\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage{verbatim,upref,amsxtra} \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}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} %%%%TBC: need the eps file \end{center} \begin{center} \vskip 1cm{\LARGE\rm New Upper Bounds for\\[2mm] Taxicab and Cabtaxi Numbers} \vskip 1cm \large Christian Boyer\\ 53, rue de Mora \\ FR-95880 Enghien-les-Bains \\ France\\ \href{mailto:cboyer@club-internet.fr}{cboyer@club-internet.fr}\\ \end{center} \vskip .2in \begin{abstract} Hardy was surprised by Ramanujan's remark about a London taxi numbered 1729: ``it is a very interesting number, it is the smallest number expressible as a sum of two cubes in two different ways''. In memory of this story, this number is now called Taxicab(2) $= 1729 = 9^3 + 10^3 = 1^3 + 12^3$, Taxicab$(n)$ being the smallest number expressible in $n$ ways as a sum of two cubes. We can generalize the problem by also allowing differences of cubes: Cabtaxi$(n)$ is the smallest number expressible in $n$ ways as a sum or difference of two cubes. For example, Cabtaxi$(2)$ $= 91 = 3^3 + 4^3 = 6^3 - 5^3.$ Results were only known up to Taxicab$(6)$ and Cabtaxi$(9)$. %by the end of the 17th century. This paper presents a history of the two problems since Fermat, Frenicle and Vi\`ete, and gives new upper bounds for Taxicab$(7)$ to Taxicab$(19)$, and for Cabtaxi$(10)$ to Cabtaxi$(30)$. Decompositions are explicitly given up to Taxicab$(12)$ and Cabtaxi$(20)$. \end{abstract} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}{Proposition}[section] \newtheorem{corollary}{Corollary}[section] \newtheorem{lemma}{Lemma}[section] \newtheorem{definition}{Definition}[section] \section{A Fermat problem solved by Frenicle} Our story starts 350 years ago, with letters exchanged between France and England during the reign of Louis XIV and the protectorate of Oliver Cromwell. On August 15th 1657, from Castres (in the south of France), Pierre de Fermat sent to Kenelm Digby some mathematical problems. Translated into English, two of them were: \begin{enumerate} \item Find two cube numbers of which the sum is equal to two other cube numbers. \item Find two cube numbers of which the sum is a cube. \end{enumerate} \noindent These two statements can be written algebraically as follows: \begin{align} x^3 + y^3 &= z^3 + w^3\label{1}\\[1mm] x^3 + y^3 &= z^3.\label{2} \end{align} \noindent Fermat asked Digby, who was living in Paris at that time, to pass the problems on to William Brouncker, John Wallis, and Bernard Frenicle de Bessy, defying them to find solutions. Frenicle succeeded in finding several numerical solutions to (\ref{1}), as announced in October 1657 in a letter sent by Brouncker to Wallis. The first solutions by Frenicle are: \begin{align} 1729 &= 9^3 + 10^3 = 1^3 + 12^3\nonumber\\[1mm] 4104 &= 9^3 + 15^3 = 2^3 + 16^3\nonumber\\[1mm] \ldots\nonumber \end{align} %\bigskip %%%TBC \begin{center} \includegraphics[width=16.5cm]{TaxicabFig1x.eps} %0.2 %TBC %\includegraphics[width=16.5cm]{TaxicabFig1x.pdf} \end{center} %\smallskip \begin{center} {\footnotesize {\sc Figure} 1: Colbert presenting the founding members of the Acad\'emie Royale des Sciences to Louis XIV, in 1666. Bernard Frenicle de Bessy (Paris circa 1605 -- Paris 1675), one of the founding members, is probably among the people on the left. [Painting by Henri Testelin, Mus\'ee du Ch\^ateau de Versailles, MV 2074].\\ } \end{center} \bigskip \begin{center} \includegraphics[width=16.5cm]{TaxicabFig2x.eps} %0.2 %TBC %\includegraphics[width=16.5cm]{TaxicabFig2x.pdf} %0.2 \end{center} %\smallskip \begin{center} {\footnotesize {\sc Figure} 2: The two smallest of Frenicle's solutions found in 1657,\\ as published in Wallis's Commercium Epistolicum, Epistola X, Oxford, 1693.\\ } \end{center} \newpage %%%TBC \bigskip Brouncker added that Frenicle said nothing about equation (\ref{2}). Slightly later, in February 1658, Frenicle sent numerous other solutions of (\ref{1}) to Digby without any explanation of the method used. Fermat himself worked on numbers which are sums of two cubes in more than two ways. Intelligently reusing Vi\`ete's formulae for solving $x^3 = y^3 + z^3 + w^3$, he proved in his famous comments on Diophantus that it is possible to construct a number expressible as a sum of two cubes in $n$ different ways, for any $n$, but his method generates huge numbers. We know now that Fermat's method essentially uses the addition law on an elliptic curve. See also Theorem 412 of Hardy \& Wright, using Fermat's method \cite[pp. 333--334 \& 339]{HW} . It was unknown at the time whether equation (\ref{2}) was soluble; we recognize Fermat's famous last theorem $x^n + y^n = z^n$, when $n = 3$. This particular case was said to be impossible by Fermat in a letter sent to Digby in April 1658, and proved impossible more than one century later by Euler, in 1770. The general case for any $n$ was also said to be impossible by Fermat in his famous note written in the margin of the {\it Arithmetica} by Diophantus, and proved impossible by Andrew Wiles in 1993--1994. For more details on the Fermat /Frenicle/Digby/Brouncker/Wallis letters, see \cite{Bee}, \cite[pp.