Equable parallelograms on the Eisenstein lattice

Christian Aebi; Grant Cairns

doi:10.1515/ms-2024-0071

Article Open Access

Equable parallelograms on the Eisenstein lattice

Christian Aebi and Grant Cairns

Published/Copyright: August 14, 2024

Published by

Become an author with De Gruyter Brill

Author Information Explore this Subject

From the journal Mathematica Slovaca Volume 74 Issue 4

Abstract

This paper studies equable parallelograms whose vertices lie on the Eisenstein lattice. Using Rosenberger’s Theorem on generalised Markov equations, we show that the set of these parallelograms forms naturally an infinite tree, all of whose vertices have degree 4, bar the root which has degree 3. This study naturally complements the authors’ previous study of equable parallelograms whose vertices lie on the integer lattice.

Keywords: Eisenstein lattice; equable; parallelogram

MSC 2010: 52C05; 11D25

1 Introduction

A planar polygon is said to be equable if its perimeter equals its area.

Definition 1

An Eisenstein lattice equable parallelogram (or ELEP, for short) is an equable parallelogram whose vertices lie on the Eisenstein lattice ℤ[ω], where ω=−12+i32.

In [2, 3, 4], we investigated equable quadrilaterals with vertices on the integer lattice. This present paper begins a project of replicating this investigation for the Eisenstein lattice, with the goal of comparing the results. In particular, in [2] we studied equable parallelograms with vertices on the integer lattice. This present paper follows the general approach adopted in [2], but it can be read independently of [2]. We find that while the mathematics in this paper is very similar to that of [2], the results are somewhat simpler. We saw in [2] that the equable parallelograms with vertices on the integer lattice form a forest of three trees, corresponding to three of Rosenberger’s six generalized Markov equations; these are equations M, R1 and R3 in the notation of [9]. In the present work, we show that the ELEPs form a single tree corresponding to the Rosenberger equation R2. It is known (see [13]) that the other two equations, R4 and R5, do not have coprime solutions and so do not appear in the type of study we undertake here or in [2]. So, in that sense, the tree of ELEPs forms the case that was curiously missing in [2].

We should also remark that it is not surprising that there are relatively fewer equable parallelograms on the Eisenstein lattice than there are on the integer lattice. It was already observed that for triangles, up to Euclidean transformations, there are only two equable triangles on the Eisenstein lattice, while there are five on the integer lattice; see [5, 6, 7] and the Appendix in [2].

Let us now describe our main results. We first show in Lemma 1 below that the sides of an ELEP are necessarily of the form n 3 with n ∈ ℕ. Throughout this paper, we will denote the side lengths a 3 and b 3 . Note that an ELEP is completely determined, up to a Euclidean motion, by the integers a, b. Indeed, if θ denotes one of the angles between the sides, then the area is 3ab sinθ and so by equability, sinθ = 2(a + b)/ 3 ab, which is determined by a and b. Notice, incidentally, that from sinθ = 2(a + b)/ 3 ab, there is no ELEP for which sinθ is rational. In particular, there is no rectangular ELEP, nor any ELEP with angle π/6.

Our main aim in this paper is to study the values of a, b for which an ELEP exists with sides a 3 , b 3 . In Section 2, we prove the following criterion, which is the exact analogue of [2: Theorem 1] in that it can be rephrased as follows: the square of the product of the sides minus the square of the perimeter is a square.

Theorem 1

Given positive integers a, b, an Eisenstein lattice equable parallelogram with sides a 3 , b 3 exists if and only if 9a² b² − 12(a + b)² is a square.

Corollary 1

There is no Eisenstein lattice equable rhombus.

In Section 3, we use Rosenberger’s Theorem on generalised Markov equations to prove the following result.

Theorem 2

The set 𝓣 of ordered pairs (a, b) of positive integers a < b, for which 9a² b² − 12(a + b)² is a square, is the set

T={2(q,r)∈N2|there exist unique odd, coprime, positive integerss,tsuch thats2+3t2+2q2=6stqandq≤r=3st−q}.

Furthermore, if (a, b) = 2(q, r) ∈ 𝓣, then ab = 2(s² + 3t²) and a + b = 6st. Moreover, q, r are coprime; in particular, 3 ∤ ab.

Theorems 1 and 2 provide a classification of ELEPs up to Euclidean motions. In Section 5, we describe how all ELEPs can be derived from a root example, (a, b) = (2, 4), by successive applications of four functions. This gives the tree of ELEPs; see Figure 3.

Figure 1

The root ELEP, (a, b) = (2, 4)

The paper [2] included a study of equable parallelograms, with vertices on the integer lattice, that have a pair of horizontal sides. No such parallelogram exists on the Eisenstein lattice, because of Lemma 1. Instead, in Section 7, we classify the ELEPs that have a horizontal diagonal. We show in Section 8 that there is no ELEP with a vertical diagonal. Finally, in Section 9, we classify the ELEPs with a vertical side, which turns out to be the same as the class of ELEPs with a side of length 2 3 or 4 3 .

Notation

In this paper, we employ the term positive in the strict sense. So ℕ = {n ∈ ℤ | n > 0}.

2 Proof of Theorem 1

Recall that the square of the distance between any pair of points in the Eisenstein lattice is an integer. Indeed, if z = x + yω ∈ ℤ[ω], then the square of the distance from the origin to z is x² − xy + y². If a triangle has its vertices on the Eisenstein lattice, then its area is of the form 34 n, where n ∈ ℕ. Indeed, for the triangle with vertices 0, x + yω, z + wω, with x, y, z, w ∈ ℤ, the area is 34 |xw − yz|.

The following result was proved for triangles in [6].

Lemma 1

If P is an equable polygon with vertices in ℤ[ω], then the side lengths of P are each of the form 3 n for some n ∈ ℕ.

Proof

Since the area of any triangle with vertices in ℤ[ω] is of the form 34 n for some n ∈ ℕ, the same is true for the area of P. Suppose P has sides s₁, …, s_n, whose squares are therefore integers. The equable hypothesis gives ∑i=1nsi∈34N, which implies

3s12+⋯+3sn2=3∑i=1nsi∈Q.

