\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \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,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://www.research.att.com/cgi-bin/access.cgi/as/~njas/sequences/eisA.cgi?Anum=#1}{\underline{#1}}} \newcommand{\R}{\mbox{$\mathbb{R}$}} \newcommand{\K}{\mbox{$\mathbb{K}$}} \newcommand{\C}{\mbox{$\mathbb{C}$}} \newcommand{\N}{\mbox{$\mathbb{N}$}} \newcommand{\nb}{\mbox{$\mathbb{N}$}} \newcommand{\Z}{\mbox{$\mathbb{Z}$}} \newcommand{\zb}{\mbox{$\mathbb{Z}$}} \newcommand{\Q}{\mbox{$\mathbb{Q}$}} \newcommand{\F}{\mbox{$\mathbb{F}$}} \def\ds{\displaystyle} \def\inf{\infty} \def\be{\beta} \def\al{\alpha} \def\ga{\gamma} \def\Ga{\Gamma} \def\de{\delta} \def\De{\Delta} \def\la{\lambda} \def\te{\theta} \def\vep{\varepsilon} \def\vfi{\varphi} \def\om{\omega} \def\Om{\Omega} \def\si{\sigma} \def\ca{\mathcal{A}} \def\cb{\mathcal{B}} \def\cf{\mathcal{F}} \def\cn{\mathcal{N}} \def\cg{\mathcal{G}} \def\cp{\mathcal{P}} \def\cc{\mathcal{C}} \def\cd{\mathcal{D}} \def\aca{\mathcal{|A|}} \def\acb{\mathcal{|B|}} \def\sark{S\'ark\"ozy } \def\ni{\noindent} \def \dv{\,|\,} \def\lcm{\textrm{lcm}} \newcommand{\leg}[2]{\left(\frac{#1}{#2}\right)} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \begin{center} \vskip 1cm{\LARGE\bf Sets with Even Partition Functions and \\ \vskip .1in 2-adic Integers, II } \vskip 1cm \large N. Baccar\footnote{Research supported DGRST of Tunisia, UR 99/15-18, Facult\'e des Sciences de Tunis.} \\ Universit\'e de Sousse \\ ISITCOM Hammam Sousse \\ D\'ep. de Math Inf. \\ 5 Bis Rue 1 Juin 1955 \\ 4011 Hammam Sousse \\ Tunisie\\ \href{mailto:naceurbaccar@yahoo.fr}{\tt naceurbaccar@yahoo.fr}\\ \ \\ A. Zekraoui\\ Universit\'e de Monastir\\ F. S. M. \\ D\'ep. de Math. \\ Avenue de l'environnement \\ 5000 Monastir \\ Tunisie \\ \href{mailto:ahlemzekraoui@yahoo.fr}{\tt ahlemzekraoui@yahoo.fr} \end{center} \vskip .2 in \begin{abstract} For $P\in \mathbb{F}_2[z]$ with $P(0)=1$ and $\deg(P)\geq 1$, let ${\cal A}={\cal A}(P)$ be the unique subset of $\mathbb{N}$ such that $\sum_{n\geq 0}p({\cal A},n)z^n\equiv P(z)$ (mod $2$), where $p({\cal A},n)$ is the number of partitions of $n$ with parts in ${\cal A}$. Let $p$ be an odd prime number, and let $P$ be irreducible of order $p$ ; i.e., $p$ is the smallest positive integer such that $P$ divides $1+z^p$ in $\F_2[z]$. N. Baccar proved that the elements of ${\cal A}(P)$ of the form $2^km$, where $k\geq 0$ and $m$ is odd, are given by the $2$-adic expansion of a zero of some polynomial $R_m$ with integer coefficients. Let $s_p$ be the order of $2$ modulo $p$, i.e., the smallest positive integer such that $2^{s_p}\equiv 1$ (mod $p$). Improving on the method with which $R_m$ was obtained explicitly only when $s_p=\frac{p-1}{2}$, here we make explicit $R_m$ when $s_p=\frac{p-1}{3}$. For that, we have used the number of points of the elliptic curve $x^3+ay^3 =1 $ modulo $p$. \end{abstract} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{cor}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{lem}[theorem]{Lemma} \newtheorem{thm}[theorem]{Theorem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{defin}[theorem]{Definition} \newenvironment{definition}{\begin{defin}\normalfont\quad}{\end{defin}} \newtheorem{examp}[theorem]{Example} \newenvironment{example}{\begin{examp}\normalfont\quad}{\end{examp}} \newtheorem{rema}[theorem]{Remark} \newenvironment{remark}{\begin{rema}\normalfont\quad}{\end{rema}} \newcommand{\thmref}[1]{Theorem~\ref{#1}} \newcommand{\secref}[1]{\S\ref{#1}} \newcommand{\lemref}[1]{Lemma~\ref{#1}} \section{Introduction.} Let $\N$ denote the set of positive integers, and let ${\cal A}=\left\{a_1,a_2,\ldots\right\}$ be a non-empty subset of $\N$. For $n \in \N$, let $p({\cal A},n)$ be the number of partitions of $n$ with parts in ${\cal A}$, i.e., the number of solutions of the diophantine equation \begin{equation}\label{eq1} a_1x_1+a_2x_2+\cdots=n \end{equation} in non-negative integers $x_1,~x_2,\ldots$. By convention, $ p({\cal A},0)=1$ and $p({\cal A},n)=0$ for all $n<0$. The generating series of $p({\cal A},n)$ is \begin{equation}\label{eq3} F_{\cal A}(z):=\sum_{n=0}^{\infty}p({\cal A},n)z^n=\prod_{a \in {\cal A}}\frac{1}{1-z^a}. \end{equation} Let $\F_2$ be the field with two elements and $P(z)=1+\epsilon_1z+\cdots +\epsilon_Nz^N \in \F_2\left[z\right]$, $N\geq 1$. J.-L. Nicolas, I. Z. Ruzsa and A. S\'ark\"ozy \cite{NRS} proved that there exists a unique set ${\cal A}={\cal A}(P)$ satisfying \begin{equation}\label{eq4} F_{\cal A}(z)\equiv P(z) \pmod 2, \end{equation} which means that \begin{equation}\label{hhhfffrrr} \textrm{$p({\cal A},n) \equiv \epsilon_n \pmod 2$ for $1\leq n \leq N$ } \end{equation} and $p({\cal A},n)$ is even for all $n> N$. Indeed, for $n=1$, \begin{displaymath} p({\cal A},1)=\left\{ \begin{array}{ll} 1, & \textrm{if $1 \in {\cal A};$}\\ 0, & \textrm{if $1\notin {\cal A}$}. \end{array} \right. \end{displaymath} and so, by (\ref{hhhfffrrr}), $$1 \in {\cal A}\Leftrightarrow \epsilon_1=1.