1 Introduction
For arbitrary integers x and y, we denote by $(x,y)$ the greatest common divisor of x and y and by $[x,y]$ their least common multiple. Let $a, b$ and n be positive integers. Let $S = \{x_1, \ldots , x_n\}$ be a set of n distinct positive integers. Let $\xi _a$ be the arithmetic function defined by $\xi _a=x^a$ for any positive integer x. Let $(S^a)$ and $[S^a]$ stand for the $n\times n$ matrices whose $(i,j)$ -entry is $\xi _a((x_i, x_j))$ and $\xi _a([x_i, x_j])$ respectively. We call $(S^a)$ the $ath$ power GCD matrix and $[S^a]$ the $ath$ power LCM matrix. The set S is factor closed (FC) if $(x\in S, d\mid x)\Rightarrow d\in S$ and gcd closed if $(x_i, x_j)\in S$ for all integers i and j with $1\le i,j\le n$ . Obviously, an FC set must be gcd closed but the converse is not true. Nearly 150 years ago, Smith [Reference Smith15] proved that
if S is FC, where $\varphi $ is Euler’s totient function and $\pi $ is the multiplicative function defined for the prime power $p^r$ by $\pi (p^r)=-p$ . There are many generalisations of Smith’s determinant (1.1) and related results (see, for instance, [Reference Altinisik, Yildiz and Keskin1–Reference Korkee and Haukkanen14, Reference Tan and Li16–Reference Zhu, Cheng and Zhao21]). In particular, an elegant result was achieved by Hong et al. [Reference Hong, Hu and Lin8] stating that for any integer $n\ge 2$ ,
where $\mu $ is the M $\ddot {\textrm {o}}$ bius function and an integer $x\ge 1$ is called square free if x is not divisible by the square of any prime number.
As usual, $\mathbb {Z}$ and $|S|$ denote the ring of integers and the cardinality of the set S. Hong [Reference Hong9] introduced the concept of greatest-type divisor when he solved the Bourque–Ligh conjecture. For any integer $x\in S$ , y is called a greatest-type divisor of x if
Let $ G_S(x):=\{y\in S: y \ \text {is a greatest-type divisor of} \ x \ \mathrm {in} \ S\} $ and let $M_n(\mathbb {Z})$ stand for the ring of $n\times n$ matrices over the integers. Bourque and Ligh [Reference Bourque and Ligh4] proved that $(S)$ divides $[S]$ in the ring $M_n(\mathbb {Z})$ (that is, $[S]=B(S)$ or $[S]=(S)B$ for some $B\in M_n(\mathbb {Z})$ ) if S is FC. Hong [Reference Hong10] showed that such a factorisation is not true when S is gcd closed and $\max _{x \in S}\{|{G_S(x)}|\}=2$ . The results of Bourque–Ligh and Hong were generalised by Korkee and Haukkanen [Reference Korkee and Haukkanen14] and by Chen et al. [Reference Chen, Feng, Hong and Qiu6]. Feng et al. [Reference Feng, Hong and Zhao7], Zhao [Reference Zhao17], Altinisik et al. [Reference Altinisik, Yildiz and Keskin1] and Zhao et al. [Reference Zhao, Chen and Hong18] used the concept of greatest-type divisor to characterise the gcd-closed sets S with $\max _{x \in S}\{|{G_S(x)}|\}\le 3$ such that $(S^a)\mid [S^a]$ which partially solved an open problem of Hong [Reference Hong10].