~551--552]{Dick}, %[12, pp.551-552], \cite{Rash, Tan1, Tan2, Wall}.% [31], We will now focus our paper on equation (\ref{1}). Euler worked on it \cite{Eul}, but the first to have worded it exactly as the problem of the ``smallest'' solution, which is the true Taxicab problem, seems to have been Edward B. Escott. It was published in 1897 in {\it L'Interm\'ediaire des Math\'ematiciens} \cite{Esc}: \begin{quote} Quel est le plus petit nombre entier qui soit, de deux fa\c{c}ons diff\'erentes, la somme de deux cubes? [In English: What is the smallest integer which is, in two different ways, the sum of two cubes?] \end{quote} Several authors responded \cite{Mor1} to Escott, stating that Frenicle had found 1729 a long time before. A more complete answer was given by C. Moreau \cite{Mor2}, listing all the solutions less than 100,000. C.~E. Britton \cite{Brit} listed all the solutions less than 5,000,000. These two lists are given in the Appendix, figures A1a and A1b. \section{Why is 1729 called a ``Taxicab'' number?} The problem about the number 1729 is now often called the ``Taxicab problem'', e.g., \cite[p.~212]{Guy}, %[18, p.~212], \cite{Mer, Slo1, EW}, in view of an anecdote, often mentioned in mathematical books, involving the Indian mathematician Srinivasa Ramanujan (1887--1920) and the British mathematician Godfrey Harold Hardy (1877--1947). Here is the story as related by Hardy and given, for example, in \cite[p.~xxxv]{Hardy-Rama}: %[19, p. xxxv]: \begin{quote} I remember once [probably in 1919] going to see him [Ramanujan] when he was lying ill in Putney [in the south-west of London]. I had ridden in taxicab No. 1729, and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavourable omen. ``No,'' he replied, ``it is a very interesting number; it is the smallest number expressible as a sum of two cubes in two different ways.'' \end{quote} As Euler did, Ramanujan worked on parametric solutions of (\ref{1}). For example, even if a similar formula had been previously found by Werebrusow \cite{Wer}, Ramanujan found \cite[p.~107]{Berndt}, %[2, p.~107], \cite[p.~387]{Ram} %[29, p.~387] the very nice condition \begin{align} \mbox{\rm If}\quad m^2 + mn + n^2 = 3a^2b,\quad \mbox{\rm then} \quad (m + ab^2)^3 + (bn + a)^3 = (bm + a)^3 + (n + ab^2)^3.\label{3} \end{align} This equation gives only a small proportion of the solutions. However, with $m=3, n=0, a=1$, and $b=3$, the equation yields $12^3 + 1^3 = 10^3 + 9^3 = 1729.$ %\end{document} %TBC \bigskip \begin{center} \includegraphics[width=7cm]{TaxicabFig3ax.eps}\hspace{1cm}\includegraphics[width=7cm]{TaxicabFig3bx.eps} \end{center} %\smallskip \begin{center} {\footnotesize {\sc Figure} 3. Equations handwritten by Ramanujan in two different notebooks:\\ \cite[p.~225]{Ram} (left panel), and \cite[p.~341]{Ram2} (right panel).\\ } \end{center} \bigskip Euler had published the complete parametric solution in rationals of (\ref{1}), but as Hardy and Wright \cite[p.~200]{HW} pointed out, %[20,p.200]: ``The problem of finding all integral solutions is more difficult''. In 1998, Ajai Choudhry published an interesting paper \cite{Chou} on the parametric solution in integers of (\ref{1}). \section{Notation used in this paper} In this paper, \underline{Taxicab$(n)$} %\footnote{How about underlining? A reader (like me) might think boldface {$\mathbf T(\mathbf n, \mathbf k)$} to have a different meaning than { $T(n,k)$} ....} denotes the smallest integer that can be written in $n$ ways as a sum of two cubes of positive integers. Example: \begin{align} \mbox{\rm Taxicab}(2) = 1729 = 12^3 + 1^3 = 10^3 + 9^3.\nonumber\ \end{align} Fermat proved that {\rm Taxicab}$(n)$ exists for any $n$. We let \underline{$T(n, k)$} denote the $k$th smallest primitive solution that can be written in $n$ ways as a sum of two cubes of positive integers, so that \begin{align} \mbox{\rm Taxicab}(n) = T(n, 1) \label{4} \end{align} Examples: \begin{align} T(2, 1) = 1729 = \mbox{\rm Taxicab}(2), \quad T(2, 2) = 4104.