But it is well known that if ∑i=1nmi is rational for integers m₁, …, m_n, then mi is rational for each i; see, for example, [15] or [16]. Thus 3si2 is rational for each i. Hence, as 3s12,…,3sn2 are integers, it follows that 3s12,…,3sn2 are also integers. So the side lengths are each of the form n3 for some n ∈ ℕ. Thus, since the squares of the side lengths are integers, the required result follows. □

Lemma 2

Suppose an ELEP P has sides a 3 , b 3 , where a, b ∈ ℕ. Then a + b ≡ 0 (mod 3) and the lengths of the diagonals of P are given by the following formula:

d2=3(a2+b2)±29a2b2−12(a+b)2.

In particular, 9a² b² − 12(a + b)² is a square.

Proof

Consider a diagonal of length d, so d² ∈ ℕ. By Heron’s formula [12: Chap. 6.7], the triangle with sides a 3 , b 3 , d, has area

s(s−a3)(s−b3)(s−d),

where s=3a+3b+d2 is the semi-perimeter. Hence, the equability hypothesis is

(3a+3b+d)(−3a+3b+d)(3a−3b+d)(3a+3b−d)=48(a+b)2.

Expanding and rearranging the left hand side gives

−9(a2−b2)2+6(a2+b2)d2−d4=48(a+b)2. (1)

In particular, the integer d² is divisible by 3, say d² = 3D. Then (1) gives

−3(a2−b2)2+6(a2+b2)D−3D2=16(a+b)2,

and hence, a + b ≡ 0 (mod 3), as required. Furthermore, solving (1) for d² gives

d2=3(a2+b2)±29a2b2−12(a+b)2,

as required. In particular, as a, b, d² are integers, 9a² b² − 12(a + b)² is a square. □

Remark 1

Suppose an ELEP P has sides a 3 , b 3 and diagonals d₁, d₂. Then the above lemma gives d1d2=(a+b)48+9(a−b)2. Indeed, by Lemma 2,

(d1d2)2=9(a2+b2)2−36a2b2+48(a+b)2=9(a2−b2)2+48(a+b)2=(a+b)2(48+9(a−b)2), (2)

as required. We remark that the same formula can be deduced from Bretschneider’s formula and the equable hypothesis, without the use of Lemma 2. Notice that the 3-adic order, ν₃((a + b)²(48 + 9(a − b)²)), of the right-hand-side of (2) is odd. Hence, ν3(d12)≢ν3(d22) (mod 2).

Proof of Theorem 1

The necessity of the condition was shown in Lemma 2. Therefore, assume that a, b are positive integers such that 9a² b² − 12(a + b)² is a square. Consider a triangle T with sides a 3 , b 3 and

d:=3(a2+b2)+29a2b2−12(a+b)2.

Notice that such a triangle exists because a 3 , b 3 < d < (a + b) 3 , where the latter inequality holds as 3(a2+b2)+29a2b2−12(a+b)2<3(a+b)2 since 29a2b2−12(a+b)2<6ab. Let θ denote the angle between sides of length a 3 , b 3 and note that θ is obtuse since d² ≥ 3a² + 3b². Therefore,

d2=3a2+3b2−6abcos⁡θ=3a2+3b2+29a2b2−9a2b2sin2⁡θ.

So, from the definition of d, we have 9a² b² sin² θ = 12 (a + b)², thus 3ab sinθ = 2 3 (a + b). But since twice the area of T is 3 ab sinθ, the area of T is then 3 (a + b). Now consider the parallelogram P made from two copies of T. From what we have just seen, P is equable. It remains to show that P can be realized on the Eisenstein lattice, or equivalently, that T can be realized on the Eisenstein lattice. To do this, we employ the following result.

Theorem 3

([5]). A planar triangle T is realizable on the Eisenstein lattice if and only if the following three conditions hold:

the area of T is of the form 34 n, where n ∈ ℕ,
the squares of the side lengths of T are integers,
one of the side lengths of T is of the form r t , where r, t ∈ ℕ and t has no prime divisors congruent to 2 (mod 3).

Note that firstly, we saw above that the area of T is 3 (a + b) ∈ 3 ℕ. Secondly, the squares of the side lengths are 3a, 3b, 3(a² + b²)+ 29a2b2−12(a+b)2 which are all integers. Thirdly, the length 3 a has the form r t , where r, t ∈ ℕ and t has no prime divisors congruent to 2 (mod 3). So by Theorem 3, T has a realization on the Eisenstein lattice. □

Proof of Corollary 1

Suppose we had an Eisenstein lattice equable rhombus with side length a. Then by Lemma 2, 9a⁴ − 48a² is a square and hence 9a² − 48 is a square. But this is impossible as 6 is not a quadratic residue modulo 9. □

3 Proof of Theorem 2

The aim of this section is to prove Theorem 2. We first show that the elements of the set 𝓣 have the required property. Suppose that (a, b) = 2(q, r) ∈ 𝓣, where for coprime positive integers s, t, we have s² + 3t² + 2q² = 6stq and q < r = 3st − q. Hence, q, r are the two solutions of the quadratic equation

2u2−6stu+(s2+3t2)=0, (3)

in u, and so, q + r = 3st and 2qr = s² + 3t². Thus we have

9a2b2−12(a+b)2=4(9⋅4q2r2−12(q+r)2)=4(9(s2+3t2)2−12⋅32s2t2)=4⋅9(s2−3t2)2,

which is a square, as required.

We now show that every solution is in 𝓣. The main tool we use is Rosenberger’s Theorem on generalised Markov equations. Recall that in [13] Rosenberger considered equations of the form

ax2+by2+cz2=dxyz, (4)

where a, b, c are pairwise coprime positive integers with a ≤ b ≤ c such that a, b, c all divide d. We are only interested in positive integer solutions, that is, x, y, z ∈ ℕ, so we use the word solution to mean positive integer solution. Rosenberger’s remarkable result is that only 6 such equations have a solution and when such a solution exists, there are infinitely many solutions. We use the R1–R5 notation of [9].

Rosenberger’s Theorem

([13]). Equation (4) only has a solution in the following 6 cases:

x² + y² + z² = 3xyz (Markov’s equation),
x² + y² + 2z² = 4xyz,
x² + 2y² + 3z² = 6xyz,
x² + y² + 5z² = 5xyz,
x² + y² + z² = xyz,
x² + y² + 2z² = 2xyz.

We will also require the following classical result. There are several proofs of this result; a direct, elementary proof is given in [1].

Lemma 3

Suppose that positive integers x, y, z satisfy the equation x² + 3y² = z². Then there exist k ∈ ℕ and coprime s, t ∈ ℕ such that

x=k2|s2−3t2|,y=kst,z=k2(s2+3t2),

where k is even if s, t have different parity.