Hong [Reference Hong12] investigated divisibility among power GCD matrices and among power LCM matrices. It was proved in [Reference Hong12] that $(S^a)\mid (S^b), (S^a) \mid [S^b]$ and $[S^a] \mid [S^b]$ if $a \mid b$ and S is a divisor chain (that is, $x_{\sigma (1)}| \cdots |x_{\sigma (n)}$ for a permutation $\sigma $ of $\{1,\ldots , n\}$ ), and such factorisations are no longer true if $a\nmid b$ and $|S|\ge 2$ . Evidently, a divisor chain is gcd closed but not conversely. Recently, Zhu [Reference Zhu19] confirmed two conjectures of Hong raised in [Reference Hong12] stating that if $a\mid b$ and S is a gcd-closed set with $\max _{x\in S}\{|G_S(x)|\}=1$ , then both the bth power GCD matrix $(S^b)$ and the bth power LCM matrix $[S^b]$ are divisible by the ath power GCD matrix $(S^a)$ . At the end of [Reference Hong12], Hong also conjectured that if $a \mid b$ and $S=\{x_1,\ldots , x_n \}$ is gcd closed and $\max _{x\in S}\{|G_S(x)|\}=1$ , then $[S^a] \mid [S^b]$ in the ring $M_n(\mathbb {Z})$ . Tan and Li [Reference Tan and Li16] partially confirmed this conjecture by proving that $[S^a] \mid [S^b]$ in the ring $M_{|S|}(\mathbb {Z})$ if $a \mid b$ and S consists of finitely many coprime divisor chains with $1\in S$ and that such a divisibility relation is not true if $a\nmid b$ . However, the conjecture still remains open.
Our goal is to present a proof of Hong’s conjecture. The main result of the paper is the following theorem.
Theorem 1.1. If a and b are positive integers such that $a \mid b$ and S is a gcd-closed set such that $\max _{x\in S}\{|G_S(x)|\}=1$ , then the ath power LCM matrix $[S^a]$ divides the bth power LCM matrix $[S^b]$ in the ring $M_{|S|}(\mathbb {Z})$ .
The proof of Theorem 1.1 is similar to that of Feng et al. [Reference Feng, Hong and Zhao7] in character, but it is more complicated. This paper is organised as follows. In Section 2, we supply several preliminary lemmas needed in the proof of Theorem 1.1. Section 3 is devoted to the proof of Theorem 1.1.
One can easily check that for any permutation $\sigma $ on the set $\{1, \ldots , n \}$ , $[S^a]\mid [S^b]\Leftrightarrow [S_{\sigma }^a]\mid [S_{\sigma } ^b]$ , where $S_\sigma :=\{x_{\sigma (1)},\ldots , x_{\sigma (n)}\}$ . Without loss of any generality, we can always assume that the set $S=\{x_1,\ldots , x_n \}$ satisfies $x_1<\cdots <x_n$ .
2 Auxiliary results
In this section, we provide several lemmas that will be needed in the proof of Theorem 1.1. We begin with a result due to Hong which gives the formula for the determinant of the power LCM matrix on a gcd-closed set.
Lemma 2.1 [Reference Hong11, Lemma 2.1].
If S is gcd closed, then
where
and ${1}/{\xi _a}$ is the arithmetic function defined for any positive integer x by ${({1}/{\xi _a})(x):= x^{-a}}$ .
Lemma 2.2 [Reference Bourque and Ligh5, Theorem 3].
If S is a gcd-closed set and $(f((x_i,x_j)))$ is invertible, then $ (f((x_i,x_j)))^{-1}=(a_{ij}), $ where
with
Lemma 2.3 [Reference Hong11, Lemma 2.3].
Let m be a positive integer. Then
Lemma 2.4 [Reference Feng, Hong and Zhao7, Lemma 2.2].
Let S be gcd closed and $\max _{x\in S}\{|G_S(x)|\}=1$ . Let $\alpha _{a,k}$ be defined as in (2.2). If $G_S(x_k)=\{x_{k_1}\}$ for $2\le k\le |S|$ , then $\alpha _{a,k}=x_k^{-a}-x_{k_1}^{-a}.$
Lemma 2.5. Let S be gcd closed and $\max _{x\in S}\{|G_S(x)|\}=1$ . Let $\alpha _{a,k}$ and $c_{ij}$ be defined as in (2.2) and (2.3), respectively. Then $[S^a]$ is nonsingular and $[S^a]^{-1}=(s_{ij})_{1\le i,j\le n}$ with
Proof. Since ${[x_i,x_j]}^a={x_i^ax_j^a}/{{(x_i,x_j)}^a}$ ,
where $D:=\mathrm {diag}(x_1^a,\ldots ,x_n^a)$ . By (2.1) and (2.4),
By Lemma 2.3, $\alpha _{a,1}=x_1^{-a}$ . For $2\le k\le n$ , since $\max _{x\in S}\{|G_S(x)|\}=1$ , one may let $G_S(x_k)=\{x_{k_1}\}$ . By Lemma 2.4, $\alpha _{a,k}=x_k^{-a}-x_{k_1}^{-a}\neq 0$ . So the matrix $(({1/\xi _a})((x_i,x_j)))$ is nonsingular. Now applying Lemma 2.2 gives
where
The desired result follows immediately from (2.4) and (2.5).
