09.7.3.

[2] A. Bremner, On diagonal cubic surfaces, Manuscripta Math. 62 (1988), 21–32.

[3] A. Bremner, Arithmetic progression on elliptic curves, Experiment. Math. 8 (1999), 409–413.

[4] A. Bremner, Arithmetic progressions on Edwards curves, J. Integer Seq. 16 (2013), Article 13.8.5.

[5] G. Campbell, A note on arithmetic progressions on elliptic curves, J. Integer Seq. 6 (2003), Article 03.1.3.

[6] A. Choudhry, Arithmetic progressions on Huff curves, J. Integer Seq. 18 (2015), Article 15.5.2.

[7] I. García-Selfa and J. M. Tornero, Searching for simultaneous arithmetic progressions on elliptic curves, Bull. Aust. Math. Soc. 71 (2005), 417–424.

[8] E. González-Jiménez, On arithmetic progressions on Edwards curves, Acta Arith. 167 (2015), 117–132.

[9] J. B. Lee and W. Y. Vélez, Integral solutions in arithmetic progression for y^{2} = x^{3} + k, Period. Math. Hungar. 25 (1992), 31–49.

[10] A. MacLeod, 14-term arithmetic progressions on quartic elliptic curves, J. Integer Seq. 9 (2006), Article 06.1.2.

[11] S. P. Mohanty, On consecutive integral solutions for y^{2} = x^{3} + k, Proc. Amer. Math. Soc. 48 (1975), 281–285.

[12] S. P. Mohanty, Integral solutions in arithmetic progression for y^{2} = x^{3} + k, Acta Math. Hungar. 34 (1980), 261–265.

[13] D. Moody, Arithmetic progressions on Edward curves, J. Integer Seq. 14 (2011), Article 11.1.7.

[14] D. Moody, Arithmetic progressions on Huff curves, Ann. Math. Inform. 38 (2011), 111–116.

[15] M. Ulas, A note on arithmetic progressions on quartic elliptic curves, J. Integer Seq. 8 (2005), Article 05.3.1.

2010 Mathematics Subject Classification: Primary 11G05; Secondary 11B25. Keywords: arithmetic progression, elliptic curve.

Received September 19 2015; revised versions received September 24 2015; May 8 2016; September 3 2016. Published in Journal of Integer Sequences, September 4 2016.

Return to Journal of Integer Sequences home page.