\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{amscd} \usepackage{graphicx} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics} \usepackage{latexsym} \usepackage{epsf} \usepackage{breakurl} \def\lcm{\mathrm{lcm}} \newcommand{\R}{{\mathbb R}} \newcommand{\Q}{{\mathbb Q}} \newcommand{\C}{{\mathbb C}} \newcommand{\N}{{\mathbb N}} \newcommand{\Z}{{\mathbb Z}} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.1in} \setlength{\textheight}{8.4in} \newcommand{\seqnum}[1]{\href{https://oeis.org/#1}{\rm \underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \begin{center} \vskip 1cm{\LARGE\bf Nontrivial Effective Lower Bounds for the \\ \vskip .02in Least Common Multiple of Some \\ \vskip .1in Quadratic Sequences } \vskip 1cm \large Sid Ali Bousla and Bakir Farhi \\ Laboratoire de Math\'ematiques appliqu\'ees \\ Facult\'e des Sciences Exactes \\ Universit\'e de Bejaia \\ 06000 Bejaia \\ Algeria \\ \href{mailto:bouslasidali@gmail.com}{\tt bouslasidali@gmail.com} \\ \href{mailto:bakir.farhi@gmail.com}{\tt bakir.farhi@gmail.com} \end{center} \vskip .2in \begin{abstract} This paper is devoted to studying the numbers $$L_{c,m,n} := \lcm\left(m^2 + c , (m+1)^2 + c ,\right. \left.\dots , n^2 + c\right),$$ where $c,m,n$ are positive integers such that $m \leq n$. More precisely, we determine a nontrivial rational divisor of $L_{c,m,n}$ and then we derive (as consequences) some nontrivial lower bounds for $L_{c,m,n}$. Our approach (focusing on commutative algebra) is new and different from those using previously by Farhi, Oon, and Hong. \end{abstract} \section{Introduction and Notation} Throughout this paper, we let $\mathbb{N^*}$ denote the set $\mathbb{N}\setminus\lbrace 0\rbrace$ of positive integers. For $t\in\mathbb{R}$, we let $\lfloor t\rfloor$ and $\lceil t\rceil$ respectively denote the floor and the ceiling functions. We say that an integer $a$ is a multiple of a non-zero rational number $r$ (or equivalently, $r$ is a divisor of $a$) if the quotient $a / r$ is an integer. If $m,n,c$ are positive integers such that $m\leq n$, we set $L_{c,m,n}:=\lcm\left\lbrace m^2+c,(m+1)^2+c,\dots,n^2+c \right\rbrace$. For a given polynomial $P \in \C[X]$, we let $\overline{P}$ denote the polynomial conjugate of $P$ in $\mathbb{C}[X]$, that is the polynomial we get by replacing each coefficient of $P$ with its complex conjugate. It is well-known that the conjugation of polynomials in $\mathbb{C}[X]$ is compatible with addition and multiplication, in the sense that for every $P,Q\in\mathbb{C}[X]$, we have $\overline{P+Q}=\overline{P}+\overline{Q}$ and $\overline{P\cdot Q}=\overline{P}\cdot\overline{Q}$. Further, we let $I$, $E_h$ $(h\in\mathbb{R})$, and $\Delta$ denote the linear operators on $\mathbb{C}[X]$ which respectively represent the identity, the shift operator with step $h$ ($E_hP\left(X\right)=P\left(X+h\right)$, $\forall P\in\mathbb{C}[X]$), and the forward difference ($\Delta P\left(X\right)=P\left(X+1\right)-P\left(X\right)$, $\forall P\in\mathbb{C}[X]$). For $n\in\mathbb{N}$, the expression of $\Delta^n$ in terms of the $E_h$'s is easily obtained from the binomial formula, as follows \begin{equation}\label{dece} \Delta^n=\left(E_1-I\right)^n=\sum_{m=0}^{n}(-1)^{n-m}\binom{n}{m}E_{1}^m=\sum_{m=0}^{n}(-1)^{n-m}\binom{n}{m}E_{m}. \end{equation} For falling factorial powers, we use Knuth's notation: \[X^{\underline{n}}:=X\left(X-1\right)\left(X-2\right)\cdots\left(X-n+1\right)~~~~(\forall n\in\mathbb{N}).\] The study of the least common multiple of the first $n$ consecutive positive integers $(n\in\mathbb{N^*})$ began with Chebyshev's work \cite{cheb} in his attempts to prove the prime number theorem. The latter showed that the prime number theorem is equivalent to stating that $\log \lcm\left(1,2,\dots,n\right)\sim_{+\infty} n$. More recently, many authors are interested in the effective estimates of the least common multiple of consecutive terms of some integer sequences. In 1972, Hanson \cite{han} showed (by using the expansion of the number 1 in Sylvester series) that $\lcm\left(1,2,\dots,n\right)\leq 3^n$ $(\forall n\in\mathbb{N^*})$. In 1982, investigating the integral $\int_{0}^{1}x^n(1-x)^n\mathrm{d}x$, Nair \cite{nair} gave a simple proof that $\lcm\left(1,2,\dots,n\right)\geq 2^n$ $(\forall n\geq 7)$. Later, the second author \cite{far} obtained nontrivial lower bounds for the least common multiple of consecutive terms in an arithmetic progression. In particular, he proved that for any $u_0,r,n\in \mathbb{N^*}$ such that $\gcd\left(u_0,r\right)=1$, we have \begin{equation}\label{01} \lcm\left(u_0,u_0+r,\dots,u_0+nr\right)\geq u_0\left(r+1\right)^{n-1}, \end{equation} and conjectured that the exponent $(n-1)$ appearing in the right-hand side of \eqref{01} can be replaced by $n$, which is the optimal exponent that can be obtained. That conjecture was confirmed by Hong and Feng \cite{hong1}. Furthermore, several authors obtained improvements of \eqref{01} for $n$ sufficiently large in terms of $u_{0}$ and $r$ (see, e.g., \cite{hong1,hongkom,kane}). The second author \cite{far} also obtained nontrivial lower bounds for the least common multiple of some quadratic sequences. In particular, he proved that for any positive integer $n$, we have \begin{equation}\label{02+} \lcm\left(1^2+1,2^2+1,\dots,n^2+1\right)\geq 0.32(1.442)^n. \end{equation} In 2013, Oon \cite{oon} managed to improve \eqref{02+} by proving that for any positive integers $c$ and $n$, we have \begin{equation}\label{02} \lcm\left(1^2+c,2^2+c,\dots,n^2+c\right)\geq 2^n. \end{equation} Actually, we have something a little stronger: \begin{theorem}[Oon \cite{oon}]\label{03} Let $c,n,m$ be positive integers such that $m\leq \left\lceil \frac{n}{2}\right\rceil$. Then we have \[L_{c,m,n}\geq 2^n.\] \end{theorem} Later, Hong et al.\ \cite{hong2} managed to generalize Theorem \ref{03} for polynomial sequences $\left(f(n)\right)_{n\geq 1}$, with $f\in\mathbb{Z}[X]$ and the coefficients of $f$ are all nonnegative. In another direction, various asymptotic estimates have been obtained by several authors. For example, Bateman et al.\ \cite{bat} proved that for any $h,k\in\mathbb{Z}$ with $k>0$, $h+k>0$, and $\gcd(h,k)=1$, we have \begin{equation}\label{04} \log\lcm\lbrace h+k,h+2k,\dots,h+nk\rbrace ~\sim_{+ \infty}~ \left(\frac{k}{\varphi(k)} \sum_{\begin{subarray}{c} 1\leq m\leq k \\ \gcd(m , k)=1 \end{subarray}}\frac{1}{m}\right) n , \end{equation} where $\varphi$ denotes the Euler totient function. Another asymptotic estimate a little harder to prove is due to Cilleruelo \cite{cil} and states that for every irreducible quadratic polynomial $f\in\mathbb{Z}[X]$, we have \begin{equation}\label{05} \log \lcm\lbrace f(1),\dots,f(n)\rbrace = n\log n +Bn+o(n), \end{equation} where $B$ is a constant depending on $f$. In this paper, we use arguments of commutative algebra and complex analysis to find a nontrivial rational divisor of $L_{c,m,n}$ ($c , m , n \in \N^*$). As a consequence, we derive some new nontrivial lower bounds for $L_{c,m,n}$. The rest of the paper is organized in five parts (subsections). In the first part, we give an algebraic lemma which allows us, on the one hand to re-demonstrate Theorem \ref{03} of Oon by an easy and purely algebraic method, and on the other hand to reformulate the problem of bounding the number $L_{c,m,n}$ from below. In this reformulation, we are led to introduce a vital arithmetic function, noted $h_c$, whose multiple provides a divisor for $L_{c,m,n}$. In the next two parts, we study the arithmetic function $h_c$ and we find for $h_c$ a simple multiple. In the fourth part, we use the obtained multiple of $h_c$ to deduce a nontrivial divisor for $L_{c,m,n}$. Our new nontrivial lower bounds for $L_{c,m,n}$ then follow from this divisor. We conclude the paper with a short last part which presents comparisons between our results and those of Oon (cf.\ Theorem \ref{03}). \section{The results and the proofs} \subsection{An algebraic method} Although the method used by Oon \cite{oon} to obtain his result (i.e., Theorem \ref{03}) is analytic, the ingredients for its success are algebraic in depth, as we will show it below by applying the following fundamental algebraic lemma: \begin{lemma}\label{l1} Let $\mathcal{A}$ be an integral domain and $n$ be a positive integer. Let also $u_0,u_1,\dots,u_n$, $a,b$ be elements of $\mathcal{A}$. Suppose that $a$ and $b$ satisfy the following conditions: \begin{enumerate} \item Each of the elements $u_0,u_1,\dots,u_n$ of $\mathcal{A}$ divides $a$. \item Each of the elements $\prod_{\begin{subarray}{c}0\leq j\leq n \\ j\neq i\end{subarray}}\left(u_i-u_j\right)$ $(i=0,1,\dots,n)$ of $\mathcal{A}$ divides $b$. \end{enumerate} Then the product $ab$ is a multiple of the product $u_0u_1\cdots u_n$. \end{lemma} \begin{proof} If the elements $u_0,u_1,\dots,u_n$ of $\mathcal{A}$ are not pairwise distinct, the result of the lemma is trivial, since by its second condition we have $b=0_{\mathcal{A}}$. Suppose in what follows that the $u_i$'s $(i=0,1,\dots,n)$ are pairwise distinct. We use the well-known result that if a polynomial in one indeterminate, with coefficients in an integral domain, has a number of zeros (in that domain) greater than its degree then it is zero. Since $a$ is a multiple of each of the elements $u_0,u_1,\dots,u_n$ of $\mathcal{A}$, there exist $k_0,k_1,\dots,k_n\in \mathcal{A}$ such that: \begin{equation}\label{eq1} a=k_0u_0=k_1u_1=\dots=k_nu_n. \end{equation} Similarly, since $b$ is a multiple of each of the elements $\prod_{\begin{subarray}{c}0\leq j\leq n \\ j\neq i\end{subarray}}\left(u_i-u_j\right) (i=0,1,\dots,n)$, then there exist $\ell_0,\ell_1,\dots,\ell_n\in \mathcal{A}$ such that: \begin{equation}\label{eq2} b=\ell_i\prod_{\begin{subarray}{c}0\leq j\leq n \\ j\neq i\end{subarray}}\left(u_i-u_j\right)~~\left(\forall i\in\left\lbrace0,1,\dots,n\right\rbrace\right). \end{equation} Now, consider the following polynomial of $\mathcal{A}[X]$: \[P\left(X\right):=\sum_{i=0}^{n}\left(\ell_i\prod_{\begin{subarray}{c}0\leq j\leq n \\ j\neq i\end{subarray}}\left(X-u_j\right)\right)-b.