On arithmetical structures on $K_{9}$

Egyptian fractions appear in an ancient Egyptian text, namely the Rhind Mathematical Papyrus, dating back to around $1650$ B.C. [28]. The fact that any positive rational number $r$ has an Egyptian fraction representation of the form

$$r=\frac{1}{x_{1}}+\dots+\frac{1}{x_{n}}\qquad\text{with $1\leq x_{1}<\dots<x_{n}$}$$

is due to Fibonacci in $1202$, using a greedy algorithm (at each step the algorithm chooses greedily the largest possible unit fraction that can be used in any representation of the remaining fraction) and later rediscovered by Sylvester in [30]. Fibonacci’s greedy algorithm consists of repeatedly performing:

$$\frac{x}{y}=\frac{1}{\lceil\frac{y}{x}\rceil}+\frac{(-y)\,\text{mod}\,x}{y\lceil\frac{y}{x}\rceil}$$

Later on, several authors tried to improve the Fibonacci-Sylvester algorithm, for instance by minimizing the number of unit fractions. A non exhaustive list is [15][18][4][5]. Recently, Louwsma and Martino studied the rational numbers with odd greedy expansion of fixed length. Their work appears in [26].

Problems and results concerning Egyptian fractions appear in [15] and [4].
The present paper is concerned with representations of the unit, that is setting $r=1$ and the $x_{i}$’s are not necessarily distinct. We present a new viewpoint for the arithmetic study of the diophantine equation, based on elementary properties on $p$-adic numbers. One interest in the solutions to this diophantine equation relies in the fact that finding such solutions to the diophantine equation is equivalent to finding $n$ positive integers $k_{i}$’s with $1\leq i\leq n$ with no common factor such that

$$k_{j}\bigg{|}\sum_{i=1}^{n}k_{i}\qquad\text{for all $j$}$$

This problem is itself equivalent to finding an arithmetical structure on the complete graph $K_{n}$, see for instance Theorem $4.4$ of [23] proven in a more general setting.
Arithmetical structures were originally introduced by Lorenzini in [25] in terms of matrices as a generalization of the Laplacian matrix in order to study intersections of degenerating curves in algebraic geometry. If $G$ is a finite connected graph with $n$ vertices and $A$ is the adjacency matrix of $G$, then an arithmetical structure on $G$ is a pair of vectors $(\mathbf{d},\mathbf{r})$ whose entries are nonnegative and positive integers respectively and the entries of $\mathbf{r}$ have no nontrivial common factor, such that:

$$\big{(}diag(\mathbf{d})-A\big{)}\mathbf{r}=\mathbf{0}$$

The critical group associated with an arithmetical structure [22] generalizes the sandpile group in the special case of the Laplacian matrix which is by definition the difference of the diagonal degree matrix with the adjacency matrix. The sandpile group itself relates to the chip-firing game, see e.g. [16] and [3]. Lorenzini showed in his Lemma $1.6$ of [25] that any finite, connected graph has a finite number of arithmetical structures. In [6], the authors show that the number of arithmetical structures on the path graph $P_{n}$ is given by the Catalan number $C(n-1)$ and that the number of arithmetical structures on the cycle graph $C_{n}$ is given by the binomial coefficient $\binom{2n-1}{n-1}$. In [2], the authors treat the case of graphs corresponding to Dynkin diagrams of type $D_{n}$. For a given $n\geq 4$, they establish the recursive formula:

$$|Arith(D_{n})|=2C(n-2)+\sum_{m=4}^{n}B(n-3,n-m)|SArith(D_{m})|,$$

where $C(n)=\frac{1}{n+1}\binom{2n}{n}$ is the Catalan number, $B(n,k)=\frac{n-k+1}{n+1}\binom{n+k}{n}$ is the ballot number and $|SArithm(D_{m})|$ is the number of smooth arithmetical structures on $D_{m}$. The authors define a smooth arithmetical structure on $D_{n}$ by imposing that $d_{i}\geq 2$ at each vertex $i$, except possibly at the trivalent vertex say $i=t$ (in fact, they show in their Proposition $2.3$ that these conditions on the $d_{i}$’s imply $d_{t}=1$). They then come up with bounds for the number of smooth arithmetical structures and ultimately deduce the following bounds for the arithmetical structures on $D_{n}$ with $n\geq 4$:

$$2C(n-2)+C(n-3)\leq|Arith(D_{n})|<2\,C(n-2)+702\,C(n-3)$$

In [17], the authors deal with arithmetical structures on paths with a doubled edge and conjecture how the number of arithmetical structures grows depending on the path length and the location of the doubled edge.
The importance of arithmetical structures on complete graphs gets reinforced by Conjecture $6.10$ of [14] which asserts that for any connected simple graph $G$ with $n$ vertices, the number of arithmetical structures on $G$ is at most the number of arithmetical structures on $K_{n}$. The number of arithmetical structures on $K_{n}$ appears in the online encyclopedia of integer sequences [27] for $n\leq 8$. In the present paper, we investigate arithmetical structures on $K_{9}$ from the perspective of the Diophantine equation, that is we look for solutions to $1/x_{1}+\dots+1/x_{9}=1$ in positive integers $x_{i}$’s, $i=1,\dots,9$. It is shown in [8] that there are exactly five solutions in distinct and odd integers. The solutions in possibly non distinct and possibly even integers have not been investigated so far.
Our paper is structured as follows. We will do some general $p$-adic analysis on the Diophantine equation in $\S\,2$ and then apply it specifically to the case of $9$ positive integer variables in $\S\,3$. Like other authors also did it, we will restrict the study to some specific cases described in $\S\,3$. Within these specific cases, the method that we use allows to count the number of solutions for the general complete graph $K_{n}$ with $n\geq 9$. We wrote a Mathematica program which given the number $n$ in entry returns this number of solutions. Last, still in the framework of these specific cases, we wrote a program in CAML allowing to obtain the complete list of solutions and applied our program in the cases of $K_{9}-K_{13}$.

In the statements below and throughout the paper, $(E)_{n}$ denotes the Diophantine equation

$$\frac{1}{x_{1}}+\dots+\frac{1}{x_{n}}=1$$

$(E)_{n}$

Our main results are the following. Theorem $1$ deals with $n=9$ specifically. Theorem $2$ provides the count of the solutions for a general $n\geq 9$ using a Mathematica program. Theorem $3$ exhibits all the solutions when $n=10,11,12,13$ using a CAML program.

The goal of this part is to study for a given prime $p$ the $p$-adic valuations of the positive integers $x_{i}$’s when these constitute a solution to the equation $(E)_{n}$. We begin our study with the fundamental property stated right below.

Proof of Lemma $1$. By assumption, the $x_{i}$’s satisfy to

$$\sum_{i=1}^{r}x_{1}\dots\hat{x_{i}}\dots x_{n}-x_{1}\dots x_{n}=0$$

(1)

The $p$-adic absolute value being ultrametric, for any $p$-adic numbers $a_{1},\dots,a_{s}$, we have:

$$v_{p}(a_{1}+\dots+a_{s})\geq Min_{1\leq i\leq s}v_{p}(a_{i}),$$

with the equality holding when the minimum is unique.
By convention, we have $v_{p}(0)=+\infty$. Then $(1)$ implies that:

$$v_{p}(x_{1}\dots x_{n})=v_{p}(x_{1}\dots\hat{x_{i}}\dots x_{n}),\qquad\text{some $1\leq i\leq n$}$$

(2)

or

$$v_{p}(x_{1}\dots\hat{x_{i}}\dots x_{n})=v_{p}(x_{1}\dots\hat{x_{j}}\dots x_{n}),\;\;\;\,\text{some $1\leq i<j\leq n$}$$