\nonumber\ \end{align} When {\rm Taxicab}$(n)$ is unknown, however, we let \underline{$T'(n, k)$} denote the $k$th smallest known primitive solution (at the time of the article) written in $n$ ways as a sum of two cubes of positive integers, and $T'(n, 1)$ is an upper bound of {\rm Taxicab}$(n)$: %(5) \begin{align} \mbox{\rm Taxicab}(n) \leq T'(n, 1) \label{5} \end{align} We let \underline{Cabtaxi$(n)$} denote the smallest integer that can be written in $n$ ways as a sum of two cubes of positive or negative integers. Example: \begin{align} \mbox{\rm Cabtaxi}(2) = 91 = 3^3 + 4^3 = 6^3 - 5^3.\nonumber\ \end{align} We let \underline{$C(n, k)$} denote the $k$th smallest primitive solution that can be written in $n$ ways as a sum of two cubes of positive or negative integers. \begin{align} %(6) \mbox{\rm Cabtaxi}(n) = C(n, 1) \label{6} \end{align} When {\rm Cabtaxi}$(n)$ is unknown, however, we let \underline{$C'(n, k)$} denote the $k$th smallest known primitive solution %(known at the time the article is being written) written in $n$ ways as a sum of two cubes of positive or negative integers. $C'(n, 1)$ is an upper bound of {\rm Cabtaxi}$(n)$: %(7) \begin{align} \mbox{\rm Cabtaxi}(n) \leq C'(n, 1). \label{7} \end{align} \bigskip% \section{1902--2002: from Taxicab$(3)$ to Taxicab$(6)$} After having asked the question above on Taxicab$(2)$, Escott asked about Taxicab$(3)$ in 1902 \cite{Esc15}. Find the smallest solution of the equation: \begin{align} %(8) u^3 + v^3 = w^3 + x^3 = y^3 + z^3. \label{8} \end{align} \begin{center} \includegraphics[height=8.5cm,width=16cm]{TaxicabFig4a.eps} \end{center} %\smallskip \begin{center} {\footnotesize {\sc Figure} 4. History of Taxicab numbers.\\ } \end{center} \newpage%\bigskip The Euler and Werebrusow \cite{Wer2} parametric solutions of (\ref{1}) and (\ref{8}) do not help us find the smallest solution. In 1908 Andr\'e G\'erardin \cite{Ger} suggested that the solution was probably \begin{align} 175959000 = 70^3 + 560^3 = 198^3 + 552^3 = 315^3 + 525^3.\nonumber\ \end{align} An important observation for our study and our future ``splitting factors'' is that G\'erardin's solution is equal to $35^3 * T(2, 2)$. Two out of its three sums come from the second solution $4104$ found by Frenicle as \begin{align} 70 = 2*35,\quad 560 &= 16*35,\nonumber\\[1mm] 315 = 9*35,\quad 525 &= 15*35.\nonumber \end{align} But $198^3 + 552^3$ is not a multiple of $35^3$ and can be considered as a ``new'' decomposition. The true Taxicab$(3)$ was discovered more than 50 years after Escott's question, and exactly 300 years after Frenicle's discovery of Taxicab$(2)$. Using an EDSAC machine, John Leech found, and published in 1957 \cite{Leech}, the five smallest 3-way solutions, the smallest of these five being \begin{align} \mbox{\rm Taxicab}(3) = 87539319 = 167^3 + 436^3 = 228^3 + 423^3 = 255^3 + 414^3.\nonumber \end{align} His results indicated that G\'erardin's solution was not Taxicab$(3)$, but $T(3, 4)$ = the fourth smallest primitive 3-way solution. %%%%%%%%%%%%%%%%%%%%%%%%%%%%% E. Rosenstiel, J.~A. Dardis \& C.~R. Rosenstiel found Taxicab$(4) = 6963472309248$, and first announced it in 1989 \cite{Mudge27}. They gave more detailed results in \cite{Rosen}, along with the next three smallest 4-way solutions. Until now, David W. Wilson was considered to have been the first to have discovered, in 1997, Taxicab$(5) = 48988659276962496$, see \cite{Wilson}, \cite[p.~391]{Bernstein}, \cite[p.~212]{Guy}. %[47] [3, p.391] [18, p.212]. But, as kindly communicated to me by Duncan Moore, this number had been previously found three years earlier in 1994 by John A. Dardis, one of the co-discoverers of Taxicab$(4)$, and published in the February 1995 ``Numbers count'' column of {\it Personal Computer World} \cite{Mudge28}. After Dardis in 1994 and Wilson in 1997, this same number was again found independently by Daniel J. Bernstein \cite{Bernstein} in 1998. Bill Butler also confirmed \cite{Butler} this number in 2006, while computing the fifteen 5-way solutions $< 1.024*10^{21}$. From 1997 to 2002, the best known upper bound of Taxicab$(6)$ was a 6-way solution found by David W. Wilson. In July 2002, Randall L. Rathbun found \cite{Rathbun} a better upper bound of Taxicab$(6), 2.42 *10^{22}$, adding: ``I don't believe that this is the lowest value sum for 6 positive cube pairs of equal value''. But it seems today that it probably is the lowest value! Calude, Calude \& Dinneen claimed in 2003 \cite{Cal9} that this upper bound is the true Taxicab$(6)$ with probability greater than $99\%$, and further claimed that results in 2005 \cite{Cal10} gave a probability greater than $99.8\%$, but these claims are not accepted by many mathematicians. And the computations done by Bill Butler proved that Taxicab$(6) > 1.024*10^{21}.