\] We have $\deg P \leq n$. On the other hand, we have (according to \eqref{eq2}) \[P\left(u_i\right)=0~~\left(\forall i\in\left\lbrace0,1,\dots,n\right\rbrace\right),\] showing that the number of zeros of $P$ in $\mathcal{A}$ is greater than its degree. So, according to the elementary result of commutative algebra announced above, the polynomial $P$ is zero. In particular, we have $P\left(0\right)=0$; that is \[b=(-1)^n\sum_{i=0}^{n}\ell_i\left(\prod_{\begin{subarray}{c}0\leq j\leq n \\ j\neq i\end{subarray}}u_j\right).\] By multiplying the two sides of this last equality by $a$, we get (according to \eqref{eq1}) \begin{align*} ab&=(-1)^n\sum_{i=0}^{n}\ell_ia\left(\prod_{\begin{subarray}{c}0\leq j\leq n \\ j\neq i\end{subarray}}u_j\right)\\&=(-1)^n\sum_{i=0}^{n}\ell_ik_iu_i\left(\prod_{\begin{subarray}{c}0\leq j\leq n \\ j\neq i\end{subarray}}u_j\right)\\&=(-1)^n\left(\sum_{i=0}^{n}\ell_ik_i\right)u_0u_1\cdots u_n, \end{align*} showing that $ab$ is a multiple of $u_0u_1\cdots u_n$, as required. This completes the proof. \end{proof} \begin{remark} Lemma \ref{l1} is inspired by Farhi's result \cite[Theorem 2]{far} which becomes a special case for $\mathcal{A} = \Z$, $a = \lcm(u_0 , u_1 , \dots , u_n)$, and $b = \lcm(\prod_{ \begin{subarray}{c} 0 \leq j \leq n \\ j \neq i \end{subarray} } (u_i - u_j) ; i = 0 , 1 , \dots , n)$. It was precisely this special case which led the second author \cite{far} to establish the first nontrivial lower bounds for the least common multiples of arithmetic progressions. \end{remark} Now, we use Lemma \ref{l1} to establish a new proof of Theorem \ref{03}, which is purely algebraic. \begin{proof}[A new proof of Theorem \ref{03}] Since $L_{c,m,n}$ is non-increasing relative to $m$, then it suffices to prove the result of the theorem for $m=\left\lceil\frac{n}{2}\right\rceil$, that is $L_{c,\left\lceil\frac{n}{2}\right\rceil,n}\geq 2^n$. For simplicity, put $m_0=\left\lceil\frac{n}{2}\right\rceil$. So, we have to show that $L_{c,m_0,n}\geq 2^n$. For $n\in\left\lbrace 1,2,\dots,6\right\rbrace$, this can be easily checked by hand (as was done by Oon). Suppose in what follows that $n\geq 7$. It is well-known and easily proved that for any integer $r\geq 7$, we have $\left\lceil \frac{r}{2}\right\rceil\binom{r}{\left\lceil \frac{r}{2}\right\rceil}\geq 2^{r}$. According to this inequality for $r=n$, it suffices to show that $L_{c,m_0,n}\geq m_0\binom{n}{m_0}$. More generally, we shall show that: \begin{equation}\label{cor2} L_{c,m',n}\geq m'\binom{n}{m'}~~~~(\forall m'\in\mathbb{N^*},~m'\leq n). \end{equation} Let $m'\in\mathbb{N^*}$ such that $m'\leq n$. To prove \eqref{cor2}, we apply Lemma \ref{l1} for $\mathcal{A}=\mathbb{Z}[\sqrt{-c}]$ by taking for the $u_i$'s the elements $m'+\sqrt{-c},m'+1+\sqrt{-c},\dots,n+\sqrt{-c}$ of $\mathcal{A}$ and for $a$ and $b$ the integers $a=L_{c,m',n}$ and $b=(n-m')!$. For any $k\in\left\lbrace m',m'+1,\dots,n\right\rbrace$, Since $L_{c,m',n}$ is a multiple (in $\mathbb{Z}$, so also in $\mathcal{A}=\mathbb{Z}[\sqrt{-c}]$) of $(k^2+c)$ and $k^2+c=\left(k+\sqrt{-c}\right)\left(k-\sqrt{-c}\right)$ is a multiple (in $\mathbb{Z}[\sqrt{-c}]$) of $k+\sqrt{-c}$, then $L_{c,m',n}$ is a multiple (in $\mathbb{Z}[\sqrt{-c}]$) of $k+\sqrt{-c}$. This shows that the first condition of Lemma \ref{l1} is satisfied. On the other hand, we have for all $k\in\left\lbrace m',m'+1,\dots,n\right\rbrace$: \[ \prod_{\begin{subarray}{c}m'\leq \ell\leq n \\ \ell\neq k\end{subarray}}\left(\left(k+\sqrt{-c}\right)-\left(\ell+\sqrt{-c}\right)\right) = \prod_{\begin{subarray}{c}m'\leq \ell\leq n\\ \ell\neq k\end{subarray}}(k-\ell)=(-1)^{n-k}(k-m')!(n-k)!,\] which divides (in $\mathbb{Z}$, so also in $\mathbb{Z}[\sqrt{-c}]$) the integer $(n-m')!$ (since $\frac{(n-m')!}{(k-m')!(n-k)!}=\binom{n-m'}{k-m'}\in\mathbb{Z}$). This shows that the second condition of Lemma \ref{l1} is also satisfied. We thus deduce (by applying Lemma \ref{l1}) that $L_{c,m',n}(n-m')!$ is a multiple (in $\mathbb{Z}[\sqrt{-c}]$) of $\prod_{k=m'}^{n}\left(k+\sqrt{-c}\right)$. So, there exist $x,y\in\mathbb{Z}$ such that: \begin{equation}\label{**} L_{c,m',n}(n-m')!=\left(x+y\sqrt{-c}\right)\prod_{k=m'}^{n}\left(k+\sqrt{-c}\right).\end{equation} Therefore, by taking the absolute value in $\mathbb{C}$ on both sides, we get \[L_{c,m',n}(n-m')!=\sqrt{x^2+cy^{2}}\prod_{k=m'}^{n}\sqrt{k^2+c}.\] Next, since $x^2+cy^2\in\mathbb{N}$ and $x^2+cy^2\neq 0$ (because $x^2+cy^2=0$ $\Longrightarrow$ $L_{c,m',n}=0$, which is false) then $x^2+cy^2\geq 1$. Hence \[L_{c,m',n}=\frac{\sqrt{x^2+cy^2}\prod_{k=m'}^{n}\sqrt{k^2+c}}{(n-m')!}\geq \frac{\prod_{k=m'}^{n}\sqrt{k^2+c}}{(n-m')!}\geq \frac{\prod_{k=m'}^{n}k}{(n-m')!}=m'\binom{n}{m'},\] as required. This completes the proof of the theorem. \end{proof} Naturally, we have the following question: \begin{quote} \textit{How could we improve the lower bound $L_{c,m,n}\geq \frac{\prod_{k=m}^{n}\sqrt{k^2+c}}{(n-m)!