$$ Further, assume that we Know ${\cal A}_{n-1}={\cal A}\cap \{1,\ldots,n-1\}$; since there exists only one partition of $n$ containing the part $n$, then $$p({\cal A},n)=p({\cal A}_{n-1},n)+\chi({\cal A},n),$$ where $\chi ({\cal A},.)$ is the characteristic function of the set ${\cal A}$, i.e., \begin{displaymath} \chi ({\cal A},n)=\left\{ \begin{array}{ll} 1, & \textrm{if $n \in {\cal A};$}\\ 0, & \textrm{if $n\notin {\cal A}$}, \end{array} \right. \end{displaymath} which with (\ref{eq4}) allow one to decide whether $n$ belongs to ${\cal A}$.\\ Let $p$ be an odd prime number, and let $s_p$ be the order of $2$ modulo $p$, i.e., $s_p$ is the smallest positive integer such that $p$ divides $2^{s_p}-1$. Let $P\in \F_2[z]$ be irreducible of order $p$ ($\text {ord}(P)=p$); in other words, $p$ is the smallest positive integer such that $P$ divides $1+z^p$ in $\F_2[z]$. N. Baccar and F. Ben Sa\"{\i}d \cite{NF2} determined the sets ${\cal A}(P)$ for all $p$ such that $s_p=\frac{p-1}{2}$. Moreover, they proved that if $k \geq 0$ and $m$ is an odd positive integer, then the elements of ${\cal A}(P)$ of the form $2^km$ are given by the $2$-adic expansion of some zero of a polynomial $R_m$ with integer coefficients. N. Baccar \cite{NF3} extended this last result to any odd prime number $p$. Unfortunately, the method used in that paper can make explicit $R_m$ only when $s_p=\frac{p-1}{2}$. In this paper, we will improve on the method given by N. Baccar \cite{NF3}, by introducing elliptic curves, to make $R_m$ explicit when $s_p=\frac{p-1}{3}$. In Section 2, some properties of the polynomial $R_m$ are exposed. In Section 3, we introduce elliptic curves to compute some cardinalities used in Section 4 to make $R_1$ explicit, and in Section 5 to get $R_m$ explicitly for any odd integer $m\geq 3$.\\ Throughout this paper, $p$ is an odd prime number and $P$ is some irreducible polynomial in $\F_2[z]$ of order $p$. We also denote by $s_p$ the order of $2$ modulo $p$. For $a\in \Z$ and $b\in \N $, we should write $a \bmod b$ for the remainder of the euclidean division of $a$ by $b$. \section{Some results on the polynomial $R_m$} Let $p$ be an odd prime. We denote by $(\mathbb{Z}/{p \mathbb{Z}})^*$ the group of invertible elements modulo $p$ and by $<2>$ its subgroup generated by $2$. We consider the action $\star $ of $<2>$ on the set $\mathbb{Z}/{p \mathbb{Z}}$ given by $a\star n=an$ for all $a\in <2>$ and all $n\in \mathbb{Z}/{p \mathbb{Z}}$. The quotient set will be denoted by $(\mathbb{Z}/{p \mathbb{Z}})/_{<2>}$ and the orbit of some $n\in \mathbb{Z}/{p \mathbb{Z}}$ by ${\it O}(n)$. So, we can write \begin{equation*} \mathbb{Z}/{p \mathbb{Z}}= {\it O}(1)\cup{\it O}(g)\cup\cdots\cup{\it O}(g^{r-1})\cup{\it O}(p), \end{equation*} where $g$ is some generator of $(\mathbb{Z}/{p \mathbb{Z}})^*$, $r=\frac{p-1}{s_p}$ is the number of invertible orbits of $\mathbb{Z}/{p \mathbb{Z}}$, \begin{equation}\label{hdnbcvf} {\it O}(g^i)=\left\{2^jg^i \bmod p:~0\leq j \leq s_p -1\right\},~~0\leq i \leq r -1, \end{equation} \begin{equation*} {\it O}(p)=\left\{0\right\}. \end{equation*} Note that for any integer $t$, \begin{equation}\label{114nnbv1} {\it O}(g^t)={\it O}(g^{t \bmod r}). \end{equation} The orbits ${\it O}(n)$ are defined as parts of $\mathbb{Z}/{p \mathbb{Z}}$; however, by extension, they are also considered as parts of $\N$.\\ If $\phi_p$ is the cyclotomic polynomial over $\F_2$ of index $p$, then \begin{equation*} 1+z^p =(1+z)\phi_p(z). \end{equation*} Moreover, one has \begin{equation*} \phi_p(z)=P_0(z)P_1(z)\cdots P_{r-1}(z), \end{equation*} where $P_0,P_1,\ldots$ and $P_{r-1}$ are the only distinct irreducible polynomials in $\F_2\left[z\right]$ of the same degree $s_p$ and all of which are of order $p$. For all $l,$ $0\leq l \leq r-1$, let ${\cal A}_l={\cal A}(P_l)$ be the set obtained from (\ref{eq4}). If $m$ is an odd positive integer, we define the $2-$adic integer $y_l(m) $ by \begin{equation}\label{bndcse} y_l(m)=\chi({\cal A}_l,m)+2\chi({\cal A}_l,2m)+4\chi({\cal A}_l,4m)+ \cdots=\sum_{k=0}^{\infty}\chi({\cal A}_l,2^km)2^k. \end{equation} By computing $y_l(m) \bmod 2^{k+1}$, one can deduce $\chi({\cal A}_l,2^jm)$ for all $j,$ $0\leq j\leq k$, and obtain all the elements of ${\cal A}_l$ of the form $2^jm$. In \cite{NFA}, some necessary conditions on integers to be in ${\cal A}_l$ were given. For instance: \begin{equation}\label{bacbszek1} p^2n\notin {\cal A}_l,~\forall ~n\in \N , \end{equation} \begin{equation}\label{bacbszek2} \text {if}~ q ~\text {is an odd prime in}~ {\it O}(1), \text {then}~ qn\notin {\cal A}_l,~\forall n\in \N . \end{equation} Let ${\K}$ be some field, and let $u(z)=\sum_{j=0}^{n}u_jz^j$ and $v(z)=\sum_{j=0}^{t}v_jz^j$ be polynomials in ${\K}[z]$. We denote the resultant of $u$ and $v$ with respect to $z$ by $\textrm{res}_z(u(z),v(z))$, and recall the following well known result \begin{lem}\label{lemee} (i) The resultant $\textrm{res}_z(u(z),v(z))$ is a homogeneous multivariate polynomial with integer coefficients, of degree $n+t$ in the $n+t+2$ variables $u_i,$ $v_j$.