$ %5 \section{Splitting factors} We have remarked that G\'erardin's solution was equal to $35^3 * T(2, 2)$. It is important to note that $T'(6, 1)$ is equal to $79^3 * $Taxicab$(5)$, as computed by Rathbun. Among the 6 decompositions, only one (underlined in Fig. 4) is a ``new'' decomposition: the others are $79^3$ multiples of the $5$ decompositions of Taxicab$(5)$. Rathbun also remarked that other multiples of Taxicab(5) are able to produce 6-way solutions: $127^3, 139^3$ and $727^3$. I add that they are not the only multiples of Taxicab$(5)$ producing 6-way solutions. The next one is $4622^3$, which indicates again, as for G\'erardin's solution, that non-prime numbers do not have to be skipped as we might initially assume: $79, 127, 139$ and $727$ were primes, but $4622 = 2*2311$ is not prime. {\bf {If $N$ is an $n$-way sum of two cubes, and if $k^3 N$ is an $(n+1)$-way sum of two cubes, then $k$ is called a ``splitting factor'' of $N$.}} This means that this $k$ factor ``splits'' $k^3 N$ into a new $(n+1)$th-way sum of two cubes, the $n$ other sums being directly the $k^3$ multiples of the already known $n$ ways of $N$. It was called the ``magnification technique'' by David W. Wilson. It is possible that other known 5-way solutions, if they have small splitting factors, may produce smaller 6-way solutions than Rathbun's upper bound. Using the list of 5-way solutions computed by Bill Butler \cite{Butler}, I have computed their splitting factors (Appendix, figure A3). These splitting factors give the smallest known 6-way solutions $< 10^{26}$ (Appendix, figure A4): the first one remains $79^3 * $Taxicab$(5)$, which means that it is impossible to do best with this method. We will use this Taxicab$(5)$ number as a basis for our search of upper bounds of Taxicab$(n)$, for larger $n$. The method used to find all our decompositions of $N$ into a sum of two cubes is as follows. We first factorize $N$, then build a list of all its possible pair of factors $(r, s)$ solving $N=rs,$ with $r < s$. Because any sum of two cubes can be written as %(9) \begin{align} N = rs = x^3 + y^3 = (x + y)(x^2 - xy + y^2), \label{9} \end{align} any possible sum of two cubes is an integer solution of the system (\ref{10}) for one of the possible pairs $(r, s)$: %(10) \begin{align} x + y = r, \quad x^2 - xy + y^2 = s. \label{10} \end{align} We search for integer solutions of this system by solving the resulting quadratic equation. Of course, most of the pairs $(r, s)$ do not give an integer solution $(x, y)$. %6 \section{Taxicab$(7)$ and beyond} The first idea is to use several of the existing splitting factors together. When we use $n$ factors together, we add $n$ new ways. For example, $127^3 * $Taxicab$(5)$ gives $5+1 = 6$ way-solutions, and $127^3 * 727^3 * $Taxicab$(5)$ gives $5+2 = 7$ way-solutions. Directly using this idea, the smallest 7-way solution is $79^3 * 127^3 * $Taxicab$(5)$. The second idea is to check, once a splitting factor is used, if a completely new splitting factor is possible on the new number. In our case, yes it is! A very pleasant surprise: $79^3 * $Taxicab$(5)$ has a new splitting factor $101$, called a ``secondary'' factor. And because $101$ is smaller than $127$, we have found a better 7-way solution $ 79^3 * 101^3 *\mbox{\rm Taxicab}(5)$ smaller than $ 79^3 * 127^3 * \mbox{\rm Taxicab}(5). $ It is possible that some other $T(5, i)$ could produce a smaller 7-way solution if it has a small secondary factor. This is not the case. For example, using $T(5, 6)$, the smallest possible 7-way solution is $25^3 * 367^3 * T(5,6)$, bigger than $79^3 * 101^3 * $Taxicab$(5)$. \bigskip \begin{center} \includegraphics[width=16cm]{TaxicabFig5x.eps} \end{center} %\smallskip \begin{center} {\footnotesize {\sc Figure} 5. Detailed list of splitting factors of Taxicab$(5)$.\\ } \end{center} \bigskip %%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{center} \includegraphics[width=16cm]{TaxicabFig6x.eps} \end{center} %\smallskip \begin{center} {\footnotesize {\sc Figure} 6. Best upper bounds, for Taxicab$(n), n=7,8,\ldots,12.