}$, obtained during the proof of Theorem \ref{03} and initially established by Oon \cite{oon}?} \end{quote} To simplify, suppose that $c=1$ and let $m,n\in\mathbb{N^*}$ such that $m\leq n$. According to Formula \eqref{**}, the positive integer $L_{1,m,n}(n-m)!$ is a multiple (in $\mathbb{Z}[i]$) of the Gauss integer \mbox{$\prod_{k=m}^{n}(k+i)$}. Next, by taking the conjugates (in $\mathbb{C}$) of both sides of \eqref{**}, we obtain that $L_{1,m,n}(n-m)!$ is also a multiple (in $\mathbb{Z}[i]$) of the Gauss integer $\prod_{k=m}^{n}(k-i)$. It follows from those two facts that $L_{1,m,n}(n-m)!$ is a multiple (in $\mathbb{Z}[i]$) of\footnote{Let $\mathcal{A}$ be an integral domain and let $a , b \in \mathcal{A}$. Then an element $d$ of $\mathcal{A}$ is called a greatest common divisor of $a$ and $b$ (and denoted by $\mathrm{gcd}_{\mathcal{A}}(a , b)$) if $d$ divides both $a$ and $b$ and if any other $d' \in \mathcal{A}$, which divides both $a$ and $b$, also divides $d$. Similarly, an element $m$ of $\mathcal{A}$ is called a least common multiple of $a$ and $b$ (and denoted by $\mathrm{lcm}_{\mathcal{A}}(a , b)$) if $m$ is a multiple of both $a$ and $b$ and if any other $m' \in \mathcal{A}$, which is a multiple of both $a$ and $b$, is also a multiple of $m$. Note that $\mathrm{gcd}_{\mathcal{A}}(a , b)$ and $\mathrm{lcm}_{\mathcal{A}}(a , b)$ exist at least when $\mathcal{A}$ is a unique factorization domain (which is the case of $\mathbb{Z}[i]$) and they are unique up to a multiplication by a unit.} \begin{align*} {\lcm}_{\mathbb{Z}[i]}\left(\prod_{k=m}^{n}(k+i),\prod_{k=m}^{n}(k-i)\right) &=\frac{\prod_{k=m}^{n}(k+i)\cdot\prod_{k=m}^{n}(k-i)}{{\gcd}_{\mathbb{Z}[i]}\left(\prod_{k=m}^{n}(k+i),\prod_{k=m}^{n}(k-i)\right)} \\ &= \frac{\prod_{k=m}^{n}(k^2+1)}{{\gcd}_{\mathbb{Z}[i]}\left(\prod_{k=m}^{n}(k+i),\prod_{k=m}^{n}(k-i)\right)}. \end{align*} Consequently \begin{equation}\label{eq4} L_{1,m,n}\geq \frac{\prod_{k=m}^{n}(k^2+1)}{(n-m)!\left|{\gcd}_{\mathbb{Z}[i]}\left(\prod_{k=m}^{n}(k+i),\prod_{k=m}^{n}(k-i)\right)\right|}. \end{equation} Remarkably, the trivial upper bound \[ \left|{\gcd}_{\mathbb{Z}[i]}\left(\prod_{k=m}^{n}(k+i),\prod_{k=m}^{n}(k-i)\right)\right|\leq \left|\prod_{k=m}^{n}(k+i)\right|\leq \prod_{k=m}^{n}\sqrt{k^2+1} \] suffices to establish the Oon lower bound $L_{1,m,n}\geq \frac{\prod_{k=m}^{n}\sqrt{k^2+1}}{(n-m)!}$. So, a nontrivial upper bound for the number $\left|{\gcd}_{\mathbb{Z}[i]}\left(\prod_{k=m}^{n}(k+i),\prod_{k=m}^{n}(k-i)\right)\right|$ certainly gives an improvement of the Oon theorem. On the other hand, for $a,b\in\mathbb{Z}$ such that $(a,b)\neq (0,0)$, we can easily check that ${\gcd}_{\mathbb{Z}[i]}\left(a+bi,a-bi\right)$ is not far from ${\gcd}_{\mathbb{Z}}(a,b)$. Precisely, we have \[{\gcd}_{\mathbb{Z}[i]}\left(a+bi,a-bi\right)=\left(\sigma + i \tau\right){\gcd}_{\mathbb{Z}}(a,b),\] where $\sigma , \tau \in \left\lbrace -1,0,1\right\rbrace$ and $(\sigma , \tau) \neq (0,0)$. So, for the case $c=1$, we are led to study the arithmetic function: \[\begin{array}{rcl}h :~\mathbb{Z}[i]\setminus\left\lbrace 0\right\rbrace &\longrightarrow &\mathbb{N^{*}}\\a+bi ~~& \longmapsto &\gcd (a,b)\end{array},\] and precisely to find nontrivial upper bounds for the quantities $h\left(\prod_{k=m}^{n}(k+i)\right)$ $(m,n\in\mathbb{N^*},~m\leq n)$. For the general case $(c\in\mathbb{N^*})$, the arithmetic function we need to study is given by: \[\begin{array}{rcl}h_c :~\mathbb{Z}[\sqrt{-c}]\setminus\left\lbrace 0\right\rbrace &\longrightarrow &\mathbb{N^{*}}\\a+b\sqrt{-c}~~& \longmapsto &\gcd (a,b)\end{array}\] and the quantities we need to bound from above are $h_c\left(\prod_{k=m}^{n}(k+\sqrt{-c})\right)$ $(m,n\in\mathbb{N^*},$ $m \leq n)$. The following proposition has as objective to replace a specific arithmetic language of the ring $\mathbb{Z}[\sqrt{-c}]$ by its simpler analog in $\mathbb{Z}$. \begin{proposition}\label{p1} Let $c\in\mathbb{N^*}$ and $N,a,b\in\mathbb{Z}$, with $(a,b)\neq (0,0)$. Then $N$ is a multiple $($in $\mathbb{Z}[\sqrt{-c}])$ of $\left(a+b\sqrt{-c}\right)$ if and only if $N$ is a multiple $($in $\mathbb{Z})$ of $\frac{a^2+cb^2}{\gcd(a,b)}$. \end{proposition} \begin{proof} The result of the proposition is trivial for $b=0$. Suppose in what follows that $b\neq 0$. Suppose that $N$ is a multiple (in $\mathbb{Z}[\sqrt{-c}]$) of $\left(a+b\sqrt{-c}\right)$; that is there exist $x,y\in\mathbb{Z}$ such that: \[N=\left(x+y\sqrt{-c}\right)\left(a+b\sqrt{-c}\right).