\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}{-.1in} \setlength{\textheight}{8.4in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \DeclareMathOperator{\des}{des} \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\perm}{perm} \DeclareMathOperator{\Sym}{Sym} \newcommand{\wh}{\widetilde{h}} \newcommand{\wlm}{\overline{F}} \DeclareMathOperator{\lcm}{lcm} \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} \newtheorem{question}[theorem]{Question} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \begin{center} \vskip 1cm{\LARGE\bf A Diophantine System Concerning \\ \vskip .11in Sums of Cubes } \vskip 1cm \large Zhi Ren\\ Mission San Jose High School\\ 41717 Palm Avenue\\ Fremont, CA 94539\\ USA\\ \href{renzhistc69@163.com}{\tt renzhistc69@163.com} \end{center} \vskip .2 in \begin{abstract} We study the Diophantine system \[\begin{cases} x_{1}+\cdots+x_{n}=a,\\ x_{1}^3+\cdots+x_{n}^3=b, \end{cases} \] where $a,b \in \mathbb{Q},ab\neq0,n\geq4$, and prove, using the theory of elliptic curves, that it has infinitely many rational parametric solutions depending on $n-3$ free parameters. Moreover, this Diophantine system has infinitely many positive rational solutions with no common element for $n=4$, which partially answers a question in our earlier paper. \end{abstract} \section{Introduction} Ren and Yang \cite{ry} considered the positive integer solutions of the Diophantine chains \begin{equation}\label{Eq1} \begin{cases} \begin{split} &\sum_{j=1}^nx_{1j}=\sum_{j=1}^nx_{2j}=\cdots=\sum_{j=1}^nx_{kj}=a,\\ &\sum_{j=1}^nx_{1j}^3=\sum_{j=1}^nx_{2j}^3=\cdots=\sum_{j=1}^nx_{kj}^3=b,\\ &n\geq2,~k\geq2, \end{split} \end{cases} \end{equation} where $a,b$ are positive integers and determined by $k$ $n$-tuples $(x_{i1},x_{i2},\ldots,x_{in}),i=1,\ldots,k$. For $n=2,k=2,$ Eq.~(\ref{Eq1}) has no nontrivial integer solutions \cite{s}, so we consider $n\geq3$. For $n=3,k=2$, Eq.~(\ref{Eq1}) reduces to \begin{equation}\label{Eq2} \begin{cases} x_{1}+x_{2}+x_{3}=y_{1}+y_{2}+y_{3},\\ x_{1}^3+x_{2}^3+x_{3}^3=y_{1}^3+y_{2}^3+y_{3}^3. \end{cases} \end{equation} Systems like \eqref{Eq2} has been investigated by many authors, at least since 1915 \cite[p.\ 713]{d}; see \cite{b,bb,bl,c1,c2,l}. Eq.~(\ref{Eq2}) is interesting because it reveals the relation between all of the nontrivial zeros of weight-$1$ $6j$ Racah coefficients and all of its non-negative integer solutions. More recently, Moreland and Zieve \cite{mz} showed that ``for triples $(a,b,c)$ of pairwise distinct rational numbers such that for every permutation $(A,B,C)$ of $(a,b,c)$, the conditions $(A+B)(A-B)^3\neq(B+C)(B-C)^3$ and $AB^2+BC^2+CA^2\neq A^3+B^3+C^3$ hold, then the Diophantine system \[\begin{cases} x+y+z=a+b+c,\\ x^3+y^3+z^3=a^3+b^3+c^3 \end{cases}\] has infinitely many rational solutions $(x,y,z)$." This gives a complete answer to Question 5 in an earlier paper of the author \cite{ry}. For $n=3,k\geq3$, Choudhry \cite{c2} proved that Eq.~(\ref{Eq1}) has a parametric solution in rational numbers, but the solutions are not all positive. There are arbitrarily long Diophantine chains of the form Eq.