$\\ } \end{center} \bigskip \begin{center} \includegraphics[height=4cm,width=16cm]{TaxicabFig7x.eps} \end{center} %\smallskip \begin{center} {\footnotesize {\sc Figure} 7. Upper bound of Taxicab$(7)$ and its $7$ decompositions\\ (2 more decompositions are differences of cubes)\\ } \end{center} \newpage %%%%%%%%%%%%%%%%%%%%%%%%%%%%% The best upper bounds using this method were computed in November--December 2006, and are listed in Fig. 6. This search needed some hours on a Pentium IV. They are the current records for the upper bounds of the Taxicab numbers. Fig. 7 gives the full decomposition of the new upper bound of Taxicab(7). Its new 7th decomposition, which is not 101 times one of the 6 decompositions of $T'(6, 1)$ from Fig. 4, is underlined. The other decompositions of upper bounds up to Taxicab(12) are in the detailed lists of the Appendix. We may continue with the other unused splitting factors of Fig. 5, giving (without explicitly stating their decompositions): \begin{align} \mbox{\rm Taxicab}(13) \leq T'(13, 1) &= 4327^3*T'(12, 1) \simeq 5.99*10^{78} \nonumber\\ \mbox{\rm Taxicab}(14) \leq T'(14, 1) &= 4622^3*T'(13, 1) \simeq 5.91*10^{89} \nonumber\\ \mbox{\rm Taxicab}(15) \leq T'(15, 1) &= 7549^3*T'(14, 1) \simeq 2.54*10^{101} \nonumber\\ \mbox{\rm Taxicab}(16) \leq T'(16, 1) &= 8063^3*T'(15, 1) \simeq 1.33*10^{113} \nonumber\\ \mbox{\rm Taxicab}(17) \leq T'(17, 1) &= 14309^3*T'(16, 1) \simeq 3.91*10^{125} \nonumber\\ \mbox{\rm Taxicab}(18) \leq T'(18, 1) &= 16227^3*T'(17, 1) \simeq 1.67*10^{138} \nonumber\\ \mbox{\rm Taxicab}(19) \leq T'(19, 1) &= 23035^3*T'(18, 1) \simeq 2.04*10^{151}. \nonumber \end{align} \section{Cabtaxi numbers} But why should we be limited to sums of {positive cubes}? We can generalize the problem, allowing sums of positive \underline{or negative} cubes, these are known as Cabtaxi numbers. Their story starts before that of the Taxicab numbers. \begin{center} \includegraphics[height=8cm,width=6cm]{TaxicabFig8x.eps} \end{center} \begin{center} {\footnotesize {\sc Figure} 8. Fran\c{c}ois Vi\`ete (Fontenay-le-Comte 1540 -- Paris 1603)\\ } \end{center} \newpage% \begin{center} \includegraphics[width=16cm]{TaxicabFig9x.eps} \end{center} \begin{center} {\footnotesize {\sc Figure} 9. Formula ``$6^3 = 3^3 + 4^3 + 5^3$'' by Fran\c{c}ois Vi\`ete, as republished in 1646 \cite[p.~75]{Viete}.\\% [41, p.~75]\\ } \end{center} \bigskip On 31 July 1589, the French king Henri III was killed by the monk Jacques Cl\'ement and was succeeded on the throne by Henri IV. In 1591, Fran\c{c}ois Vi\`ete, ``one of the most influential men at the court'' of Henri IV \cite[p.~3]{Viete2} published this very nice formula about his problem XVIII, fourth book of {\it Zetetica} \cite[p.~75]{Viete} \cite[p.~146]{Viete2}: % [41, p.~75] [42, p.~146]: $$6^3 = 3^3 + 4^3 + 5^3.$$ %Fig. 8. Fran\c{c}ois Vi\`ete (Fontenay-le-Comte 1540 - Paris 1603) Moving only one term, we can consider that Vi\`ete knew Cabtaxi(2): $$91 = 3^3 + 4^3 = 6^3 - 5^3.$$ In exactly the same year, 1591, Father Pietro Bongo (``Petrus Bungus'' in Latin), canon of Bergamo, independently published this same formula in {\it Numerorum Mysteria} \cite[p.~550]{Dick}. % [12, p.~550]. Bongo is also known to have ``demonstrated'' that the Antichrist was Martin Luther by using the Hebrew alphabet, the sum of the letters being 666: the number of the Beast. It was an answer to the German mathematician Michael Stifel (1487--1567) who previously proved, using the Latin alphabet, that Pope Leo X was the Antichrist. So strange and mystic the reasoning by some mathematicians at that time $\ldots$ Back to mathematics! Vi\`ete and Euler worked on parametric solution in rationals of: \begin{align} %(11) x^3 = y^3 + z^3 + w^3. \label{11} \end{align} In 1756, Euler published \cite{Eul} the same $x=6$ solution of Vi\`ete and Bongo, and some other solutions. In 1920 H.~W. Richmond published \cite{{Richmond}} a list of $C(2, i)$ numbers, with a solving method. Euler was probably the first to have mentioned some 3-way solutions, his smallest being \begin{align} 87^3 - 79^3 = 20^3 + 54^3 = 38^3 + 48^3. \nonumber\end{align} But the first mention of the true Cabtaxi(3) that I have found was by Escott in 1902 \cite{Esc14}: \begin{align} 728 = 12^3 - 10^3 = 9^3 - 1^3 = 8^3 + 6^3. \nonumber\end{align} Answering Escott's problem in 1904, Werebrusow published \cite{Wer2}, \cite[p.~562]{Dick} % [46] [12, p.~562] this 3-way formula: %(12) \begin{align} \hspace*{1cm}\mbox{\rm If}\; \; m^2+mn+n^2 = 3a^2bc,\; \mbox{\rm then}\hspace*{15cm}\nonumber \end{align} \vspace*{-1cm}\begin{align} \bigl((m+n)c+ab^2\bigr)^3 + \bigl(-(m+n)b-ac^2\bigr)^3 &= (-mc+ab^2)^3 + (mb-ac^2)^3 \nonumber\\[1mm] &= (-nc+ab^2)^3 + (nb-ac^2)^3. \label{12} \end{align} \bigskip \begin{center} \includegraphics[width=16cm]{TaxicabFig10a.eps} \end{center} %\smallskip \begin{center} {\footnotesize {\sc Figure} 10. History of Cabtaxi numbers.