\] By identifying the real and imaginary parts of the two hand-sides of this equality, we get \begin{align} N&=ax-byc, \label{eq5} \\ 0&=bx+ay. \label{eq6} \end{align} Next, putting $d:=\gcd(a,b)$, there exist $a',b'\in\mathbb{Z}$, with $b'\neq 0$ and $\gcd(a',b')=1$, such that $a=da'$ and $b=db'$. By substituting these in \eqref{eq6}, we obtain (after simplifying) \begin{equation}\label{eq7} b'x=-a'y. \end{equation} This last equality shows that $b'$ divides $a'y$. But since $\gcd(a',b')=1$, then (according to the Gauss lemma) $b'$ divides $y$. So there exists $k\in\mathbb{Z}$ such that $y=kb'$. By reporting this in \eqref{eq7}, we get $x=-ka'$. Then by substituting $x=-ka'=-k\frac{a}{d}$ and $y=kb'=k\frac{b}{d}$ in \eqref{eq5}, we finally obtain \[N=-k\frac{a^2+cb^2}{d}=-k\frac{a^2+cb^2}{\gcd(a,b)},\] showing that $N$ is a multiple (in $\mathbb{Z}$) of $\frac{a^2+cb^2}{\gcd(a,b)}$, as required. Conversely, suppose that $N$ is a multiple (in $\mathbb{Z}$) of $\frac{a^2+cb^2}{\gcd(a,b)}$. Then there exists $k\in\mathbb{Z}$ such that: \[N=k\frac{a^2+cb^2}{\gcd(a,b)}=k\frac{a-b\sqrt{-c}}{\gcd(a,b)}\left(a+b\sqrt{-c}\right)=\left(k\frac{a}{\gcd(a,b)}-k\frac{b}{\gcd(a,b)}\sqrt{-c}\right)\left(a+b\sqrt{-c}\right).\] Since $\left(k\frac{a}{\gcd(a,b)}-k\frac{b}{\gcd(a,b)}\sqrt{-c}\right)\in\mathbb{Z}[\sqrt{-c}]$, the last equality shows that $N$ is a multiple (in $\mathbb{Z}[\sqrt{-c}]$) of $\left(a+b\sqrt{-c}\right)$, as required. This completes the proof of the proposition. \end{proof} From Proposition \ref{p1}, we derive the following corollary, which is the first key step to obtaining the results of this paper. \begin{corollary}\label{p2} Let $c,m,n\in\mathbb{N^*}$ such that $m\leq n$. Then the positive integer $L_{c,m,n}(n-m)!$ is a multiple $($in $\mathbb{Z})$ of the positive integer: \[\frac{\prod_{k=m}^{n}\left(k^2+c\right)}{h_c\left(\prod_{k=m}^{n}\left(k+\sqrt{-c}\right)\right)}.\] \end{corollary} \begin{proof} Formula \eqref{**} (obtained during our new proof of Theorem \ref{03}) shows that $L_{c,m,n}(n-m)!$ is a multiple (in $\mathbb{Z}[\sqrt{-c}]$) of $\prod_{k=m}^{n}\left(k+\sqrt{-c}\right)$. But, according to Proposition \ref{p1}, this last property is equivalent to the statement of the corollary. \end{proof} In view of Corollary \ref{p2}, to bound from below $L_{c,m,n}$ $(c,m,n\in\mathbb{N^*},~m\leq n)$, it suffices to bound from above $h_c\left(\prod_{k=m}^{n}(k+\sqrt{-c})\right)$. Likewise, to find a nontrivial (rational) divisor of $L_{c,m,n}$, it suffices to find a nontrivial multiple of $h_c\left(\prod_{k=m}^{n}(k+\sqrt{-c})\right)$. This is what we will do in what follows. \subsection{An explicit B\'ezout identity}\label{sub22} In the following, let $c\in\mathbb{N^*}$ and $k\in\mathbb{N}$ be fixed and define \begin{align*} P_{k}\left(X\right)&:=\left(X+\sqrt{-c}\right)\left(X-1+\sqrt{-c}\right)\cdots \left(X-k+\sqrt{-c}\right):=A_{k}\left(X\right)+B_{k}\left(X\right)\sqrt{-c}, \\ \overline{P_{k}}\left(X\right)&:=\left(X-\sqrt{-c}\right)\left(X-1-\sqrt{-c}\right)\cdots \left(X-k-\sqrt{-c}\right):=A_{k}\left(X\right)-B_{k}\left(X\right)\sqrt{-c}, \end{align*} where we understand that $A_{k},B_{k}\in\mathbb{Z}[X]$. In what follows, we find nontrivial multiples for the positive integers $h_{c}\left(P_{k}(n)\right)=\gcd\left(A_{k}(n),B_{k}(n)\right)$ $(n\geq 1)$. To do so, we look for two polynomial sequences $\left(a_{k}(n)\right)_{n}$ and $\left(b_{k}(n)\right)_{n}$ so that the polynomial sequence $\left(a_{k}(n)A_{k}(n)+b_{k}(n)B_{k}(n)\right)_{n}$ be independent on $n$. This leads to looking for two polynomials $U_{k},V_{k}\in\mathbb{Q}[X]$ which satisfy the B\'ezout identity: \[U_{k}\left(X\right)A_{k}\left(X\right)+V_{k}\left(X\right)B_{k}\left(X\right)=1.\] Next, since $A_{k}=\frac{P_{k}+\overline{P_{k}}}{2}$ and $B_{k}=\frac{P_{k}-\overline{P_{k}}}{2\sqrt{-c}}$, the problem is equivalent to looking for $\sigma_{k},\tau_{k} \in \mathbb{Q}\left(\sqrt{-c}\right)[X]$ such that: \[\sigma_{k}\left(X\right)P_{k}\left(X\right)+\tau_{k}\left(X\right)\overline{P_{k}}\left(X\right)=1.\] Let us first justify the existence of such $\sigma_k$ and $\tau_k$. Denoting by $Z\left(P\right)$ the set of all the complex zeros of a polynomial $P\in\mathbb{C}[X]$, we have \[Z\left(P_{k}\right)=\left\lbrace -\sqrt{-c},1-\sqrt{-c},\dots,k-\sqrt{-c}\right\rbrace ~\text{and}~Z\left(\overline{P_{k}}\right)=\left\lbrace \sqrt{-c},1+\sqrt{-c},\dots,k+\sqrt{-c}\right\rbrace,\] showing that $Z\left(P_{k}\right)\cap Z\left(\overline{P_{k}}\right)=\emptyset$; that is $P_{k}$ and $\overline{P_{k}}$ do not have a common zero in $\mathbb{C}$. This implies that $P_{k}$ and $\overline{P_{k}}$ are coprime in $\mathbb{C}[X]$; so coprime also in $\mathbb{Q}\left(\sqrt{-c}\right)[X]$ (since $P_{k},\overline{P_{k}}\in\mathbb{Q}\left(\sqrt{-c}\right)[X]$). It follows (according to B\'ezout's theorem) that there exist $\sigma_{k},\tau_{k}\in\mathbb{Q}\left(\sqrt{-c}\right)[X]$ such that: $\sigma_{k}P_{k}+\tau_{k}\overline{P_{k}}=1$, as required. Now, to find explicitly such $\sigma_k$ and $\tau_k$, we need the following more precise version of B\'ezout's theorem: \begin{theorem}\label{p3} Let $\mathbb{K}$ be a field and $P$ and $Q$ be two non-constant polynomials of $\mathbb{K}[X]$ such that ${\gcd}_{\mathbb{K}[X]}\left(P,Q\right)=1$. Then there exists a unique pair $\left(U,V\right)$ of polynomials of $\mathbb{K}[X]$, with $\deg U<\deg Q$ and $\deg V<\deg P$, such that: \[PU+QV=1.