# Equable Parallelograms on the Eisenstein Lattice

Christian Aebi* and Grant Cairns  * Collège Calvin
Geneva
Switzerland 1211<EMAIL_ADDRESS>** Department of Mathematical and
Physical Sciences
La Trobe University
Melbourne
Australia 3086<EMAIL_ADDRESS>
###### 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.

###### Key words and phrases:

Eisenstein lattice, equable, parallelogram

###### 2020 Mathematics Subject Classification:

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
$\mathbb{Z}[\omega]$, where $\omega=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$.

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 $n\sqrt{3}$ with
$n\in\mathbb{N}$. Throughout this paper we will denote the side lengths
$a\sqrt{3}$ and $b\sqrt{3}$. Note that an ELEP is completely determined, up to
a Euclidean motion, by the integers $a,b$. Indeed, if $\theta$ denotes one of
the angles between the sides, then the area is $3ab\sin\theta$ and so by
equability, $\sin\theta=2(a+b)/\sqrt{3}ab$, which is determined by $a$ and
$b$. Notice, incidentally, that from $\sin\theta=2(a+b)/\sqrt{3}ab$, there is
no ELEP for which $\sin\theta$ is rational. In particular, there is no
rectangular ELEP, nor any ELEP with angle $\pi/6$.

Our main aim in this paper is to study the values of $a,b$ for which an ELEP
exists with sides $a\sqrt{3},b\sqrt{3}$. 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 $a\sqrt{3},b\sqrt{3}$ exists if and only if
$9a^{2}b^{2}-12(a+b)^{2}$ 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 $\mathcal{T}$ of ordered pairs $(a,b)$ of positive integers $a<b$, for
which $9a^{2}b^{2}-12(a+b)^{2}$ is a square, is the set

$\displaystyle\mathcal{T}=\\{2(q,r)\in\mathbb{N}^{2}\ |\ $
$\displaystyle\text{there exist unique odd, coprime, positive integers}\ s,t$
$\displaystyle\text{such that}\ \ s^{2}+3t^{2}+2q^{2}=6stq\ \ \text{and}\ \
q\leq r=3st-q\\}.$

Furthermore, if $(a,b)=2(q,r)\in\mathcal{T}$, then $ab=2(s^{2}+3t^{2})$ and
$a+b=6st$. Moreover, $q,r$ are coprime; in particular, $3\nmid ab$.

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 $2\sqrt{3}$ or $4\sqrt{3}$.

######  Notation.

In this paper, we employ the term _positive_ in the strict sense. So
$\mathbb{N}=\\{n\in\mathbb{Z}\ |\ n>0\\}$.

##  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
$z=x+y\omega\in\mathbb{Z}[\omega]$, then the square of the distance from the
origin to $z$ is $x^{2}-xy+y^{2}$. If a triangle has its vertices on the
Eisenstein lattice, then its area is of the form $\frac{\sqrt{3}}{4}n$, where
$n\in\mathbb{N}$. Indeed, for the triangle with vertices
$0,x+y\omega,z+w\omega$, with $x,y,z,w\in\mathbb{Z}$, the area is
$\frac{\sqrt{3}}{4}|xw-yz|$.

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

######  Lemma 1.

If $P$ is an equable polygon with vertices in $\mathbb{Z}[\omega]$, then the
side lengths of $P$ are each of the form $\sqrt{3}n$, for some
$n\in\mathbb{N}$.

###### Proof.

Since the area of any triangle with vertices in $\mathbb{Z}[\omega]$ is of the
form $\frac{\sqrt{3}}{4}n$, for some $n\in\mathbb{N}$, the same is true for
the area of $P$. Suppose $P$ has sides $s_{1},\dots,s_{n}$, whose squares are
therefore integers. The equable hypothesis gives
$\sum_{i=1}^{n}s_{i}\in\frac{\sqrt{3}}{4}\mathbb{N}$, which implies

$\sqrt{3s_{1}^{2}}+\dots+\sqrt{3s_{n}^{2}}=\sqrt{3}\sum_{i=1}^{n}s_{i}\in\mathbb{Q}.$

But it is well known that if $\sum_{i=1}^{n}\sqrt{m_{i}}$ is rational for
integers $m_{1},\dots,m_{n}$, then $\sqrt{m_{i}}$ is rational for each $i$;
see for example [15] or [16]. Thus $\sqrt{3s_{i}^{2}}$ is rational for each
$i$. Hence, as $3s_{1}^{2},\dots,3s_{n}^{2}$ are integers, it follows that
$\sqrt{3s_{1}^{2}},\dots,\sqrt{3s_{n}^{2}}$ are also integers. So the side
lengths are each of the form $\frac{n}{\sqrt{3}}$, for some $n\in\mathbb{N}$.
Thus, since the squares of the side lengths are integers, the required result
follows. ∎

######  Lemma 2.

Suppose an ELEP $P$ has sides $a\sqrt{3},b\sqrt{3}$, where $a,b\in\mathbb{N}$.
Then $a+b\equiv 0\pmod{3}$ and the lengths of the diagonals of $P$ are given
by the following formula:

$d^{2}=3(a^{2}+b^{2})\pm 2\sqrt{9a^{2}b^{2}-12(a+b)^{2}}.$

In particular, $9a^{2}b^{2}-12(a+b)^{2}$ is a square.

###### Proof.

Consider a diagonal of length $d$, so $d^{2}\in\mathbb{N}$. By Heron’s formula
[12, Chap. 6.7], the triangle with sides $a\sqrt{3},b\sqrt{3},d$ has area

$\sqrt{s(s-a\sqrt{3})(s-b\sqrt{3})(s-d)},$

where $s=\frac{\sqrt{3}a+\sqrt{3}b+d}{2}$ is the semi-perimeter. Hence the
equability hypothesis is

$(\sqrt{3}a+\sqrt{3}b+d)(-\sqrt{3}a+\sqrt{3}b+d)(\sqrt{3}a-\sqrt{3}b+d)(\sqrt{3}a+\sqrt{3}b-d)=48(a+b)^{2}.$

Expanding and rearranging the left hand side gives

$-9(a^{2}-b^{2})^{2}+6(a^{2}+b^{2})d^{2}-d^{4}=48(a+b)^{2}.$ (1)

In particular, the integer $d^{2}$ is divisible by 3, say $d^{2}=3D$. Then (1)
gives

$-3(a^{2}-b^{2})^{2}+6(a^{2}+b^{2})D-3D^{2}=16(a+b)^{2},$

and hence $a+b\equiv 0\pmod{3}$, as required. Furthermore, solving (1) for
$d^{2}$ gives

$d^{2}=3(a^{2}+b^{2})\pm 2\sqrt{9a^{2}b^{2}-12(a+b)^{2}},$

as required. In particular, as $a,b,d^{2}$ are integers,
$9a^{2}b^{2}-12(a+b)^{2}$ is a square. ∎

######  Remark 1.

Suppose an ELEP $P$ has sides $a\sqrt{3},b\sqrt{3}$ and diagonals
$d_{1},d_{2}$. Then the above lemma gives
$d_{1}d_{2}=(a+b)\sqrt{48+9(a-b)^{2}}$. Indeed, by Lemma 2,

$\displaystyle(d_{1}d_{2})^{2}$
$\displaystyle=9(a^{2}+b^{2})^{2}-36a^{2}b^{2}+48(a+b)^{2}$
$\displaystyle=9(a^{2}-b^{2})^{2}+48(a+b)^{2}=(a+b)^{2}(48+9(a-b)^{2}),$ (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, $\nu_{3}((a+b)^{2}(48+9(a-b)^{2}))$, of the
right-hand-side of (2) is odd. Hence
$\nu_{3}(d_{1}^{2})\not\equiv\nu_{3}(d_{2}^{2})\pmod{2}$.

###### Proof of Theorem 1.

The necessity of the condition was shown in Lemma 2. Therefore, assume that
$a,b$ are positive integers such that $9a^{2}b^{2}-12(a+b)^{2}$ is a square.
Consider a triangle $T$ with sides $a\sqrt{3},b\sqrt{3}$ and

$d:=\sqrt{3(a^{2}+b^{2})+2\sqrt{9a^{2}b^{2}-12(a+b)^{2}}}.$

Notice that such a triangle exists because
$a\sqrt{3},b\sqrt{3}<d<(a+b)\sqrt{3}$, where the latter inequality holds as
$3(a^{2}+b^{2})+2\sqrt{9a^{2}b^{2}-12(a+b)^{2}}<3(a+b)^{2}$ since
$2\sqrt{9a^{2}b^{2}-12(a+b)^{2}}<6ab$. Let $\theta$ denote the angle between
sides of length $a\sqrt{3},b\sqrt{3}$ and note that $\theta$ is obtuse since
$d^{2}\geq 3a^{2}+3b^{2}$. Therefore

$d^{2}=3a^{2}+3b^{2}-6ab\cos\theta=3a^{2}+3b^{2}+2\sqrt{9a^{2}b^{2}-9a^{2}b^{2}\sin^{2}\theta}.$

So, from the definition of $d$, we have
$9a^{2}b^{2}\sin^{2}\theta=12(a+b)^{2}$, thus $3ab\sin\theta=2\sqrt{3}(a+b)$.
But since twice the area of $T$ is $3ab\sin\theta$, the area of $T$ is then
$\sqrt{3}(a+b)$. 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:

  1. (1)

the area of $T$ is of the form $\frac{\sqrt{3}}{4}n$, where $n\in\mathbb{N}$,

  2. (2)

the squares of the side lengths of $T$ are integers,

  3. (3)

one of the side lengths of $T$ is of the form $r\sqrt{t}$, where
$r,t\in\mathbb{N}$ and $t$ has no prime divisors congruent to $2\pmod{3}$.