\\ } \end{center} \bigskip Werebrusow needed the condition $a^3 = 1$, but his formula is true without this condition. This 3-way formula (\ref{12}) reuses his previous 2-way formula (\ref{3}). No example was given by Werebrusow, but we can remark that Cabtaxi$(3)$ can be found, applying $(m, n, a, b, c) = (0, 3, 1, 3, 1)$. Another observation is that Cabtaxi(3) can be deduced from Cabtaxi(2), using a splitting factor $2$, which adds one new decomposition $9^3 - 1^3$. The two other decompositions of Cabtaxi$(2)$ are $2^3$ multiples of Cabtaxi$(2)$. Cabtaxi$(4)$, Cabtaxi$(5)$, Cabtaxi$(6)$, Cabtaxi$(7)$ were found by Randall L. Rathbun in the beginning of the 1990s \cite[p.~211]{Guy}, %[18, p.~211], while Cabtaxi$(8)$ was discovered by Daniel J. Bernstein in 1998 \cite{Bernstein}. In the same month, January 2005, there were two nice results on Cabtaxi$(9)$ from two different people: on the 24th, Jaroslaw Wroblewski found an upper bound of Cabtaxi$(9)$ \cite{Mer}, and one week later, on the 31st January 2005, Duncan Moore found the true Cabtaxi$(9)$ \cite{Moore} Moore's search also proved that Cabtaxi$(10) > 4.6*10^{17}$. %%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%% Just as Taxicab$(5)$ was a strong basis for Taxicab numbers, we observe in Fig. 10 that Cabtaxi$(6)$ is a strong basis used by bigger Cabtaxi numbers. These interesting relations were never published, and show the strength of splitting factors: \begin{align} \mbox{\rm Cabtaxi}(7) &= 2^3 * \mbox{\rm Cabtaxi}(6)\nonumber\\ \mbox{\rm Cabtaxi}(8) &= 23^3 * \mbox{\rm Cabtaxi}(7)\nonumber\\ \mbox{\rm Cabtaxi}(9) &= (5*67)^3 * \mbox{\rm Cabtaxi}(7).\nonumber \end{align} Our method is similar to Taxicab numbers, and uses the splitting factors of Cabtaxi$(9)$ given in Fig. 11a. However, because Jaroslaw Wroblewki's number $C'(9, 2) = 8.25 * 10^{17}$ is close to $C(9, 1) =$ Cabtaxi$(9) = 4.25 * 10^{17}$, it is interesting also to analyze its splitting factors, as shown in Fig.~11b. The best upper bounds up to $C'(20, 1)$ using the splitting factors of Cabtaxi(9) were computed in November--December 2006. Three better upper bounds $C'(11, 1), C'(17, 1), C'(18, 1)$ are possible, coming from $C'(9, 2)$: they were found later, in February 2007. All these numbers are listed in Fig.~12 and are the current records for the upper bounds of the Cabtaxi numbers. Fig.~13 gives the full decomposition of the new upper bound of Cabtaxi(10). Its new 10th decomposition, which is not 13 times one of the 9 decompositions of Cabtaxi(9) from Fig 10, is underlined. %%%%%%%%%%%%%%%%%%%%%%%%%%%%% \bigskip \begin{center} \includegraphics[width=11cm]{TaxicabFig11ax.eps} \end{center} %\smallskip \begin{center} {\footnotesize {\sc Figure} 11a. Detailed list of splitting factors of Cabtaxi$(9) = 424910390480793000$.\\ } \end{center} \bigskip \begin{center} \includegraphics[width=11cm]{TaxicabFig11bx.eps} \end{center} \begin{center} {\footnotesize {\sc Figure} 11b. Detailed list of splitting factors of $C'(9, 2) = 825001442051661504$.\\ } \end{center} \bigskip \bigskip \begin{center} \includegraphics[width=16cm]{TaxicabFig12a.eps} \end{center} %\smallskip \begin{center} {\footnotesize {\sc Figure} 12. Best upper bounds for Cabtaxi(10) to Cabtaxi(20).\\ } \end{center} \bigskip \bigskip \bigskip \begin{center} \includegraphics[width=14cm]{TaxicabFig13x.eps} \end{center} %\smallskip \begin{center} {\footnotesize {\sc Figure} 13. Upper bound of Cabtaxi(10) and its 10 decompositions.\\ } \end{center} The other decompositions of upper bounds up to Cabtaxi(20) are presented in the detailed lists of the Appendix. We may continue with the other unused splitting factors of Fig. 11a, giving (without explicitly stating their decompositions): \begin{align} \mbox{\rm Cabtaxi}(21) \leq C'(21, 1) &= 349^3 * C'(20,1) \simeq 9.14*10^{76} \nonumber\\ \mbox{\rm Cabtaxi}(22) \leq C'(22, 1) &= 436^3 * C'(21,1) \simeq 7.57*10^{84}\nonumber\\ \mbox{\rm Cabtaxi}(23) \leq C'(23, 2) &= 661^3 * C'(22,1) \simeq 2.19*10^{93}\nonumber\\ \mbox{\rm Cabtaxi}(24) \leq C'(24, 2) &= 859^3 * C'(23,1) \simeq 1.39*10^{102}\nonumber\\ \mbox{\rm Cabtaxi}(25) \leq C'(25, 2) &= 1009^3 * C'(24,1) \simeq 1.42*10^{111}\nonumber\\ \mbox{\rm Cabtaxi}(26) \leq C'(26, 2) &= (4367*439)^3 * C'(24,1) \simeq 9.77*10^{120}\nonumber\\ \mbox{\rm Cabtaxi}(27) \leq C'(27, 