On the solutions of parametrized families of elliptic equations

Attila Pethő, University of Debrecen

Debrecen, October 25, 2001.

Parametrized families of Thue equations

Let n ≥ 3, m ∈ Z and f_{1}, …, f_{n} ∈ Z[X]. Then

x^{n} + f_{1}(k)x^{n−1}y + ⋯ + f_{n}(k)y^{n} = m

is called a parametrized family of Thue equations.

The solutions of parametrized families of Thue equations behave not always nicely. Consider for example

x^{3} − ky^{3} = 1.

But there may exist subfamilies, which are nice. If k = b^{3} ± 1 then (x, y) = (∓1, 1).

Parametrized families of elliptic equations - introduction

Let a_{2}, a_{4}, a_{6} ∈ Z[X]. Then

y^{2} = x^{3} + a_{2}(k)x^{2} + a_{4}(k)x + a_{6}(k)

is called a parametrized family of elliptic equations. Classical examples:

\begin{aligned}y^{2} & = & x^{3} + k \\ y^{2} & = & x^{3} + kx \\ y^{2} & = & x(x + k)(x + 2k) \end{aligned}

Modern examples in cronological order:

a.) R. Stroeker (1995), A. Bremner, R. Stroeker and N. Tzanakis (1995)

Y^{2} = X^{2} + ⋯ + (X + k − 1)^{2}

E_{1} : y^{2} = x^{3} − 36x − 864k(k − 1)(2k − 1). b.) A. Dujella (2000) Y^{2} = ((k − 1)X + 1)((k + 1)X + 1)(4kX + 1)

\begin{aligned}E_{2} : y^{2} & = & x^{3} + (9k^{2} − 1)x^{2} \\ & + & 24k^{2}(k^{2} − 1)x + 16k^{2}(k^{2} − 1)^{2}. \end{aligned}

c.) A. Dujella and A.Pethő (2000)

Y^{2} = (X + 1)(3X + 1)(c_{k}X + 1) E_{3} : y^{2} = x^{3}+(4c_{k}+3)x^{2}+(3c_{k}^{2}+12c_{k})x+9c_{k}^{2}.

d.) S. Duquesne (2001)

E_{4} : y^{2} = x^{3} + kx^{2} − (k + 3)x + 1.

Modern examples in logical order:

a.) There are two trivial integral point on

E_{1} : y^{2} = x^{3} − 36x − 864k(k − 1)(2k − 1) :

T_{k} = (6(2k − 1), 0), the only rational torsion point on E_{1}(Q) provided k ≥ 2 and Q_{k} = (12k + 6, 72k).

Theorem 1. Q_{k} is of infinite order. If

E_{1}(Q)/E_{1}(Q)_{tors} =< Q_{k} >

then the only integral points on E_{1}(Q) are T_{k}, Q_{k}.

Conjecture 1. If rank(E_{1}(Q)) = 1 then E_{1}(Q)/E_{1}(Q)_{tors} =< Q_{k} > and Theorem 1 gives all integral points on E_{1}(Q).

This is true if k ≤ 100. b.)

E_{4} : y^{2} = x^{3} + kx^{2} − (k + 3)x + 1.

Assume that Δ = k^{2} + 3k + 9 is square-free, which is true for infinitely many k by a result of P. Erdős. The point P_{4} = (0, 1) is of infinite order, and 2P_{4} = (b^{2}+2b+3, b^{3}+4b^{2}+7b+5) ∈ E_{4}(Z) provided k = 2b + 1.

Theorem 2. There exists no P ∈ E_{4}(Q) such that P_{4} = nP, n ∈ Z, n ≠ ±1. If rank(E_{4}(Q)) = 1 then the only integral points on E_{4}(Q) are ±P_{4} if k is even and ±P_{4}, ±2P_{4} if k is odd.

Problem: Does there exist infinitely many k such that rank(E_{4}(Q)) = 1?

Main steps of the proof. Let P = (\frac{α}{d^{2}}, \frac{β}{d^{3}}) ∈ E_{4}(Q), gcd (α, d) = gcd (β, d) = 1 and h(P) = log max {|α|, d^{2}}. Then

ĥ(P) ≤ \frac{h(P)}{2} + 1.57 + \frac{log Δ}{4} + \frac{log k}{2}

ĥ(P) ≥ \frac{log k}{16} + \frac{9.47}{32 log k} + \frac{log (βd)}{4}
+ \frac{log (0.99 + 1.91/ log k)}{8} − \frac{log log k}{8}
+ 0.02.

Using these estimates we obtain n ≤ 4.

E_{4}(Q) = E_{egg}(Q) ∪ E_{0}(Q), where E_{egg}(Q) denotes the compact and E_{0}(Q) the noncompact part of E_{4}(Q). As P_{4} ∈ E_{egg}(Q) we have P ∈ E_{egg}(Q). As 2P_{4} and 4P_{4} ∈ E_{0}(Q) it remains to study n = 3, which is excluded by elementary considerations.