(3)

Moreover, if $(2)$ holds, then we get $v_{p}(x_{i})=0$ and if $(3)$ holds, we get $v_{p}(x_{i})=v_{p}(x_{j})$. $\square$

Proof of Proposition $1$. If $p$ divides none of the $x_{i}$’s, then for all $i$ with $1\leq i\leq n$, we have $v_{p}(x_{i})=0$ and so we are done. From now on, suppose that $p$ divides at least one of the $x_{i}$’s. The fact that $(x_{1},\dots,x_{n})$ is a solution forces $v_{p}(x_{i})=0$ for some $i$ or $v_{p}(x_{k})=v_{p}(x_{l})$ for some $k\neq l$. In the latter case, we are done. Otherwise, assume without loss of generality that $v_{p}(x_{1})=0$ and factor the diophantine equation as follows.

$$x_{1}(\sum_{i=2}^{n}x_{2}\dots\hat{x_{i}}\dots x_{n}-x_{2}\dots x_{n})+x_{2}\dots x_{n}=0$$

(4)

It comes:

$$v_{p}(x_{2}\dots x_{n})=v_{p}(\sum_{i=2}^{n}x_{2}\dots\hat{x_{i}}\dots x_{n}-x_{2}\dots x_{n})$$

(5)

By assumption, there exists an integer $s$ amongst $\{2,\dots,n\}$ such that $v_{p}(x_{s})>0$. Then,

$$v_{p}(x_{2}\dots x_{n})>Min_{2\leq i\leq r}v_{p}(x_{2}\dots\hat{x_{i}}\dots x_{r})$$

(6)

In turn, there must exist two integers $k$ and $l$ with $2\leq k<l\leq n$ such that:

$$v_{p}(x_{2}\dots\hat{x_{k}}\dots x_{n})=v_{p}(x_{2}\dots\hat{x_{l}}\dots x_{n})$$

(7)

Then, for these two distinct integers $k$ and $l$, we have $v_{p}(x_{k})=v_{p}(x_{l})$ and moreover, these two valuations are non-zero. Hence the straightforward corollary.

Proof of Corollary $2$. If $p$ divides none of the $x_{i}$’s, the result is trivial. Otherwise, by Corollary $1$, there exist distinct integers $k$ and $l$ with $v_{p}(x_{k})=v_{p}(x_{l})=\alpha>0$. Either $\alpha$ is the highest integer such that $p^{\alpha}$ occurs as a factor amongst the $x_{i}$’s, in which case we are done or we have:

$$Max_{j\not\in\{k,l\}}v_{p}(x_{j})>\alpha$$

(8)

Suppose for a contradiction that this maximum is unique and attained at $x_{s}$ with $s\not\in\{k,l\}$.
Write:

$$\begin{array}[]{cccc}x_{k}&=&p^{\alpha}x^{{}^{\prime}}_{k}&\text{with $p\not|x^{{}^{\prime}}_{k}$}\\ x_{l}&=&p^{\alpha}x^{{}^{\prime}}_{l}&\text{with $p\not|x^{{}^{\prime}}_{l}$}\end{array}$$

After division by $p^{\alpha}$, the diophantine equation reads:

$$p^{\alpha}x^{{}^{\prime}}_{k}x^{{}^{\prime}}_{l}\prod_{j\not\in\{k,l\}}x_{j}=x^{{}^{\prime}}_{k}\prod_{j\not\in\{k,l\}}x_{j}+x^{{}^{\prime}}_{l}\prod_{j\not\in\{k,l\}}x_{j}+\sum_{m\not\in\{k,l\}}p^{\alpha}x^{{}^{\prime}}_{k}x^{{}^{\prime}}_{l}\prod_{j\not\in\{m,k,l\}}x_{j}$$

(9)

It follows that:

$$v_{p}\big{(}\prod_{j\not\in\{k,l\}}x_{j}\big{)}\leq\alpha+\sum_{j\not\in\{k,l,s\}}v_{p}(x_{j})$$