~(\ref{Eq1}) with $n=3$. For $n\geq 3$, Ren and Yang \cite{ry} obtained a special result of Eq.~(\ref{Eq1}) with $(x_1,x_2,\ldots,x_{n-3})=(1,2,\ldots,n-3)$, which leads to Eq.~(\ref{Eq1}) has infinitely many coprime positive integer solutions for $n\geq3$. Now we study the case of Eq.~(\ref{Eq1}) for $n\geq 4$ with the greatest possible generality. For convenience, let us consider the non-zero rational solutions of the Diophantine system \begin{equation}\label{Eq3} \begin{cases} x_{1}+\cdots+x_{n}=a,\\ x_{1}^3+\cdots+x_{n}^3=b, \end{cases} \end{equation} where $a,b \in \mathbb{Q},ab\neq0,n\geq4$. Using the theory of elliptic curves, we prove the following theorems: \begin{theorem}\label{Thm1} For $n\geq 4$, the Diophantine system (\ref{Eq3}) has infinitely many rational parametric solutions depending on $n-3$ free parameters. \end{theorem} \begin{theorem}\label{Thm2} For $n=4$, the Diophantine system (\ref{Eq3}) has infinitely many positive rational solutions. \end{theorem} From these two theorems, we have \begin{corollary}\label{Cor3} For $n\geq 4$ and every positive integer $k$, there are infinitely many primitive sets of $k$ $n$-tuples of polynomials in $\mathbb{Z}[t_1,t_2,\ldots,t_{n-3}]$ with the same sum and the same sum of cubes. \end{corollary} \begin{corollary}\label{Cor4} For $n=4$ and every positive integer $k$, there are infinitely many primitive sets of $k$ $4$-tuples of positive integers with the same sum and the same sum of cubes. \end{corollary} \section{The proofs of the theorems}\label{TPT} In this section, we give the proofs of our theorems, which are related to the rational points of some elliptic curves. The proof of Theorem~\ref{Thm1} is inspired by the method of \cite{u}. \begin{proof} In view of the homogeneity of Eq.~(\ref{Eq3}), we let $a,b \in \mathbb{Z},ab\neq0$. First, we prove it for $n=4$ and then deduce the solution of Eq.~(\ref{Eq3}) for all $n\geq 5$. In the following Diophantine system \begin{equation}\label{Eq4} x_{1}+x_2+x_3+x_{4}=a,x_{1}^3+x_2^3+x_3^3+x_{4}^3=b, \end{equation} eliminating $x_4$ from the first equation and letting $x_3=tx_2$, we get \begin{equation}\label{Eq5} \begin{split} &3(tx_2+x_2-a)x_1^2+3(tx_2+x_2-a)^2x_1+3t(t+1)x_2^3\\ &-3a(t+1)^2x_2^2+3a^2(t+1)x_2+b-a^3=0. \end{split} \end{equation} To prove Theorem \ref{Thm1} for $n=4$, it is enough to show that the set of $x_2\in \mathbb{Q}(t)$, such that Eq.~(\ref{Eq5}) has a solution (with respect to $x_1$), is infinite. Then we need to show that there are infinitely many $x_2\in \mathbb{Q}(t)$ such that the discriminant of Eq.~(\ref{Eq5}) is a square, which leads to the problem of finding infinitely many rational parametric solutions on the following curve \[\begin{split} C:~y^2=&9(t^2-1)^2x_2^4+36at(t+1)x_2^3\\ &-18a^2(t+1)^2x_2^2+12(a^3-b)(t+1)x_2-3a(a^3-4b). \end{split}\] The discriminant of $C$ is \[\begin{split} \Delta(t)=&-5038848(t+1)^4\big((-b+a^3)t^2+(-2b-a^3)t-b+a^3\big)^2\\ &\big((9b^2+a^6-10a^3b)t^4+(-36b^2+14a^3b-2a^6)t^3+(54b^2-24a^3b+3a^6)t^2\\ &+(-36b^2+14a^3b-2a^6)t+9b^2+a^6-10a^3b\big),\end{split}\] and is non-zero as an element of $\mathbb{Q}(t)$. Then $C$ is smooth. By \cite[Prop.