We next recall some basic results on gcd-closed sets.
Lemma 2.6 [Reference Feng, Hong and Zhao7, Lemma 2.3].
Let S be a gcd-closed set with $|S|\ge 2$ . Let $c_{ij}$ be defined as in (2.3). Then
Further, if $G_S(x_m)=\{x_{m_1}\}$ for $2\le m\le |S|$ , then
Lemma 2.7 [Reference Feng, Hong and Zhao7, Lemma 3.1].
Let S be gcd closed and $x,z\in S$ such that $x\nmid z$ . If ${G_S(x)=\{y\}}$ , then $(x, z)=(y, z)$ .
Lemma 2.8. Let S be gcd closed and $x, y\in S$ with $G_S(x)=\{y\}$ . If $a\mid b$ , then for any $z, r\in S$ with $r\mid x$ , $y^a[z,x]^b-x^a[z,y]^b$ is divisible by each of $x^a(y^a-x^a)$ and $r^a(y^a-x^a)$ .
Proof. We divide the proof into two cases.
Case 1: $x\nmid z$ . By Lemma 2.7, $(x, z)=(y, z)$ , which implies
Since $a\mid b$ ,
Hence, $(x^a-y^a)\mid (x^{b-a}-y^{b-a})$ . Then by (2.6), we deduce that $y^a[z,x]^b-x^a[z,y]^b$ is divisible by each of $x^a(y^a-x^a)$ and $r^a(y^a-x^a)$ .
Case 2: $x\mid z$ . Then $[x, z]=[y, z]=z$ . It follows that
Since $a\mid b$ , the desired results follow immediately.
Lemma 2.9. Let S be gcd closed and $\max _{x\in S}\{|G_S(x)|\}=1$ . If $a\mid b$ , then all the elements of the nth column and the nth row of $[S^b][S^a]^{-1}$ are integers.
Proof. The proof of Lemma 2.9 is divided into two cases.
Case 1: $1\le i\le n$ and $j=n$ . By Lemmas 2.5 and 2.6,
Since $\max _{x\in S}\{|G_S(x)|\}=1$ , we may let $G_S(x_n)=\{x_{n_1}\}$ . Then by Lemmas 2.4, 2.6 and 2.8,
as required.
Case 2: $i=n,\ 1\le j\le n-1$ . Then
We claim that
for any positive integer k with $x_j\mid x_k$ .
If $k=1$ , then $m=j=1$ . In this case,
Now let $k>1$ . We can set $G_S(x_k)=\{x_{k_1}\}$ since $|G_S(x_k)|=1$ . By Lemmas 2.4, 2.6 and 2.8,
as desired. This concludes the proof of the claim and of Lemma 2.9.
Finally, we can use Lemma 2.9 to establish the main result of this section.
Lemma 2.10. Let S be gcd closed and $\max _{x\in S}\{|G_S(x)|\}=1$ . Let $S_1:=S\setminus \{x_n\}=\{x_1, \ldots , x_{n-1}\}$ . If $a\mid b$ , then $[S^b][S^a]^{-1}\in M_n(\mathbb {Z})$ if and only if $[S_1^b][S_1^a]^{-1}\in M_{n-1}(\mathbb {Z}).$
Proof. First, it follows from the hypothesis and Lemma 2.9 that all the elements of the nth column and the nth row of $[S^b][S^a]^{-1}$ are integers. So it suffices to show that
for all integers i and j with $1\le i,j\le n-1$ .