Note that firstly, we saw above that the area of $T$ is
$\sqrt{3}(a+b)\in\sqrt{3}\mathbb{N}$. Secondly, the squares of the side
lengths are $3a,3b,3(a^{2}+b^{2})+2\sqrt{9a^{2}b^{2}-12(a+b)^{2}}$ which are
all integers. Thirdly, the length $\sqrt{3}a$ has the form $r\sqrt{t}$, where
$r,t\in\mathbb{N}$ and $t$ has no prime divisors congruent to $2\pmod{3}$. 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, $9a^{4}-48a^{2}$ is a square and hence $9a^{2}-48$ 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 $\mathcal{T}$ have the required property. Suppose that
$(a,b)=2(q,r)\in\mathcal{T}$, where for coprime positive integers $s,t$ we
have $s^{2}+3t^{2}+2q^{2}=6stq$ and $q<r=3st-q$. Hence $q,r$ are the two
solutions of the quadratic equation

$2u^{2}-6stu+(s^{2}+3t^{2})=0,$ (3)

in $u$, and so $q+r=3st$ and $2qr=s^{2}+3t^{2}$. Thus we have

$\displaystyle 9a^{2}b^{2}-12(a+b)^{2}$ $\displaystyle=4(9\cdot
4q^{2}r^{2}-12(q+r)^{2})$ $\displaystyle=4(9(s^{2}+3t^{2})^{2}-12\cdot
3^{2}s^{2}t^{2})=4\cdot 9(s^{2}-3t^{2})^{2},$

which is a square, as required.

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

$ax^{2}+by^{2}+cz^{2}=dxyz,$ (4)

where $a,b,c$ are pairwise coprime positive integers with $a\leq b\leq c$ such
that $a,b,c$ all divide $d$. We are only interested in positive integer
solutions, that is, $x,y,z\in\mathbb{N}$, 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:

  1. M:

$x^{2}+y^{2}+z^{2}=3xyz$ (Markov’s equation),

  2. R1:

$x^{2}+y^{2}+2z^{2}=4xyz$,

  3. R2:

$x^{2}+2y^{2}+3z^{2}=6xyz$,

  4. R3:

$x^{2}+y^{2}+5z^{2}=5xyz$,

  5. R4:

$x^{2}+y^{2}+z^{2}=xyz$,

  6. R5:

$x^{2}+y^{2}+2z^{2}=2xyz$.

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
$x^{2}+3y^{2}=z^{2}$. Then there exist $k\in\mathbb{N}$ and coprime
$s,t\in\mathbb{N}$ such that

$x=\frac{k}{2}\,|s^{2}-3t^{2}|,\ y=kst,\ z=\frac{k}{2}(s^{2}+3t^{2}),$

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 $9a^{2}b^{2}-12(a+b)^{2}$ is a square. So $a+b\equiv
0\pmod{3}$ and $a^{2}b^{2}-12\left(\frac{a+b}{3}\right)^{2}$ is a square
integer. Let
$z=ab,y=\frac{2(a+b)}{3},x=\sqrt{a^{2}b^{2}-12\left(\frac{a+b}{3}\right)^{2}}$,
so that $x^{2}+3y^{2}=z^{2}$, and $y$ is even. So by Lemma 3, there exist
$k^{\prime}\in\mathbb{N}$ and coprime $s,t\in\mathbb{N}$ such that

$x=\frac{k^{\prime}}{2}\,|s^{2}-3t^{2}|,\ y=k^{\prime}st,\
z=\frac{k^{\prime}}{2}(s^{2}+3t^{2}),$

where $k^{\prime}$ is even if $s,t$ have different parity. In particular, from
$z$ and $y$, we have

$ab=\frac{k^{\prime}}{2}(s^{2}+3t^{2}),\qquad a+b=\frac{3k^{\prime}}{2}st.$
(5)

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

$ab=k(s^{2}+3t^{2}),\qquad a+b=3kst.$ (6)

Hence $a,b$ are solutions to the quadratic equation
$X^{2}-3kstX+k(s^{2}+3t^{2})=0$. In particular, we have

$ks^{2}+3kt^{2}+a^{2}=3kast.$ (7)

Let $k=fg^{2}$, where $f$ is square-free. From (7), $f$ divides $a^{2}$ and
hence $f$ divides $a$. Let $a=f\alpha$. Dividing (7) by $f$ gives

$g^{2}s^{2}+3g^{2}t^{2}+f\alpha^{2}=3fg^{2}st\alpha.$ (8)

From (8), $g^{2}$ divides $f\alpha^{2}$ and hence $g^{2}$ divides
$\alpha^{2}$, and thus $g$ divides $\alpha$. Let $\alpha=gq$. Dividing (8) by
$g^{2}$ gives

$s^{2}+3t^{2}+fq^{2}=3fg\,stq.$ (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

$s^{2}+3t^{2}+2q^{2}=6stq.$ (10)

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

Notice that as $s^{2}+3t^{2}+2q^{2}=6stq$, then modulo $2$, this gives
$s+t\equiv 0$, 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

$ab=2(s^{2}+3t^{2}),\qquad a+b=6st.$ (11)

Suppose $p$ is an odd prime divisor of $\gcd(a,b)$. Then $p^{2}$ divides
$s^{2}+3t^{2}$ 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 $s^{2}+3t^{2}$, from the first
Equation in (11). But this is impossible since $9$ divides $s^{2}$ 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 $s^{\prime},t^{\prime}$
with

$ab=2(s^{\prime 2}+3t^{\prime 2}),\qquad a+b=6s^{\prime}t^{\prime}.$ (12)

From the second equations in (11) and (12), we have $s^{\prime}t^{\prime}=st$.
From (12), we also have $s^{\prime 2}+3t^{\prime
2}+2q^{2}=6s^{\prime}t^{\prime}q$, where $a=2q$, as before. Hence, by (11) and
(10), $s^{\prime 2}+3t^{\prime 2}=s^{2}+3t^{2}$, so as
$t^{\prime}=st/s^{\prime}$,

$s^{\prime 4}+3s^{2}t^{2}=s^{\prime 2}(s^{2}+3t^{2}),$

and hence $(s^{\prime 2}-s^{2})(s^{\prime 2}-3t^{2})=0$. As $s^{\prime}$ is a
positive integer, it follows that $s^{\prime}=s$ and thus $t^{\prime}=t$. 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 $\mathcal{T}$ of ordered pairs $(a,b)$ of
possible side lengths of ELEPs, divided by $\sqrt{3}$, 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

$s^{2}+3t^{2}+2u^{2}=6stu$

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

$a=3st-\sqrt{9s^{2}t^{2}-2(s^{2}+3t^{2})},\qquad
b=3st+\sqrt{9s^{2}t^{2}-2(s^{2}+3t^{2})}.$ (13)

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

$s^{2}+3t^{2}=\frac{ab}{2}\ \text{and}\ s^{2}\cdot
3t^{2}=\frac{(a+b)^{2}}{12}$

and so

$\displaystyle s^{2}$
$\displaystyle=\frac{3ab+\sigma(a,b)\sqrt{9a^{2}b^{2}-12(a+b)^{2}}}{12},$ (14)
$\displaystyle 3t^{2}$
$\displaystyle=\frac{3ab-\sigma(a,b)\sqrt{9a^{2}b^{2}-12(a+b)^{2}}}{12},$ (15)

where $\sigma(a,b)=\pm 1$, and will now be explained. By Remark 2,
$s\not\equiv 0\pmod{3}$, so $s^{2}\equiv 1\pmod{3}$. Notice that

$s^{2}=\frac{3ab+\sigma(a,b)\sqrt{9a^{2}b^{2}-12(a+b)^{2}}}{12}=qr+\sigma(a,b)\sqrt{q^{2}r^{2}-\frac{(q+r)^{2}}{3}}.$

By Theorem 2, $a,b$ are not divisible by $3$. So, as $q+r=3st$, we have
$qr\equiv-1\pmod{3}$. Hence, since $\sigma(a,b)=1$ if and only if
$s^{2}>3t^{2}$, we have the following result.

