Rational Points of Some Elliptic Curves Related to the Tilings of the Equilateral Triangle
- PDF / 266,506 Bytes
- 10 Pages / 439.37 x 666.142 pts Page_size
- 45 Downloads / 246 Views
Rational Points of Some Elliptic Curves Related to the Tilings of the Equilateral Triangle Miklós Laczkovich1 Received: 3 February 2019 / Revised: 7 October 2019 / Accepted: 9 October 2019 © The Author(s) 2019
Abstract Let n be a positive and squarefree integer. We show that the equilateral triangle can be dissected into n · k 2 congruent triangles for some k if and only if n ≤ 3, or at least one of the curves Cn : y 2 = x(x − n)(x + 3n) and C−n : y 2 = x(x + n)(x − 3n) has a rational point with y = 0. We prove that if p is a positive prime such that p ≡ 7 (mod 24), then C p and C− p do not have such points. Consequently, for these primes the equilateral triangle cannot be dissected into p · k 2 congruent triangles for any k. Keywords Tilings of the equilateral triangle · Rank of some elliptic curves over the rationals
1 Introduction and Main Results Let Cn denote the elliptic curve y 2 = x(x − n)(x + 3n), where n is an integer. The group of rational points of Cn will be denoted by n . We say that (x, y) ∈ Cn is a nontrivial rational point of Cn if x, y are nonzero rational numbers; that is, if the order of (x, y) as an element of the group n is greater than two. Our first result shows that the existence of nontrivial rational points of Cn is closely related to the number of pieces in certain tilings of the equilateral triangle. Theorem 1.1 For every positive and squarefree integer n the following are equivalent. (i) There is a positive integer k such that the equilateral triangle can be dissected into n · k 2 congruent triangles. (ii) Either n ≤ 3, or at least one of the curves Cn and C−n has a nontrivial rational point.
Dedicated to the memory of Ricky Pollack. Miklós Laczkovich [email protected] 1
Eötvös Loránd University, Budapest, Hungary
123
Discrete & Computational Geometry
The proof of Theorem 1.1 is based on the fact that the congruent copies of a triangle with sides a, b, c and corresponding angles α, β, γ tile an equilateral triangle if and only if either α, β, γ are multiples of π/6, or γ ∈ {π/3, 2π/3} and a, b, c are pairwise commensurable (see [4, Thm. 3.3]). By the law of cosines, we have γ = π/3 or 2π/3 if and only if c2 = a 2 + b2 ± ab. Such triples are, e.g., (a, b, c) = (7, 8, 13) or (a, b, c) = (3, 5, 7). Suppose that a, b, c are positive integers with c2 = a 2 + b2 ± ab. Then the triangle with sides a, b, c tiles an equilateral triangle T . If the side length of T is m and the tiling has N pieces, then, comparing the areas we get m 2 = N · ab, and thus the square free part of N is the same as that of ab. For example, if (a, b, c) = (7, 8, 13), then the construction described in [3, Thm. 3.1] produces a tiling with 2, 469, 600 = 14 · 4202 pieces. For the triangle with sides 3, 5, 7, a tiling with 10, 935 = 15 · 272 pieces was found by Michael Beeson (see [2, Fig. 22, p. 28]). As we shall see, a simple transformation maps these triples into nontrivial rational points of one of the corresponding curves Cn or C−n . Thus the triple (7, 8, 13) gives the point (−6, 48) of C−14 , a
Data Loading...