\ 7.2.1, p.\ 476]{co}, we can transform the curve $C$ into a family of elliptic curves \[\begin{split} E:~&Y^2=X^3-18a^2(1+t)^2X^2\\ &+108a(1+t)^2((a^3-4b)t^2+(2a^3+4b)t+a^3-4b)X\\ &-648(1+t)^2((a^6-8ba^3-2b^2)t^4+(-8ba^3+4b^2+4a^6)t^2+a^6-8ba^3-2b^2), \end{split}\] by the inverse birational map $\phi:~(x_2,y)\longrightarrow(X,Y)$. Because the coordinates of this map are quite complicated, we omit these equations. An easy calculation shows that the point \[\begin{split}P=&\bigg(18a^2(t^4+1)/(t-1)^2,36\big((a^3-b)t^6+(2b+a^3)t^5\\ &+(b-a^3)t^4+(4a^3-4b)t^3+(b-a^3)t^2+(2b+a^3)t-b+a^3\big)/(t-1)^3\bigg)\end{split}\] lies on $E$. To prove that the group $E(\mathbb{Q}(t))$ is infinite, it is enough to find a point on $E$ with infinite order. By the group law of the elliptic curves, we can get $[2]P$. Let $[2]P_2$ be the point of specialization at $t=2$ of $[2]P$. The $X$-coordinate of $[2]P_2$ is \[\frac{ 18a^2(-567b^2+2322ba^3+80937a^6)}{(-9b+111a^3)^2}.\] Let $E_2$ be the specialization of $E$ at $t=2$, i.e., \[E_2:~Y^2=X^3-162a^2X^2+972a(9a^3-12b)X-192456a^6+979776ba^3+104976b^2.\] There are two cases we need to discuss. 1. For $b=37a^3/3$, the curve $E_2$ becomes \[Y^2=X^3-162a^2X^2-135108a^4X+27859464a^6.\] Now $[2]P_2$ is the point at infinity on $E_2$, and we need find a point of infinite order. Let $Y'=Y/a^3,X'=X/a^2$. We have an elliptic curve \[E'_2:~Y'^2=X'^3-162X'^2-135108X'+27859464.\] It is easy to show that $Q=(234,-432)$ is a point of infinite order on $E'_2$. Then there are infinitely many rational points on $E'_2$ and $E$. 2. For $b\neq 37a^3/3$, when the numerator of the $X-$coordinate of $[2]P_2$ is divided by the denominator with respect to $b$, the remainder equals \[r=69984a^5(-3b+43a^3).\] 1. For $a\neq 0$ and $b\neq 43a^3/3$, we see that $r$ is not zero. By the Nagell-Lutz theorem (\cite[p.\ 56]{st}), $[2]P_2$ is a point of infinite order on $E_2$. Thus $P$ is a point of infinite order on $E$. 2. For $a\neq 0$ and $b=43a^3/3$, the curve $E_2$ becomes \[Y^2=X^3-162a^2X^2-158436a^4X+35417736a^6.\] Let $Y'=Y/a^3,X'=X/a^2$. We have an elliptic curve \[E'_2:~Y'^2=X'^3-162X'^2-158436X'+35417736.\] It is easy to show that $R=(306,-648)$ is a point of infinite order on $E'_2$. Then there are infinitely many rational points on $E'_2$ and $E$. In summary, for $a,b \in \mathbb{Z},ab\neq0$, there are infinitely many rational points on $E$. By the birational map $\phi$, we can get infinitely many rational solutions of Eqs.~(\ref{Eq5}) and (\ref{Eq4}). This completes the proof of Theorem~\ref{Thm1} for $n=4$. Next, we will deal with Eq.~(\ref{Eq3}) for $n\geq 5$. Let $x_5',x_6',\ldots,x_n'$ be rational parameters and set \[a'=\sum_{i=5}^{n}x_i',~b'=\sum_{i=5}^{n}x_i'^3.\] From the proof of the previous part, we know that Eq.~(\ref{Eq4}) has infinitely many rational solutions \[(x'_{1j},x'_{2j},x'_{3j},x'_{4j}),j\geq1,\] depending on one parameter $t$ for $A=a-a'$ and $B=b-b'$. This leads to the conclusion that for each $j\geq 1$, the $n$-tuple of the following form \[x_1=x'_{1j},x_2=x'_{2j},x_3=x'_{3j},x_4=x'_{4j},x_i=x'_i,i\geq 5\] satisfies Eq.