######  Lemma 4.

Suppose $(a,b)\in\mathcal{T}$ with $a=2q$ and $b=2r$. Then

$\sigma(a,b)\equiv-\sqrt{q^{2}r^{2}-\frac{(q+r)^{2}}{3}}\pmod{3},$

and the following conditions are equivalent:

  1. (a)

$\sigma(a,b)=1$,

  2. (b)

$s^{2}>3t^{2}$,

  3. (c)

$\sqrt{q^{2}r^{2}-\frac{(q+r)^{2}}{3}}\equiv-1\pmod{3}$.

######  Remark 3.

Suppose that an element $z=x^{\prime}+y^{\prime}\omega\in\mathbb{Z}[\omega]$
has length $\sqrt{3}n$, for some $n\in\mathbb{N}$. Then in complex numbers,
$z=x^{\prime}-\frac{y^{\prime}}{2}+\frac{\sqrt{3}y^{\prime}}{2}i$ and

$3n^{2}=x^{\prime 2}-x^{\prime}y^{\prime}+y^{\prime 2}.$

Thus, if $n$ is even, then $x^{\prime},y^{\prime}$ are necessarily both even.
Let $y^{\prime}=2y$ and $x=x^{\prime}-y$, so $z=x+y\sqrt{3}i$, where
$x,y\in\mathbb{Z}$. In particular, this is the case for the sides of an ELEP,
by Theorem 2.

##  5\. The tree of ELEPs

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

$s^{2}+3t^{2}+2u^{2}=6stu.$ (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:

$\displaystyle\phi_{1}(x)$ $\displaystyle=(6tu-s,t,u)$
$\displaystyle\phi_{2}(x)$ $\displaystyle=(s,2su-t,u)$
$\displaystyle\phi_{3}(x)$ $\displaystyle=(s,t,3st-u).$

The group of transformations of $\mathcal{S}$ generated by the maps $\phi_{i}$
is the free product of three copies of $\mathbb{Z}_{2}$, and this group acts
transitively on $\mathcal{S}$. Moreover, the maps $\phi_{i}$ give the set
$\mathcal{S}$ 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 $\phi_{i}$ 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 $\phi_{1},\phi_{3}$ are empty, and $(1,1,1)$
is the unique fixed point of $\phi_{2}$.

###### Proof.

If $(s,t,u)$ is a fixed point of $\phi_{1}$, then $s=3tu$. Hence $t=1$, as
$s,t$ are coprime. Replacing in (16) gives the contradiction $3=7u^{2}$.
Similarly, a fixed point $(s,t,u)$ of $\phi_{3}$ would have $2u=3st$, which is
impossible as $s,t$ are odd. If $(s,t,u)$ is a fixed point of $\phi_{2}$, we
have $t=su$, so $s=1$, as $s,t$ are coprime. Then (16) gives $u^{2}=1$ and so
$t^{2}=1$. Thus finally $s=t=u=1$. ∎

In summary so far: the group
$\mathbb{Z}_{2}\ast\mathbb{Z}_{2}\ast\mathbb{Z}_{2}$ generated by
$\phi_{1},\phi_{2},\phi_{3}$ acts freely on $\mathcal{S}$ except at the
fundamental solution $(s,t,u)=(1,1,1)$, which we can take as the root of the
tree $\mathcal{S}$.

Having recalled Rosenberger’s theory, we now describe how the solution tree
$\mathcal{S}$ of the Markov-Rosenberger equation determine the induced
structure on the set $\mathcal{T}$ of ELEPs. Motivated by (13), we define the
map $\pi:\mathcal{S}\to\mathcal{T}$ by

$\pi(s,t,u)=\left(3st-\sqrt{9s^{2}t^{2}-2(s^{2}+3t^{2})},\
3st+\sqrt{9s^{2}t^{2}-2(s^{2}+3t^{2})}\right).$ (17)

The map $\pi$ is well defined and surjective by Theorem 2. Note that if
$\pi(s,t,u)=(a,b)$, then by (13) and (16), either $a=2u$ or $b=2u$.
Furthermore, trivially, $\pi\circ\phi_{3}(s,t,u)=(a,b)$; that is, $(s,t,u)$
and $\phi_{3}(s,t,u)$ correspond to the same ELEP. In other words, for each
$(a,b)\in\mathcal{T}$, the pre-image $\pi^{-1}(a,b)$ consist of two points,
which are interchanged by the involution $\phi_{3}$.

Consequently, as contraction of edges of a tree produces another tree, we can
form a tree structure on $\mathcal{T}$ by contracting each of the edges of
$\mathcal{S}$ that are given by the map $\phi_{3}$; see Figure 2. 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_ $(2,4)\in\mathcal{T}$. The tree $\mathcal{T}$
is shown in Figure 3. Here the elements $(a,b)$ are shown above the
corresponding pairs $(s,t)$.

$\phi_{3}$ Figure 2. Contraction of the $\phi_{3}$ edges

We now see how the maps $\phi_{1},\phi_{2},\phi_{3}$ on $\mathcal{S}$ generate
maps on $\mathcal{T}$. For each $i=1,2$, maps
$\varphi_{i},\psi_{i}:\mathcal{T}\to\mathcal{T}$ can be naturally defined as
follows. If $(a,b)\in\mathcal{T}$, with $a=2q,b=2r$, let
$(s,t,q)\in\mathcal{S}$ such that $\pi(s,t,q)=(a,b)$. Then set

$\varphi_{i}(a,b):=\pi\circ\phi_{i}(s,t,q),\qquad\psi_{i}(a,b):=\pi\circ\phi_{i}(s,t,r).$
(18)

Note that from (11) we have $a+b=6st$, so $(a,b)=(2q,6st-2q)$. From the
definition of $\phi_{1}$, the map $\varphi_{1}$ leaves $a$ and $t$ unchanged
and $s$ is changed to $s^{\prime}=6qt-s=3at-s$. Then under $\varphi_{1}$, the
value of $b$ is changed to

$\displaystyle b^{\prime}$
$\displaystyle=6s^{\prime}t-a=18at^{2}-6st-a=18at^{2}-(a+b)-a$
$\displaystyle=\frac{3a^{2}b-\sigma(a,b)a\sqrt{9a^{2}b^{2}-12(a+b)^{2}}}{2}-2a-b,$
(19)

using (15). Note that a priori, we don’t know which of the resulting
components, $a$ or $b^{\prime}$, is the larger. So we set

$\varphi_{1}:(a,b)\mapsto\left(\min(a,b^{\prime}),\max(a,b^{\prime})\right).$

Similarly, using $\phi_{2}$, we set

$\varphi_{2}:(a,b)\mapsto\left(\min(a,b^{\prime\prime}),\max(a,b^{\prime\prime})\right),$

where

$b^{\prime\prime}=\frac{3a^{2}b+\sigma(a,b)a\sqrt{9a^{2}b^{2}-12(a+b)^{2}}}{2}-2a-b.$
(20)

Similarly, analogous to $\varphi_{1},\varphi_{2}$, interchanging the roles of
$a$ and $b$, we have two further maps:

$\displaystyle\psi_{1}:(a,b)$
$\displaystyle\mapsto\left(\min(a^{\prime},b),\max(a^{\prime},b)\right),$
$\displaystyle\psi_{2}:(a,b)$
$\displaystyle\mapsto\left(\min(a^{\prime\prime},b),\max(a^{\prime\prime},b)\right),$

where

$\displaystyle a^{\prime}$
$\displaystyle=\frac{3ab^{2}+\sigma(a,b)b\sqrt{9a^{2}b^{2}-12(a+b)^{2}}}{2}-a-2b,$
$\displaystyle a^{\prime\prime}$
$\displaystyle=\frac{3ab^{2}-\sigma(a,b)b\sqrt{9a^{2}b^{2}-12(a+b)^{2}}}{2}-a-2b.$

These maps are obtained from the maps $\phi_{1}\circ\phi_{3}$ and
$\phi_{2}\circ\phi_{3}$, respectively.