\\ (ii) If $u(z)$ is written as $u(z)= u_n(z-\alpha_1)(z-\alpha_2)\cdots(z-\alpha_n)$ in the splitting field of $u$ over ${\K}$ then \begin{equation}\label{hgfd} \textrm{res}_z(u(z),v(z))=u_n^t\prod_{i=1}^{n}v(\alpha_i). \end{equation} \end{lem} N. Baccar proved \cite{NF3} that, for all $l$, $0\leq l\leq r-1$, the $2$-adic integers $y_l(m)$ defined by (\ref{bndcse}) are the zeros of some polynomial $R_m$ with integer coefficients and which can be written as the resultant of two polynomials. We mention here that the expressions given in that paper to $R_m$, for $m=1$ and $m\geq 3$, can be encoded in only one. So that we have \begin{thm}\label{sfj14jjjj}(\cite{NF3}) 1) Let $m $ be an odd positive integer such that $m \notin {\it O}(p)$ (i.e., $\gcd(m,p)=1$), and let $\delta =\delta (m)$ be the unique integer in $ \left\{0,1,\ldots,r-1\right\}$ such that $m\in {\it O}(g^\delta)$. We define the polynomial $A_m$ by \begin{equation}\label{deee} A_m(z)=\sum_{h=0}^{r-1}\alpha_h(m)B_h(z), \end{equation} where for all $h,$ $0\leq h \leq r-1$, \begin{equation}\label{vfrd} \alpha_h(m)= \sum_{\substack{d\dv \widetilde{m},~d \in {\it O}(g^h)}} \mu(d), \end{equation} $\widetilde{m}=\prod_{q \ \textrm{prime} \ q\mid m}q$ is the radical of $m$ with $\widetilde{1}=1$, $\mu$ is the M\"obius function and $B_h$ is the polynomial \begin{equation}\label{dee} B_h(z)=B_{h,m}(z)= \sum_{j=0}^{s_p-1}z^{\left(2^jg^{(\delta-h) \bmod r}\right) \bmod p }. \end{equation} Then, the $2$-adic integers $y_0(m),y_1(m),\ldots $ and $y_{r-1}(m)$ are the zeros of the polynomial $R_m(y)$ of $\Z[y]$ defined by the resultant \begin{equation}\label{sfre} R_m(y)=\textrm{res}_z(\phi_p(z),my+A_m(z)) \end{equation} and we have \begin{equation}\label{sfre22} R_m(y)=m^{p-1}\bigl(( y-y_0(m) )(y-y_1(m))\cdots(y-y_{r-1}(m))\bigr)^{s_p}. \end{equation} 2) The $2$-adic integers $y_0(p),y_1(p),\ldots $ and $y_{r-1}(p)$ are the zeros of the polynomial $R_1(-py-s_p)$; while if $m=pm'$, $m'\geq 3$ and $\gcd (m',p)=1$, then $y_0(m),y_1(m),\ldots$ and $y_{r-1}(m)$ are the zeros of the polynomial $R_{m'}(-py)$ defined by (\ref{sfre}).\\ 3) If $ m$ is divisible by $p^2$ or by some prime $q$ belonging to ${\it O}(1)$ then we extend the definition (\ref{sfre}) to $R_m(y)=m^{p-1}y^{s_p}$; so that $y_0(m),y_1(m),\ldots $ and $ y_{r-1}(m)$ remain zeros of $R_m$ since, from (\ref {bacbszek1}) and (\ref{bacbszek2}), they all vanish.\\ \end{thm} \begin{remark}\label{hhuuggg} Explicit formulas to the polynomials $R_m$ defined by (\ref{sfre}), when $s_p=\frac {p-1}{2}$, are given in \cite{NF3}. Moreover in that paper, it is shown that if $\theta$ is a certain primitive $p$-th root of unity over the $2$-adic field $\Q_2$, then for all $l$, $0\leq l \leq r-1$, \begin{equation}\label{nbsqz} y_l(1)=-T_l , \end{equation} where, for all $l \in \Z$, \begin{equation}\label{hgtrvcdes} T_l=T_{l \bmod 3}=\sum_{k=0}^{s_p-1}\theta^{2^kg^l}=\sum_{j\in {\it O}(g^l)}\theta^j. \end{equation} We also mention here that N. Baccar \cite{NF3} proved that for all $m\in \N$, \begin{equation}\label{Rm(yy)} R_m(y)=\prod_{l=0}^{r-1}\left( my+A_m(\theta^{g^l})\right)^{s_p}. \end{equation} \end{remark} \section{ Orbits and elliptic curves.} From now on, we keep the above notation and assume that the prime number $p$ is such that $s_p=\frac{p-1}{3}$ (the first ones up to $1000$ are: $p=43,109,157,229,277,283,307,499,643,691,733,739,\\811,997$ ). So the number of invertible orbits is $r=3$ and \begin{equation}\label{orbital} \mathbb{Z}/{p \mathbb{Z}}= {\it O}(1)\cup{\it O}(g)\cup{\it O}(g^2)\cup{\it O}(p), \end{equation} where $g$ is some generator of the cyclic group $(\mathbb{Z}/{p \mathbb{Z}})^*$. The order of $2$ is $s_p=\frac{p-1}{3}$; if $2$ were a square modulo $p$, its order should divide $\frac{p-1}{2}$, which is impossible. Hence $2$ cannot be a square modulo $p$, and by Euler criterion, $p$ has to satisfy $p\equiv \pm 3 $ (mod $8$), and, as $p\equiv 1$ (mod $3$), $p\equiv 13,~19 $ (mod $24$). \begin{lem}\label{lem orbff} For all $i,$ $0 \leq i \leq 2$, let ${\it O}(g^i)$ be the orbit of $g^i$ defined by (\ref{hdnbcvf}). Then \begin{equation}\label{lknb} {\it O}(g^i)=\left\{-g^i,-2g^i,\ldots,-2^{s_p-1}g^i\right\}=\left\{g^i,g^{i+3}. \ldots,g^{i+3(s_p-1)}\right\}. \end{equation} In particular, $2$ is a cube modulo $p$ and the sub-group generated by $2$ is the sub-group of cubes (generated by $g^3$) and contains $-1$. \end{lem} \begin{proof}[Proof] To get the first equality of (\ref{lknb}), it suffices to show that $-1 \in {\it O}(1)$. This follows from $-1=(\frac {2}{p})\equiv 2^{\frac {p-1}{2}}$ (mod $p$).