2) &= (4367*439)^3 * C'(25,1) \simeq 1.00*10^{130}\nonumber \end{align} and of Fig 11b, giving: \begin{align} \mbox{\rm Cabtaxi}(21) \leq C'(21, 2) &= (139*283*291)^3 * C'(18,1) \simeq 2.72*10^{77}\nonumber\\ \mbox{\rm Cabtaxi}(22) \leq C'(22, 2) &= 307^3 * C'(21,1) \simeq 7.88*10^{84}\nonumber\\ \mbox{\rm Cabtaxi}(23) \leq C'(23, 1) &= 379^3 * C'(22,1) \simeq 4.29*10^{92}\nonumber\\ \mbox{\rm Cabtaxi}(24) \leq C'(24, 1) &= 409^3 * C'(23,1) \simeq 2.94*10^{100}\nonumber\\ \mbox{\rm Cabtaxi}(25) \leq C'(25, 1) &= 1021^3 * C'(24,1) \simeq 3.12*10^{109}\nonumber\\ \mbox{\rm Cabtaxi}(26) \leq C'(26, 1) &= 1153^3 * C'(25,1) \simeq 4.79*10^{118}\nonumber\\ \mbox{\rm Cabtaxi}(27) \leq C'(27, 1) &= 1693^3 * C'(26,1) \simeq 2.32*10^{128}\nonumber\\ \mbox{\rm Cabtaxi}(28) \leq C'(28, 1) &= 1829^3 * C'(27,1) \simeq 1.42*10^{138}\nonumber\\ \mbox{\rm Cabtaxi}(29) \leq C'(29, 1) &= 2307^3 * C'(28,1) \simeq 1.75*10^{148}\nonumber\\ \mbox{\rm Cabtaxi}(30) \leq C'(30, 1) &= 5543^3 * C'(29,1) \simeq 2.97*10^{159}.\nonumber \end{align} %8. \section{Unsolved problems} %8.1. \subsection{Are these the true Taxicab and Cabtaxi numbers?} The new upper bounds of Taxicab(7) and Cabtaxi(10) announced in this paper, and detailed in Fig 7 and 13, may have a chance of being the correct Taxicab and Cabtaxi numbers. But the probability decreases as $n$ increases, and is close to 0 for Taxicab(19) and Cabtaxi(30). Who can check if some of these upper bounds are the correct Taxicab and Cabtaxi numbers? Or who will find smaller upper bounds? This is a good subject for mathematical computation. %8.2. \subsection{Prime versions of Taxicab and Cabtaxi numbers} Our construction with splitting factors generates sums of cubes of non-prime integers: at least $n-1$ decompositions are $k^3$ multiples. What about sums of two cubes of primes? The 2-way solutions using only sums of cubed primes are rare. For what we can call ``the prime version of Taxicab numbers'', the smallest 2-way solutions are \begin{subequations}\begin{align} 6058655748 &= 61^3 + 1823^3 = 1049^3 + 1699^3\label{13a}\\[1mm] 6507811154 &= 31^3 + 1867^3 = 397^3 + 1861^3.\label{13b} \end{align}\end{subequations} For the prime version of Cabtaxi numbers, the smallest 2-way solutions are \begin{subequations}\begin{align} 62540982 &= 397^3 - 31^3 = 1867^3 - 1861^3\label{14a}\\[1mm] 105161238 &= 193^3 + 461^3 = 709^3 - 631^3.\label{14b} \end{align}\end{subequations} The solution (\ref{14a}) is just a different arrangement of (\ref{13b}). %\smallskip But nobody has succeeded yet (as far as we know) in constructing a 3-way solution using only sums, or sums and differences, of cubed primes. Who will be the first, or who can prove that it is impossible? An ``easier'' question: instead of directly searching for a 3-way solution using 6 cubed primes, is there another 3-way solution using at least 4 cubed primes, different from this one $$68913 = 40^3 + 17^3 = 41^3 - 2^3 = 89^3 - 86^3$$ (the 4 primes used are $17, 41, 2, 89$). See puzzles 90 \cite{Rivera34} and 386 \cite{Rivera35} of Carlos Rivera. A supplemental remark: our 3-way problems are unsolved, but are solved for a long time if only coprime pairs are used instead of primes. Several 3-way and 4-way solutions using sums of two coprime cubes are known. The smallest 3-way solution was found by Paul Vojta \cite[p.~211]{Guy} %[18, p.~211] in 1983: $$15170835645 = 517^3 + 2468^3 = 709^3 + 2456^3 = 1733^3 + 2152^3.$$ And 3-way, 4-way and 5-way solutions using sums or differences of two coprime cubes are known. It is easy to find the smallest 3-way solution: $$3367 = 15^3 - 2^3 = 16^3 - 9^3 = 34^3 - 33^3.$$ %8.3. \subsection{Who can construct a $4\times 4$ magic square of cubes?} A $3\times 3$ magic square of cubes, using $9$ distinct cubed integers, has been proved impossible \cite[p.~270]{Guy}, \cite[p.~59]{Boyer5}: %[18, p.~270] [4, p.~59]: if $z^3$ is the number in the centre cell, then any line going through the center should have $x^3 + y^3 = 2z^3.$ Euler and Legendre demonstrated that such an equation is impossible with distinct integers. But the question of $4\times 4$ magic squares of cubes, using $16$ distinct positive cubed integers, is still open. A breakthrough was made in 2006 by Lee Morgenstern \cite{Boyer7} who found a very nice construction method using Taxicab numbers. If \begin{align} a^3 + b^3 &= c^3 + d^3 = u\label{15}\\[1mm]%\quad {\rm and} \quad {\rm and} \quad e^3 + f^3 &= g^3 + h^3 = v,\label{16} \end{align} then the $4\times 4$ square of cubes in Fig. 14 is semi-magic, its 4 rows and 4 columns having the same magic sum $S=uv.