~(\ref{Eq3}). \end{proof} \begin{example} For $n=4$, from the point $[2]P$, we get \[\begin{split} x_1=&-\frac{q(t)}{3a^2t(t+1)(t^2-t+1)(t-1)^2p(t)},\\ x_2=&\frac{ah(t)}{(t+1)(t-1)^2p(t)},\\ x_3=&~tx_2,\\ x_4=&a-x_1-x_2-x_3=\frac{s(t)}{3a^2t(t+1)(t^2-t+1)(t-1)^2p(t)}, \end{split}\] where $q(t)$ and $s(t)$ have degree 13 as a polynomial of $\mathbb{Q}(t)$, $h(t)$ has degree 8, and $p(t)$ has degree 6. \end{example} From the above example, it seems too difficult to prove that these rational parametric solutions are positive, so we need a new idea to prove Theorem~\ref{Thm2}. \begin{proof} In the proof of Theorem \ref{Thm1}, for $n=4$ we get the curve \[\begin{split} C:~y^2=&9(t^2-1)^2x_2^4+36at(t+1)x_2^3\\ &-18a^2(t+1)^2x_2^2+12(a^3-b)(t+1)x_2-3a(a^3-4b). \end{split}\] The discriminant of $C$ is \[\begin{split} \Delta(t)=&-5038848(t+1)^4\big((-b+a^3)t^2+(-2b-a^3)t-b+a^3\big)^2\\ &\big((9b^2+a^6-10a^3b)t^4+(-36b^2+14a^3b-2a^6)t^3+(54b^2-24a^3b+3a^6)t^2\\ &+(-36b^2+14a^3b-2a^6)t+9b^2+a^6-10a^3b\big).\end{split}\] Let us consider $\Delta(t)=0$, so that $C$ has multiple roots. Put \[(-b+a^3)t^2+(-2b-a^3)t-b+a^3=0,\] and solving for $t$, we get\[t=\frac{2b+a^3\pm \sqrt{12ba^3-3a^6}}{-b+a^3}.\] In order to make $t$ be a rational number, take \[12ba^3-3a^6=c^2,\] where $c$ is a rational parameter. Then we have \[b=\frac{3a^6+c^2}{12a^3},~t=\frac{3a^3+c}{3a^3-c},~or~\frac{3a^3-c}{3a^3+c}.\] According to the symmetry of $t$, consider \[t=\frac{3a^3+c}{3a^3-c}.\] Let\[Y_1=Y+\frac{6atX}{t-1}+36(a^3-b)(t-1)(t+1)^2,\] we get \[\begin{split} E':~&Y_1^2=X^3-18a^2(t+1)^2X^2-108a(t+1)^2((a^3-4b)t^2+(4b+2a^3)t+a^3-4b)X\\ &-648(t+1)^2((a^6-8ba^3-2b^2)t^4+(-8ba^3+4b^2+4a^6)t^2+a^6-8ba^3-2b^2). \end{split}\] Substituting \[b=\frac{3a^6+c^2}{12a^3},~t=\frac{3a^3+c}{3a^3-c}\]into $E'$, we get \[Y_1^2=\frac{((3a^3-c)^2X+72a^2c^2)((3a^3-c)^2X-36a^2(c^2+9a^6))^2}{(3a^3-c)^6}.\] To get infinitely many solutions of $(Y_1,X)$, put \[(3a^3-c)^2X+72a^2c^2=d^2,\] which leads to \[X=\frac{d^2-72a^2c^2}{(3a^3-c)^2}.\] Then \[Y=-\frac{d(d+12ca)(27a^7+9ac^2-dc)}{c(3a^3-c)^3}.\] Tracing back, we get \[\begin{split} x_1=&\frac{(-3a^3+c)d^2+(54a^7+18ac^2)d+108a^2(3a^3+c)(3a^6+c^2)}{72a^3(dc+27a^7+9c^2a)},\\ x_2=&\frac{d(d+12ca)(3a^3-c)}{72a^3(dc+27a^7+9c^2a)},\\ x_3=&\frac{d(d+12ca)(3a^3+c)}{72a^3(dc+27a^7+9c^2a)},\\ x_4=&\frac{(-3a^3-c)d^2+(-54a^7-18ac^2)d+108a^2(3a^3-c)(3a^6+c^2)}{72a^3(dc+27a^7+9c^2a)}. \end{split}\] To prove $x_i>0,i=1,2,3,4$, assume that $a>0,c>0,d>0$. Then we have \[72a^3(dc+27a^7+9c^2a)>0,x_3>0,\] so we just need to consider the numerators of $x_1,x_2,x_4$. Moreover, set $3a^3-c>0$, we have $x_2>0$, and the discriminants of \[(-3a^3+c)d^2+(54a^7+18ac^2)d+108a^2(3a^3+c)(3a^6+c^2)\]and \[(-3a^3-c)d^2+(-54a^7-18ac^2)d+108a^2(3a^3-c)(3a^6+c^2)\] are $108(-c^2+45a^6)(3a^6+c^2)a^2>0$. We see that the intervals of $d$ such that $x_1>0,x_4>0$ are given by \[\bigg(\frac{3(9a^6+3c^2-\sqrt{\delta})a}{3a^3-c},\frac{3(9a^6+3c^2+\sqrt{\delta})a}{3a^3-c}\bigg)\] and\[\bigg(\frac{3(-9a^6-3c^2-\sqrt{\delta})a}{3a^3+c},\frac{3(-9a^6-3c^2+\sqrt{\delta})a}{3a^3+c}\bigg),\] respectively, where $\delta=405a^{12}+126a^6c^2-3c^4$. It is easy to show that \[\frac{3(-9a^6-3c^2+\sqrt{\delta})a}{3a^3+c}>0,\frac{3(9a^6+3c^2-\sqrt{\delta})a}{3a^3-c}<0,\] and \[\frac{3(9a^6+3c^2+\sqrt{\delta})a}{3a^3-c}>\frac{3(-9a^6-3c^2+\sqrt{\delta})a}{3a^3+c}.