\\ To prove the second equality of (\ref{lknb}), one just use the fact that (cf. (\ref{114nnbv1})) $g^3 \in {\it O}(1)$. \end{proof} Let us define the integers $\ell_{i,j},~0\leq i,~j\leq 2$, by \begin{equation}\label{vfer} \ell_{i,j}=\left| \left\{t:~0\leq t\leq s_p-1, ~1+g^{j+3t} \in {\it O}(g^i) \right\} \right|. \end{equation} \begin{remark}\label{remllll} As shown just above, $-1\in {\it O}(1)$, so that there exists one and only one $t\in \left\{0,1,\ldots,s_p-1\right\}$ such that $1+g^{3t}\in {\it O}(p)$. Moreover, for all $t\in \left\{0,1,\ldots,s_p-1\right\}$ and $j \in \{1,2\}$, $1+g^{j+3t}\notin {\it O}(p)$. Hence the integers $\ell_{i,j}$ defined by (\ref {vfer}) satisfy \begin{equation}\label{sumlij} \sum_{i=0}^2\ell_{i,j}=s_p-\delta_{0,j}, \end{equation} where $\delta_{i,j}$ is the Kronecker symbol given by \begin{equation*} \delta_{i,j}=\left\{ \begin{array}{ll} 1, & \textrm{if $i=j$};\\ 0, & \textrm{otherwise}. \end{array} \right. \end{equation*} \end{remark} ~\\ The integers $\ell_{i,j}$ defined above are cardinalities of some curves. Indeed, let us consider the curve over $\F_p$, \begin{equation*} {\cal C}_{i,j}:~~1+g^jX^3=g^iY^3 \end{equation*} and denote by $c_{i,j}$ its cardinality, $c_{i,j}=\mid {\cal C}_{i,j}\mid $. Since $-1$ is a cube modulo $p$, it is clear that \begin{equation*} c_{j,i}=c_{i,j}. \end{equation*} Using (\ref{vfer}) it follows that \begin{equation*} \ell_{i,j}=|\{(X^3,Y^3):~X\neq0,~Y\neq 0~\textrm{and}~(X,Y) \in {\cal C}_{i,j} \}|. \end{equation*} Therefore, \begin{equation}\label{lij=lji} \ell_{j,i}=\ell_{i,j}. \end{equation} Note that, $(X^3,Y^3)=(X'^3,Y'^3)$ if and only if $X'=Xg^{vs_p}$ and $Y'=Yg^{ws_p}$ for some $v,w\in\{0,1,2\}$. Moreover, if $i\neq 0$ (resp. $j\neq 0$), no point on the curve ${\cal C}_{i,j}$ can be of the form $(0,Y)$ (resp. $(X,0)$). But if $i=0$ (resp. $j=0$), we obtain three points on the curve ${\cal C}_{i,j}$ with $X=0$ (resp. $Y=0$). Consequently, we obtain the relation \begin{equation}\label{cijlij} c_{i,j}=9\ell_{i,j}+3\delta_{i,0}+3\delta_{0,j}. \end{equation} Now, let us consider the projective plane cubic curve \begin{equation*} {\cal E}_{i,j}:~~Z^3+g^jX^3=g^iY^3 \end{equation*} and $e_{i,j}=\mid {\cal E}_{i,j}\mid$ its cardinality. If $i\not= j$, ${\cal E}_{i,j}$ has no points at infinity; whereas if $i=j$, it has three points at infinity. Hence \begin{equation}\label{projectiveeij=cij} e_{i,j}=c_{i,j}+3\delta_{i,j}. \end{equation} By multiplying the equation $Z^3+gX^3=gY^3$ by $g^2$, we get the curve $g^2Z^3+{X'}^3={Y'}^3$. So, by permuting the variables, we deduce that $e_{1,1}=e_{2,0}$. Similarly, we obtain $e_{2,2}=e_{1,0}$. Hence, by (\ref{projectiveeij=cij}) and (\ref{cijlij}) we find that \begin{equation}\label{ttrfecd} \ell_{1,1}=\ell_{2,0}, \end{equation} \begin{equation}\label{ttrfecd1} \ell_{2,2}=\ell_{1,0}. \end{equation} Therefore, from (\ref{sumlij}), it follows that \begin{equation}\label{ttrfeceeed} \ell_{2,1}=\ell_{0,0}+1. \end{equation} Furthermore, from (\ref{projectiveeij=cij}) and (\ref{cijlij}) we have for all $i$, $0\leq i \leq 2$, \begin{eqnarray}\label{eijcij} % \nonumber to remove numbering (before each equation) 9\ell_{i,0} &=& c_{i,0}-3\delta_{i,0}-3 \nonumber \\ &=& e_{i,0}-6\delta_{i,0}-3. \end{eqnarray} Hence, to get all the numbers $\ell_{i,j}$, $0\leq i,j \leq 2$, it suffices to know the values of $e_{i,0}, $ $0 \leq i \leq 2$.\\ \\ {\bf Computation of $e_{i,0},~i\in \{0,1,2\}$.}\\ Here, we are interested with the curve ${\cal E}_{i,0}:~~Z^3+X^3=g^iY^3$. By setting $X=9g^iz+2y$, $Y=6x$ and $Z=9g^iz-2y$, we get the Weierstrass's form \begin{equation*} zy^2=x^3-(27/4)g^{2i}z^3, \end{equation*} which, when divided by $z^3$, gives the form \begin{equation*} y^2=x^3-(27/4)g^{2i}. \end{equation*} Let \begin{equation*} y^2=x^3+\alpha x+\beta \end{equation*} be the equation of an elliptic curve ${\cal E}$ defined over $\F_p$. It is well known that the number of points of ${\cal E}$ is equal to \begin{equation}\label{numberpoints} \mid {\cal E}\mid =p+1+\sum_{x\in \F_p}\left(\frac {x^3+\alpha x+\beta}{p}\right), \end{equation} where $(\frac {.}{p})$ is the Legendre's symbol. For $\alpha =0$, the sum $\sum_{x\in \F_p}\left(\frac {x^3+\alpha x+\beta}{p}\right)$ was investigated by S. A. Katre \cite{Kat85}. He obtained: \begin{lem}\label{Katre} Let $p$ be a prime number such that $p\equiv 1$ (mod $3$). Then there exist a unique $L$, $L\equiv 1 $ (mod $3$) and a unique $M$ up to a sign such that $4p=L^2+27M^2$. Moreover, if $\beta $ is an integer $\not= 0$ then \begin{displaymath} \sum_{x\in \F_p}\left(\frac {x^3+\beta }{p}\right)=\left\{ \begin{array}{ll} (\frac {\beta }{p})L, & \textrm{if} ~4\beta \textrm~{is~ a~ cube~ modulo~ p};\\-\frac {1}{2}(\frac {\beta }{p})(L+9M),& \textrm{otherwise, where $M$ is chosen uniquely }\\ & \textrm{by $\left(4 \beta \right)^{\frac{p-1}{3}}\equiv \frac{L+9M}{L-9M} \pmod p$ }. \end{array} \right. \end{displaymath} \end{lem} ~\\ Thanks to Lemma \ref{Katre}, we can give the values of $e_{i,0}$ for $0 \leq i \leq 2.$\\ {\bf Computation of $e_{0,0}$.