(10)

This implies $v_{p}(x_{s})\leq\alpha$, a contradiction. $\square$

In what follows, we denote the elementary symmetric functions of $a_{1},\dots,a_{n}$ by $\sigma_{i}(a_{1},\dots,a_{n})$ with $1\leq i\leq n$. And so by definition,

$$\sigma_{i}(a_{1},\dots,a_{n}):=\sum_{1\leq k_{1}<\dots<k_{i}\leq n}a_{k_{1}}\dots a_{k_{i}}$$

Using this notation, the following result holds.

Proof of Corollary $3$. We will write a similar equation as the one in $(9)$ with a larger number of $x^{{}^{\prime}}_{i}$’s involved. We have, after dividing the diophantine equation by $p^{(s-1)\alpha}$:

$$\begin{split}\sigma_{s-1}(x^{{}^{\prime}}_{i_{1}},\dots,x^{{}^{\prime}}_{i_{s}})\prod_{j\not\in\{i_{1},\dots,i_{s}\}}x_{j}&+p^{\alpha}x^{{}^{\prime}}_{i_{1}}\dots x^{{}^{\prime}}_{i_{s}}\sum_{m\not\in\{i_{1},\dots,i_{s}\}}\prod_{j\not\in\{m,i_{1},\dots,i_{s}\}}x_{j}\\ &=p^{\alpha}x^{{}^{\prime}}_{i_{1}}\dots x^{{}^{\prime}}_{i_{s}}\prod_{j\not\in\{i_{1},\dots,i_{s}\}}x_{j}\end{split}$$

(11)

It follows immediately that:

$$v_{p}(\sigma_{s-1}(x^{{}^{\prime}}_{i_{1}},\dots,x^{{}^{\prime}}_{i_{s}}))\geq\alpha-Max_{m\not\in\{i_{1},\dots,i_{s}\}}v_{p}(x_{m}),$$

(12)

with equality holding when the maximum is unique and positive. $\square$

Proof of Corollary $4$. Eq. $(11)$ still holds with a smaller number of $x^{{}^{\prime}}_{i_{t}}$’s involved. In particular, by writing $(11)$ with only $(s-1)$ of the $s$ $x^{{}^{\prime}}_{i_{t}}$’s, we see that the maximum that is involved is unique and equal to $\alpha$. It yields the result. $\square$

The latter three corollaries are best illustrated on an example.
When $n=11$, Burshtein proves that there are exactly $17$ solutions $\mathbf{B_{1}}-\mathbf{B_{17}}$ in distinct integers of the form $3^{\alpha}5^{\beta}7^{\gamma}$ and exhibits these solutions in [9]. Consider for instance

$$\mathbf{B_{12}}=\{3,5,7,3^{2},3^{2}.5.7,5^{2},3^{3},7^{2},7^{2}.3^{3},7.5^{2}.3^{3}\},$$

written in terms of prime numbers. Set $p=3$ and so $\alpha=3$ using the previous notations. Visibly $3^{3}$ occurs $3$ times as a factor. Obviously, $3$ does not divide $\sigma_{1}(7.5^{2},1)=176$, nor $\sigma_{1}(7^{2},1)=50$, nor $\sigma_{1}(7^{2},7.5^{2})=224$, just like implied by Corollary $4$. The highest other $3$-adic valuation is $2$ and we have $v_{3}(\sigma_{2}(7^{2},7.5^{2},1))=v_{3}(8799)\geq 1$, just like implied by Corollary $3$. Set now $p=5$ and so $\alpha=2$. Visibly, $5^{2}$ occurs exactly twice as a factor and indeed, we have $5|1+7.27\,(=190)$