Note that while $\phi_{1},\phi_{2},\phi_{3}$ are involutions and generate a
group of transformations of $\mathcal{S}$, the same is not true of the maps
$\varphi_{1},\varphi_{2},\psi_{1},\psi_{2}$ of $\mathcal{T}$. Indeed, they are
not even all bijections. For example, $\phi_{2}(2,4)=(2,4)=\phi_{2}(4,14)$.
Moreover, the study of the maps $\phi_{1},\phi_{2},\psi_{1},\psi_{2}$ is
considerably complicated by the term $\sigma(a,b)$. So it is often convenient
to work instead with the pairs $(s,t)$. From the definition of the maps
$\phi_{i}$, we have

$\phi_{1}(s,t,u/2)=(3ut-s,t,u/2),\quad\phi_{2}(s,t,u/2)=(s,us-t,u/2),$

where $u=a$ or $b$, and $a,b$ are given by (13). We define the linear
involutions $\phi_{1,u},\phi_{2,u}$ on the set
$\Sigma:=\\{(s,t)\in\mathbb{N}^{2}:s,t\ \text{odd and coprime}\\}$, by
$\phi_{1,u}(s,t)=(3ut-s,t),\phi_{2,u}(s,t)=(s,us-t)$, or equivalently in
matrix form,

$\phi_{1,u}=\begin{pmatrix}-1&3u\\\
0&1\end{pmatrix},\qquad\phi_{2,u}=\begin{pmatrix}1&0\\\ u&-1\end{pmatrix}.$
(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
$\varphi_{1},\varphi_{2},\psi_{1},\psi_{2}$, 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
$\phi_{i,u}$, for $i\in\\{1,2\\}$ and $u\in\\{a,b\\}$.

$(2,4)$$(1,1)$$\phi_{1,2}$$(2,28)$$(5,1)$$\phi_{2,2}$$(2,268)$$(5,9)$$\phi_{1,2}$$(2,2644)$$(49,9)$$\phi_{2,2}$$(2,26164)$$(49,89)$$\phi_{1,2}$$(2,258988)$$(485,89)$$(28,446)$$(79,1)$$\phi_{1,28}$$(28,4142)$$(5,139)$$\phi_{2,28}$$(268,390206)$$(7231,9)$$\phi_{1,268}$$(268,39662)$$(5,1331)$$\phi_{2,268}$$(4,62)$$(11,1)$$\phi_{1,4}$$(62,988)$$(175,1)$$\phi_{1,62}$$(988,15746)$$(2789,1)$$\phi_{1,988}$$(4,14)$$(1,3)$$\phi_{2,4}$$(14,52)$$(1,11)$$\phi_{2,14}$$(52,194)$$(1,41)$$\phi_{2,52}$
Figure 3. The tree of ELEPs; the elements $(a,b)$ are shown above the
corresponding pairs $(s,t)$.

##  6\. Diagonals, heights and altitudes

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

######  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 $\eta_{l}$ (resp. $\eta_{s}$).

$h_{l}$$h_{s}$$d_{s}$$d_{l}$$\eta_{l}$$\eta_{s}$ Figure 4. Diagonals, heights
and altitudes

######  Lemma 6.

Suppose an ELEP $P$ has sides $a\sqrt{3},b\sqrt{3}$ with $a,b\in\mathbb{N}$
and $a\leq b$. Then

  1. (a)

$\displaystyle h_{l}=\frac{2(a+b)}{a},\qquad h_{s}=\frac{2(a+b)}{b}$,

  2. (b)

$\displaystyle\eta_{l}=\frac{2\sqrt{3}(a+b)}{d_{s}},\qquad\eta_{s}=\frac{2\sqrt{3}(a+b)}{d_{l}}$.

###### Proof.

By equability, the area of $P$ is $2\sqrt{3}(a+b)$, but the area is obviously
also $a\sqrt{3}h_{l}$ and $b\sqrt{3}h_{s}$. 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 $\eta_{l}d_{s}$ and $\eta_{s}d_{l}$. This gives (b). ∎

######  Remark 4.

If $h_{s}$ is an integer, then $h_{s}=2+\frac{2a}{b}$ by Lemma 6(a), and so as
$b>a$, the only possibility is $b=2a$. But then
$9a^{2}b^{2}-12(a+b)^{2}=36a^{2}(a^{2}-3)$, 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 $h_{l}=2+\frac{2b}{a}$ by Lemma
6(a), and so $2b=ak$, for some positive integer $k$. In this case, $h_{l}=2+k$
and $h_{s}=2+\frac{4}{k}$.

######  Theorem 4.

For every ELEP the altitudes satisfy $2<\eta_{s}\leq\frac{6}{\sqrt{7}}$ and
$2<\eta_{l}\leq 2\sqrt{3}$.

######  Remark 5.

Suppose an ELEP $P$ has sides $a\sqrt{3},b\sqrt{3}$ with $a,b\in\mathbb{N}$
and $a\leq b$. By Theorem 2, we have $ab=2(s^{2}+3t^{2})$ and $a+b=6st$, so
$\sqrt{9a^{2}b^{2}-12(a+b)^{2}}=6|s^{2}-3t^{2}|$, for some odd, coprime
integers $s,t$. If $s^{2}>3t^{2}$, then from Lemma 2, the diagonals are given
by

$\displaystyle d_{l}^{2}$
$\displaystyle=3(a^{2}+b^{2})+2\sqrt{9a^{2}b^{2}-12(a+b)^{2}}$
$\displaystyle=3\cdot
36s^{2}t^{2}-12(s^{2}+3t^{2})+12|s^{2}-3t^{2}|=12(9s^{2}t^{2}-6t^{2}),$
$\displaystyle d_{s}^{2}$ $\displaystyle=3\cdot
36s^{2}t^{2}-12(s^{2}+3t^{2})-12|s^{2}-3t^{2}|=12(9s^{2}t^{2}-2s^{2}).$

Similarly, if $s^{2}<3t^{2}$, the diagonals are given by
$d_{l}^{2}=12(9s^{2}t^{2}-2s^{2})$ and $d_{s}^{2}=12(9s^{2}t^{2}-6t^{2})$.
Putting the cases together, we have

$d_{l}^{2}=12(9s^{2}t^{2}-2\min(s^{2},3t^{2})),\qquad
d_{s}^{2}=12(9s^{2}t^{2}-2\max(s^{2},3t^{2})).$ (22)

Notice that when a diagonal $d$ satisfies $d^{2}=12(9s^{2}t^{2}-6t^{2})$, one
has $d=6t\sqrt{3s^{2}-2}$, so $d$ is an integer if and only if $3s^{2}-2$ is a
square. When $d^{2}=12(9s^{2}t^{2}-2s^{2})$, one has $d=2s\sqrt{27t^{2}-6}$.
In this case, $d$ is never an integer, since $27t^{2}-6\equiv 3\pmod{9}$ and
$3$ is not a quadratic residue modulo 9.

###### Proof.

Suppose an ELEP $P$ has sides $a\sqrt{3},b\sqrt{3}$ with $a,b\in\mathbb{N}$
and $a\leq b$. By Theorem 2, we have $ab=2(s^{2}+3t^{2})$ and $a+b=6st$, so
$\sqrt{9a^{2}b^{2}-12(a+b)^{2}}=6|s^{2}-3t^{2}|$, for some odd, coprime
integers $s,t$. Using Lemma 6 and Remark 5, the altitudes are given by

$\displaystyle\eta_{l}^{2}$
$\displaystyle=\frac{12(a+b)^{2}}{d_{s}^{2}}=\frac{36s^{2}t^{2}}{9s^{2}t^{2}-2\max(s^{2},3t^{2})},$
$\displaystyle\eta_{s}^{2}$
$\displaystyle=\frac{12(a+b)^{2}}{d_{l}^{2}}=\frac{36s^{2}t^{2}}{9s^{2}t^{2}-2\min(s^{2},3t^{2})}.$

We consider the two cases according to whether $s^{2}>3t^{2}$ or
$s^{2}<3t^{2}$. First suppose $s^{2}>3t^{2}$. Then

$\displaystyle\eta_{l}^{2}$
$\displaystyle=\frac{36s^{2}t^{2}}{9s^{2}t^{2}-2s^{2}}=\frac{36t^{2}}{9t^{2}-2}=4+\frac{8}{9t^{2}-2},$
$\displaystyle\eta_{s}^{2}$
$\displaystyle=\frac{36s^{2}t^{2}}{9s^{2}t^{2}-6t^{2}}=\frac{12s^{2}}{3s^{2}-2}=4+\frac{8}{3s^{2}-2}.$

Hence, as $t\geq 1$, we have $4<\eta_{l}^{2}\leq 4+\frac{8}{7}$, so
$2<\eta_{l}\leq\frac{6}{\sqrt{7}}$. Similarly, as $s\geq 1$, we have
$4<\eta_{s}^{2}\leq 12$, so $2<\eta_{s}\leq 2\sqrt{3}$. Thus, as
$\eta_{s}<\eta_{l}$, we have $2<\eta_{s}<\eta_{l}\leq\frac{6}{\sqrt{7}}$.