To see this, define
for all integers u and v between 1 and n. Then $e_{nj}=1$ if $x_j\mid x_n$ and $e_{nj}=0$ otherwise. Furthermore, for any integer m with $1\le m\le n-1$ , one has $e_{nm}=1$ if $x_m\mid x_n$ and $e_{nm}=0$ otherwise. We then deduce that
Let us now show that $A_{ij}\in \mathbb {Z}$ . Since $\max _{x\in S}\{|G_S(x)|\}=1$ , one may let $G_S(x_n)=\{x_{n_1}\}$ . From Lemma 2.4, $\alpha _{a,n}=x_n^{-a}-x_{n_1}^{-a}.$ However, by Lemma 2.6, for any integer m with $1\le m\le n-1$ , $c_{mn}=-1$ if $m=n_1$ and $c_{mn}=0$ otherwise. It follows from (2.8) and Lemma 2.8 that
Since $e_{nj}\in \{0,1\}$ , (2.8) and (2.9) yield (2.7).
The proof of Lemma 2.10 is complete.
3 Proof of Theorem 1.1
We prove Theorem 1.1 by using induction on $n=|S|$ .
For $n=1$ , the statement is clearly true.
Let $n=2$ . Since $S=\{x_1, x_2\}$ is gcd closed, $(x_1, x_2)=x_1$ and $x_1\mid x_2$ . It follows that
where
Since $a\mid b$ , implying that $a\mid (b-a)$ , it follows that $\mathcal {B}\in \mathbb {Z}$ and $\mathcal {C}\in \mathbb {Z}$ , that is, ${[S^b]}[S^a]^{-1}\in M_2(\mathbb {Z})$ . The statement is true for this case.
Let $n=3$ . Since $S=\{x_1,x_2,x_3\}$ is gcd closed, we have $x_1\mid x_i\ (i=2, 3)$ and $(x_2,x_3)=x_1$ or $x_2$ . Consider the following two cases.
Case 1: $(x_2,x_3)=x_1$ . Then one computes
where $\mathcal {B}$ and $\mathcal {C}$ are as given earlier in this section, $\mathcal {D}:={x_2^b}/{x_1^b}$ , $\mathcal {E}:={x_3^b}/{x_1^b}$ and $\mathcal {F}:=({x_3^{b-a}-x_1^{b-a})}/({x_3^a-x_1^a}).$ Since $x_1\mid x_2$ , $x_1\mid x_3$ and $a\mid (b-a)$ , all of $\mathcal {B}, \mathcal {C}, \mathcal {D}, \mathcal {E}$ and $\mathcal {F}$ are integers. Hence, ${[S^b]}[S^a]^{-1}\in M_3(\mathbb {Z})$ . The statement holds in this case.
Case 2: $(x_2,x_3)=x_2$ . Then $x_2\mid x_3$ . We compute
where $\mathcal {B}$ is as before, $\mathcal {G}:={({x_3^b-x_2^b)}/{(x_3^a-x_2^a})}$ and $\mathcal {H}:={(x_3^{b-a}-x_2^{b-a})}/{(x_3^a-x_2^a)}$ . Since $a\mid b$ and $a\mid (b-a)$ imply that $\mathcal {G}\in \mathbb {Z}$ and $\mathcal {H}\in \mathbb {Z}$ , it follows immediately that ${[S^b]}[S^a]^{-1}\in M_3(\mathbb {Z})$ . The statement is true for this case.
Now let $n\ge 4$ . Assume that the statement is true for the $n-1$ case. In what follows, we show that the statement is true for the n case. Since S is gcd closed and $\max _{x\in S}\{|G_S(x)|\}=1$ , it follows that $S_1:=\{x_1,\ldots , x_{n-1}\}$ is also gcd closed and $\max _{x\in S_1}\{|G_{S_1}(x)|\}=1$ . Hence by the inductive hypothesis, $[S_1^b][S_1^a]^{-1}\in M_{n-1}( \mathbb {Z})$ . Finally, from Lemma 2.10, $[S^b][S^a]^{-1}\in M_n(\mathbb {Z})$ as desired.
This finishes the proof of Theorem 1.1. $\Box $
Acknowledgement
The authors would like to thank the anonymous referee for careful reading of the paper and helpful suggestions that improved its presentation.