\] Hence if \[d\in \bigg(0,\frac{3(-9a^6-3c^2+\sqrt{\delta})a}{3a^3+c}\bigg),\] we have $x_1,x_4>0$. This completes the proof of Theorem~\ref{Thm2}. \end{proof} \begin{example}\label{Eg6} If we take $a=c=1$, then $t=2$, $b=1/3$, and \[ x_1=\frac{-d^2+36d+864}{36(d+36)}, x_2=\frac{d(d+12)}{36(d+36)}, x_3=\frac{d(d+12)}{18(d+36)}, x_4=\frac{-d^2-18d+216}{18(d+36)},\] where $d\in (0,-9+3\sqrt{33}\approx 8.233687940)$ and $d$ is a rational number. Taking $d=1,2,3,4,5,6$,\ $7,8$, we get eight $4$-tuples of positive rational solutions with the same sum $1$ and the same sums of cubes $1/3$, which are as follows: \[\begin{split}(x_1,x_2,x_3,x_4)=&\bigg(\frac{899}{1332},\frac{13}{1332},\frac{13}{666},\frac{197}{666}\bigg),\bigg(\frac{233}{342},\frac{7}{342},\frac{7}{171},\frac{44}{171}\bigg),\\ &\bigg(\frac{107}{156},\frac{5}{156},\frac{5}{78},\frac{17}{78}\bigg),\bigg(\frac{31}{45},\frac{2}{45},\frac{4}{45},\frac{8}{45}\bigg),\bigg(\frac{1019}{1476},\frac{85}{1476},\frac{85}{738},\frac{101}{738}\bigg),\\ &\bigg(\frac{29}{42},\frac{1}{14},\frac{1}{7},\frac{2}{21}\bigg),\bigg(\frac{1067}{1548},\frac{133}{1548},\frac{133}{774},\frac{41}{774}\bigg),\bigg(\frac{68}{99},\frac{10}{99},\frac{20}{99},\frac{1}{99}\bigg).\end{split}\] \end{example} \section{The proofs of the corollaries}\label{TPC} In this section, we give the proofs of the corollaries and two examples. \begin{proof} Take any $k$ rational parametric solutions $(x_{i1},\ldots,x_{i,n}),i\leq k$ of Eq.~(\ref{Eq3}), where $x_{i5}=t_2,\ldots,x_{in}=t_{n-3},i\leq k$ are parameters. Let $c=\lcm_{i,j}(x_{ij},j=1,\ldots,n,i\leq k)$, and write \[x_{ij}=\frac{y_{ij}}{c},y_{ij}\in \mathbb{Z}[t_1,t_2,\ldots,t_{n-3}],\]with $(\gcd_{i,j}(y_{ij},c))=1$ and $c \in \mathbb{Z}[t_1,t_2,\ldots,t_{n-3}]$, where $t_1=t$. Then \[\sum_{j=1}^{n}y_{ij}=ac,\sum_{j=1}^{n}y_{ij}^3=bc^3.\] Hence \[\gcd_{i,j}~(y_{ij})=1.\] For two sets of solutions $\{(x_{i1},\ldots,x_{in}),i\leq k\}$ and $\{(x'_{i1},\ldots,x'_{in}),i\leq k\}$, if the sets of $n$-tuples $\{(y_{i1},\ldots,y_{in}),i\leq k\}$ and $\{(y'_{i1},\ldots,y'_{in}),i\leq k\}$ coincide, then $d=d'$ and the $n$-tuples coincide. Since there are infinitely many choices of $k$ elements, for every $k$ there are infinitely many primitive sets of $k$ $n$-tuples of polynomials with the same sum and the same sum of cubes. This finishes the proof of Corollary~\ref{Cor3}. \end{proof} \begin{example} For $n=4$, we have the rational parametric solutions \[\begin{split} x_1=&-\frac{q(t)}{3a^2t(t+1)(t^2-t+1)(t-1)^2p(t)},\\ x_2=&\frac{ah(t)}{(t+1)(t-1)^2p(t)},\\ x_3=&~tx_2,\\ x_4=&a-x_1-x_2-x_3=\frac{s(t)}{3a^2t(t+1)(t^2-t+1)(t-1)^2p(t)}. \end{split}\] Multiply the least common multiple of the denominator of $x_i,i=1,\ldots,4$. When $a,b \in \mathbb{Z}$, we get that \[x_1=-q(t),x_2=3a^3t(t^2-t+1)h(t), x_3=3a^3t^2(t^2-t+1)h(t),x_4=s(t)\] are the $4$-tuples of polynomials in $\mathbb{Z}[t]$ satisfying Eq.