Now suppose $s^{2}<3t^{2}$. Then

$\displaystyle\eta_{l}^{2}$
$\displaystyle=\frac{36s^{2}t^{2}}{9s^{2}t^{2}-6t^{2}}=\frac{12s^{2}}{3s^{2}-2}=4+\frac{8}{3s^{2}-2}$
$\displaystyle\eta_{s}^{2}$
$\displaystyle=\frac{36s^{2}t^{2}}{9s^{2}t^{2}-2s^{2}}=\frac{36t^{2}}{9t^{2}-2}=4+\frac{8}{9t^{2}-2}.$

Hence, as $s\geq 1$, we have $4<\eta_{l}^{2}\leq 12$, so $2<\eta_{l}\leq
2\sqrt{3}$. Similarly, as $t\geq 1$, we have $4<\eta_{s}^{2}\leq
4+\frac{8}{7}$, so $2<\eta_{s}\leq\frac{6}{\sqrt{7}}$. Thus, as
$\eta_{s}<\eta_{l}$, we have $2<\eta_{s}<\eta_{l}\leq 2\sqrt{3}$.

Combining the two cases gives the required bounds on $\eta_{s}$ and
$\eta_{l}$. ∎

##  7\. ELEPs with horizontal diagonal

######  Example 1.

Here we give examples of ELEPs with a horizontal diagonal. Consider the
sequence $(u_{n})$ defined by $u_{n}=4u_{n-1}-u_{n-2}$, with $u_{0}=0$ and
$u_{1}=1$. 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 $n\geq 1$, one has

  1. (a)

$u_{n+1}u_{n-1}=u_{n}^{2}-1$,  (b) $3u_{n}^{2}+1=(2u_{n}-u_{n-1})^{2}$.

###### Proof.

(a) One has $u_{2}u_{0}=0=u_{1}^{2}-1$. Then for $n\geq 2$, by induction,

$\displaystyle u_{n+1}u_{n-1}$ $\displaystyle-
u_{n}^{2}=(4u_{n}-u_{n-1})u_{n-1}-u_{n}(4u_{n-1}-u_{n-2})$
$\displaystyle=u_{n}u_{n-2}-u_{n-1}^{2}=-1.$

(b). For all $n\geq 1$, expanding $(2u_{n}-u_{n-1})^{2}$, one has, using (a),

$\displaystyle 3u_{n}^{2}+1-(2u_{n}-u_{n-1})^{2}$
$\displaystyle=1-u_{n}^{2}+u_{n-1}(4u_{n}-u_{n-1})$
$\displaystyle=1-u_{n}^{2}+u_{n-1}u_{n+1}=0.\qed$

Let

$A_{n}=(-6u_{n},-2\sqrt{3}),\quad B_{n}=(6u_{n+1}-6u_{n},0),\quad
C_{n}=(6u_{n+1},2\sqrt{3}).$

The vertices $OA_{n}B_{n}C_{n}$ form a parallelogram on the Eisenstein lattice
with diagonal $OB_{n}$ on the $x$-axis. The parallelogram $OA_{n}B_{n}C_{n}$
has side lengths

$\displaystyle\overline{OA_{n}}$
$\displaystyle=\sqrt{36u_{n}^{2}+12}=2\sqrt{3}\sqrt{3u_{n}^{2}+1}=2\sqrt{3}(2u_{n}-u_{n-1}),$
$\displaystyle\overline{OC_{n}}$
$\displaystyle=\sqrt{36u_{n+1}^{2}+12}=2\sqrt{3}\sqrt{3u_{n+1}^{2}+1}=2\sqrt{3}(2u_{n+1}-u_{n}),$

by Lemma 7(b), and area

$\text{area}(OA_{n}B_{n}C_{n})=\det\begin{pmatrix}-6u_{n}&-2\sqrt{3}\\\
6u_{n+1}&2\sqrt{3}\end{pmatrix}=12\sqrt{3}(u_{n+1}-u_{n}).$

Hence

$\displaystyle 2(\overline{OA_{n}}+\overline{OC_{n}})$
$\displaystyle=4\sqrt{3}(2u_{n+1}+u_{n}-u_{n-1})$
$\displaystyle=4\sqrt{3}(2u_{n+1}+u_{n}+(u_{n+1}-4u_{n}))$
$\displaystyle=12\sqrt{3}(u_{n+1}-u_{n})=\text{area}(OA_{n}B_{n}C_{n}),$

so $OA_{n}B_{n}C_{n}$ 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.

$n$ | $(q,u)$ | $a$ | $b$ | $A$ | $B$ | $(s,t)$
---|---|---|---|---|---|---
$0$ | $(1,0)$ | $2$ | $4$ | $-2-4\omega$ | $6$ | $(1,1)$
$1$ | $(2,1)$ | $4$ | $14$ | $-8-4\omega$ | $18$ | $(1,3)$
$2$ | $(7,4)$ | $14$ | $52$ | $-26-4\omega$ | $66$ | $(1,11)$
$3$ | $(26,15)$ | $52$ | $194$ | $-92-4\omega$ | $246$ | $(1,41)$
$4$ | $(97,56)$ | $194$ | $724$ | $-338-4\omega$ | $918$ | $(1,153)$
$5$ | $(362,209)$ | $724$ | $2702$ | $-1256-4\omega$ | $3426$ | $(1,571)$
$6$ | $(1351,780)$ | $2702$ | $10084$ | $-4682-4\omega$ | $12786$ | $(1,2131)$
$7$ | $(5042,2911)$ | $10084$ | $37634$ | $-17468-4\omega$ | $47718$ | $(1,7953)$
$8$ | $(18817,10864)$ | $37634$ | $140452$ | $-65186-4\omega$ | $178086$ | $(1,29681)$
Table 1. Examples of ELEPs with horizontal diagonal Figure 5. Two ELEPs with
horizontal diagonal

######  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 $A=x-y\omega,B=z,C=(z-x)+y\omega,O=0$, where $x\in\mathbb{Z}$
and $y,z\in\mathbb{N}$. Let $a\sqrt{3},b\sqrt{3}$ denote the lengths of $OA$
and $AB$ respectively, with $a,b\in\mathbb{N}$ and $a<b$. In particular, we
have

$3a^{2}=x^{2}+xy+y^{2}.$ (23)

The altitudes from $A$ and $C$ have length $\eta:=\frac{y\sqrt{3}}{2}$. By
Theorem 4, $2<\eta\leq 2\sqrt{3}$, which gives $\frac{4}{\sqrt{3}}<y\leq 4$.
So, as $y\in\mathbb{N}$, we have $y=3$ or $4$. First suppose that $y=3$. Then
(23) gives

$3a^{2}=x^{2}+3x+9.$ (24)

So $x$ is divisible by $3$, say $x=3x^{\prime}$. But then by (24), $a$ is
divisible by $3$, contrary to Theorem 2. So we have $y=4$.

As $y=4$, we have $\eta=2\sqrt{3}>\frac{6}{\sqrt{7}}$. So $\eta=\eta_{l}$ by
Theorem 4, and from the proof of Theorem 4 we have $s^{2}<3t^{2}$ and
$\eta_{l}^{2}=4+\frac{8}{3s^{2}-2}$. Hence
$(2\sqrt{3})^{2}=4+\frac{8}{3s^{2}-2}$, giving $s=1$. Then by Theorem 2,
$(a,b)=2(q,r)\in\mathbb{N}^{2}$ where

$1+3t^{2}+2q^{2}=6tq$ (25)

and $q\leq r=3t-q$. Let $u:=t-q$. From (25), we have

$q^{2}-3u^{2}=1.$ (26)

Notice that $u\geq 0$. Indeed, from (25), we have
$q=\frac{3t-\sqrt{3t^{2}-2}}{2}$, so $u=\frac{-t+\sqrt{3t^{2}-2}}{2}\geq 0$,
for $t\geq 1$. Notice also that $a^{2}=4q^{2}$ and (23) gives
$3a^{2}=x^{2}+4x+16$, so $12q^{2}=x^{2}+4x+16$. Thus $x$ is even, say
$x=2x^{\prime}$, and so $3q^{2}=x^{\prime 2}+2x^{\prime}+4$. Hence, by (26),
$9u^{2}=x^{\prime 2}+2x^{\prime}+1=(x^{\prime}+1)^{2}$, so

$x=2x^{\prime}=\pm 6u-2.$ (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

$(q_{n},u_{n})=4(q_{n-1},u_{n-1})-(q_{n-2},u_{n-2}),\qquad(q_{1},u_{1})=(2,1),\
(q_{0},u_{0})=(1,0).$

In particular, $u_{n}$ agrees with the sequence $u_{n}$ of Example 1. Notice
also that

$q_{n}+2u_{n}=u_{n+1},\quad\text{for all }n\geq 0.$ (28)

Indeed, both sides of the equation satisfy the same recurrence relation, with
the initial conditions $q_{0}+2u_{0}=1=u_{1}$ and $q_{1}+2u_{1}=4=u_{2}$.