Returning to the proof of Theorem 2, suppose that a, b are positive integers with a < b such that 9a² b² − 12(a + b)² is a square. So a + b ≡ 0(mod 3) and a2b2−12a+b32 is a square integer. Let z=ab,y=2(a+b)3,x=a2b2−12a+b32, so that x² + 3y² = z², and y is even. So by Lemma 3, there exist k′ ∈ ℕ and coprime s, t ∈ ℕ such that

x=k′2|s2−3t2|,y=k′st,z=k′2(s2+3t2),

where k′ is even if s, t have different parity. In particular, from z and y, we have

ab=k′2(s2+3t2),a+b=3k′2st. (5)

Since s, t are coprime, s, t are both odd if they have the same parity. But in this case, k′ must be even, by the second equation in (5). So k′ is even in all cases; set k′ = 2k. Then from equation (5), we have

ab=k(s2+3t2),a+b=3kst. (6)

Hence, a, b are solutions to the quadratic equation X² − 3kst X + k(s² + 3t²) = 0. In particular, we have

ks2+3kt2+a2=3kast. (7)

Let k = fg², where f is square-free. From (7), f divides a² and hence f divides a. Let a = fα. Dividing (7) by f gives

g2s2+3g2t2+fα2=3fg2stα. (8)

From (8), g² divides fα² and hence g² divides α², and thus g divides α. Let α = gq. Dividing (8) by g² gives

s2+3t2+fq2=3fgstq. (9)

Thus by Rosenberger’s Theorem, using x = s, y = q, z = t, there is only one possibility for (9) to have a solution, namely it is the equation R2, with f = 2 and g = 1. Consequently, we have k = 2 and a = 2q. So equation (9) gives

s2+3t2+2q2=6stq. (10)

In particular, a is even and hence b is even, say b = 2r, by the second equation in (6).

Notice that as s² + 3 t² + 2 q² = 6 s t q, then modulo 2, this gives s + t ≡ 0, and so s, t are both odd.

We now show that if (a, b) = (2q, 2r), then q, r are coprime. Rewriting (6), we have

ab=2(s2+3t2),a+b=6st. (11)

Suppose p is an odd prime divisor of gcd(a, b). Then p² divides s² + 3t² and p divides 3st, by equation (11). Thus p divides (s + 3t)² and (s − 3t)², and so p divides s + 3t and s − 3t. Hence, p divides 2s and 6t. Consequently, as p is odd and s, t are coprime, p = 3 and furthermore, 3 divides s and 3 doesn’t divide t. Since 3 divides a and b, we have 9 divides s² + 3t², from the first equation in (11). But this is impossible since 9 divides s² and 3 doesn’t divide t. Hence gcd(a, b) is a power of 2. Then, as s, t are both odd, the second equation in (11) gives gcd(a, b) = 2, as required.

Since a + b = 6st and gcd(a, b) = 2, neither a nor b is divisible by 3. It remains to show that, given a, b, the integers s, t are unique. Suppose that (11) holds and that one has coprime positive integers s′, t′ with

ab=2(s′2+3t′2),a+b=6s′t′. (12)

From the second equations in (11) and (12), we have s′t′ = s t. From (12), we also have s^′² + 3t^′² + 2q² = 6s′ t′ q, where a = 2q, as before. Hence, by (11) and (10), s^′² + 3t^′² = s² + 3t², so as t′ = st/s′,

s′4+3s2t2=s′2(s2+3t2),

and hence (s^′² − s²)(s^′² − 3t²) = 0. As s′ is a positive integer, it follows that s′ = s and thus t′ = t. This completes the proof of Theorem 2. □

Remark 2

Since a = 2q is not divisible by 3, neither is q. So by (10), s is not divisible by 3.

4 Comments on Theorem 2

Theorem 2 identifies the set 𝓣 of ordered pairs (a, b) of possible side lengths of ELEPs, divided by 3 , with a < b and a, b even; a = 2q and b = 2r. It relates these pairs (a, b) to pairs of coprime positive integers s, t, for which the equation

s2+3t2+2u2=6stu

is satisfied for both u = q and u = r. From given s, t, we have, from (3),

a=3st−9s2t2−2(s2+3t2),b=3st+9s2t2−2(s2+3t2). (13)

Conversely, given a, b, we have, from (11),

s2+3t2=ab2ands2⋅3t2=(a+b)212

and so

s2=3ab+σ(a,b)9a2b2−12(a+b)212, (14)

3t2=3ab−σ(a,b)9a2b2−12(a+b)212, (15)

where σ(a, b) = ± 1, and will now be explained. By Remark 2, s ≢ 0(mod 3), so s² ≡ 1(mod 3). Notice that

s2=3ab+σ(a,b)9a2b2−12(a+b)212=qr+σ(a,b)q2r2−(q+r)23.

By Theorem 2, a, b are not divisible by 3. So, as q + r = 3st, we have qr ≡ − 1(mod 3). Hence, since σ(a, b) = 1 if and only if s² > 3t², we have the following result.

Lemma 4

Suppose (a, b) ∈ 𝓣 with a = 2q and b = 2r. Then

σ(a,b)≡−q2r2−(q+r)23(mod3),

and the following conditions are equivalent:

σ(a, b) = 1,
s² > 3t²,
q2r2−(q+r)23≡−1(mod3).

Remark 3

Suppose that an element z = x′ + y′ω ∈ ℤ[ω] has length 3 n for some n ∈ ℕ. Then in complex numbers, z=x′−y′2+3y′2i and

3n2=x′2−x′y′+y′2.

Thus, if n is even, then x′, y′ are necessarily both even. Let y′ = 2y and x = x′ − y, so z = x + y 3 i, where x, y ∈ ℤ. In particular, this is the case for the sides of an ELEP, by Theorem 2.

5 The tree of ELEPs

Consider the set 𝓢 of solutions (s, t, u), with s, t coprime, of the Markov-Rosenberger equation given in (10) with q = u:

s2+3t2+2u2=6stu. (16)

Note that we are not assuming that u < 3st. Following the presentation given in [9], from a solution x = (s, t, u) to (10), one can generate three new solutions by applying the involutions:

ϕ1(x)=(6tu−s,t,u),ϕ2(x)=(s,2su−t,u),ϕ3(x)=(s,t,3st−u).

