\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{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}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} %\newtheorem{theorem}{Theorem}[section] %\newtheorem{proposition}{Proposition}[section] %\newtheorem{corollary}{Corollary}[section] %\newtheorem{lemma}{Lemma}[section] \def\NN{{\mathbb N}} \def\ZZ{{\mathbb Z}} \def\QQ{{\mathbb Q}} \begin{center} \vskip 1cm{\LARGE\bf {Perfect Powers With All Equal Digits But One}} \vskip 1cm \large Omar~ Kihel \\ Department of Mathematics \\ Brock University \\ St. Catharines, Ontario L2S 3A1 \\ Canada \\ \href{mailto:okihel@brocku.ca}{\tt okihel@brocku.ca} \\ \ \\ Florian~Luca\\ Instituto de Matem{\'a}ticas\\ Universidad Nacional Aut\'onoma de M{\'e}xico\\ C.P.~58089 \\ Morelia, Michoac{\'a}n \\ M{\'e}xico\\ \href{mailto:fluca@matmor.unam.mx}{\tt fluca@matmor.unam.mx}\\ \end{center} \vskip .2 in \begin{abstract} In this paper, among other results, we show that for any fixed integer $l \geq 3$, there are only finitely many perfect $l$-th powers all of whose digits are equal but one, except for the trivial families $10^{ln}$ when $l \ge 3$ and $8\cdot 10^{3n}$ if $l = 3$. \end{abstract} \vskip .2in \renewcommand{\theequation}{\arabic{equation}} \renewcommand{\theequation}{\arabic{equation}} \newtheorem{prop}{Proposition} \newtheorem{lemma}[prop]{Lemma} \newtheorem{cor}{Corollary} \newtheorem{corollary}[cor]{Corollary} \newtheorem{conj}[prop]{Conjecture} \newtheorem{defi}[prop]{Definition} \newtheorem{theorem}[prop]{Theorem} \newtheorem{fac}[prop]{Fact} \newtheorem{facs}[prop]{Facts} \newtheorem{com}[prop]{Comments} \newtheorem{prob}{Problem} \newtheorem{problem}[prob]{Problem} \newtheorem{ques}{Question} \newtheorem{question}[ques]{Question} \newtheorem{remark}[prop]{Remark} \section{Introduction} Obl\'ath \cite{Ob} proved that the only perfect powers all of whose digits are equal to a fixed one $ a \neq 1$ in decimal representation are 4, 8 and 9. This is equivalent to saying that the diophantine equation \begin{equation} \label{eq:(1)} a\frac{x^{n}-1}{x - 1} = y^{q}, \; \mbox{in integers}\;~ n\geq 3,~ x \geq 2,~ 1 \leq a \leq x,~y \geq 2,~ q \geq 2 \end{equation} has no solution when $x = 10$ and $a \neq 1$. Inkeri \cite{In} extended Obl\'ath's result by proving that when $x \in \{3, \ldots, 10 \}$ and $a \neq 1$, equation (\ref{eq:(1)}) has the unique solution $(a,x,n,y,q) = (4, 7, 4, 40, 2)$. Thanks to results of Bugeaud and Mignotte \cite{BM}, we now know that equation (\ref{eq:(1)}) has only the following three solutions: $$ \frac{3^5 - 1}{3 - 1} = 11^2, \;~ \frac{7^4 - 1}{7 - 1} = 20^2\;~ \mbox{and} \;~ \frac{18^3 - 1}{18 - 1} = 7^3, $$ when $a = 1$ and $x \in \{2, \ldots, 10\}$. Gica and Panaitopol \cite{GP} studied a variant on Obl\'ath's problem. Namely, they found all squares of $k$ decimal digits having $k-1$ of their digits equal to each other. They asked to solve the analogous problem for higher powers. In the first part of this paper, we prove the following result. \begin{theorem} \label{thm:1} For a fixed integer $l \geq 3$, there are only finitely many perfect $l$-th powers all whose digits are equal but one, except for the trivial families $10^{ln}$ for $l\ge 3$ and $8\cdot 10^{3n}$ for $l = 3$. \end{theorem} Our main tool for the proof of Theorem \ref{thm:1} is the following result of Corvaja and Zannier from \cite{CZ}. \begin{theorem} \label{thm:2} Let $f(X,Y)=a_0(X)Y^d+\cdots+a_d(X)$ be a polynomial in $\QQ[X,Y]$ with $d\geq 