\] \end{theorem} \begin{proof} Since ${\gcd}_{\mathbb{K}[X]}\left(P,Q\right)=1$, then (according to B\'ezout's theorem) there exist $U_0,V_0\in\mathbb{K}[X]$ such that: \[PU_0+QV_0=1.\] Next, consider in $\mathbb{K}[X]$ the euclidean division of $U_0$ by $Q$ and the euclidean division of $V_0$ by $\left(-P\right)$: \begin{align*} U_0 &= U_1Q+U , \\ V_0 &= V_1\left(-P\right)+V, \end{align*} where $U_1,V_1,U,V\in\mathbb{K}[X]$, $\deg U<\deg Q$, and $\deg V<\deg \left(-P\right)=\deg P$. So, we have $$ PU+QV = P\left(U_0-U_1Q\right)+Q\left(V_0+V_1P\right) = PQ\left(V_1-U_1\right)+PU_0+QV_0 = PQ\left(V_1-U_1\right)+1 . $$ If $V_1-U_1\neq 0$, then the last equality implies that $\deg\left(PU+QV\right)\geq \deg\left(PQ\right)$, which is impossible, since $\deg U<\deg Q$ and $\deg V<\deg P$. Thus $V_1-U_1=0$, which gives $PU+QV=1$. The existence of the pair $\left(U,V\right)$ as required by the theorem is proved. It remains to prove the uniqueness of $\left(U,V\right)$. Let $\left(U_*,V_*\right)$ another pair of polynomials of $\mathbb{K}[X]$, with $\deg U_* <\deg Q$, $\deg V_* <\deg P$, and $PU_*+QV_*=1$, and let us prove that $\left(U_*,V_*\right)=\left(U,V\right)$. We have \begin{align*} P\left(UV_*-U_*V\right)=\left(PU\right)V_*-\left(PU_*\right)V=\left(1-QV\right)V_*-\left(1-QV_*\right)V=V_*-V, \end{align*} showing that the polynomial $\left(V_*-V\right)$ is a multiple of $P$ in $\mathbb{K}[X]$. But since $\deg\left(V_*-V\right)<\deg P$ (because $\deg V<\deg P$ and $\deg V_*<\deg P$), we have inevitably $V_*-V=0$; hence $V_*=V$. Using this, we get $PU_*=1-QV_*=1-QV=PU$. Thus $U_*=U$. Consequently, we have $\left(U_*,V_*\right)=\left(U,V\right)$, as required. This completes the proof of the theorem. \end{proof} In our context, the application of Theorem \ref{p3} gives the following corollary: \begin{corollary}\label{c1} There exists a unique polynomial $\sigma_k\in\mathbb{C}[X]$, with degree $\leq k$, such that: \[\sigma_{k}P_{k}+\overline{\sigma_{k}}\overline{P_{k}}=1.\] \end{corollary} \begin{proof} According to Theorem \ref{p3} (applied for $\mathbb{K}=\mathbb{C}$ and $\left(P,Q\right)=\left(P_k,\overline{P_k}\right)$), there exists a unique pair $\left(\sigma_k,\tau_k\right)$ of polynomials of $\mathbb{C}[X]$, with $\deg \sigma_k < \deg \overline{P_k}=k+1$ and $\deg \tau_k < \deg P_k = k+1$, such that $\sigma_{k}P_{k}+\tau_{k}\overline{P_{k}}=1$. By taking the conjugates in $\mathbb{C}[X]$ of both sides of the last equality, we derive that $\overline{\sigma_{k}}\overline{P_{k}}+\overline{\tau_{k}}P_{k}=1$, that is $\overline{\tau_{k}}P_{k}+\overline{\sigma_{k}}\overline{P_{k}}=1$. Since $\deg \overline{\tau_k}=\deg \tau_k < k+1$ and $\deg \overline{\sigma_k}=\deg \sigma_k < k+1$, this shows that the pair $\left(\overline{\tau_k},\overline{\sigma_k}\right)$ satisfies the property that characterizes $\left(\sigma_k,\tau_k\right)$. Thus $\left(\overline{\tau_k},\overline{\sigma_k}\right)=\left(\sigma_k,\tau_k\right)$, that is $\tau_k=\overline{\sigma_k}$. Consequently, we have $\sigma_{k}P_{k}+\overline{\sigma_{k}}\overline{P_{k}}=1$. This completes the proof of the corollary. \end{proof} Now, we are going to determine the explicit expression of the polynomial $\sigma_k$ announced by Corollary \ref{c1}. By replacing, in the identity $\sigma_{k}\left(X\right)P_{k}\left(X\right)+\overline{\sigma_{k}}\left(X\right)\overline{P_{k}}\left(X\right)=1$, the indeterminate $X$ by the numbers $s+\sqrt{-c}$ $(s=0,1,\dots,k)$, we get \begin{equation}\label{dec1} \sigma_k\left(s+\sqrt{-c}\right)=\frac{1}{P_k\left(s+\sqrt{-c}\right)}~~(\forall s\in\left\lbrace 0,1,\dots,k\right\rbrace). \end{equation} (since $\overline{P_{k}}\left(s+\sqrt{-c}\right)=0$ for $s=0,1,\dots,k$). So the values of $\sigma_{k}$ are known for $(k+1)$ equidistant points with distance $1$. Since $\deg\sigma_k\leq k$, this is sufficient to determine the expression of $\sigma_{k}\left(X\right)$ by using for example the Newton forward interpolation formula. Doing so, we obtain that: \[\sigma_k\left(X\right)=\sum_{\ell=0}^{k}\frac{\left(\Delta^{\ell}\sigma_k\right)\left(\sqrt{-c}\right)}{\ell!}\left(X-\sqrt{-c}\right)^{\underline{\ell}}.\] Then by using \eqref{dece}, we derive that: \begin{align*} \sigma_{k}\left(X\right)&=\sum_{\ell=0}^{k}\sum_{j=0}^{\ell}\frac{(-1)^{\ell-j}}{\ell!}\binom{\ell}{j}\sigma_{k}\left(j+\sqrt{-c}\right)\left(X-\sqrt{-c}\right)^{\underline{\ell}}\\ &=\sum_{\ell=0}^{k}\left(\frac{1}{\ell!}\sum_{j=0}^{\ell}(-1)^{\ell-j}\binom{\ell}{j}\sigma_{k}\left(j+\sqrt{-c}\right)\right)\left(X-\sqrt{-c}\right)^{\underline{\ell}}\\ &=\sum_{\ell=0}^{k}\left(\frac{1}{\ell!}\sum_{j=0}^{\ell}(-1)^{\ell-j}\binom{\ell}{j}\frac{1}{P_{k}\left(j+\sqrt{-c}\right)}\right)\left(X-\sqrt{-c}\right)^{\underline{\ell}} \end{align*} (according to \eqref{dec1}). So, by setting for all $\ell \in\left\lbrace 0,1,\dots,k\right\rbrace$: \begin{equation}\label{dec2} \Theta_{k,\ell}:=\frac{1}{\ell!