The group of transformations of 𝓢 generated by the maps ϕ_i is the free product of three copies of ℤ₂, and this group acts transitively on 𝓢. Moreover, the maps ϕ_i give the set 𝓢 of solutions the structure of an infinite binary tree: each solution is a vertex and two distinct solutions are connected by an edge if one of the maps ϕ_i sends one solution to the other. The fundamental solution has the smallest values of s + t + u; it is (s, t, u) = (1, 1, 1).

Lemma 5

The fixed point sets of the maps ϕ₁, ϕ₃ are empty, and (1, 1, 1) is the unique fixed point of ϕ₂.

Proof

If (s, t, u) is a fixed point of ϕ₁, then s = 3tu. Hence t = 1, as s, t are coprime. Replacing in (16) gives the contradiction 3 = 7u². Similarly, a fixed point (s, t, u) of ϕ₃ would have 2u = 3st, which is impossible as s, t are odd. If (s, t, u) is a fixed point of ϕ₂, we have t = su, so s = 1, as s, t are coprime. Then (16) gives u² = 1 and so t² = 1. Thus finally s = t = u = 1. □

In summary so far: the group ℤ₂ ∗ ℤ₂ ∗ ℤ₂ generated by ϕ₁, ϕ₂, ϕ₃ acts freely on 𝓢 except at the fundamental solution (s, t, u) = (1, 1, 1), which we can take as the root of the tree 𝓢.

Having recalled Rosenberger’s theory, we now describe how the solution tree 𝓢 of the Markov-Rosenberger equation determine the induced structure on the set 𝓣 of ELEPs. Motivated by (13), we define the map π : 𝓢 → 𝓣 by

π(s,t,u)=3st−9s2t2−2(s2+3t2),3st+9s2t2−2(s2+3t2). (17)

The map π is well defined and surjective by Theorem 2. Note that if π (s, t, u) = (a, b), then by (13) and (16), either a = 2u or b = 2u. Furthermore, trivially, π ∘ ϕ₃(s, t, u) = (a, b); that is, (s, t, u) and ϕ₃(s, t, u) correspond to the same ELEP. In other words, for each (a, b) ∈ 𝓣, the pre-image π⁻¹(a, b) consist of two points, which are interchanged by the involution ϕ₃.

Consequently, as contraction of edges of a tree produces another tree, we can form a tree structure on 𝓣 by contracting each of the edges of 𝓢 that are given by the map ϕ₃; see Figure 2.

Figure 2

Contraction of the ϕ₃ edges

This contraction turns vertices of degree 3 into vertices of degree 4. This is the case for all vertices except for the image of the fundamental solution, which has degree 3. This is the root (2, 4) ∈ 𝓣. The tree 𝓣 is shown in Figure 3. Here the elements (a, b) are shown above the corresponding pairs (s, t).

Figure 3

The tree of ELEPs; the elements (a, b) are shown above the corresponding pairs (s, t).

We now see how the maps ϕ₁, ϕ₂, ϕ₃ on 𝓢 generate maps on 𝓣. For each i = 1, 2, maps φ_i, ψ_i : 𝓣 → 𝓣 can be naturally defined as follows. If (a, b) ∈ 𝓣, with a = 2q, b = 2r, let (s, t, q) ∈ 𝓢 such that π (s, t, q) = (a, b). Then set

φi(a,b):=π∘ϕi(s,t,q),ψi(a,b):=π∘ϕi(s,t,r). (18)

Note that from (11) we have a + b = 6st, so (a, b) = (2q, 6st − 2q). From the definition of ϕ₁, the map φ₁ leaves a and t unchanged and s is changed to s′ = 6qt − s = 3at − s. Then under φ₁, the value of b is changed to

b′=6s′t−a=18at2−6st−a=18at2−(a+b)−a=3a2b−σ(a,b)a9a2b2−12(a+b)22−2a−b, (19)

using (15). Note that a priori, we don't know which of the resulting components, a or b′, is the larger. So we set

φ1:(a,b)↦min(a,b′),max(a,b′).

Similarly, using ϕ₂, we set

φ2:(a,b)↦min(a,b″),max(a,b″),

where

b″=3a2b+σ(a,b)a9a2b2−12(a+b)22−2a−b. (20)

Similarly, analogous to φ₁, φ₂, interchanging the roles of a and b, we have two further maps:

ψ1:(a,b)↦min(a′,b),max(a′,b),ψ2:(a,b)↦min(a″,b),max(a″,b),

where

a′=3ab2+σ(a,b)b9a2b2−12(a+b)22−a−2b,a″=3ab2−σ(a,b)b9a2b2−12(a+b)22−a−2b.

These maps are obtained from the maps ϕ₁ ∘ ϕ₃ and ϕ₂ ∘ ϕ₃, respectively.

Note that while ϕ₁, ϕ₂, ϕ₃ are involutions and generate a group of transformations of 𝓢, the same is not true of the maps φ₁, φ₂, ψ₁, ψ₂ of 𝓣. Indeed, they are not even all bijections. For example, ϕ₂(2, 4) = (2, 4) = ϕ₂(4, 14). Moreover, the study of the maps ϕ₁, ϕ₂, ψ₁, ψ₂ is considerably complicated by the term σ(a, b). So it is often convenient to work instead with the pairs (s, t). From the definition of the maps ϕ_i, we have

ϕ1(s,t,u/2)=(3ut−s,t,u/2),ϕ2(s,t,u/2)=(s,us−t,u/2),

where u = a or b, and a, b are given by (13). We define the linear involutions ϕ_1,u, ϕ_2,u on the set Σ := {(s, t) ∈ ℕ² : s, t odd and coprime}, by ϕ_1,u(s, t) = (3u t − s, t), ϕ_2,u(s, t) = (s, us − t), or equivalently in matrix form,

ϕ1,u=−13u01,ϕ2,u=10u−1. (21)

By taking u = a or b, (21) effectively gives four functions. For given (a, b), it is often easier to employ these functions than the functions φ₁, φ₂, ψ₁, ψ₂, and then use (13) to determine the resulting corresponding values of a, b, where necessary. In Figure 3, and later in Figure 7, the edges are labelled with the corresponding maps ϕ_i,u for i ∈ {1, 2} and u ∈ {a, b}.

6 Diagonals, heights and altitudes

Let us first fix some terminology and notation; see Figure 4.