~(\ref{Eq3}). \end{example} \begin{proof} The proof of Corollary \ref{Cor4} is similar to the proof of Corollary \ref{Cor3}, so we omit it. \end{proof} \begin{example} From the eight $4$-tuples of positive rational solutions of Example~\ref{Eg6}, we get the following eight $4$-tuples of positive integers \[\begin{split}(y_1,y_2,y_3,y_4)=&(150719584015,2179482305,4358964610,66055079090),\\ &(152140218230,4570736170,9141472340,57460683280),\\ &(153169889565,7157471475,14314942950,48670806030),\\ &(153837920236,9925027112,19850054224,39700108448),\\ &(154170771755,12860172325,25720344650,30561821290),\\ &(154192385490,15950936430,31901872860,21267915240),\\ &(153924475705,19186462295,38372924590,11829247430),\\ &(153386782640,22556879800,45113759600,2255687980)\end{split}\]with the same sum 223313110020 and the same sum of cubes 3712114854198399246457100577\ 336000. \end{example} \section{A remaining question}\label{ARQ} Ren and Yang \cite[Ques.\ 4]{ry} raised the following question: \begin{question}\label{ques9} Are there infinitely many $n$-tuples of positive integers, having no common element, with the same sum and the same sum of their cubes for $n\geq4$? \end{question} It's easy to calculate that any $4$-tuples $(x_1,x_2,x_3,x_4)$, given by the same method from Example \ref{Eg6}, have no common element for $d\in \mathbb{Q}\bigcap(0,-9+3\sqrt{33})$. This gives a positive answer to Question~\ref{ques9} for $n=4$. When $n\geq 5$, it seems out of our reach. However, we conjecture that the answer to Question~\ref{ques9} is yes. \section{Acknowledgment} The author would like to thank the referee for his valuable comments and suggestions. \begin{thebibliography}{99} \bibitem {b} A. Bremner, Diophantine equations and nontrivial Racah coefficients, \textit{J. Math. Phys.} \textbf{27} (1986), 1181--1184. \bibitem {bb} A. Bremner and S. Brudno, A complete determination of the zeros of weigh-$1$ $6j$ coefficients, \textit{J. Math. Phys.} \textbf{27} (1986), 2613--2615. \bibitem {bl} S. Brudno and J. D. Louck, Nontrivial zeros of weight-$1$ $3j$ and $6j$ coefficients: Relation to Diophantine equations of equal sums of like powers, \textit{J. Math. Phys.} \textbf{26} (1985), 2092--2095. \bibitem {c1} A. Choudhry, Symmetric Diophantine systems, \textit{Acta Arith.} \textbf{59} (1991), 291--307. \bibitem {c2} A. 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Sinha, A relation between the coefficients and the roots of two equations and its applications to diophantine problems, \textit{J. Res. Nat. Bur. Standards, Sect. B} \textbf{74B} (1970), 31--36. \bibitem {u} M. Ulas, On some Diophantine systems involving symmetric polynomials, to appear, \textit{Math. Comp.}. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 11D25; Secondary 11D72, 11G05. \noindent \emph{Keywords: } Diophantine system, $n$-tuple, elliptic curve. \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received August 4 2013; revised version received September 4 2013. Published in {\it Journal of Integer Sequences}, September 8 2013. Minor revision, November 1 2013. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document} .