       The case $n=9$ studied here is special in that when the integers are odd, the minimal length of representation of the unit as an Egyptian fraction is $9$. This was first predicted by J. Leech, see [21], pp. $89$. In [8], Burshtein proved that the diophantine equation has exactly $5$ solutions that were first identified by S. Yamashita in $1976$, see the online The prime puzzles and problems connection, Problem $35$ entitled ”More wrong turns…” Ronald Graham had once asked André Weil why the Egyptians did this, that is to represent fractions as a sum of unit fractions, and Weil’s answer was ”They took a wrong turn…”
The five solutions in odd and distinct integers copied from [8] are:

$$\left\{\begin{array}[]{l}\mathbf{B_{1}}=\{3,5,7,9,11,15,21,231,315\}\\ \mathbf{B_{2}}=\{3,5,7,9,11,15,35,45,231\}\\ \mathbf{B_{3}}=\{3,5,7,9,11,15,21,135,10395\}\\ \mathbf{B_{4}}=\{3,5,7,9,11,15,33,45,385\}\\ \mathbf{B_{5}}=\{3,5,7,9,11,15,21,165,693\}\end{array}\right.$$

We are now concerned with searching for solutions when the $x_{i}$’s are not necessarily distinct and are possibly even. First and foremost, we will search for solutions with $2$ occurring at least twice as a factor.

Proof. Using our usual notations, set $\alpha$ the highest positive $2$-adic valuation amongst the $x_{i}$’s and let $s$ denote the number of occurrences. By Corollary $2$, we know that $2^{\alpha}$ occurs at least twice as a factor amongst the $x_{i}$’s. Without loss of generality, $x_{1}=2^{\alpha}x^{{}^{\prime}}_{1}$ and $x_{2}=2^{\alpha}x^{{}^{\prime}}_{2}$ with $x^{{}^{\prime}}_{1}$ and $x^{{}^{\prime}}_{2}$ both odd. The sum of two odd numbers being even, we have $2|x^{{}^{\prime}}_{1}+x^{{}^{\prime}}_{2}$, which by Corollary $5$ implies $s=2$ or $s\geq 4$. Suppose now $s\neq 2$. Without loss of generality, $x_{i}=2^{\alpha}x^{{}^{\prime}}_{i}$ for $i=1,\dots,4$ and each $x^{{}^{\prime}}_{i}$ is odd. Now, $2$ divides $\sigma_{3}(x^{{}^{\prime}}_{1},x^{{}^{\prime}}_{2},x^{{}^{\prime}}_{3},x^{{}^{\prime}}_{4})$ since the latter elementary symmetric function is the sum of $4$ odd terms. A new application of Corollary $5$ yields $s=4$ or $s\geq 6$. The result follows from proceeding by induction, since there is an even number of ways of picking $2l-1$ elements amongst $2l$ elements and a sum of an even number of odd integers is an even integer. $\square$

For the remainder of our discussion, it will be convenient to denote by $\alpha_{p}$ the highest $p$-adic valuation of the $x_{i}$’s and by $s_{p}$ the number of occurrence(s) amongst the $x_{i}$’s. For instance, we have just shown that $s_{2}\in\{0,2,4,6,8\}$.
Set also $X:=\{x_{1},\dots,x_{9}\}$ as the set of constituents of a solution to the diophantine equation. If along the proofs, we need to refer to solutions with $k$ vertices with $k$ not necessarily equal to $9$, we will specify it by writing $X_{n=k}$ instead of $X$.

In what follows, we assume that the $x_{i}$’s are distinct and $2$ is a divisor.

We will first prove Lemma $4$ and then prove Lemma $3$ by using some elements in the proof of Lemma $4$.

Proof of Lemma $4$. In the proof, there is no distinction between $2\in X$ or $2\not\in X$. Having $2\not\in X$ is more rare, hence the distinction in the statement itself. We first lay out some general considerations based on $\alpha_{2}=2$ and our results of $\S\,2$. Let $q$ be an odd divisor. Our assumptions imply that the highest power of $q$ occurs twice or thrice. We apply Corollary $3$ to $q$.