Figure 4

Diagonals, heights and altitudes

Definition 2

Consider a non-square ELEP P. We denote the length of its long (resp. short) diagonal d_l (resp. d_s). The heights of P are the distances between opposite sides; we denote the long (resp. short) height h_l (resp. h_s). Notice that the long (resp. short) height connects short (resp. long) sides. Each diagonal d partitions P into two congruent triangles T. We will call the distance from d to the third vertex of T an altitude of P. We call the altitude from d_s (resp. d_l) the long (resp. short) altitude and denote it η_l (resp. η_s).

Lemma 6

Suppose an ELEP P has sides a 3 , b 3 with a, b ∈ ℕ and a ≤ b. Then

hl=2(a+b)a,hs=2(a+b)b,
ηl=23(a+b)ds,ηs=23(a+b)dl.

Proof

By equability, the area of P is 2 3 (a + b), but the area is obviously also a 3 h_l and b 3 h_s. This gives (a). But the area of P is also twice the area of the triangle determined by each diagonal. So the area of P is both η_l d_s and η_s d_l. This gives (b). □

Remark 4

If h_s is an integer, then h_s = 2 + 2ab by Lemma 6(a), and so as b > a, the only possibility is b = 2a. But then 9a² b² − 12(a + b)² = 36 a² (a² − 3), which is a square only when a = 2. This is the root parallelogram (a, b) = (2, 4). It has h_s = 3 and h_l = 6. Similarly, if h_l is an integer, then h_l = 2 + 2ba by Lemma 6(a), and so 2b = ak for some positive integer k. In this case, h_l = 2 + k and h_s = 2 + 4k .

Theorem 4

For every ELEP the altitudes satisfy 2 < η_s ≤ 67 and 2 < η_l ≤ 2 3 .

Remark 5

Suppose an ELEP P has sides a 3 , b 3 with a, b ∈ ℕ and a ≤ b. By Theorem 2, we have ab = 2(s² + 3t²) and a + b = 6st, so 9a2b2−12(a+b)2=6|s2−3t2| for some odd, coprime integers s, t. If s² > 3t², then from Lemma 2, the diagonals are given by

dl2=3(a2+b2)+29a2b2−12(a+b)2=3⋅36s2t2−12(s2+3t2)+12|s2−3t2|=12(9s2t2−6t2),ds2=3⋅36s2t2−12(s2+3t2)−12|s2−3t2|=12(9s2t2−2s2).

Similarly, if s² < 3t², the diagonals are given by dl2=12(9s2t2−2s2) and ds2=12(9s2t2−6t2). Putting the cases together, we have

dl2=12(9s2t2−2min(s2,3t2)),ds2=12(9s2t2−2max(s2,3t2)). (22)

Notice that when a diagonal d satisfies d² = 12(9 s² t² − 6t²), one has d=6t3s2−2, so d is an integer if and only if 3s² − 2 is a square. When d² = 12(9 s² t² − 2s²), one has d=2s27t2−6. In this case, d is never an integer, since 27t² − 6 ≡ 3(mod 9) and 3 is not a quadratic residue modulo 9.

Proof

Suppose an ELEP P has sides a 3 , b 3 with a, b ∈ ℕ and a ≤ b. By Theorem 2, we have ab = 2(s² + 3t²) and a + b = 6st, so 9a2b2−12(a+b)2=6|s2−3t2| for some odd, coprime integers s, t. Using Lemma 6 and Remark 5, the altitudes are given by

ηl2=12(a+b)2ds2=36s2t29s2t2−2max(s2,3t2),ηs2=12(a+b)2dl2=36s2t29s2t2−2min(s2,3t2).

We consider the two cases according to whether s² > 3t² or s² < 3t². First suppose s² > 3t². Then

ηl2=36s2t29s2t2−2s2=36t29t2−2=4+89t2−2,ηs2=36s2t29s2t2−6t2=12s23s2−2=4+83s2−2.

Hence, as t ≥ 1, we have 4<ηl2≤4+87, so 2 < η_l ≤ 67 . Similarly, as s ≥ 1, we have 4 < ηs2 ≤ 12, so 2 < η_s ≤ 2 3 . Thus, as η_s < η_l, we have 2 < η_s < η_l ≤ 67 .

Now suppose s² < 3t². Then

ηl2=36s2t29s2t2−6t2=12s23s2−2=4+83s2−2ηs2=36s2t29s2t2−2s2=36t29t2−2=4+89t2−2.

Hence, as s ≥ 1, we have 4 < ηl2 ≤ 12, so 2 < η_l ≤ 2 3 . Similarly, as t ≥ 1, we have 4 < ηs2 ≤ 4 + 87 , so 2 < η_s ≤ 67 . Thus, as η_s < η_l, we have 2 < η_s < η_l ≤ 2 3 .

Combining the two cases gives the required bounds on η_s and η_l. □

7 ELEPs with horizontal diagonal

Example 1

Here we give examples of ELEPs with a horizontal diagonal. Consider the sequence (u_n) defined by u_n = 4 u_n−1 − u_n−2, with u₀ = 0 and u₁ = 1. The following result is well known; see [10: Chap. 5, Ex. 2.1] and sequence A001353 in [14]. We provide a proof for completeness.

Lemma 7

For all n ≥ 1, one has

u_n+1 u_n−1 = un2 − 1,
3 un2 + 1 = (2u_n − u_n−1)².

Proof

One has u₂ u₀ = 0 = u12 − 1. Then for n ≥ 2, by induction,

un+1un−1−un2=(4un−un−1)un−1−un(4un−1−un−2)=unun−2−un−12=−1.
For all n ≥ 1, expanding (2u_n − u_n−1)², one has, using (a),

3un2+1−(2un−un−1)2=1−un2+un−1(4un−un−1)=1−un2+un−1un+1=0.

□

Let

An=(−6un,−23),Bn=(6un+1−6un,0),Cn=(6un+1,23).

The vertices OA_nB_nC_n form a parallelogram on the Eisenstein lattice with diagonal OB_n on the x-axis. The parallelogram OA_nB_nC_n has side lengths

OAn¯=36un2+12=233un2+1=23(2un−un−1),OCn¯=36un+12+12=233un+12+1=23(2un+1−un),

by Lemma 7(b), and area

area(OAnBnCn)=det−6un−236un+123=123(un+1−un).