}\sum_{j=0}^{\ell}(-1)^{\ell-j}\binom{\ell}{j}\frac{1}{P_{k}(j+\sqrt{-c})}, \end{equation} we get \begin{equation}\label{dec3} \sigma_{k}\left(X\right)=\sum_{\ell=0}^{k}\Theta_{k,\ell}\left(X-\sqrt{-c}\right)^{\underline{\ell}}. \end{equation} It remains to simplify the expressions of the numbers $\Theta_{k,\ell}$ $(0\leq\ell\leq k)$. To do so, we introduce the functions $R_{k,\ell}$ $(0\leq\ell\leq k)$, defined by: \begin{equation}\label{dec4} R_{k,\ell}(z):=\frac{1}{\ell!}\sum_{j=0}^{\ell}(-1)^{\ell-j}\binom{\ell}{j}\frac{1}{P_{k}(z+j+\sqrt{-c})}, \end{equation} so that we have \begin{equation}\label{dec5} \Theta_{k,\ell}=R_{k,\ell}(0)~~~~(\forall \ell\in\left\lbrace 0,1,\dots,k\right\rbrace). \end{equation} Since the $R_{k , \ell}$'s ($0 \leq \ell \leq k$) are all rational functions, each of them is holomorphic on its domain of definition, that is each $R_{k , \ell}$ ($0 \leq \ell \leq k$) is holomorphic on $$ D_{\ell} := \C \setminus \left\{j - 2 \sqrt{- c} ~;~ j \in \Z \text{ and} - \ell \leq j \leq k\right\} . $$ Consequently, the common domain of holomorphy of the functions $R_{k,\ell}$ $(0\leq\ell\leq k)$ is the open connected region $D$ of $\mathbb{C}$, given by: \[D := \bigcap_{0 \leq \ell \leq k} D_{\ell} = \mathbb{C}\setminus\lbrace j - 2 \sqrt{-c} ~;~ j \in \Z \text{ and} -k\leq j\leq k\rbrace.\] Using the principle of analytical continuation together with the theory of the gamma and beta functions, we can find another expression of $R_{k,\ell}$ $(0\leq\ell\leq k)$, which is simpler than the above. We have the following proposition: \begin{proposition}\label{pp} For all $\ell \in \mathbb{N}$, with $\ell\leq k$, and all $z\in D$, we have \begin{equation}\label{dec6} R_{k,\ell}(z)=\frac{(-1)^{k+\ell}}{z+2\sqrt{-c}}\binom{k+\ell}{\ell}\frac{1}{\left(k-2\sqrt{-c}-z\right)^{\underline{k}}\left(\ell+2\sqrt{-c}+z\right)^{\underline{\ell}}}. \end{equation} \end{proposition} \begin{proof} Let $\ell \in \mathbb{N}$ such that $\ell \leq k$. According to the principle of analytical continuation, it suffices to prove Formula \eqref{dec6} for $z\in\mathbb{C}$, such that $\Re(z)>k$. For a such $z$, we have \begin{align*} R_{k,\ell}(z) &:= \frac{1}{\ell!}\sum_{j=0}^{\ell}(-1)^{\ell-j}\binom{\ell}{j}\frac{1}{P_{k}(z+j+\sqrt{-c})} \\ &= \frac{1}{\ell!}\sum_{j=0}^{\ell}(-1)^{\ell-j}\binom{\ell}{j}\frac{1}{\left(z+j+2\sqrt{-c}\right)\left(z+j-1+2\sqrt{-c}\right)\cdots\left(z+j-k+2\sqrt{-c}\right)} \\ &= \frac{1}{\ell!}\sum_{j=0}^{\ell}(-1)^{\ell-j}\binom{\ell}{j}\frac{\Gamma\left(z+j-k+2\sqrt{-c}\right)}{\Gamma\left(z+j+1+2\sqrt{-c}\right)} \\ &= \frac{1}{\ell!}\sum_{j=0}^{\ell}(-1)^{\ell-j}\binom{\ell}{j}\frac{1}{k!}\beta\left(z+j-k+2\sqrt{-c},k+1\right) \\ &= \frac{1}{k!\ell!}\sum_{j=0}^{\ell}\left((-1)^{\ell-j}\binom{\ell}{j}\int_{0}^{1}t^{z+j-k-1+2\sqrt{-c}}(1-t)^{k}\mathrm{d}t\right) \\ &= \frac{1}{k!\ell!}\int_{0}^{1} t^{z-k-1+2\sqrt{-c}}(1-t)^{k}\left(\sum_{j=0}^{\ell}(-1)^{\ell-j}\binom{\ell}{j}t^{j}\right) \mathrm{d}t \\ &=\frac{1}{k!\ell!}\int_{0}^{1} t^{z-k-1+2\sqrt{-c}}(1-t)^{k}\left(t-1\right)^{\ell} \mathrm{d}t \\ &=\frac{(-1)^{\ell}}{k!\ell!}\int_{0}^{1} t^{z-k-1+2\sqrt{-c}}(1-t)^{k+\ell}\mathrm{d}t \\ &=\frac{(-1)^\ell}{k!\ell!}\beta\left(z-k+2\sqrt{-c},k+\ell+1\right)\\ &=\frac{(-1)^\ell}{k!\ell!}\frac{\Gamma\left(z-k+2\sqrt{-c}\right)\Gamma\left(k+\ell+1\right)}{\Gamma\left(z+\ell+1+2\sqrt{-c}\right)}\\ &=(-1)^\ell\binom{k+\ell}{\ell}\frac{1}{\left(z+\ell+2\sqrt{-c}\right)\left(z+\ell-1+2\sqrt{-c}\right)\cdots\left(z-k+2\sqrt{-c}\right)} \\ &= \frac{(-1)^{k+\ell}}{z+2\sqrt{-c}}\binom{k+\ell}{\ell}\frac{1}{\left(k-2\sqrt{-c}-z\right)^{\underline{k}}\left(\ell+2\sqrt{-c}+z\right)^{\underline{\ell}}} , \end{align*} as required. This completes the proof. \end{proof} From Proposition \ref{pp}, we immediately derive a simpler explicit expression of $\sigma_k\left(X\right)$. We have the following corollary: \begin{corollary}\label{jan1} We have \[\sigma_{k}\left(X\right)=\frac{1}{2\sqrt{-c}\left(k-2\sqrt{-c}\right)^{\underline{k}}}\sum_{\ell=0}^{k}\frac{(-1)^{k+\ell}\binom{k+\ell}{\ell}}{\left(\ell+2\sqrt{-c}\right)^{\underline{\ell}}}\left(X-\sqrt{-c}\right)^{\underline{\ell}}.\] \end{corollary} \begin{proof} This immediately follows from Formulas \eqref{dec3}, \eqref{dec5}, and \eqref{dec6}. \end{proof} \subsection{Nontrivial multiples of some values of $h_c$} In this subsection, we preserve the notation of Subsection \ref{sub22}. From Corollary \ref{jan1}, we derive the following theorem: \begin{theorem}\label{jan2} For all $c,n,m\in\mathbb{N^*}$, with $m\leq n$, we have \[h_c\left(\prod_{\ell=m}^{n}\left(\ell+\sqrt{-c}\right)\right)~~\text{divides}~~c\prod_{\ell=1}^{n-m}(\ell^2+4c).\] \end{theorem} \begin{proof} Let $c,n,m\in\mathbb{N^*}$, with $m\leq n$. Putting $k:=n-m\in\mathbb{N}$ and $d:=c\prod_{\ell=1}^{n-m}(\ell^2+4c)\in\mathbb{N^*}$, we have $\prod_{\ell=m}^{n}\left(\ell+\sqrt{-c}\right)=P_k(n)$; so, we have to show that $h_c\left(P_k(n)\right)$ divides $d$. By noting that $2d=\sqrt{-c}\cdot 2\sqrt{-c}\left(k-2\sqrt{-c}\right)^{\underline{k}}\left(k+2\sqrt{-c}\right)^{\underline{k}}$, we derive from Corollary \ref{jan1} that $2d\sigma_k\in\mathbb{Z}[\sqrt{-c}][X]$. So, there exist $r_k,s_k\in\mathbb{Z}[X]$ such that: \[2d\sigma_k\left(X\right)=r_k\left(X\right)+s_k\left(X\right)\sqrt{-c}.