} From (\ref{numberpoints}), since $-27$ is a cube , by using Lemma \ref{Katre} with $\beta=-27/4$, we obtain \begin{eqnarray}\label{mmppoo} % \nonumber to remove numbering (before each equation) e_{0,0} &=& p + 1 +\left(\frac{-27/4}{p}\right)L \nonumber \\ &=& p+1+\left(\frac{-27}{p}\right)L \nonumber\\ &=& p+1+\left(\frac{-3}{p}\right)^3L\nonumber. \end{eqnarray} Since $p\equiv 1$ (mod $3$) then, by the quadratic reciprocity law, $-3$ is a quadratic residue modulo $p$. Hence, \begin{equation}\label{mmddppoo} e_{0,0}=p + 1 +L. \end{equation} {\bf Computation of $e_{1,0}$. } If $\beta=-27g^2/4$ then $4 \beta=-27g^2$ is not a cube modulo $p$. Hence, by using Lemma \ref{Katre} again, it follows that \begin{eqnarray}\label{mmlpoiu} % \nonumber to remove numbering (before each equation) e_{1,0} &=& p + 1 -\frac{1}{2}\left(\frac{-27g^2/4 }{p}\right)(L+9M) \nonumber \\ &=& p + 1 -\frac{1}{2}(L+9M), \end{eqnarray} where the sign of $M$ is given by \begin{eqnarray*} % \nonumber to remove numbering (before each equation) \left(-27g^{2}\right)^{(p-1)/3} &\equiv & (g^{2})^{(p-1)/3} \pmod p \nonumber \\ & \equiv & \frac{L+9M}{L-9M} \pmod p. \end{eqnarray*} {\bf Computation of $e_{2,0}$.} Let $M$ be fixed by last congruence, and let $\beta=-27g^4/4$. Since $\left(g^{4}\right)^{(p-1)/3}\not\equiv \left(g^{2}\right)^{(p-1)/3} $ (mod $p$), Lemma \ref{Katre} implies that \begin{equation}\label{vkjhgfer} e_{2,0}=p + 1 -\frac{1}{2}(L-9M). \end{equation} \section{The explicit form of $R_1$ when $s_p=\frac {p-1}{3}$.} First, we remark that the polynomial $R_1(y)$ given by (\ref{sfre}) can also be defined (cf. (\ref{sfre22}), (\ref{nbsqz})) by \begin{equation}\label{R1*1/sp(y)} \left(R_1(y)\right)^{1/s_p}=\prod_{i \in \{0,1,2\}}~\biggl( y+T_i \biggr), \end{equation} where $\theta\neq 1 $ is some $p$-th root of unity. \begin{thm}\label{jhgvcsd} Let $p$ be an odd prime such that $s_p=\frac {p-1}{3}$. Then the polynomial $R_1$ given by (\ref{sfre}) or (\ref{R1*1/sp(y)}) is equal to \begin{equation*} R_1(y)=\left(y^3-y^2-s_py+\lambda_p\right)^{s_p}, \end{equation*} with \begin{equation*} \lambda_p=\frac{p(L+3)-1}{27}, \end{equation*} where $L$ is the unique integer satisfying $4p=L^2+27M^2$ and $L\equiv 1$ (mod $3$). \end{thm} \begin{remark} $p(L+3) \equiv 1$ (mod $27$) follows easily from the congruences $L\equiv 1$ (mod $3$) and $p \equiv L^2 $ (mod $27$). \end{remark} \begin{proof}[Proof of Theorem \ref{jhgvcsd}] From (\ref{R1*1/sp(y)}), we have \begin{equation*} R_1(y)= \biggl( (y+T_0)(y+T_1)(y+T_2)\biggr)^{s_p}. \end{equation*} This can be written as \begin{equation*} R_1(y)= \biggl( y^3+\lambda''_py^2+\lambda'_py+\lambda_p\biggr)^{s_p}, \end{equation*} with \begin{equation}\label{lambdap} \lambda_p=T_0T_1T_2, \end{equation} \begin{equation}\label{lambdap'} \lambda'_p=T_0T_1+T_0T_2+T_1T_2, \end{equation} \begin{equation*} \lambda''_p=T_0+T_1+T_2. \end{equation*} We begin by calculating $\lambda''_p$. It follows immediately from (\ref{hgtrvcdes}) and (\ref{orbital}) that \begin{eqnarray}\label{trefc} % \nonumber to remove numbering (before each equation) \lambda''_p &=& T_0+T_1+T_2 \nonumber\\ &=& \sum_{i=0}^2\sum_{k=0}^{s_p-1}\theta^{2^kg^i} \nonumber \\ &=& \sum_{j=1}^{p-1}\theta^{j} \nonumber \\ &=& -1, \end{eqnarray} since cf. Remark \ref{hhuuggg}, $\theta $ is a primitive $p$-th root of unity.\\ Now, let us prove that $\lambda_p'=-s_p$. From (\ref{lambdap'}) and (\ref{hgtrvcdes}), we have \begin{equation}\label{dscss} \lambda'_p=\sum_{0 \leq k,k' \leq s_p-1}\theta^{2^kg+2^{k'}g^2}+ \sum_{0 \leq k,k' \leq s_p-1}\theta^{2^kg^2+2^{k'}}+\sum_{0 \leq k,k' \leq s_p-1}\theta^{2^kg+2^{k'}}. \end{equation} To treat the last sum in (\ref{dscss}), let us fix $k$ and $k'$ in $\left\{0,1,\ldots,s_p-1\right\}$. We have $\theta^{2^kg+2^{k'}}=\theta^{2^{k'}(1+2^{k-k'}g)}$. Since $2^{k-k'}g\in {\it O}(g)$ then, from the second equality in (\ref{lknb}), there exists a unique $t\in \{0,1,\ldots,s_p-1\}$ such that $2^{k-k'}g=g^{1+3t}$. Hence, the last sum in (\ref{dscss}) becomes \begin{equation}\label{jhgfv 3} \sum_{0 \leq k,k' \leq s_p-1}\theta^{2^kg+2^{k'}}=\sum_{0 \leq k',t \leq s_p-1}\theta^{2^{k'}(1+g^{1+3t})}. \end{equation} For the first and second sums in (\ref{dscss}), arguing as above, we get \begin{eqnarray}\label{jhgfv 2} % \nonumber to remove numbering (before each equation) \sum_{0 \leq k,k' \leq s_p-1}\theta^{2^kg^2+2^{k'}} &=& \sum_{0 \leq k,k' \leq s_p-1}\theta^{2^{k'}(1+2^{k-k'}g^2)} \nonumber \\ &=& \sum_{0 \leq {k'},t \leq s_p-1}\theta^{2^{k'}(1+g^{2+3t})} \end{eqnarray} and \begin{eqnarray}\label{jhgfv} % \nonumber to remove numbering (before each equation) \sum_{0 \leq k,k' \leq s_p-1}\theta^{2^kg+2^{k'}g^2} &=& \sum_{0 \leq k,k' \leq s_p-1}\theta^{2^k(g+2^{k'-k}g^2)} \nonumber \\ &=& \sum_{0 \leq k,t \leq s_p-1}\theta^{2^k(g+g^{2+3t})}. \end{eqnarray} Now, from (\ref{vfer}), (\ref{ttrfecd}), (\ref{ttrfecd1}), (\ref{hgtrvcdes}) and Remark \ref{remllll}, we obtain \begin{eqnarray}\label{uytrtg} % \nonumber to remove numbering (before each equation) \sum_{0 \leq k',t \leq s_p-1}\theta^{2^{k'}(1+g^{1+3t})} &=&\sum_{i=0}^{2}\ell_{i,1}T_i \nonumber\\ &=& \ell_{1,0}T_0+ \ell_{2,0}T_1+\ell_{2,1}T_2 \end{eqnarray} and \begin{eqnarray}\label{uytrtg 