Hence,

2(OAn¯+OCn¯)=43(2un+1+un−un−1)=43(2un+1+un+(un+1−4un))=123(un+1−un)=area(OAnBnCn),

so OA_nB_nC_n is equable. Table 1 lists the first 9 examples of ELEPs with horizontal diagonal. Note that the first four of these ELEPs appear in Figure 3 in the branch that starts at the root (2, 4) and descends vertically. Figure 5 shows the first two examples, with the first one translated 6 units to the left.

Figure 5

Two ELEPs with horizontal diagonal

Table 1

Examples of ELEPs with horizontal diagonal

n	(q, u)	a	b	A	B	(s, t)
0	(1, 0)	2	4	−2 − 4ω	6	(1, 1)
1	(2, 1)	4	14	−8 − 4ω	18	(1, 3)
2	(7, 4)	14	52	−26 − 4ω	66	(1, 11)
3	(26, 15)	52	194	−92 − 4ω	246	(1, 41)
4	(97, 56)	194	724	−338 − 4ω	918	(1, 153)
5	(362, 209)	724	2702	−1256 − 4ω	3426	(1, 571)
6	(1351, 780)	2702	10084	−4682 − 4ω	12786	(1, 2131)
7	(5042, 2911)	10084	37634	−17468 − 4ω	47718	(1, 7953)
8	(18817, 10864)	37634	140452	−65186 − 4ω	178086	(1, 29681)

Theorem 5

Up to Euclidean isometry, the only ELEPs with horizontal diagonal are those of Example 1.

Proof

By translating and reflecting in the x and/or y axes if necessary, we may assume that the horizontal diagonal lies on the positive x-axis, starting at the origin 0, and that the side starting at 0, and lying in the 3rd or 4th quadrants, is the shorter of the two sides. Therefore, suppose we have a ELEP with vertices A = x − yω, B = z, C = (z − x) + yω, O = 0, where x ∈ ℤ and y, z ∈ ℕ. Let a 3 , b 3 denote the lengths of OA and AB, respectively, with a, b ∈ ℕ and a < b. In particular, we have

3a2=x2+xy+y2. (23)

The altitudes from A and C have length η:=y32. By Theorem 4, 2 < η ≤ 2 3 , which gives 43 < y ≤ 4. So, as y ∈ ℕ, we have y = 3 or 4. First suppose that y = 3. Then (23) gives

3a2=x2+3x+9. (24)

So x is divisible by 3, say x = 3x′. But then by (24), a is divisible by 3, contrary to Theorem 2. So we have y = 4.

As y = 4, we have η = 2 3 > 67 . So η = η_l by Theorem 4, and from the proof of Theorem 4 we have s² < 3t² and ηl2=4+83s2−2. Hence, (23)2=4+83s2−2, giving s = 1. Then by Theorem 2, (a, b) = 2(q, r) ∈ ℕ² where

1+3t2+2q2=6tq (25)

and q ≤ r = 3t − q. Let u := t − q. From(25), we have

q2−3u2=1. (26)

Notice that u ≥ 0. Indeed, from (25), we have q=3t−3t2−22, so u=−t+3t2−22≥0 for t ≥ 1. Notice also that a² = 4q² and (23) gives 3a² = x² + 4x + 16, so 12q² = x² + 4x + 16. Thus x is even, say x = 2x′, and so 3q² = x′² + 2x′ + 4. Hence, by (26), 9u² = x′² + 2x′ + 1 = (x′ + 1)², so

x=2x′=±6u−2. (27)

Now (26) is one of Pell’s equations and it is well known (see [11]) that the solutions (q_n, u_n) to (26) are given by the recurrence relation

(qn,un)=4(qn−1,un−1)−(qn−2,un−2),(q1,u1)=(2,1),(q0,u0)=(1,0).

In particular, u_n agrees with the sequence u_n of Example 1. Notice also that

qn+2un=un+1for all n≥0. (28)

Indeed, both sides of the equation satisfy the same recurrence relation, with the initial conditions q₀ + 2u₀ = 1 = u₁ and q₁ + 2u₁ = 4 = u₂.

For each n ≥ 0, we denote by a_n, b_n, x_n, z_n, t_n the values of a, b, x, z, t, respectively, determined by q_n and u_n, and we denote the corresponding parallelogram OA_nB_nC_n. Since there are so many variables, we recall for the reader’s convenience that A_n = x_n − 4ω, B_n = z_n, C = (z_n − x_n) + 4ω, the lengths of OA_n, OC_n are 3 a_n, 3 b_n respectively, and from the definition of u, we also have a_n + b_n = 6t_n and t_n = u_n + q_n. Furthermore, a_n = 2q_n and

bn=6tn−an=6tn−2qn=6un+4qn. (29)

Since A_n has y-coordinate −2 3 , the area of OA_nB_nC_n is 2 3 z_n, so the equability condition is 2 3 z_n = 2(a_n + b_n) 3 ; that is, z_n = a_n + b_n = 6t_n. So z_n = 6(u_n + q_n).

We now consider the two cases given by (27):

x_n = 6u_n − 2 for u_n > 0,
x_n = − 6u_n − 2 for u_n ≥ 0.

We show that this case leads to a contradiction. Suppose x_n = 6u_n − 2 for u_n > 0, so A_n = 6u_n − 2 − 4ω, B_n = z_n = 6(u_n + q_n) and C_n = 6q_n + 2 + 4ω. So, from (29),

3⋅(6un+4qn)2=3bn2=OCn¯2=(6qn+2)2−(6qn+2)4+16,

where the expression on the right comes for the square length of the element C_n = 6q_n + 2 + 4 ω. Expanding, rearranging and dividing by 12 gives qn2+12qnun+9un2=1. But this is impossible for positive integers u_n and q_n.
Now suppose x_n = − 6u_n − 2 for u_n ≥ 0. Using (28), we have A_n = − 6u_n − 2 − 4 ω, B_n = z_n = 6(u_n + q_n) = 6u_n+1 − 6u_n, and C_n = 6u_n+1 + 2 + 4ω, which is one of the parallelograms of Example 1. □

Since the parallelograms of Example 1 have horizontal short diagonal, we have the following conclusion.

Corollary 2

No ELEP has a horizontal long diagonal.

Remark 6

The upper bounds η_s = 67 and η_l = 2 3 of Theorem 4 are both attained by the first of the parallelograms of Example 1; that is, the root parallelogram (a, b) = (2, 4) given by n = 0 in Table 5.

8 ELEPs with a vertical diagonal

Theorem 6

There is no ELEP having a vertical diagonal.