For each $n\geq 0$, 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 $OA_{n}B_{n}C_{n}$. Since there are so many
variables, we recall for the reader’s convenience that
$A_{n}=x_{n}-4\omega,B_{n}=z_{n},C=(z_{n}-x_{n})+4\omega$, the lengths of
$OA_{n},OC_{n}$ are $\sqrt{3}a_{n},\sqrt{3}b_{n}$ 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

$b_{n}=6t_{n}-a_{n}=6t_{n}-2q_{n}=6u_{n}+4q_{n}.$ (29)

Since $A_{n}$ has $y$-coordinate $-2\sqrt{3}$, the area of $OA_{n}B_{n}C_{n}$
is $2\sqrt{3}z_{n}$, so the equability condition is
$2\sqrt{3}z_{n}=2(a_{n}+b_{n})\sqrt{3}$; 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): (i) $x_{n}=6u_{n}-2$ for
$u_{n}>0$ and (ii) $x_{n}=-6u_{n}-2$ for $u_{n}\geq 0$.

(i). We show that this case leads to a contradiction. Suppose $x_{n}=6u_{n}-2$
for $u_{n}>0$, so $A_{n}=6u_{n}-2-4\omega,B_{n}=z_{n}=6(u_{n}+q_{n})$ and
$C_{n}=6q_{n}+2+4\omega$. So, from (29),

$3\cdot(6u_{n}+4q_{n})^{2}=3b_{n}^{2}=\overline{OC_{n}}^{2}=(6q_{n}+2)^{2}-(6q_{n}+2)4+16,$

where the expression on the right comes for the square length of the element
$C_{n}=6q_{n}+2+4\omega$. Expanding, rearranging and dividing by 12 gives
$q_{n}^{2}+12q_{n}u_{n}+9u_{n}^{2}=1$. But this is impossible for positive
integers $u_{n}$ and $q_{n}$.

(ii). Now suppose $x_{n}=-6u_{n}-2$ for $u_{n}\geq 0$. Using (28), we have
$A_{n}=-6u_{n}-2-4\omega,B_{n}=z_{n}=6(u_{n}+q_{n})=6u_{n+1}-6u_{n}$, and
$C_{n}=6u_{n+1}+2+4\omega$, 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 $\eta_{s}=\frac{6}{\sqrt{7}}$ and $\eta_{l}=2\sqrt{3}$ 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
$A=x+y\omega,B=z\omega,C=-x+(z-y)\omega,O=0$, where $x,z\in\mathbb{N}$ and
$y\in\mathbb{Z}$. Let $a\sqrt{3},b\sqrt{3}$ denote the lengths of $OA$ and
$AB$ respectively, with $a,b\in\mathbb{N}$ and $a<b$. In particular, we have

$3a^{2}=x^{2}-xy+y^{2}.$ (30)

The altitudes from $A$ and $C$ have length $\eta:=x$. By Theorem 4,
$2<\eta\leq 2\sqrt{3}$, which gives $x=3$. Then (30) gives

$3a^{2}=9-3y+y^{2}.$ (31)

So $y$ is divisible by $3$, say $y=3y^{\prime}$. 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

$3u^{2}=2m^{2}+1.$ (32)

This equation is well known; see entry A072256 in [14]. Its solutions
$(u_{n},m_{n})$ satisfy the recurrence relation

$(u_{n},m_{n})=10(u_{n-1},m_{n-1})-(u_{n-2},m_{n-2}),$

with $(u_{1},m_{1})=(1,1),(u_{2},m_{2})=(9,11)$.

Now, using complex numbers, consider the parallelogram with vertices
$A_{n}=2\sqrt{3}i,B_{n}=-x_{n}+(2+y_{n})\sqrt{3}i,C_{n}=-x_{n}+y_{n}\sqrt{3}i,O=0$,
where $x_{n}=3+3u_{n}$ and $y_{n}=2m_{n}$. Note that the vertical side
$OA_{n}$ has length $a\sqrt{3}$, where $a:=2$, while the side $OC_{n}$ has
length

$\displaystyle b_{n}\sqrt{3}=\sqrt{x_{n}^{2}+3y_{n}^{2}}$
$\displaystyle=\sqrt{(3+3u_{n})^{2}+12m_{n}^{2}}$
$\displaystyle=\sqrt{9+18u_{n}+9u_{n}^{2}+6(3u_{n}^{2}-1)}\quad(\text{by
}\eqref{E:Pell})$ $\displaystyle=\sqrt{3}(1+3u_{n}).$

So $OA_{n}B_{n}C_{n}$ has perimeter $2(a+b_{n})\sqrt{3}=6\sqrt{3}(1+u_{n})$.
Furthermore, $OA_{n}B_{n}C_{n}$ has area
$a\sqrt{3}x_{n}=2x_{n}\sqrt{3}=6\sqrt{3}(1+u_{n})$. So $OA_{n}B_{n}C_{n}$ 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.

$n$ | $(u,m)$ | $b$ | $x$ | $y$ | $(s,t)$
---|---|---|---|---|---
$1$ | $(1,1)$ | $4$ | $6$ | $2$ | $(1,1)$
$2$ | $(9,11)$ | $28$ | $30$ | $22$ | $(5,1)$
$3$ | $(89,109)$ | $268$ | $270$ | $218$ | $(5,9)$
$4$ | $(881,1079)$ | $2644$ | $2646$ | $2158$ | $(49,9)$
$5$ | $(8721,10681)$ | $26164$ | $26166$ | $21362$ | $(49,89)$
$6$ | $(86329,105731)$ | $258988$ | $258990$ | $211462$ | $(485,89)$
$7$ | $(854569,1046629)$ | $2563708$ | $2563710$ | $2093258$ | $(485,881)$
$8$ | $(8459361,10360559)$ | $25378084$ | $25378086$ | $20721118$ | $(4801,881)$
$9$ | $(83739041,102558961)$ | $2486793150$ | $2030458102$ | $205117922$ | $(4801,8721)$
Table 2. Examples of ELEPs with vertical side of length $2\sqrt{3}$ and vertex
$C=-x+y\sqrt{3}i$ Figure 6. Two ELEPs having a vertical side of length
$2\sqrt{3}$.

$(2,4)$$(1,1)$$\phi_{1,2}$$(2,28)$$(5,1)$$\phi_{2,2}$$(2,268)$$(5,9)$$\phi_{1,2}$$(2,2644)$$(49,9)$$\phi_{2,2}$$(2,26164)$$(49,89)$$(4,62)$$(11,1)$$\phi_{1,4}$$(4,14)$$(1,3)$$\phi_{2,4}$$\phi_{1,4}$$(4,626)$$(35,3)$$\phi_{2,4}$$(4,28766)$$(35,137)$$\phi_{1,4}$$(4,1322594)$$(1609,137)$$\phi_{2,4}$$(4,60810542)$$(1609,6299)$$\phi_{2,4}$$(4,2834)$$(11,43)$$\phi_{1,4}$$(4,130286)$$(505,43)$$\phi_{2,4}$$(4,5990306)$$(505,1977)$$\phi_{1,4}$$(4,275423774)$$(23219,1977)$
Figure 7. The paths of ELEPs having a side of length $2$ or $4$.

######  Example 3.

We now give two sequences of ELEPs having a vertical side of length
$4\sqrt{3}$. Consider the equation

$132w^{2}+36w+1=y^{2}.$ (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

$\begin{pmatrix}w_{n}\\\ y_{n}\end{pmatrix}=\begin{pmatrix}23&2\\\
264&23\end{pmatrix}\begin{pmatrix}w_{n-1}\\\
y_{n-1}\end{pmatrix}+\begin{pmatrix}3\\\ 36\end{pmatrix},$

but different initial conditions. One sequence has $(w_{1},y_{1})=(1,13)$,
while the other has $(w_{1},y_{1})=(5,59)$.

Now, for each of the two sequences, consider the parallelogram with vertices
$A_{n}=4\sqrt{3}i,B_{n}=-x_{n}+(4+y_{n})\sqrt{3}i,C_{n}=-x_{n}+y_{n}\sqrt{3}i,O=0$,
where $x_{n}=6w_{n}+3$. Note that the vertical side $OA_{n}$ has length
$a:=4\sqrt{3}$, while the side $OC_{n}$ has length

$\displaystyle b_{n}\sqrt{3}=\sqrt{x_{n}^{2}+3y_{n}^{2}}$
$\displaystyle=\sqrt{(6w_{n}+3)^{2}+3y_{n}^{2}}$
$\displaystyle=\sqrt{(6w_{n}+3)^{2}+3(132w_{n}^{2}+36w_{n}+1)}\quad(\text{by
}\eqref{E:notPell})$ $\displaystyle=2\sqrt{3}(1+6w_{n}).$

So $OA_{n}B_{n}C_{n}$ has perimeter $2(a+b_{n})=12\sqrt{3}(1+2w_{n})$.
Furthermore, $OA_{n}B_{n}C_{n}$ has area
$ax_{n}=4x_{n}\sqrt{3}=12\sqrt{3}(1+2w_{n})$. So $OA_{n}B_{n}C_{n}$ is
equable.

