A Note on Arithmetic Progressions on Quartic Elliptic Curves

Maciej Ulas Jagiellonian University Institute of Mathematics Reymonta 4 30-059 Kraków Poland Maciej.Ulas@im.uj.edu.pl

Abstract G. Campbell described a technique for producing infinite families of quartic elliptic curves containing a length-9 arithmetic progression. He also gave an example of a quartic elliptic curve containing a length-12 arithmetic progression. In this note we give a construction of an infinite family of quartics on which there is an arithmetic progression of length 10. Then we show that there exists an infinite family of quartics containing a sequence of length 12.

1 Introduction

Let us consider a curve E : y^{2} = f(x), where f ∈ ℚ[x] and f is not a square of a polynomial. We say that points P_{i} = (x_{i}, y_{i}), i = 1, …, k on the curve E form an arithmetic progression of length k if the sequence x_{1}, x_{2}, …, x_{k} form an arithmetic progression.

G. Campbell [2] proved the following theorems:

Theorem 1.1. There are infinitely many elliptic curves of the form y^{2} = w(x), with w(x) a quartic, containing 9 points in arithmetic progression.

Theorem 1.2. There exists an elliptic curve in the form y^{2} = w(x), with w(x) a quartic, containing 12 points in arithmetic progression.

In this paper we propose two different ways to obtain an infinite family of quartic elliptic curves with a sequence of length 10. Then we present a family of quartics (parametrized by the rational points on an elliptic curve with a non zero rank) with an arithmetic progression of length 12.

2 Arithmetic Progressions of Length 10

The proof of the first theorem is similar to the one given by G. Campbell [2].

Theorem 2.1. There are infinitely many quartic elliptic curves y^{2} = f(x) containing 10 points in arithmetic progression.

Proof. Let us consider the following polynomial

P_{t}(x) = (x^{2} − 9x − 4t) ∏_{i=0}^{9}(x − i) ∈ ℚ(t)[x].  (2.1)

Then, we have

P_{t}(x) = Q_{t}(x)^{2} − F_{t}(x),

where Q_{t} is the unique monic polynomial defined over ℚ(t) such that F_{t} has degree 4. The discriminant of the polynomial F_{t}(x) is non zero for t ∈ ℚ \ S, where

S = {±1, ±2, ±4, −5, −6, −8, −11}.

Hence, for such parameters t, the quartic elliptic curve

E_{t} : y^{2} = F_{t}(x)  (2.2)

contain the points P_{i} = (i, Q_{t}(i)) for i = 0, …, 9 which form an arithmetic progression of length 10 on the curve E_{t}.

3 The sequence of length 12

There doesn’t seem to be a way to construct an infinite family of quartics with an arithmetic progression of length 12 from family of curves (2.2). In this section we shall construct another family which will be better for our purposes. First let us consider the polynomial f ∈ ℚ[p, q, r, s, t][x]

f(x) = ∑_{i=0}^{4} a_{i}x^{i},

where

a_{0} = 5p^{2} − 10q^{2} + 10r^{2} − 5s^{2} + t^{2},
a_{1} = \frac{1}{12}(−77p^{2} + 214q^{2} − 234r^{2} + 122s^{2} − 25t^{2}),
a_{2} = \frac{1}{24}(71p^{2} − 236q^{2} + 294r^{2} − 164s^{2} + 35t^{2}),
a_{3} = \frac{1}{12}(−7p^{2} + 26q^{2} − 36r^{2} + 22s^{2} − 5t^{2}),
a_{4} = \frac{1}{24}(p^{2} − 4q^{2} + 6r^{2} − 4s^{2} + t^{2}).

We have

f(1) = p^{2}, f(2) = q^{2}, f(3) = r^{2}, f(4) = s^{2}, f(5) = t^{2},

so we see that

E : y^{2} = f(x)

is a five parameter family of quartics containing an arithmetic progression of length 5. In order to obtain a family with a sequence of length 10 we have to consider the following system of equations

\begin{cases}f(6) = p^{2} − 5q^{2} + 10r^{2} − 10s^{2} + 5t^{2} = P^{2}, \\ f(7) = 5p^{2} − 24q^{2} + 45r^{2} − 40s^{2} + 15t^{2} = Q^{2}, \\ f(8) = 15p^{2} − 70q^{2} + 126r^{2} − 105s^{2} + 35t^{2} = R^{2}, \\ f(9) = 35p^{2} − 160q^{2} + 280r^{2} − 224s^{2} + 70t^{2} = S^{2}, \\ f(10) = 70p^{2} − 315q^{2} + 540r^{2} − 420s^{2} + 126t^{2} = T^{2}, \end{cases}  (3.1)

in integers p, q, r, s, t, P, Q, R, S, T. Since the general solution is hard to obtain we will look for particular solutions with P = t, Q = s, R = r, S = q, T = p. Then, in this case it is easy to realize that (3.1) is equivalent to

\begin{cases}p^{2} = 15r^{2} − 35s^{2} + 21t^{2}, \\ q^{2} = 5r^{2} − 9s^{2} + 5t^{2}. \end{cases}  (3.2)