Proof

Suppose we have an ELEP with a vertical diagonal. By translating and reflecting in the x and/or y axes if necessary, we may assume that the vertical diagonal lies on the positive y-axis, starting at the origin 0, and that the side starting at 0, and lying in the 1st or 2nd quadrants, is the shorter of the two sides. Therefore, suppose we have an ELEP with vertices A = x + yω, B = zω, C = − x + (z − y)ω, O = 0, where x, z ∈ ℕ and y ∈ ℤ. Let a 3 , b 3 denote the lengths of OA and AB respectively, with a, b ∈ ℕ and a < b. In particular, we have

3a2=x2−xy+y2. (30)

The altitudes from A and C have length η := x. By Theorem 4, 2 < η ≤ 2 3 , which gives x = 3. Then (30) gives

3a2=9−3y+y2. (31)

So y is divisible by 3, say y = 3y′. But then by (31), a is divisible by 3, contrary to Theorem 2. □

9 ELEPs with a vertical side

Example 2

Consider the Pell-like equation

3u2=2m2+1. (32)

This equation is well known; see entry A072256 in [14]. Its solutions (u_n, m_n) satisfy the recurrence relation

(un,mn)=10(un−1,mn−1)−(un−2,mn−2),

with (u₁, m₁) = (1, 1), (u₂, m₂) = (9, 11).

Now, using complex numbers, consider the parallelogram with vertices A_n = 2 3 i, B_n = − x_n + (2 + y_n) 3 i, C_n = − x_n + y_n 3 i, O = 0, where x_n = 3 + 3u_n and y_n = 2m_n. Note that the vertical side OA_n has length a 3 , where a := 2, while the side OC_n has length

bn3=xn2+3yn2=(3+3un)2+12mn2=9+18un+9un2+6(3un2−1)(by (32))=3(1+3un).

So OA_nB_nC_n has perimeter 2(a + b_n) 3 = 6 3 (1 + u_n). Furthermore, OA_nB_nC_n has area a 3 x_n = 2x_n 3 = 6 3 (1 + u_n). So OA_nB_nC_n is equable.

Table 2 lists the first 9 of these examples. The first five of these ELEPs appear in Figure 7, and in Figure 3, in the branch that starts at the root (2, 4) and proceeds horizontally to the right. The values of (s, t) were computed using (21). Figure 6 shows the first two examples, with the first one reflected in the y-axis.

$Figure 6 Two ELEPs having a vertical side of length 3 $\begin{array}{} \displaystyle \sqrt3 \end{array}$.$

Figure 6

Two ELEPs having a vertical side of length 3 .

Figure 7

The paths of ELEPs having a side of length 2 or 4.

Table 2

Examples of ELEPs with vertical side of length 2 3 and vertex C = −x + y 3 i

n	(u, m)	b	x	y	(s, t)
1	(1, 1)	4	6	2	(1, 1)
2	(9, 11)	28	30	22	(5, 1)
3	(89, 109)	268	270	218	(5, 9)
4	(881, 1079)	2644	2646	2158	(49, 9)
5	(8721, 10681)	26164	26166	21362	(49, 89)
6	(86329, 105731)	258988	258990	211462	(485, 89)
7	(854569, 1046629)	2563708	2563710	2093258	(485, 881)
8	(8459361, 10360559)	25378084	25378086	20721118	(4801, 881)
9	(83739041, 102558961)	2486793150	2030458102	205117922	(4801, 8721)

Example 3

We now give two sequences of ELEPs having a vertical side of length 4 3 . Consider the equation

132w2+36w+1=y2. (33)

This equation is not particularly well known. The first six solutions for (w, y) are:

{1,13},{5,59},{52,599},{236,2713},{2397,27541},{10857,124739}.

In fact, using Alpern’s integer equation solver [8], one finds that the sequence of solutions is composed of two interspersed sequences, having the same recurrence relation

wnyn=23226423wn−1yn−1+336,

but different initial conditions. One sequence has (w₁, y₁) = (1, 13), while the other has (w₁, y₁) = (5, 59).

Now, for each of the two sequences, consider the parallelogram with vertices A_n = 4 3 i, B_n = − x_n + (4 + y_n) 3 i, C_n = − x_n + y_n 3 i, O = 0, where x_n = 6w_n + 3. Note that the vertical side OA_n has length a := 4 3 , while the side OC_n has length

bn3=xn2+3yn2=(6wn+3)2+3yn2=(6wn+3)2+3(132wn2+36wn+1)(by (33))=23(1+6wn).

So OA_nB_nC_n has perimeter 2(a + b_n) = 12 3 (1 + 2 w_n). Furthermore, OA_nB_nC_n has area ax_n = 4x_n 3 = 12 3 (1 + 2w_n). So OA_nB_nC_n is equable.

Table 3 lists the first six examples of ELEPs corresponding to the sequence with initial condition (w₁, y₁) = (1, 13). The first five of these ELEPs appear in Figure 7 in the branch that starts at the element (a, b) = (4, 14) and proceeds horizontally to the right. Table 4 lists the first six examples of ELEPs corresponding to the sequence with initial condition (w₁, y₁) = (5, 59). The first five of these ELEPs appear in Figure 7 in the branch that starts at the element (a, b) = (4, 62) and proceeds horizontally to the right. Notice that the two sequences are actually the two ends of a bi-infinite path, and are connected by the vertical path of length 2 that passes from (4, 14) through the root (2, 4) to (4, 62).

Table 3

First sequence of ELEPs with vertical side of length 4 3 and vertex C = − x + y 3 i

n	(w, y)	b	x	(s, t)
1	(1, 13)	14	9	(1, 3)
2	(52, 599)	626	315	(35, 3)
3	(2397, 27541)	28766	14385	(35, 137)
4	(110216, 1266287)	1322594	661299	(1609, 137)
5	(5067545, 58221661)	60810542	30405273	(1609, 6299)
6	(232996860, 2676930119)	2795962322	1397981163	(73979, 6299)

Table 4

Second sequence of ELEPs with vertical side of length 4 3 and vertex C = −x + y 3 i

n	(w, y)	b	x	(s, t)
1	(5, 59)	62	18	(11, 1)
2	(236, 2713)	2834	1419	(11, 43)
3	(10857, 124739)	130286	65145	(505, 43)
4	(499192, 5735281)	5990306	2995155	(505, 1977)
5	(22951981, 263698187)	275423774	137711889	(23219, 1977)
6	(1055291940, 12124381321)	12663503282	6331751643	(23219, 90899)

Theorem 7

Up to Euclidean transformations, the only ELEPs having a side of length 2 3 or 4 3 are those of Example 2 and Example 3, respectively. Moreover, all ELEPs with a vertical side have a side of length 2 3 or 4 3 .