Table 3 lists the first six examples of ELEPs corresponding to the sequence
with initial condition $(w_{1},y_{1})=(1,13)$. 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
$(w_{1},y_{1})=(5,59)$. 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 $(4,62)$.

$n$ | $(w,y)$ | $b$ | $x$ | $(s,t)$
---|---|---|---|---
$1$ | $(1,13)$ | $14$ | $9$ | $(1,3)$
$2$ | $(52,599)$ | $626$ | $315$ | $(35,3)$
$3$ | $(2397,27541)$ | $28766$ | $14385$ | $(35,137)$
$4$ | $(110216,1266287)$ | $1322594$ | $661299$ | $(1609,137)$
$5$ | $(5067545,58221661)$ | $60810542$ | $30405273$ | $(1609,6299)$
$6$ | $(232996860,2676930119)$ | $2795962322$ | $1397981163$ | $(73979,6299)$
Table 3. First sequence of ELEPs with vertical side of length $4\sqrt{3}$ and vertex $C=-x+y\sqrt{3}i$ $n$ | $(w,y)$ | $b$ | $x$ | $(s,t)$
---|---|---|---|---
$1$ | $(5,59)$ | $62$ | $18$ | $(11,1)$
$2$ | $(236,2713)$ | $2834$ | $1419$ | $(11,43)$
$3$ | $(10857,124739)$ | $130286$ | $65145$ | $(505,43)$
$4$ | $(499192,5735281)$ | $5990306$ | $2995155$ | $(505,1977)$
$5$ | $(22951981,263698187)$ | $275423774$ | $137711889$ | $(23219,1977)$
$6$ | $(1055291940,12124381321)$ | $12663503282$ | $6331751643$ | $(23219,90899)$
Table 4. Second sequence of ELEPs with vertical side of length $4\sqrt{3}$ and
vertex $C=-x+y\sqrt{3}i$

######  Theorem 7.

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

###### Proof.

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

$\sqrt{9a^{2}b^{2}-12(a+b)^{2}}=2\sqrt{6((b-1)^{2}-3)}\in\mathbb{N}.$

So $(b-1)^{2}-3=6m^{2}$ for some $m\in\mathbb{N}$. In particular, $b\equiv
1\pmod{3}$, say $b=3u+1$, as in Example 2. Thus $3u^{2}=2m^{2}+1$, which is
the same equation as (32). 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

$\sqrt{9a^{2}b^{2}-12(a+b)^{2}}=2\sqrt{3(11b^{2}-8b-16)}\in\mathbb{N}.$

So $11b^{2}-8b-16=3m^{2}$ for some $m\in\mathbb{N}$. Writing $b=2r$, as
before, we have that $m$ is even, say $m=2\ell$. Thus
$11r^{2}-4r-4=3\ell^{2}$. 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
$44v^{2}+36v+3=3\ell^{2}$. 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
$132w^{2}+36w+1=\ell^{2}$, 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
$A=a\sqrt{3}i,B=-x+(a+y)\sqrt{3}i,C=-x+y\sqrt{3}i,O=0$, where
$a,x,y\in\mathbb{N}$ by Remark 3. Then $OA$ has length $a\sqrt{3}$. Let
$b\sqrt{3}$ denote the length of $OC$. In particular, we have

$3b^{2}=x^{2}+3y^{2}.$ (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. ∎

## References

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# Indirect-to-direct band gap crossover of single walled MoS2 nanotubes

Kaoru Hisama  Department of Mechanical Engineering, The University of Tokyo,
Tokyo 113-8656, Japan  Mina Maruyama  Department of Physics, University of
Tsukuba, Tsukuba 305-8571, Japan  Shohei Chiashi  Department of Mechanical
Engineering, The University of Tokyo,
Tokyo 113-8656, Japan  Shigeo Maruyama<EMAIL_ADDRESS>Department of Mechanical Engineering, The University of Tokyo,
Tokyo 113-8656, Japan  Susumu Okada<EMAIL_ADDRESS>Department of Physics, University of Tsukuba, Tsukuba 305-8571, Japan

###### Abstract

Using density functional theory, the electronic structures of single walled
molybdenum disulfide nanotubes (MoS2 NTs) were investigated as a function of
diameter. Our calculations show that the electronic structure near the band
gap is sensitive to the NT diameter: armchair MoS2 NTs act as indirect gap
semiconductors for diameters up to approximately 5.0 nm, while armchair MoS2
NTs with larger diameters act as direct gap semiconductors with band edges
located in the vicinity of $k=2\pi/3$. This finding implies that MoS2 NTs with
large diameters should exhibit similar photoluminescence to 2D monolayer MoS2
sheets. This indirect-to-direct band gap crossover is ascribed to the upward
shift of the valence band peak at the $\Gamma$ point in small diameter NTs,
which is caused by the tensile strain resulting from their tubular structures.

Since the discovery of graphene, atomic layer materials have attracted much
attention due to their unique physical properties arising from chemically
inert 2D covalent networks with an unusual network topology. The constituent
elements and network structure provide these materials with a variety of
properties, ranging from metals to insulators. Transition metal
dichalcogenides (TMDCs), such as MoS2, WS2, WSe2, and MoTe2, are
representative examples of such atomic layer materials. TMDCs consist of an
atomic layer of transition metals that form a triangular lattice sandwiched by
atomic layers of chalcogens arranged in a prismatic manner, resulting in a
hexagonal network of these elements with a thickness of approximately 0.3 nm
Friend and Yoffe (1987); Wang  _et al._ (2012). According to the chemical
valence of the constituent elements in TMDCs, monolayers of TMDCs are
primarily semiconductors, with a direct band gap at the K point Roldán  _et
al._ (2014); Mak  _et al._ (2010); Splendiani  _et al._ (2010), which strongly
depends on the constituent elements; in contrast, corresponding thin films or
bulk materials are indirect band gap semi-conductors.

Due to the chemical stability of these 2D lattices with covalent bonds, TMDCs
have been employed as starting materials to construct various derivatives.
Such derivatives can provide building blocks of various layered
heterostructures, where each layer is bound via weak van der Waals
interactions Geim and Grigorieva (2013); Novoselov  _et al._ (2016); Liu  _et
al._ (2016). In these van der Waals heterostructures, one can tune the
electronic structure near the band edges by assembling and stacking them in an
appropriate manner, leading to the potential for optical and electronic
applications Lee  _et al._ (2014); Zhang  _et al._ (2016). Nanoribbons provide
another possible derivative, obtained by imposing an open boundary condition
on TMDC sheets. The formation energy and electronic structures of TMDC
nanoribbons are strongly dependent on the edge morphology Li  _et al._ (2008);
Ataca  _et al._ (2011), due to the polarity at the edges and structural
relaxation. In addition to nanoribbons, nanotubes (NTs) of TMDCs Tenne  _et
al._ (1992); Tenne and Redlich (2010) are also available, by imposing a
tubular boundary condition on nanoribbons. Indeed, researchers recently
reported the synthesis of single walled MoS2 NT wrapping walls of multi-walled
boron nitride NTs Xiang  _et al._ (2020). The authors reported
photoluminescence (PL) from these single walled MoS2 NTs, with a similar
energy to the 2D monolayer MoS2. This behavior implies that the electronic
properties of single walled MoS2 NT act as a direct band gap semiconductor,
even with the tubular structure.

Prior to the experimental synthesis of single walled MoS2 NTs, their
electronic properties were intensively studied Seifert  _et al._ (2000);
Zhang, Dumitricǎ, and Seifert (2010); Zibouche, Kuc, and Heine (2012). It has
been reported that these NTs have intrinsic semiconducting electronic
structures because their 2D counterpart is a semiconductor with a direct band
gap at the K point Mak  _et al._ (2010); Splendiani  _et al._ (2010). It has
also been revealed that the band edge structure depends on the NT chirality.
Zigzag ($n$,0) NTs have direct band gaps at the $\Gamma$ point, while armchair
($n$,$n$) and chiral NTs have indirect band gaps between the vicinity of the
$\Gamma$ point and the middle of the 1D Brillouin zone (BZ) for the valence
and conduction band edges, respectively. These theoretical insights into the
electronic band structure of single-walled MoS2 NTs are not fully accountable
for the recent experimental finding that MoS2 NTs have a direct band gap
causing PL. In addition, the results also imply the existence of a missing
link that connects the indirect band gap of the tubular form and the direct
band gap of the planar form of MoS2.