2} % \nonumber to remove numbering (before each equation) \sum_{0 \leq k',t \leq s_p-1}\theta^{2^{k'}(1+g^{2+3t})} &=& \sum_{i=0}^{2}\ell_{i,2}T_i\nonumber \\ &=& \ell_{2,0}T_0+\ell_{2,1} T_1+\ell_{1,0}T_2 . \end{eqnarray} On the other hand, since for all $v\geq 0$, $0 \leq i,j \leq 2$, \begin{equation*} 1+g^{j+3t} \in {\it O}(g^i)\Longleftrightarrow g^v+g^{v+j+3t} \in {\it O}(g^{v+i}), \end{equation*} again by (\ref{vfer}), (\ref{ttrfecd}), (\ref{hgtrvcdes}) and Remark \ref{remllll}, we get \begin{eqnarray}\label{uytrtg 3} % \nonumber to remove numbering (before each equation) \sum_{0 \leq k,t \leq s_p-1}\theta^{2^k(g+g^{2+3t})} &=& \sum_{i=0}^{2}\ell_{i,1}T_{1+i}\nonumber \\ &=& \ell_{2,1}T_0+ \ell_{1,0}T_1+\ell_{2,0}T_2. \end{eqnarray} Clearly, from (\ref{sumlij}) and (\ref{ttrfeceeed}), one can deduce that \begin{equation}\label{l0+l1+l2} \ell_{1,0}+\ell_{2,0}+\ell_{2,1}=s_p, \end{equation} which, by (\ref{dscss})-(\ref{uytrtg 3}) and (\ref{trefc}), gives \begin{eqnarray}\label{lknbv} % \nonumber to remove numbering (before each equation) \lambda'_p &=& s_p\left(T_0+T_1+T_2\right) \nonumber \\ &=& s_p\sum_{j=1}^{p-1}\theta^j \nonumber \\ &=& -s_p. \end{eqnarray} Finally, let us calculate $\lambda_p$. From (\ref{lambdap}) and (\ref{hgtrvcdes}), we have \begin{eqnarray}\label{lambdap2} % \nonumber to remove numbering (before each equation) \lambda_p &=& \sum_{0\leq k,k',k"\leq s_p-1}\theta^{2^k+2^{k'}g+2^{k"}g^2} \nonumber\\ &=& \sum_{0\leq k\leq s_p-1}\theta^{2^k}\left(\sum_{0\leq k',k"\leq s_p-1}\theta^{2^{k'}g+2^{k"}g^2}\right). \end{eqnarray} Hence, by (\ref{jhgfv}), (\ref{uytrtg 3}) and (\ref{hgtrvcdes}), we get \begin{eqnarray*} \lambda_p &=&\sum_{0\leq k\leq s_p-1}\theta^{2^k}\left(\ell_{2,1}\sum_{j \in {\it O}(1)}\theta^j+\ell_{1,0}\sum_{j \in {\it O}(g)}\theta^j+\ell_{2,0}\sum_{j \in {\it O}(g^2)}\theta^j\right)\nonumber\\ &=&\ell_{2,1}\sum_{0\leq k,k'\leq s_p-1}\theta^{2^k+2^{k'}}+\ell_{1,0}\sum_{0\leq k,k'\leq s_p-1}\theta^{2^k+2^{k'}g}+\ell_{2,0}\sum_{0\leq k,k'\leq s_p-1}\theta^{2^k+2^{k'}g^2}. \end{eqnarray*} Consequently, from (\ref{jhgfv 3}), (\ref{uytrtg}), (\ref{jhgfv 2}) and (\ref{uytrtg 2}), it happens that \begin{eqnarray}\label{fgds21} \lambda_p&=&\ell_{2,1}\sum_{0\leq k,k'\leq s_p-1}\theta^{2^k+2^{k'}}+(\ell_{1,0}^2+\ell_{2,0}^2)T_0+(\ell_{1,0}\ell_{2,0}+\ell_{2,0}\ell_{2,1})T_1\nonumber\\ & & +(\ell_{1,0}\ell_{2,1}+\ell_{1,0}\ell_{2,0})T_2. \end{eqnarray} Since $2^{k'-k}\in {\it O}(1)= {\it O}(g^3)$ then, cf. (\ref{lknb}), there exists a unique $t \in \left\{0,1,\ldots,s_p-1\right\}$ such that $2^{k'-k}=g^{3t}$. Hence \begin{eqnarray*} % \nonumber to remove numbering (before each equation) \sum_{0\leq k,k'\leq s_p-1}\theta^{2^k+2^{k'}} &=& \sum_{0\leq k,k'\leq s_p-1}\theta^{2^k(1+2^{k'-k})} \nonumber \\ &=& \sum_{0\leq k,t\leq s_p-1}\theta^{2^k(1+g^{3t})}. \end{eqnarray*} Now, we recall that cf. Remark \ref{remllll} there exists one and only one $t\in \left\{0,1,\ldots,s_p-1\right\}$ satisfying $1+g^{3t} \in {\it O}(p)$. Consequently, by (\ref{vfer}) and (\ref{hgtrvcdes}), we get \begin{eqnarray}\label{kjf11} % \nonumber to remove numbering (before each equation) \sum_{0\leq k,k'\leq s_p-1}\theta^{2^k+2^{k'}} &=& \sum_{0\leq k,t\leq s_p-1}\theta^{2^k(1+g^{3t})} \nonumber \\ &=& \ell_{0,0}T_0+ \ell_{1,0}T_1+\ell_{2,0}T_2+s_p. \end{eqnarray} Hence, (\ref{fgds21}) gives \begin{eqnarray}\label{lambdapvaleur1} \lambda_p&=&(\ell_{1,0}^2+\ell_{2,0}^2+\ell_{0,0}\ell_{2,1})T_0 +(\ell_{1,0}\ell_{2,0}+\ell_{2,0}\ell_{2,1}+\ell_{1,0}\ell_{2,1})T_1\nonumber\\ &&+(\ell_{1,0}\ell_{2,0}+\ell_{1,0}\ell_{2,1}+\ell_{2,0}\ell_{2,1})T_2+\ell_{2,1}s_p. \end{eqnarray} On the other hand, from (\ref{lambdap2}), by changing the order of summation, $\lambda_p$ can be written as \begin{equation*} \lambda_p=\sum_{0\leq k'\leq s_p-1}\theta^{2^{k'}g}\left(\sum_{0\leq k,k"\leq s_p-1}\theta^{2^{k}+2^{k"}g^2}\right) \end{equation*} and we get in the way as above \begin{eqnarray}\label{lambdapvaleur2} \lambda_p&=&(\ell_{1,0}\ell_{2,0}+\ell_{1,0}\ell_{2,1}+\ell_{2,0}\ell_{2,1})T_0 +(\ell_{1,0}^2+\ell_{2,0}^2+\ell_{0,0}\ell_{2,1})T_1\nonumber\\ & & +(\ell_{1,0}\ell_{2,0}+\ell_{2,0}\ell_{2,1}+\ell_{1,0}\ell_{2,1})T_2+\ell_{2,1}s_p \end{eqnarray} Whereas, if we write $\lambda_p$ in the form \begin{equation*} \lambda_p=\sum_{0\leq k"\leq s_p-1}\theta^{2^{k"}g^2}\left(\sum_{0\leq k,k'\leq s_p-1}\theta^{2^{k}+2^{k'}g}\right), \end{equation*} we get \begin{eqnarray}\label{lambdapvaleur3} \lambda_p&=&(\ell_{1,0}\ell_{2,0}+\ell_{2,0}\ell_{2,1}+\ell_{1,0}\ell_{2,1})T_0 +(\ell_{1,0}\ell_{2,0}+\ell_{1,0}\ell_{2,1}+\ell_{2,0}\ell_{2,1})T_1\nonumber\\ & & +(\ell_{1,0}^2+\ell_{2,0}^2+\ell_{0,0}\ell_{2,1})T_2+\ell_{2,1}s_p. \end{eqnarray} By summing (\ref{lambdapvaleur1}), (\ref{lambdapvaleur2}) and (\ref{lambdapvaleur3}), we obtain \begin{eqnarray*} % \nonumber to remove numbering (before each equation) 3\lambda_p &=& (\ell_{1,0}^2+\ell_{2,0}^2+2\ell_{1,0}\ell_{2,0}+2\ell_{1,0}\ell_{2,1}+2\ell_{2,0}\ell_{2,1}+\ell_{0,0}\ell_{2,1})\times \nonumber \\ & &(T_0+T_1+T_2)+3\ell_{2,1}s_p. \end{eqnarray*} But, according to (\ref{trefc}), $T_0+T_1+T_2=-1$. Hence, \begin{eqnarray*} % \nonumber to remove numbering (before each equation) 3\lambda_p&=& - (\ell_{1,0}^2+\ell_{2,0}^2+2\ell_{1,0}\ell_{2,0}+2\ell_{1,0}\ell_{2,1}+2\ell_{2,0}\ell_{2,1}+\ell_{0,0}\ell_{2,1})+3\ell_{2,1}s_p \\ &=& -\left((\ell_{1,0}+\ell_{2,0})^2+\ell_{1,0}\ell_{2,1}+\ell_{2,0}\ell_{2,1}+\ell_{2,1}(\ell_{0,0}+\ell_{1,0}+\ell_{2,0})\right)+3\ell_{2,1}s_p. \end{eqnarray*} So that, from (\ref{l0+l1+l2}) and (\ref{sumlij}), we obtain $$3\lambda_p=-(s_p-\ell_{2,1})^2-\ell_{2,1}(s_p-\ell_{2,1})-\ell_{2,1}(s_p-1)+3\ell_{2,1}s_p,$$ which gives $\lambda_p=\frac {(3s_p+1)\ell_{2,1}-s_p^2}{3}=\frac {p\ell_{2,1}-s_p^2}{3}$. Finally, using the value of $\ell_{2,1}$: \begin{equation}\label{1kkjjhhggfv} \ell_{2,1}=\frac{1}{9}(p+1+L) \end{equation} which follows from (\ref{ttrfeceeed}), (\ref{eijcij}) and (\ref{mmddppoo}), we complete the proof of Theorem \ref{jhgvcsd}. \end{proof} In the following table we give $L$, $g$, $M$ and $R_1(y)$ when $p\leq 1000.$ $$\begin{tabular}{|l|l|l|l|l|} \hline % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ... $p$ & $L$ &$g$ & $M$ &$R_1(y)$ \\ \hline $43$ & $$-8$$ & $$3$$& $$-2$$ & $(y^3 - y^2 - 14 y - 8)^{14}$ \\ \hline $109$ & $$-2$$ &$$6$$ & $$4$$ & $(y^3 - y^2 - 36 y +4)^{36}$ \\ \hline $157$& $$-14$$ & $$5$$& $$4$$ & $ (y^3 - y^2 - 52 y - 64)^{52}$ \\ \hline $229$ & $$22$$ & $$6$$ & $$4$$ & $(y^3 - y^2 - 76 y +212)^{76}$ \\ \hline $ 277$& $$-26$$ & $$5$$&$$-4$$ & $(y^3 - y^2 - 92 y - 236)^{ 92}$ \\ \hline $283$ & $$-32$$ & $$3$$& $$-2$$ & $(y^3 - y^2 - 94 y - 304)^{ 94}$ \\ \hline $307$ & $$16$$ & $$5$$&$$-6$$ & $(y^3 - y^2 - 102 y + 216)^{ 102}$ \\ \hline $499$ & $$-32$$ & $$7$$ &$$-6$$ & $(y^3 - y^2 - 166 y -536)^{ 166}$ \\ \hline $643$ & $$40$$ & $$11$$&$$6$$ & $(y^3 - y^2 - 214 y + 1024)^{ 214}$ \\ \hline $691$ & $$-8$$ & $$3$$&$$10$$ & $(y^3 - y^2 - 230 y - 128)^{ 230}$ \\ \hline $733$ & $$-50$$ & $$6$$&$$4$$ & $(y^3 - y^2 - 244 y - 1276)^{244}$\\ \hline $739$ & $$16$$ & $$3$$&$$10$$ & $(y^3 - y^2 - 246 y + 520)^{246}$ \\ \hline $811$ & $$-56$$ & $$3$$&$$-2$$ &$(y^3 - y^2 - 270 y - 1592)^{270}$ \\ \hline $997$ & $$10$$ & $$7$$&$$-12$$ & $(y^3 - y^2 - 332 y + 480)^{ 332}$ \\ \hline \end{tabular}$$ \begin{center} xxxx \end{center} \section{The explicit form of $R_m$ when $s_p=\frac {p-1}{3}$ and $m\geq 3$.} We remind that if $m\in {\it O}(p)$ or $m$ is divisible by some prime $q$ belonging to ${\it O}(1)$, then the polynomial $R_m$ is given by Theorem \ref{sfj14jjjj}, 2) and 3). Let $m$ be an odd integer $\geq 3$ such that all its prime divisors are in ${\it O}(g)\cup{\it O}(g^2)$. For $i \in \{1,2\}$, we denote by $\omega_i$ the arithmetic function which counts the number of distinct prime divisors belonging to ${\it O}(g^i)$ of an integer, i.e., \begin{equation}\label{arith 0} \omega_i(n)=\sum_{q~\text {prime},\ q\in{\it O}(g^i),~q \dv n}1. \end{equation} Let the decomposition of $m$ into irreducible factors be \begin{equation}\label{jhgvfre} m=q_{1,1}^{\gamma_{1,1}}q_{1,2}^{\gamma_{1,2}}\cdots q_{1,\omega_1}^{\gamma_{1,\omega_1}}q_{2,1}^{\gamma_{2,2}}q_{2,2}^{\gamma_{2,2}}\cdots q_{2,\omega_2}^{\gamma_{2,\omega_2}}, \end{equation} where $\omega_i=\omega_i(m)$, $\omega =\omega (m)=\omega_1+\omega_2$ and $q_{i,j}\in {\it O}(g^i)$.\\ We shall begin with some result concerning binomial coefficients: \begin{lem}\label{lemmas d} For all $n \in \N$ and all $j$ , $0 \leq j \leq 2$, \begin{equation}\label{cdsew10} \sum_{k\geq 0}\binom{n}{3k+j}(-1)^{k+j}=2.3^{\frac{n}{2}-1}\cos(\frac{n \pi}{6}+\frac{2j\pi}{3}). \end{equation} \end{lem} \begin{proof} Let $z_1=e^{(2i\pi)/3}$ and $z_2=e^{(4i\pi)/3}$ be the two cubic primitive roots of unity, and let $f(z)=\sum_{k\geq 0}a_kz^k$ be some convergent power series. Since for all $j$, $0 \leq j \leq 2$, \begin{equation*} \frac{1+z_1^{n-j}+z_2^{n-j}}{3}=\begin{cases} 1, ~~\text{if}~n\equiv j \pmod 3;\\ 0,~~\text{otherwise },\end{cases} \end{equation*} it follows that \begin{equation*} \frac{f(z) + \frac{1}{z_1^j}f(z_1z) +\frac{1}{z_2^j} f(z_2z)} {3}=\sum_{k\geq 0}a_{3k+j}z^{3k+j}. \end{equation*} Hence, defining $g_j(z)$, $0 \leq j \leq 2$, by \begin{equation*} g_j(z)= \sum_{k\geq 0}\binom{n}{3k+j}z^{3k+j}, \end{equation*} and taking $f(z)=(1+z)^n$, we get \begin{equation*} g_j(z)=\frac{f(z) + \frac{1}{z_1^j}f(z_1z) +\frac{1}{z_2^j} f(z_2z)} {3}. \end{equation*} By making the substitution $z=-1$, we obtain \begin{eqnarray*} g_j(-1)&=&\sum_{k\geq 0}\binom{n}{3k+j}(-1)^{k+j}\\ &=&\frac{\frac{1}{z_1^j}(1-z_1)^n +\frac{1}{z_2^j} (1-z_2)^n}{3}\nonumber \\ &=&\frac{1}{3}\left\{\frac{1}{z_1^j}\left(\frac{3-i\sqrt{3}}{2}\right)^n+\frac{1}{z_2^j} \left(\frac{3+i\sqrt{3}}{2}\right)^n\right\}. \end{eqnarray*} To get (\ref{cdsew10}), we need only transform the right hand-side of the last equality. \end{proof} \begin{cor}\label{nbcdse2} Let $m$ be an odd integer $\geq 3$ of the form (\ref{jhgvfre}), and let $\alpha_h (m)$ be the quantity defined by (\ref{vfrd}). For all $h$, $0\leq h \leq 2$, we have \begin{equation}\label{sansd 3} \alpha_h(m)=\eta(m) \cos \left((\omega_2-\omega_1)\frac{\pi}{6}+4h\frac{\pi}{3}\right) \end{equation} where \begin{equation}\label{sadsqnsd 3} \eta(m)=2.3^{\frac{\omega }{2}-1}. \end{equation} \end{cor} \begin{proof} From (\ref{vfrd}), for all $h$, $0\leq h \leq 2$, we have \begin{equation*} \alpha_h(m)= \sum_{\substack{d\dv \widetilde{m},~d \in {\it O}(g^h)}} \mu(d). \end{equation*} First, let us suppose that $\omega_1\neq 0$ and $\omega_2 \neq0$. By (\ref{arith 0}) and (\ref{jhgvfre}), we obtain that for all $h$, $0 \leq h \leq 2$, \begin{equation*} \alpha_h(m)=\sum_{i_1= 0}^{\omega_1}(-1)^{i_1}\binom{\omega_1}{i_1}\sum_{\substack{i_2=0\\i_2 \equiv i_1+2h \pmod 3}}^{\omega_2}(-1)^{i_2}\binom{\omega_2}{i_2}. \end{equation*} So that, by (\ref{cdsew10}), we get \begin{eqnarray*} \alpha_h(m)&=&\sum_{i_1=0}^{\omega_1}(-1)^{i_1}\binom{\omega_1}{i_1}2.3^{\frac{\omega_2}{2}-1}\cos\left(\frac{\omega_2 \pi}{6}+\frac{2(i_1+2h)\pi}{3}\right)\nonumber \\ &=&2.3^{\frac{\omega_2}{2}-1}\sum_{j=0}^{2}\cos\left(\frac{\omega_2 \pi}{6}+\frac{2(j+2h)\pi}{3}\right)\sum_{\substack{i_1=0\\i_1 \equiv j \pmod 3}}^{\omega_1}(-1)^{i_1}\binom{\omega_1}{i_1}, \end{eqnarray*} which, by (\ref{cdsew10}) again, gives \begin{eqnarray*} \alpha_h(m)&=&4.3^{\frac{\omega }{2}-2}\sum_{j=0}^{2}\cos\left(\frac{\omega_2 \pi}{6}+\frac{2(j+2h)\pi}{3}\right)\cos\left(\frac{\omega_1 \pi}{6}+\frac{2j\pi}{3}\right). \end{eqnarray*} Consequently, to get (\ref{sansd 3}), one need only use the elementary trigonometric formulas \begin{equation*} \cos a\cos b=\frac{1}{2}\left( \cos(a+b)+ \cos(a-b)\right)~\textrm{for all}~a ~ \textrm{and }~b ~\textrm{in} ~\R \end{equation*} and \begin{equation*} \cos c+\cos(c+\frac{2\pi}{3})+\cos(c+\frac{4\pi}{3})=0,~\textrm{for all}~c \in \R. \end{equation*} In case $\omega_1=$ or $\omega_2=0$, (\ref{sansd 3}) follows immediately from (\ref{cdsew10}). \end{proof} \begin{thm}\label{hgfdrce} Let $m$ be an odd integer $\geq 3$ of the form (\ref {jhgvfre}). Let $\eta (m)$ be as defined in (\ref{sadsqnsd 3}), and let $R_m$ be the polynomial given by (\ref{sfre}). Then \begin{equation}\label{Rm1(y)} R_m(y)=\left(m^3y^3-\frac{3}{4}pm\eta^2(m)y+\nu_p\right)^{s_p}, \end{equation} with \begin{equation}\label{1111444mmmpp} \nu_p=\left\{% \begin{array}{lll} \frac{1}{8}\left(-1\right)^{\frac{\omega_2-\omega_1}{2}}p\eta^3(m)L, & \hbox{\textrm{if $\omega_2-\omega_1$ is even }}; \\ & \\ \frac{3\sqrt{3}}{8}\left(-1\right)^{\frac{\omega_2-\omega_1-1}{2}}p\eta^3(m)M, & \hbox{\textrm{if $\omega_2-\omega_1$ is odd,}} \\ \end{array}% \right. \end{equation} where $L$ and $M$ are the unique integers satisfying $4p=L^2+27M^2$, $L\equiv 1$ (mod $3$) and $(g^{2})^{(p-1)/3} \equiv \frac{L+9M}{L-9M}$ (mod $p$). \end{thm} \begin{proof} From (\ref{Rm(yy)}), we have \begin{eqnarray}\label{Rm(y)} R_m(y)&=&\prod_{l=0}^{2}\left(my+A_m(\theta^{g^l})\right)^{s_p}\nonumber\\ &=& \left(m^3y^3+m^2\nu''_py^2+m\nu'_py+\nu_p\right)^{s_p}, \end{eqnarray} where \begin{equation}\label{dfcsve, 1} \nu''_p=A_m(\theta)+A_m(\theta^{g})+A_m(\theta^{g^2}), \end{equation} \begin{equation}\label{dfcsve, 2} \nu'_p=A_m(\theta)A_m(\theta^{g})+A_m(\theta)A_m(\theta^{g^2})+A_m(\theta^{g})A_m(\theta^{g^2}) \end{equation} and \begin{equation}\label{dfcsve, 3} \nu_p=A_m(\theta)A_m(\theta^{g})A_m(\theta^{g^2}). \end{equation} Recall that cf. Theorem \ref{sfj14jjjj}, $\delta$ is the unique integer in $ \left\{0,1,2\right\}$ such that $m\in {\it O}(g^\delta)$. So that from (\ref{deee})-(\ref{dee}) and (\ref{hgtrvcdes}), we get for $i \in\{0,1,2\}$ \begin{equation}\label{hhttrefd} A_m(\theta^{g^i})=\sum_{h=0}^{2}\alpha_h(m)T_{\delta -h+i }. \end{equation} {\bf Computation of $\nu''_p$.}\\ From (\ref{hhttrefd}) and (\ref{dfcsve, 1}) we deduce that \begin{equation*} \nu''_p=\biggl(\alpha_0(m)+\alpha_1(m)+\alpha_2(m)\biggr)\biggl(T_0+T_1+T_2\biggr). \end{equation*} Since $\gcd (m,p)=1$ and $m\neq 1$, it follows immediately from (\ref{vfrd}) and (\ref{orbital}) that \begin{equation}\label{lkytmlnb} \alpha_0(m)+\alpha_1(m)+\alpha_2(m)= \sum_{d\dv \widetilde{m}} \mu(d)=0 \end{equation} and thus $\nu''_p=0$.\\ {\bf Computation of $\nu'_p$.}\\ From (\ref{Rm(y)}), to prove (\ref{Rm1(y)}) it suffices to show (\ref{1111444mmmpp}) and that $\nu_p'=\frac {-3}{4}p\eta^2(m)$. By (\ref{dfcsve, 2}) and (\ref{hhttrefd}), we have \begin{equation*} \nu'_p=\sum_{k=0}^{2} \sum_{h=0}^{k}\alpha_h(m)\alpha_k(m) U(h,k) \end{equation*} with \begin{equation*} U(h,h)=\sum_{(i,j)\in \{(0,1),(0,2),(1,2)\}}T_{\delta-h+i}T_{\delta-h+j} \end{equation*} and, for $h< k$, \begin{equation*} U(h,k)=\sum_{(i,j)\in \{(0,1),(0,2),(1,2)\}}\left( T_{\delta-h+i}T_{\delta-k+j}+ T_{\delta-k+i}T_{\delta-h+j} \right). \end{equation*} Observing that, for $0\leq \delta, h, k\leq 2$, $U(h,k)$ does not depend on $\delta$ and is equal to $T_0T_1+T_0T_2+T_1T_2$ when $h=k$ and to $T_0T_1+T_0T_2+T_1T_2+T_0^2+T_1^2+T_2^2$ when $h