2$ such that $a_0(X)\in \QQ$ and the polynomial $f(0,Y)$ has no multiple root. Let $i, j$ be integers $>1$ which are not relatively prime. If the equation $f(i^n,y)=j^m$ has an infinite sequence of solutions $(m,n,y)\in \ZZ^3$ such that $\min\{m,n,y\}\to\infty$, then there exist $h\ge 1$ and $p(X)\in\QQ[X]$ nonconstant such that $f(X^h,p(X))$ is nonconstant and has only one term. \end{theorem} We point out that in \cite{CZ}, it was shown that the pair $h\ge 1$ and $p(X)\in \QQ[X]$ with the property that $f(X^h,p(X))$ has only one term exists only under the hypothesis that $\min\{m,n\}\to\infty$. It was not shown that $p(X)$ is nonconstant. However, a close analysis of the proof of the result from \cite{CZ} shows that if $(m,n,y)$ is any infinite family of integer solutions to the equation $f(i^n,y)=j^m$ with $\min\{m,n\}\rightarrow\infty$, then an infinite subfamily of such solutions have the property that $y$ is in the range of the polynomial $p(X)$; thus, if $y\to\infty$, then $p(X)$ cannot be a constant polynomial. Similarly, it is not specifically said in \cite{CZ} that $f(X^h,p(X))$ is nonconstant but this is also clear from the arguments from \cite{CZ}. \section{Proofs of Theorems 1 and 2} {\sc Proof of Theorem 1.} Suppose that $l \geq 3$ is a fixed integer. Consider a perfect $l$-th power with all identical digits but one of them. Writing it first as $$ x^l={\overline{a\dots aba\dots a}}_{(10)}, $$ it follows that we may also rewrite it as \begin{equation} \label{eq:(2)} x^l=a\frac{10^{n}-1}{9}+c10^m, \end{equation} where $c=b-a$. If $a=0$, we then get $x^l=b\cdot 10^m$, which easily leads to the conclusions that $m$ is a multiple of $l$, $b=1$ if $l \neq 3$, and $b\in \{1,8\}$ if $l = 3$. \medskip {From} now on, we assume that $a\ne 0$. We also assume that $c\ne 0$, otherwise we recover Obl\'ath's problem. We let $$ f(X,Y)=\frac{1}{c}Y^l-\frac{a}{9c}(X-1). $$ Since $a\ne 0$, the polynomial $f(X,Y)$ fulfills the hypothesis from Theorem \ref{thm:2}. If $(m,n,x)$ is an integer solution of equation (\ref{eq:(2)}), then $$ f(10^n,x)=10^m. $$ Thus, we may take $i = j = 10$ in the statement of Theorem \ref{thm:2}. Assume that we have infinitely many solutions for equation (\ref{eq:(2)}). Since the pair $(a,c)$ can assume only finitely many values, it follows that we may assume that $(a,c)$ is fixed in equation (\ref{eq:(2)}). If $m$, remains bounded over an infinity of solutions, it follows that we may assume that it is fixed. But then, the sequence $$ u_n=\frac{a}{9}\cdot 10^n+\left(-\frac{a}{9}+c10^m\right)=A\cdot 10^n+B, $$ with $A=a/9$, and $B=-a/9+c\cdot 10^m$ is a binary recurrent sequence. If $B\ne 0$, then it is nondegenerate, therefore it can contain only finitely many perfect $l$-th powers (see \cite{ST}), which is a contradiction. If $B=0$, then $c\cdot 10^m=a/9$. Since $c$ is an integer and $a$ is a digit, we get that $a=9,~c=1,~m=0$, therefore $b=a+c=10$, which is a contradiction. \medskip Since $a \neq 0$, $\min\{m,n\} = m$. Thus, we may assume that $\min\{m,n\}\rightarrow\infty$. Clearly, $x\rightarrow\infty$ as well. By Theorem \ref{thm:2}, it follows that there exist a positive integer $h$ and a nonconstant polynomial $p(x)\in \QQ[X]$ such that $f(X^h,p(X))$ has only one term. Write $f(X^h,p(X))=qX^k$ for some nonzero rational $q$ and positive integer $k$. This leads to $$ p(X)^l=cqX^k+\frac{a}{9}(X^h-1):=r(X). $$ Assume $k\ne h$. Then $$ r'(X)=cqkX^{k-1}+\frac{a}{9}hX^{h-1}. $$ Since $r(0)=-a/9\ne 0$, and all roots of $r$ are of multiplicity at least $l \geq 3$, it follows that all roots of $r$ are also among the nonzero roots of $r'(X)$. If $k>h$, then these are the roots of $r_1(x)=cqkX^{k-h}+ah/9$, while if $h>k$, then these are the roots of $r_2(X)=ah/9X^{h-k}+cqk$. In both cases, the roots of $r(X)$ are multiple roots of $r_1(X)$ (or $r_2(X)$, respectively). However, since $ahcqk\ne 0$, neither the polynomial $r_1(X)$ nor the polynomial $r_2(X)$ has multiple roots. Thus, we get $k=h$, and $$ p(X)^l=(cq+a/9)X^h-a/9. $$ Since $a \neq 0$, the above relation is impossible (again, the polynomial on the right is nonconstant since $p(X)$ is nonconstant and does not admit multiple roots). This shows that indeed equation (\ref{eq:(2)}) has only finitely many nontrivial solutions; i.e., solutions different from $x^l=10^{ln}$ if $l\ge 3$ and from $x^3=8\cdot 10^{3n}$ if $l = 3$. \medskip Bugeaud \cite{Bu} extended Inkeri's result from \cite{In} to other values of the basis $x$. He completely solved equation (\ref{eq:(1)}) for $x \leq 100$ and also for $x = 1000$. His result includes that positive integers of the form $$ {\overline{aaaa \ldots aa}}_{(10)},\qquad {\overline{abab \ldots ab}}_{(10)},\qquad {\overline{abcabc \ldots abc}}_{(10)} $$ cannot be perfect powers except for the integers $a$, $ab$ and $abc$, when these integers themselves are perfect powers. Here, we will consider the problem of which perfect powers have the form $$ {\overline{aa \ldots ab \ldots b}}_{(10)} $$ when written in decimal representation, where the number of $a$'s and the number of $b$'s are not necessarily equal. We prove the following theorems. \begin{theorem} \label{thm:3} The only squares of the form $${\overline{aa \ldots ab \ldots b}}_{(10)}$$ in decimal representation are the trivial infinite families $10^{2i},\; 4\cdot 10^{2i}$, $9\cdot 10^{2i}$ with $i \in \NN$ together with 16, 25, 36, 49, 64, 81, 144, 225, 441, 1444 and 7744. \end{theorem} \begin{theorem} \label{thm:4} For every fixed integer $l \geq 3$, there are only finitely many perfect $l$-th powers of the form $$ {\overline{aa \ldots ab \ldots b}}_{(10)} $$ when written in decimal representation, except for the trivial infinite families $10^{ln}$ if $ l \ge 3$ and $8\cdot 10^{3n}$ if $l = 3$. \end{theorem} {\sc Proof of theorem 3.} Suppose that $ 1 \leq a \leq 9$ and $0\leq b \leq 9$ are two integers, not necessarily equal, such that \begin{equation} \label{eq:(3)} {\overline{aa \ldots ab \ldots b}}_{(10)} = y^2, \end{equation} where the number of $a$'s in equation (\ref{eq:(3)}) is $n$ and the number of $b$'s is $m$. It is easy to verify as done by Gica and Panaitopol \cite{GP} that the last 4 digits of a square are equal only when they are equal to $0$. Thus, if $ m > 3$, then the integer $b$ is equal $0$. Hence, equation (\ref{eq:(3)}) yields $$ 10^{m} \cdot a \frac{10^{n} - 1}{9} = y^2 , $$ which easily leads to the conclusion that $m$ is an even number; i.e., $m = 2i$ for a certain integer $i \in \NN$ and $$ {\overline{aa \ldots a}}_{(10)} = Y^2 . $$ This last equation is Obl\'ath's problem for squares which is known to have only the solutions $a =1$, $a=4$, $a = 9$ and $n = 1$. \medskip We suppose now that $m \leq 3 $. Equation (\ref{eq:(3)}) can be solved using congruences for few values of $a$ and $b$ but not for all values $ 1 \leq a \leq 9$ and $0\leq b \leq 9$. So, we proceed as follows. \medskip \noindent $\bullet$ If $m = 3$, then equation (\ref{eq:(3)}) yields $$ 10^3 aa \ldots a + 111b = y^2. $$ Hence, \begin{equation} \label{eq:(4)} 10^{n+3} a - 10^3 a + 999b = (3y)^2. \end{equation} If $n\equiv 0\pmod 3$; i.e., if $n = 3N$ for a some integer $N$, then equation (\ref{eq:(4)}) yields \begin{equation} \label{eq:(5)} Y^2 = X^3 - 10^3 a^3 + 999 a^2 b, \end{equation} where $Y = 3ay$ and $X = 10^{N+1}a.