Making a substitution (p, r, s, t) = (p, a + p, b + p, c + p) we get a parametrized solution of the first equation in (3.2)

\begin{cases}p = 15a^{2} − 35b^{2} + 21c^{2}, \\ r = −15a^{2} + 70ab − 35b^{2} − 42ac + 21c^{2}, \\ s = 15a^{2} − 30ab + 35b^{2} − 42bc + 21c^{2}, \\ t = 15a^{2} − 35b^{2} − 30ac + 70bc − 21c^{2}. \end{cases}  (3.3)

Now inserting r, s, t from (3.3) to the second equation in (3.2), we obtain

q^{2} = 441c^{4} − 168(15a − 7b)c^{3} + 2(675a^{2} + 2520ab − 2303b^{2})c^{2}
+ 40(45a^{3} − 189a^{2}b + 63ab^{2} + 49b^{3})c  (3.4)
+ 25(9a^{4} − 96a^{3}b + 278a^{2}b^{2} − 224ab^{3} + 49b^{4}).

Moreover, if we take b = 3(a + 1), c = 2(a + 1), we get that q = 3(56a^{2} + 142a + 91). Finally we have

\begin{cases}p = T = −3(72a^{2} + 154a + 77), \\ q = S = 3(56a^{2} + 142a + 91), \\ r = R = −3(40a^{2} + 112a + 77), \\ s = Q = 3(24a^{2} + 68a + 49), \\ t = P = −3(8a^{2} + 6a − 7), \end{cases}  (3.5)

which is a solution of (3.1). Specializing the polynomial f as given by (3.5) we obtain

f_{a}(x) = ∑_{i=0}^{4} a_{i}x^{i},

where

a_{0} = 9(7744a^{4} + 25216a^{3} + 22544a^{2} − 784a − 5831),
a_{1} = − 66(a + 1)(16a + 21)(24a^{2} − 52a − 119),
a_{2} = 3(a + 1)(16a + 21)(48a^{2} − 709a − 1085),
a_{3} = 66(a + 1)(5a + 7)(16a + 21),
a_{4} = − 3(a + 1)(5a + 7)(16a + 21),

and the discriminant R_{a} of the polynomial f_{a} is

R_{a} = − 419904(a + 1)^{4}(2a + 3)^{2}(4a + 5)^{2}(2a + 7)^{2}(4a + 7)^{2}(5a + 7)^{3}
× (6a + 7)^{2}(12a + 17)^{2}(16a + 21)^{4}(1008a^{2} + 1831a + 623).

Then for a ∈ ℚ \ W, where

W := {−1, −3/2, −5/4, −7/2, −7/4, −7/5, −7/6, −17/12, −21/16},

we get a nontrivial quartic elliptic curve

C_{a} : y^{2} = f_{a}(x)

containing an arithmetic progression of length 10.

Now we are ready to prove the following:

Theorem 3.1. There are infinitely many quartic elliptic curves y^{2} = f(x) containing 12 points in arithmetic progression.

Proof. The curve C_{a} defined above contains the 10 points with x = 1, …, 10. Observe that

f_{a}(0) = f_{a}(11) = 9(7744a^{4} + 25216a^{3} + 22544a^{2} − 784a − 5831).

Now, consider the curve:

E : Y^{2} = 9(7744a^{4} + 25216a^{3} + 22544a^{2} − 784a − 5831).

A short computer search reveals that P = (−1, 15) is a rational point on the quartic E. Using the program APECS [3] we found that E is birationally equivalent to the elliptic curve

E′ : y^{2} = x^{3} − x^{2} − 33433x + 2213737.

For the curve E′ we have

Tors E(ℚ) = {O, (127, 0)},

and with the help of mwrank [4] we see that the free part of E′ is generated by

G_{1} = (77, −300), G_{2} = (−193, −1200), G_{3} = (−48, −1925).

Hence the curve E has an infinite number of rational points and it is clear that all but finitely many of them leads to the quartic C_{a} containing arithmetic progression of length 12.

It is natural to state the following question:

Open Question 3.2. Is there an quartic elliptic curve E containing a length 13 arithmetic progression?

4 Acknowledgment

I would like to thank anonymous referee for his/her valuable comments.

References

[1] A. Bremner, On arithmetic progressions on elliptic curves. Experiment. Math. 8 (1999), 409–413.

[2] G. Campbell, A note on arithmetic progressions on elliptic curves. Journal of Integer Sequences, Paper 03.1.3, 2003.

[3] I. Connel, APECS, available from ftp.math.mcgill.ca/pub/apecs/.

[4] J. Cremona, mwrank program, available from

    http://www.maths.nottingham.ac.uk/personal/jec/ftp/progs/.

2000 Mathematics Subject Classification: 11G05, 11B25. Keywords: elliptic curves, arithmetic progression.

Received November 11 2004; revised version received May 21 2005. Published in Journal of Integer Sequences, May 24 2005.

Return to Journal of Integer Sequences home page.