\] Next, the identity of polynomials $\sigma_kP_k+\overline{\sigma_k}\overline{P_k}=1$ (given by Corollary \ref{c1}) implies that $2d\sigma_k\cdot P_k+\overline{2d\sigma_k}\cdot \overline{P_k}=2d$. By substituting in this last equality $P_k$ by $\left(A_k+B_k\sqrt{-c}\right)$ and $2d\sigma_k$ by $\left(r_k+s_k\sqrt{-c}\right)$, we obtain (in particular) that: \[r_kA_k-cs_kB_k=d,\] implying that ${\gcd}_{\mathbb{Z}[X]}\left(A_k,B_k\right)$ divides $d$. We then conclude that ${\gcd}_{\mathbb{Z}}\left(A_k(n),B_k(n)\right)=h_c\left(P_k(n)\right)$ divides $d$, as required. \end{proof} \subsection{New estimates for the number $L_{c,m,n}$} We have the following theorem: \begin{theorem}\label{t7} Let $c,m,n\in\mathbb{N^*}$ such that $m\leq n$. Then: \begin{enumerate} \item The positive integer $L_{c,m,n}$ is a multiple of the rational number \[\frac{\displaystyle\prod_{k=m}^{n}\left(k^2+c\right)}{c\cdot (n-m)!\displaystyle\prod_{k=1}^{n-m}\left(k^2+4c\right)}.\] \item We have \[L_{c,m,n}\geq \lambda_1(c) \cdot m^2\frac{n!^2}{m!^2(n-m)!^3},\] where $\lambda_1(c):=e^{-\frac{2\pi^{2}}{3} c}/c$. \end{enumerate} \end{theorem} \begin{proof} The first point of the theorem is an immediate consequence of Corollary \ref{p2} and Theorem \ref{jan2}. Next, using the well-known inequality $1 + x \leq e^x$ ($\forall x \in \R$), we have \begin{eqnarray*} \prod_{k = 1}^{n - m} \left(k^2 + 4 c\right) & = & \prod_{k = 1}^{n - m} k^2 \left(1 + \frac{4 c}{k^2}\right) \\ & = & (n - m)!^2 \prod_{k = 1}^{n - m} \left(1 + \frac{4 c}{k^2}\right) \\ & \leq & (n - m)!^2 \prod_{k = 1}^{n - m} e^{\frac{4 c}{k^2}} \\ & \leq & (n - m)!^2 \prod_{k = 1}^{+ \infty} e^{\frac{4 c}{k^2}} \\ & = & (n - m)!^2 e^{4 c \sum_{k = 1}^{+ \infty} \frac{1}{k^2}} \\ & = & (n - m)!^2 e^{4 c (\frac{\pi^2}{6})} \\ & = & (n - m)!^2 e^{\frac{2 \pi^2}{3} c} . \end{eqnarray*} Hence \begin{eqnarray*} \frac{\prod_{k=m}^{n}\left(k^2+c\right)}{c\cdot (n-m)!\prod_{k=1}^{n-m}\left(k^2+4c\right)} & \geq & \frac{\prod_{k = m}^{n} k^2}{c \cdot (n - m)! \cdot (n - m)!^2 e^{\frac{2 \pi^2}{3} c}} \\ & = & \frac{m^2 \left(\frac{n!}{m!}\right)^2}{c \cdot (n - m)!^3 e^{\frac{2 \pi^2}{3} c}} \\ & = & \frac{e^{-\frac{2\pi^{2}}{3} c}}{c}\cdot m^2\frac{n!^2}{m!^2(n-m)!^3} . \end{eqnarray*} The second point of the theorem then follows from the first one. This completes the proof. \end{proof} We shall now impose conditions on $m$ (in terms of $n$) in order to optimize (resp., simplify) the estimate of the second point of Theorem \ref{t7}. To do so, we first need to get rid of the factorials in that estimate. We have the following: \begin{corollary}\label{t9} Let $c,n,m\in\mathbb{N^*}$ such that $m 0$ ($\forall x \in [0 , n]$), then $f$ is increasing. Next, we have \begin{align*} f\left(n-\frac{1}{2}n^{2/3}\right)&=\left(n-\frac{1}{2}n^{2/3}\right)^2-\left(\frac{1}{2}n^{2/3}\right)^3\\&=n^2-n^{5/3}+\frac{1}{4}n^{4/3}-\frac{1}{8}n^2\\&=\frac{7}{8}n^2-n^{5/3}+\frac{1}{4}n^{4/3}. \end{align*} But since $n^2\geq \frac{8}{7}n^{5/3}$ (because $n\geq 2$), it follows that $f\left(n-\frac{1}{2}n^{2/3}\right)\geq \frac{1}{4}n^{4/3}>0$. So, the increase of $f$ ensures that $f(m)>0$ (since $m\geq n-\frac{1}{2}n^{2/3}$ by hypothesis). Thus $\frac{m^2}{(n-m)^3}>1$ and $\frac{m}{(n-m)^{3/2}}>1$. By reporting these into \eqref{jan3}, we then conclude that: \[L_{c,m,n}\geq \lambda_2(c)\cdot ne^{3(n-m)},\] as required. This completes the proof. \end{proof} \subsection{Comparison with Oon's lower bound} In the application of Oon's lower bound (i.e., Theorem \ref{03}), the number of the terms included in the least common multiple $\lcm(m^2 + c , \dots , n^2 + c) = L_{c , m , n}$ must be $\geq n - \lceil\frac{n}{2}\rceil + 1 > \frac{n}{2}$; whereas when we put our theorems \ref{c5} and \ref{c6} together, this constraint is deleted. However, if the condition of application of Oon's theorem holds then we obtain a lower bound for $L_{c , m , n}$ stronger than those of our theorems. Further, our key result is rather the point 1 of Theorem \ref{t7} which provides a nontrivial rational divisor of $L_{c , m , n}$. Here, we have exploited this key result in a naive way. 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Monthly} {\bf 89} (1982), 126--129. \bibitem{oon} S. M. Oon, Note on the lower bound of least common multiple, {\it Abstr. Appl. Anal.} (2013), Article ID 218125. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 11A05; Secondary 11B83, 13G05, 30B40, 33B15. \noindent \emph{Keywords:} least common multiple, quadratic sequence. \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received January 30 2020; revised versions received May 8 2020; May 14 2020. Published in {\it Journal of Integer Sequences}, June 11 2020. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}