$ \medskip If $n \equiv 1\pmod 3$; i.e., if $n = 3N + 1$ for some integer $N$, then equation (\ref{eq:(4)}) yields \begin{equation} \label{eq:(6)} Y^2 = X^3 - 10^5 a^3 + 99900 a^2 b, \end{equation} where $Y = 30ay$ and $X = 10^{N+2}a.$ \medskip If $n \equiv 2\pmod 3$; i.e., if $n = 3N + 2$ for some integer $N$, then equation (\ref{eq:(4)}) yields \begin{equation} \label{eq:(7)} Y^2 = X^3 - 10^7 a^3 + 9990000 a^2 b, \end{equation} where $Y = 300ay$ and $X = 10^{N+3}a.$ \medskip \noindent For fixed values of $a$ and $b$, equations (\ref{eq:(5)}), (\ref{eq:(6)}) and (\ref{eq:(7)}) represent elliptic curves. We used SIMATH to find all integral points on these elliptic curves. The only solution that we found which corresponds to an integer solution to equation (\ref{eq:(1)}) is $x=1444.$ \medskip \noindent $\bullet$ If $m = 2$, then equation (\ref{eq:(3)}) yields to \begin{equation} \label{eq:(8)} 10^{n + 2}a - 10^2 a + 99b = (3y)^2. \end{equation} The same technique used above reduces equation (\ref{eq:(8)}) to finding integral points on a family of elliptic curves. We used SIMATH and found that the integer solutions that correspond to solutions to equation (\ref{eq:(3)}) are $x=144$ and $x=7744.$ \medskip \noindent $\bullet$ If $m = 1$, then we can use the same technique as above but this is already a particular case of Gica and Panaitopol's results from \cite{GP}. These solutions are then $16,~ 25,~ 36,~ 49,~ 81,~ 225$ and $441.$ \medskip \noindent {\sc Proof of Theorem 4.} Suppose that $l \geq 3$ is a fixed integer and that $1 \leq a \leq 9$ and $0 \leq b \leq 9$ are two integers such that $$ {\overline{aa \ldots ab \ldots b}}_{(10)} = x^{l} , $$ where the number of $a$'s is $n$ and the number of $b$'s is $m$. It follows that we may write rewrite this equation as \begin{equation} \label{eq:(9)} x^l = a 10^{m} \frac{10^{n} - 1}{9} + b \frac{10^{m} - 1}{9}. \end{equation} Hence, \begin{equation} \label{eq:(10)} x^l = \frac{a}{9}10^{N} + \frac{c}{9} 10^{m} - \frac{b}{9}, \end{equation} where $N = n + m$ and $c=b-a$. We suppose that $c \neq 0$, otherwise we recover Obl\'ath's problem. We let $$ f(X,Y) = \frac{9}{c} Y^l - \frac{a}{c} X + \frac{b}{c}. $$ If $b = 0$, we then get $$10^{m}\cdot a \frac{10^{n} -1}{9} = x^l, $$ which leads to the conclusion that $m$ is a multiple of $l$, $n = 1$, $a = 1$ if $l \neq 3$, and $a \in \{1,8\}$ if $l = 3$. We now assume that $b \neq 0$. The polynomial $f$ fulfills the hypothesis of Theorem \ref{thm:2}. If $(m,N,x)$ is a solution of equation (\ref{eq:(10)}), then $$ f(10^{N},x) = 10^{m}. $$ Thus, we may take $i = j = 10$ in the statement of Theorem \ref{thm:2}. Since the pair $(a, b)$ can assume only finitely many values, it follows that we may assume that the $(a, b)$ is fixed in equation (\ref{eq:(10)}). If $m$ remains bounded over an infinity of solutions, it then follows that we may assume that it is fixed. But then, the sequence $$ u_N=\frac{a}{9}\cdot 10^N+\left(-\frac{b}{9}+\frac{c}{9}10^m\right)=A\cdot 10^N+B, $$ where $A=a/9$ and $B = -{b}/{9}+({c}/{9})10^m$ is a binary recurrent sequence. If $B\ne 0$, then it is nondegenerate, therefore it can contain finitely many perfect powers (see again \cite{ST}), which is a contradiction. If $B=0$, then $c\cdot 10^m=a/9$. Since $c$ is an integer and $a$ is a digit, we get that $a=9,~c=1,~m=0$, so $b=a+c=10$, which is a contradiction. Since $N > m$, $\min\{m, N\} = m$. So, we may assume that $\min\{m,N\}\rightarrow\infty$. It is also clear that $x\to\infty$. By Theorem \ref{thm:2}, it follows that there exist a positive integer $h$ and a nonconstant polynomial $p(x)\in \QQ[X]$, such that $f(X^h,p(X))$ is nonconstant and has only one term. Write $f(X^h,p(X))=qX^k$ for some nonzero rational $q$ and positive integer $k$. This leads to $$ p(X)^l=\frac{cq}{9}X^k+\frac{a}{9}X^h - \frac{b}{9}:=r(X). $$ Assume $k\ne h$. Then $$ r'(X)=\frac{cqk}{9}X^{k-1}+\frac{ah}{9}X^{h-1}. $$ Since $r(0)=-b/9\ne 0$, and all roots of $r$ are of multiplicity at least $l \geq 3$, it follows that all roots of $r$ are also among the nonzero roots of $r'(X)$. If $k>h$, then these are the roots of $r_1(x)=cqkX^{k-h}+ah$, while if $h>k$, then these are the roots of $r_2(X)=ahX^{h-k}+cqk$. In both cases, the roots of $r(X)$ are multiple roots of $r_1(X)$ (or $r_2(X)$, respectively). However, since $ahcqk\ne 0$, neither the polynomial $r_1(X)$ nor the polynomial $r_2(X)$ has multiple roots. Thus, we get $k=h$, and $$ p(X)^l = \frac{cq + a}{9}X^h - \frac{b}{9}. $$ Since $b \neq 0$, the above relation is impossible (again, the polynomial on the right is nonconstant since $P(X)$ is nonconstant and does not admit multiple roots). This shows that indeed equation (\ref{eq:(10)}) has only finitely many nontrivial solutions; i.e., solutions different from $x^l=10^{ln}$ if $ l \ge 3$ and from $x^3=8\cdot 10^{3n}$ if $l = 3$. \medskip \noindent {\bf Remark.} The proofs of Theorem 1 and Theorem 4 can easily be generalized to other basis. \section{Acknowledgments} The first author was supported in part by a grant from NSERC. Research of F.~L. was supported in part by grants PAPIIT IN104505, SEP-CONACyT 46755 and a Guggenheim Fellowship. \begin{thebibliography}{99} \bibitem{Bu} Y. Bugeaud, On the Diophantine equation $a(x^n-1)/(x-1)=y^q$, in {\it Number Theory: Proceedings of the Turku Symposium on Number Theory in Memory of Kustaa Inkeri, Turku, Finland, May 31-June 4, 1999\/}, Berlin, de Gruyter, 2001, pp.\ 19--24. \bibitem{BM} Y. Bugeaud and M. Mignotte, On integers with identical digits, {\it Mathematika,} {\bf 46} (1999), 411--417. \bibitem{CZ} P. Corvaja and U. Zannier, On the diophantine equation $f(a^m,y)=b^n$, {\it Acta Arith.} {\bf 94} (2000), 25--40. \bibitem{GP} A. Gica and L. Panaitopol, \href{http://www.cs.uwaterloo.ca/journals/JIS/VOL6/Panaitopol/panaitopol41.html}{On Obl\'ath's problem}, {\it J. Integer Seq.\/} {\bf 6} (2003), Paper 03.3.5. \bibitem{In} K. Inkeri, On the Diophantine equation $a(x^n-1)/(x-1)=y^m$, {\it Acta Arith.} {\bf 21} (1972), 299--311. \bibitem{Ob} R. Obl\'ath, Une propriet\'e des puissances parfaites, {\it Mathesis} {\bf 65} (1956), 356--364. \bibitem{ST} T.N. Shorey and R. Tijdeman, {\it Exponential Diophantine Equations}, Cambridge University Press, 1986. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2000 {\it Mathematics Subject Classification}: Primary 11D75; Secondary 11J75. \noindent \emph{Keywords: } perfect powers, digits. \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received October 3 2005; revised version received November 9 2005. Published in {\it Journal of Integer Sequences}, November 14 2005. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document} \end{document}