$ \bigskip \begin{center} \includegraphics[width=6cm]{TaxicabFig14x.eps} \end{center} %\smallskip \begin{center} {\footnotesize {\sc Figure} 14. Parametric $4 \times 4$ magic square of cubes, Morgenstern's method.\\ } \end{center} Using $u = \mbox{\rm Taxicab}(2) = 1729$ and the second smallest 2-way solution $v = T(2, 2) = 4104,$ both found by Frenicle, which implies $(a, b, c, d, e, f, g, h) = (1, 12, 9, 10, 2, 16, 9, 15)$, we find the $4\times4$ semi-magic square of cubes shown in Fig. 15. \bigskip \begin{center} \includegraphics[width=6cm]{TaxicabFig15x.eps} \end{center} %\smallskip \begin{center} {\footnotesize {\sc Figure} 15. $4 \times 4$ semi-magic square of cubes. Magic sum $S=1729*4104 = 7,095,816.$\\ } \end{center} This is not a full solution of the problem, because this square is only ``semi-magic'', in that the diagonals each have a wrong sum. The diagonals (and the square) would be fully magic if a third equation is simultaneously true: \begin{align}(ae)^3 + (bf)^3 = (cg)^3 + (dh)^3.\label{17}\end{align} Using 2-way lists kindly provided by Jaroslaw Wroblewski, University of Wroc{\l}aw, I can say that there is no solution to the system of 3 equations (\ref{15}), (\ref{16}) and (\ref{17}), with $a, b, c, d, e, f, g, h$ $ < 500,000$ or with $a, b, c, d < 1,000,000$ and $e, f, g, h < 25,000.$ But that does not mean that the system is impossible. The first person who finds a numerical solution of this system of 3 equations will directly get a $4\times 4$ magic square of cubes! But perhaps somebody will succeed in constructing a $4\times 4$ magic square of cubes using a different method. 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Available at \href{http://www.cs.uwaterloo.ca/journals/JIS/wilson10.html}{http://www.cs.uwaterloo.ca/journals/JIS/wilson10.html} \end{thebibliography} \section{Appendix} \bigskip \begin{center} \includegraphics[width=17cm]{TaxicabFigA1aA1bx.eps} \end{center} %\smallskip \begin{center} {\footnotesize {\sc Figure} A1a. Smallest 2-way solutions\footnote{All these solutions were previously found by Frenicle.} listed by Moreau.\\ {\sc Figure} A1b. Smallest 3-way solutions listed by Leech. } \end{center} \bigskip \begin{center} %\hspace*{-8cm} \includegraphics[width=10cm]{TaxicabFigA2x.eps} \end{center} %\smallskip \begin{center} {\footnotesize {\sc Figure} A2. Splitting factors of Taxicab numbers. } \end{center} \bigskip \begin{center} \includegraphics[width=17cm]{TaxicabFigA3x.eps} \end{center} %\smallskip \begin{center} {\footnotesize {\sc Figure} A3. Splitting factors of the smallest 5-way solutions. } \end{center} \bigskip \begin{center} \includegraphics[width=17cm]{TaxicabFigA4x.eps} \end{center} %\smallskip \begin{center} {\footnotesize {\sc Figure} A4. Smallest 6-way solutions derived from 5-way solutions and splitting factors\\ (other 6-way solutions are possible, if they are not derived from 5-way solutions). } \end{center} \bigskip \begin{center} {\footnotesize {\sc List} 1. Upper bounds of Taxicab(7..12) and decompositions. } \end{center} %\bigskip \begin{center} \includegraphics[width=17cm]{TaxicabList1x.eps} \end{center} \newpage \begin{center} {\footnotesize {\sc List} 2. Upper bounds of Taxicab(10..20) and decompositions. } \end{center} \begin{center} \includegraphics[width=15cm]{TaxicabList2a.eps} \end{center} \newpage \begin{center} {\footnotesize {\sc List} 2 (cont'd). Upper bounds of Taxicab(10..20) and decompositions. } \end{center} \begin{center} \includegraphics[width=15cm]{TaxicabList2b.eps} \end{center} \newpage \begin{center} {\footnotesize {\sc List} 3. Upper bounds of Taxicab(13..19). } \end{center} \begin{center} \includegraphics[width=17cm]{TaxicabList3x.eps} \end{center} \bigskip \bigskip \begin{center} {\footnotesize {\sc List} 4. Upper bounds of Taxicab(21..30). } \end{center} \begin{center} \includegraphics[width=17cm]{TaxicabList4x.eps} \end{center} \bigskip \hrule \bigskip \noindent 2000 {\it Mathematics Subject Classification}: Primary 11D25.\\ \noindent {\it Keywords}: Taxicab number, Cabtaxi number, Hardy--Ramanujan number, Bernard Frenicle de Bessy, Fran\c{c}ois Vi\`ete, sum of two cubes, difference of two cubes, magic square of cubes. \bigskip \hrule \bigskip \noindent (Concerned with sequences \seqnum{A011541}, \seqnum{A047696}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received June 26 2007; revised version received February 22 2008. Published in {\it Journal of Integer Sequences}, March 7 2008. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}