If the highest power of $q$ occurs three times, then $q|\sigma_{2}(1,2,4)=14$. It forces $q=7$. Moreover, still by Corollary $3$, the second largest valuation must be $\alpha_{7}-1$. Further, we can show that this second largest valuation may only occur as $\{7^{\alpha_{7}-1},2.7^{\alpha_{7}-1}\}$. Indeed, in all the other cases, by contracting $\{7^{\alpha_{7}},2.7^{\alpha_{7}},2^{2}.7^{\alpha_{7}}\}$ to $2^{2}.7^{\alpha_{7}-1}$ and possibly contracting further, we would get a solution of $X_{n=k}$ for some adequate $k\leq 7$ with the highest $7$-valuation occurring less than three times. Continuing this process inductively, it forces $2\in X$ and moreover, we obtain the result of the first point of $(ii)$, with $s_{2}=2$ the only available option.

Suppose now the highest power of $q$ occurs exactly twice. Then $q|\sigma_{1}(1,2)=3$ or $q|\sigma_{1}(1,2^{2})=5$ or $q|\sigma_{1}(2,2^{2})=2.3$. This implies $q=3$ or $q=5$.
In the first case, the last two terms of the sum are either $\big{\{}\frac{1}{3^{\alpha_{3}}},\frac{1}{2.3^{\alpha_{3}}}\big{\}}$ or $\big{\{}\frac{1}{2.3^{\alpha_{3}}},\frac{1}{2^{2}.3^{\alpha_{3}}}\big{\}}$ and the second largest $3$-valuation occurring is $\alpha_{3}-1$.
In the second case, the last two terms of the sum are $\frac{1}{5^{\alpha_{5}}}$ and $\frac{1}{2^{2}.5^{\alpha_{5}}}$. Again, the second largest $5$-valuation occurring is $\alpha_{5}-1$.
We first deal with the case $q=5$. This time, it will be useful to note that $\{5^{\alpha_{5}},2^{2}.5^{\alpha_{5}}\}$ contracts to $2^{2}.5^{\alpha_{5}-1}$. By the same reasoning as for $q=7$ above, the only option is to have $5^{\alpha_{5}-1}$ in the solution. We then work inductively until getting a solution of $X_{n=3}$ of the form $\{x_{1},x_{2},2^{2}.5^{\alpha_{5}-6}\}$ forcing $x_{1}=2$, $x_{2}=4$ and $\alpha_{5}=6$. Consequently the only possible original solution is $U$, provided in point $(ii)$ of the lemma.
When $q=3$, we treat both cases evoked above simultaneously and inductively on the number of vertices.

We state below a series of facts on which our discussion will be based.

Proof of Fact $1$. Let’s treat $(i)$. Such a solution in $X_{n=9}$ would provide a solution in $X_{n=8}$ ending in $\{3^{\alpha_{3}-1},2^{2}3^{\alpha_{3}-1}\}$, which is forbidden.
Point $(ii)$ is straightforward.
Regarding $(iii)$, if such a solution existed, realize two contractions and obtain a solution in $X_{n=7}$ ending in $2.3^{\alpha_{3}-1}$ with no other occurrence with valuation $\alpha_{3}-1$. This is impossible.
Point $(iv)$ is similar to point $(ii)$ and is likewise straightforward. So is point $(vi)$.
As for $(v)$, it follows from the same argument as in $(iii)$ with $2.3^{\alpha_{3}-1}$ replaced with $3^{\alpha_{3}-1}$. $\square$