Further results: If k = 2t^{2} +2t−1 then P_{4,0} = (−1, 2t + 1), P_{4,1} = (0, 1), P_{4,2} = (2, 2t + 1) ∈ E_{4}(Q) are independent and 2P_{1}, P_{0} ± P_{1}, P_{2} ± P_{1}, P_{0} + P_{2}E_{4}(Z).

Similarly for k = 5t^{2} − 3t + 3 rank(E_{4}(Q)) ≥ 3.

For the integer solutions of 2t_{1}^{2} + 2t_{1} − 1 = 5t_{2}^{2} − 3t_{2} + 3 one obtains curves of rank at least 5. b.)

Y^{2} = ((k − 1)X + 1)((k + 1)X + 1)(4kX + 1)

\begin{aligned}E_{2} : y^{2} & = & x^{3} + (9k^{2} − 1)x^{2} \\ & + & 24k^{2}(k^{2} − 1)x + 16k^{2}(k^{2} − 1)^{2}. \end{aligned}

Let A_{k} = (−4k^{2} − 4k, 0), B_{k} = (−4k^{2} + 4k, 0), C_{k} = (−k^{2} + 1, 0), O are the only torsion points on E_{2}(Q). The point P_{2} = (0, 4k^{3} − 4k) is of infinite order.

(X, ±Y) = (16k^{3} − 4k, 126k^{6} − 112k^{4} − 2k^{2} − 1)

are trivial integral points on E_{2}(Q).

Theorem 3. If k ≥ 3 and rankE_{2}(Q)) = 1 then all integral points on E_{2}(Q)) are trivial.

Main  steps  of  the  proof. Let

E_{2}(Q)/E_{2}(Q)_{tors} =< U > and Q ∈ E_{2}(Q). Then Q = mU +T and P_{2} = m_{P} U +T_{P} , m, m_{P} ∈ Z. One can see that m_{P} is odd, hence

U ≡ P_{2} + T_{P} (mod 2E_{2}(Q))

and

Q ≡ Q_{1} (mod 2E_{2}(Q)),

where Q_{1} ∈ {O, A_{k}, B_{k}, C_{k}, P_{2}, P_{2} + A_{k}, P_{2} + B_{k}, P_{2} + C_{k}}. From here one derives simultanous equations of the form

\begin{aligned}(k − 1)x + 1 & = & αy_{1}^{2}, \\ (k + 1)x + 1 & = & βy_{2}^{2}, \\ 4kx + 1 & = & γy_{3}^{2}. \end{aligned}

The case α = β = γ = 1 was solved, Dujella (1997). The others are not solvable.

Dujella found two infinite subfamilies with k = 3t^{2} + 2t − 2, k = (3t^{2} + 5t)/2 of E_{2} with rankE_{2}(Q) ≥ 2 and one with rankE_{2}(Q) ≥ 3. He proved that in these cases and for k ≤ 1000 all integral points are trivial.

Conjecture 2. Let k ≥ 3 then all integral points on E_{2}(Q) are trivial. c.) A. Dujella and A.Pethő (2000)

Y^{2} = (X + 1)(3X + 1)(c_{k}X + 1)

E_{3} : y^{2} = x^{3}+(4c_{k}+3)x^{2}+(3c_{k}^{2}+12c_{k})x+9c_{k}^{2}.

Trivial integral points:

(X, ±Y) = {(−1, 0), (0, 1),

(c_{k−1}, s_{k−1}t_{k−1}(2c_{k} − s_{k}t_{k})), (c_{k+1} , s_{k+1} t_{k+1} (2c_{k} + s_{k}t_{k}))}

Theorem 4. If rankE_{3}(Q) = 2 or k ≤ 40 with the possible exceptions k = 23 and k = 37 then all all integral points on E_{3}(Q) are trivial.

Theorem 5 (Jacobson, Williams (2001)). If k ≤ 100 with the possible exceptions k = 37 then all integral points on E_{3}(Q) are trivial. Moreover, if the GRH is true there is no exceptions in the range k ≤ 100.

The proof is based on the study of

ax_{1}^{2} − bx_{2}^{2} = c.

Let D = ab = f^{2}D_{0}, D_{0} square-free, Δ = 4D, when 4D_{0} ≢ 1 (mod 4) and Δ = D otherwise. Let K = Q(√D) and O be an order of K. If α ∈ O there exists a compact representation

α = γ n ; ∏ ; j=1 (α_{j}/d_{j})^{2^{n−j}},

where

1. n < log_{2} log_{2}(2 max (|α|, |α′|) + 2

2. d_{j} ∈ Z^{>}, γ ∈ O, α_{j} = (a_{j} + b_{j}√Δ )/2, a_{j}, b_{j} ∈ Z.

3. 2 max (|γ|, |γ′|) ≤ |N(α)|

4. d_{j} < Δ^{1/2} , |a_{j}| < Δ^{5/2} , |b_{j}| < Δ^{2}.

Let ε = v_{1} + √abu_{1} be the fundamental unit of K, and ε^{m} = v_{m} + √abu_{m}. Lemma 1. Let m be odd.

ax_{1}^{2} − bx − 2^{2} = c.

has a solution iff

(2a/c)|v_{m} + 1 and (2b/c)|v_{m} − 1.