Proof

Let OABC be an ELEP with sides of length a 3 and b 3 . First suppose that a = 2. Then Lemma 2 gives the integer

9a2b2−12(a+b)2=26((b−1)2−3)∈N.

So, (b − 1)² − 3 = 6m² for some m ∈ ℕ. In particular, b ≡ 1(mod 3), say b = 3u + 1, as in Example 2. Thus 3u² = 2m² + 1, which is the same equation as (26). Consequently, b is one of the numbers b_n of Example 2. It follows that, up to a Euclidean transformation, the OABC is one of the ELEPs of Example 2.

Now suppose that a = 4. Then Lemma 2 gives the integer

9a2b2−12(a+b)2=23(11b2−8b−16)∈N.

So, 11b² − 8b − 16 = 3m² for some m ∈ ℕ. Writing b = 2r, as before, we have that m is even, say m = 2ℓ. Thus 11r² − 4r − 4 = 3ℓ². Note that as a = 2q = 4, we have q = 2. So r must be odd, since q, r are coprime by Theorem 2. Let r = 2v + 1. Thus 44v² + 36 v + 3 = 3ℓ². Hence, v is divisible by 3, say v = 3w. So b = 2r = 2(2v + 1) = 2(6w + 1) = 12w + 2, as in Example 3. It follows that 132 w² + 36 w + 1 = ℓ², which is the same equation as (33). Consequently, b is one of the numbers b_n of Example 3. It follows that, up to a Euclidean transformation, the OABC is one of the ELEPs of Example 3.

Finally, suppose that the ELEP OABC has a vertical side. By translating and reflecting in the x and/or y axes if necessary, we may assume that the vertical side lies on the positive y-axis, starting at the origin 0, and that the other side starting at 0 lies in the 2nd quadrant. Therefore, using complex numbers, we consider an ELEP with vertices A = a 3 i, B = − x + (a + y) 3 i, C = − x + y 3 i, O = 0, where a, x, y ∈ ℕ by Remark 3. Then OA has length a 3 . Let b 3 denote the length of OC. In particular, we have

3b2=x2+3y2. (34)

The height from C (to the y-axis) is h := x, which is an integer. So, by Remark 4, if h = h_s, then OABC is the root parallelogram and {a, b} = {2, 4}. So we may assume that h = h_l, in which case, by Remark 4, 2b = ak for some positive integer k. Let a = 2q, b = 2r, as before. Then 2b = ak gives 2r = qk. By Theorem 2, q, r are coprime, so q = 1 or 2; that is, a = 2 or 4. This completes the proof of the theorem. □

Communicated by István Gaál

References

[1] Abdelalim, S.—Dyani, H.: The solution of the diophantine equation x² + 3y² = z², Int. J. Algebra 8(5) (2014), 729–732.10.12988/ija.2014.4779Search in Google Scholar

[2] Aebi, C.—Cairns, G.: Lattice equable quadrilaterals I: parallelograms, L’Enseign. Math. 67(3/4) (2021), 369–401.10.4171/lem/1013Search in Google Scholar

[3] Aebi, C.—Cairns, G.: Lattice equable quadrilaterals II: kites, trapezoids and cyclic quadrilaterals, Int. J. Geom. 11(2) (2022), 5–27.Search in Google Scholar

[4] Aebi, C.—Cairns, G.: Lattice equable quadrilaterals III: tangential and extangential cases, Integers 23 (2023), A48.Search in Google Scholar

[5] Aebi, C.—Cairns, G.: Following in Yiu’s footsteps but on the Eisenstein lattice, Amer. Math. Monthly 131(6) (2024), 526–529.10.1080/00029890.2024.2325863Search in Google Scholar

[6] Aebi, C.—Cairns, G.: Equable triangles on the Eisenstein lattice, Math. Gazette, to appear in March 2025; preprint available at https://arxiv.org/abs/2309.04476.Search in Google Scholar

[7] Aebi, C.—Cairns, G.: Less than Equable Triangles on the Eisenstein lattice, Math. Gazette, to appear in July 2025; preprint available at https://arxiv.org/abs/2312.10866.Search in Google Scholar

[8] Alpern, D.: Generic two integer variable equation solver, https://www.alpertron.com.ar/QUAD.HTM, accessed 6 September 2023.Search in Google Scholar

[9] Baragar, A.—Umeda, K.: The asymptotic growth of integer solutions to the Rosenberger equations, Bull. Austral. Math. Soc. 69(3) (2004), 481–497.10.1017/S0004972700036261Search in Google Scholar

[10] Barbeau, E. J.: Pell’s Equation, Springer-Verlag, New York, 2003.10.1007/b97610Search in Google Scholar

[11] Deemer, B.: A Recurrence Formula Solution to dy² + 1 = x², Math. Mag. 32(1) (1958), 37–40.10.2307/3029051Search in Google Scholar

[12] Ostermann, A.—Wanner, G.: Geometry by its History, Springer, Heidelberg, 2012.10.1007/978-3-642-29163-0Search in Google Scholar

[13] Rosenberger, G.: Über die diophantische Gleichung ax² + by² + cz² = dxyz, J. Reine Angew. Math. 305 (1979), 122–125.10.1515/crll.1979.305.122Search in Google Scholar

[14] Sloane, N. J. A.: The On-line Encyclopedia of Integer Sequences, https://oeis.org.Search in Google Scholar

[15] Wang, V.: The Art of problem Solving, https://artofproblemsolving.com/community/c1461h1035155.Search in Google Scholar

[16] Yuan, Q.: Annoying Precision, https://qchu.wordpress.com/2009/07/02/square-roots-have-no-unexpected-linear-relationships/.Search in Google Scholar

[17] Zimhoni, N.: A forest of Eisensteinian triangles, Amer. Math. Monthly 127(7) (2020), 629–637.10.1080/00029890.2020.1760018Search in Google Scholar

Received: 2023-11-09

Accepted: 2024-01-03

Published Online: 2024-08-14

Published in Print: 2024-08-27

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/ms-2024-0071

Keywords for this article

Eisenstein lattice; equable; parallelogram

Creative Commons

BY 4.0