Proof of Fact $2$. If we had a solution of $X_{n=9}$ ending in this following set $\{3^{\alpha_{3}-1},2^{2}.3^{\alpha_{3}-1},2.3^{\alpha_{3}},3^{\alpha_{3}}\}$ or that following set $\{3^{\alpha_{3}-1},2.3^{\alpha_{3}-1},2^{2}.3^{\alpha_{3}},2.3^{\alpha_{3}}\}$ (resp $\{2.3^{\alpha_{3}-1},2^{2}.3^{\alpha_{3}-1},2.3^{\alpha_{3}},3^{\alpha_{3}}\}$ or the set $\{3^{\alpha_{3}-1},2.3^{\alpha_{3}-1},2.3^{\alpha_{3}},3^{\alpha_{3}}\}$, resp $\{2.3^{\alpha_{3}-1},2^{2}.3^{\alpha_{3}-1},2^{2}.3^{\alpha_{3}},2.3^{\alpha_{3}}\}$), then we would have a solution of $X_{n=8}$ (resp $X_{n=7}$, resp $X_{n=6}$) ending in $\{3^{\alpha_{3}-1},2.3^{\alpha_{3}-1},2^{2}.3^{\alpha_{3}-1}\}$ (resp $2.3^{\alpha_{3}-1}$, resp $3^{\alpha_{3}-1}$). None of these situations is permitted. $\square$

Proof of Fact $3$. Suppose first the solution ends in the following set $\{3^{\alpha_{3}-1},2.3^{\alpha_{3}-1},2^{2}.3^{\alpha_{3}-1},2^{2}.3^{\alpha_{3}},2.3^{\alpha_{3}}\}$. From there, realize two contractions leading to a solution of $X_{n=7}$ containing the set $\{2^{2}.3^{\alpha_{3}-2},3^{\alpha_{3}-1},2^{2}.3^{\alpha_{3}-1}\}$. This is impossible.
Suppose now the solution ends in the set $\{3^{\alpha_{3}-1},2.3^{\alpha_{3}-1},2^{2}.3^{\alpha_{3}-1},3^{\alpha_{3}},2.3^{\alpha_{3}}\}$. Then, by operating three contractions, we get a solution of $X_{n=6}$ which contains $\{2.3^{\alpha_{3}-2},2^{2}.3^{\alpha_{3}-2}\}$. The procedure is now the following. By arguments similar to those already used before, there are only three available options. Namely,

• $\clubsuit$

  If $\forall i\in\{0,1,2\}$, we have $2^{i}3^{\alpha_{3}-2}\not\in X_{n=9}$, then we use induction.
  

• $\clubsuit\clubsuit$

  If $\{2.3^{\alpha_{3}-2},3^{\alpha_{3}-2}\}\subseteq X_{n=9}$, then by operating successive contractions, we obtain a solution of $X_{n=3}$ containing $2^{2}3^{\alpha_{3}-4}$. The solutions (in non necessarily distinct integers) are well known from OEIS and are precisely $\{2,4,4\}$, $\{2,3,6\}$ and $\{3,3,3\}$. Then the solution of $X_{n=3}$ must be $\{2,4,4\}$ and the exponent $\alpha_{3}$ must be equal to $4$. Then, the original solution which we are inspecting is $\widehat{Z}_{1}$.
  

• $\clubsuit\clubsuit\clubsuit$

  If $(2^{2}.3^{\alpha_{3}-2},2.3^{\alpha_{3}-2}\}\subseteq X_{n=9}$, then by successive contractions, we obtain a solution of $X_{n=3}$ containing $2.3^{\alpha_{3}-3}$. Then, the solution of $X_{n=3}$ is $\{2,3,6\}$ with $\alpha_{3}=3$ or $\alpha_{3}=4$ or $\{2,4,4\}$ with $\alpha_{3}=3$. The latter possibility must be excluded as it leads to a solution of $X_{n=9}$ with non distinct integers. The first situation with $\alpha_{3}=3$ must also be excluded for the same reason (since then $2.3$ would appear twice). Thus, we necessarily have $\alpha_{3}=4$, leading to the solution which we denoted by $\widehat{Z}_{2}$. $\square$

We will finish treating Fact $3$ $(ix)$ $(b)$ using induction and the results contained in the three facts.