In this report, we aim to determine the critical NT diameter at which the
indirect-to-direct band gap crossover occurs in order to connect the
discontinuity between the indirect gaps of NTs and the direct gap of sheets.
Using density functional theory (DFT) Hohenberg and Kohn (1964); Kohn and Sham
(1965), we investigated the electronic structures of armchair MoS2 NTs, with
diameters ranging from 1.6 to 6.8 nm. The DFT calculations revealed that MoS2
NTs with diameters larger than 5.2 nm are semiconductors with an indirect band
gap in the vicinity of $k=2\pi/3$, corresponding to the valence and conduction
band valleys at the K point of an isolated MoS2 sheet. Furthermore, our
structural analysis of MoS2 NTs demonstrated that the curvature-induced
mechanical strain causes the indirect band gaps of MoS2 NTs with small
diameters.

All of the calculations in this study were performed using DFT implemented in
the Simulation Tool for Atom Technology (STATE) program package Morikawa,
Iwata, and Terakura (2001). The general gradient approximation (GGA) was
applied with the Perdew–Burke–Ernzerhof functional Perdew, Burke, and
Ernzerhof (1996, 1997) to describe the exchange correlation potential energy
among interacting electrons. The ultrasoft pseudopotential generated by the
Vanderbilt scheme Vanderbilt (1990) was used to provide the interaction
between valence electrons and nuclei. The valence wave functions and deficit
charge density were expanded in terms of the plane wave basis set, with cutoff
energies of 340 and 3061 eV (25 and 225 Ry), respectively.

In this work, we considered armchair MoS2 NTs with diameters ranging from 1.6
to 6.8 nm, which correspond to the indices (9,9) and (39,39), to determine the
critical diameter at which the band gap crossover occurs. To exclude the
effect arising from inter-tube periodic boundary conditions normal to the NT,
the NTs were separated from their periodic images by 0.9 nm vacuum spacing. BZ
integration was conducted under six equidistant k points along the tube axis.
The atomic coordinates were optimized using a fixed lattice parameter of 0.315
nm, corresponding to the experimental value for planar MoS2 Wakabayashi,
Smith, and Nicklow (1975), until the forces acting on each atom were less than
0.684 eV/nm (1.33$\times$10-3 Hartree/au).

Figure 1 shows the electronic structure of the MoS2 NTs, from the $\Gamma$
point ($k=0$) to the X point ($k=\pi$) in the Brillouin zone. The band
dispersion relation was sensitive to the tube diameter, even though these NTs
had the same armchair chirality. The (39,39) MoS2 NT was a direct band gap
semiconductor, with the valence and conduction band edges located in the
vicinity of $k=2\pi/3$ [Fig. 1(a)], corresponding to the K point of the 2D BZ.
This result indicates that the band edge positions of the NTs with large
diameters originate from the valleys of the isolated MoS2 sheet, confirming
the expectation that the electronic band structure near the gap of the NTs
should asymptotically approach that of an isolated MoS2 sheet. In contrast to
the (39,39) NT, the (9,9) NT had an indirect band gap between near $\Gamma$
and $k=2\pi/3$ for the valence and conduction band edges, respectively [Fig.
1(b)], as observed in the previous calculations. Thus, NTs with a large
diameter should exhibit PL, as observed for isolated sheets of MoS2.

Figure 1: Electronic structures of (a) (39,39) MoS2 NT and (b) (9,9) MoS2 NT.
The energies are measured with respect to that of the valence band edge at the
$\Gamma$ point.  Figure 2: Energy difference between valence band peaks,
$E_{K}^{\mathrm{HO}}-E_{0}^{\mathrm{HO}}$, as a function of diameter, $D$, for
MoS2 NTs. The dotted line indicates the value for monolayer MoS2.

To determine the critical diameter at which the indirect-to-direct band gap
crossover occurs, we investigated the energy difference between the peak of
the highest energy of the occupied state in valence band at near $\Gamma$,
$E_{0}^{\mathrm{HO}}$, and at $k=2\pi/3$, corresponding to the K point of 2D,
$E_{K}^{\mathrm{HO}}$. Figure 2 shows the energy difference,
$E_{0}^{\mathrm{HO}}-E_{K}^{\mathrm{HO}}$ as a function of NT diameter, $D$.
The energy difference, $E_{0}^{\mathrm{HO}}-E_{K}^{\mathrm{HO}}$ monotonically
increases with increasing diameter, crossing zero at a diameter of
approximately 5.0 nm. Therefore, the NTs with a diameter of 5.0 nm or larger
have a direct band gap in the vicinity of $k=2\pi/3$ and are expected to
exhibit PL at the band gap energy. Note that the asymptotic property of the
energy peak difference rises extremely slowly with increasing diameter. The
peak difference for the NT with the largest diameter is still smaller (by 0.1
eV) than that of the monolayer MoS2 sheet.

The diameter dependence of the valence band edge of the MoS2 NTs indicates
that the effect of the structural strain arises from the curvature. Figure 3
shows the bond length between Mo and S atoms as a function of tube diameter.
The tubular structure and symmetry result in four distinct Mo-S distances. Two
of the four distances, corresponding to outer Mo-S bonds, increase with
decreasing diameter, while the other two distances, corresponding to inner
Mo-S bonds, decrease with decreasing diameter. The bond deformations of the
outer Mo-S bonds are larger than those of the inner Mo-S bonds. In particular,
the outer Mo-S bond perpendicular to the tube axis is elongated by a factor of
four compared with that of the inner Mo-S bonds. Therefore, the atoms in the
MoS2 NTs are subjected to tensile and compressive strain in the outer and
inner regions, respectively. Furthermore, the tensile strain for the outer
part is dominant for the structural distortion. This structural deformation
can account for the diameter-dependent edge modulation observed for the
valence band of the MoS2 NT because the valence band edge of monolayer MoS2 is
sensitive to strain Conley  _et al._ (2013); Gong  _et al._ (2013); Ghorbani-
Asl  _et al._ (2013).

Figure 3:  (a) Schematic view of an atomic arrangement of an MoS2 NT, where
the yellow and purple balls indicate S and Mo atoms, respectively. (b) Mo-S
bond length as a function of the MoS2 NT diameter, $D$, where indices
correspond with those in (a). Figure 4:  Energy difference between valence
band peaks, $E_{K}^{\mathrm{HO}}-E_{0}^{\mathrm{HO}}$, of monolayer MoS2 as a
function of the lattice constant.

To confirm the effect of strain on the electronic structure of MoS2 sheets, we
investigated the electronic structure of isolated MoS2 sheets. Figure 4 shows
the energy difference between the K and $\Gamma$ points of the highest branch
of the valence band as a function of the lattice parameter. With increasing
lattice parameter, which corresponds to the tensile strain, the energy
difference decreased and crossed zero at a tensile strain of 1.84%, with an
Mo-S distance of 0.318 nm. Therefore, the tensile strain caused the upward
shift of the peak at the $\Gamma$ point, leading to the indirect band gap
structure for the MoS2 sheet. This upward shift is associated with the wave
function distribution of the valence band top at the $\Gamma$ point, where the
state consists of the $d_{z}$ state of Mo and the $p_{z}$ state of S atoms
with a bonding nature Gong  _et al._ (2013). Therefore, the state was shifted
upward by the tensile strain, which reduced the orbital hybridization. This
analysis clearly confirms that the band gap crossover in MoS2 NTs is induced
by the strain associated with their curvature.

Based on DFT with GGA, we investigated the electronic structure of armchair
MoS2 NTs with diameters of 1.6-6.8 nm. Our DFT calculations showed that NTs
with diameters of up to 5.0 nm are semiconductors with an indirect band gap,
whereas NTs with diameters of 5.0 nm or larger have a direct band gap between
valleys in the vicinity of $k=2\pi/3$, corresponding to the band edge valleys
of monolayer MoS2. The results confirm that the electronic structure of MoS2
exhibits asymptotical properties transitioning from an indirect to a direct
band gap based on the diameter. Therefore, MoS2 NTs with diameters of 5.2 nm
or larger may exhibit PL, as observed for monolayer MoS2 sheets, and may
represent a constituent material for designing optical, optoelectronic, and
valleytronic devices. Structural analysis of the MoS2 NTs revealed that the
tensile strain on the outer part of the NTs caused an upward shift of the
valence band peak near the $\Gamma$ point, leading to the indirect-to-direct
band gap crossover associated with the NT diameter.

Part of this work was financially supported by JSPS KAKENHI Grant Numbers
JP17K06187, JP18H05329, JP19J13818 and JP20H00220. Some of the calculations
were performed on NEC SX-Ace at the Cybermedia Center in Osaka University and
on Oakbridge-CX at Information Technology center in the University of Tokyo.

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