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\begin{document}
\vspace*{-40pt}
\centerline{\smalltt INTEGERS: \smallrm
ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 7
(2007), \#A19}
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\begin{center}
\uppercase{\bf Triangular Numbers in Geometric Progression}
\vskip 20pt
{\bf Yong-Gao Chen}\\ {\smallit Department of Mathematics, Nanjing Normal
                       University, Nanjing 210097, P. R. China}\\ {\tt
ygchen @ njnu.edu.cn}\\
\vskip 10pt
{\bf Jin-Hui Fang\footnote{Supported by the National Natural
Science Foundation of China, Grant No.10471064.}}\\ 
{\smallit Department of Mathematics, Nanjing Normal
University, Nanjing 210097, P. R. China}
\end{center}
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\centerline{\smallit Received: 1/11/07,
Accepted: 4/9/07, Published: 4/12/07}
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\centerline{\bf Abstract}

\noindent
In [R. K. Guy, Unsolved Problems in
Number Theory, 3rd ed. Springer Verlag, New York, 2004, D23], it
is stated that Sierpinski asked the question of whether or not
there exist four (distinct) triangular numbers in geometric
progression. Szymiczek  conjectured that the answer is negative.
 Recently M. A. Bennett [Integers: Electronic Journal of Combinatorial
 Number Theory {\bf 5(1)} (2005)]
 proved that there do not exist four distinct triangular
 numbers in geometric progression with the common ratio being a positive
 integer.
In this paper we prove that there do not exist four distinct
triangular numbers in geometric progression. Thus Sierpinski's
question is answered and Szymiczek's conjecture is confirmed.

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\markright{\smalltt INTEGERS: \smallrm ELECTRONIC
 JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 7 (2007), \#A19\hfill}

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In [4, D23], it is stated that Sierpinski asked the
question of whether or not there exist four (distinct) triangular
numbers in geometric progression. Szymiczek conjectured that the
answer is negative. Recall that a triangular number is one of the
form $T_n=\frac{n(n+1)}{2}$ for $n\in \mathbb{N}$. The problem
of finding three such triangular numbers is readily reduced to
finding solutions to a Pell equation(whereby, an old result of
Gerardin[3] (see also[2], [5]) implies that there are infinitely
many such triples, the smallest of which is $(T_1,T_3,T_8)$).
 Recently M. A. Bennett[1] proved that there do not exist four distinct triangular
 numbers in geometric progression with the ratio being positive
 integer.
In this paper, we extend Bennett's result to the rational common
ratio and prove that there do not exist four distinct triangular
numbers in geometric progression. Thus Sierpinski's question is
answered and Szymiczek's conjecture is
confirmed.

\noindent
{\bf Theorem} {\it There do not exist four distinct triangular
numbers in geometric progression.}

\noindent
{\it Proof.} Suppose that there  exist four distinct triangular numbers
$T_{n_1}$, $T_{n_2}$, $T_{n_3}$, $T_{n_4}$ in geometric
progression. Let $q$ be the common ratio. It is obvious that $q>0$
and $q\not= 1$. Without loss of generality,  we may assume that
$0<q<1$. Let $a=8T_{n_1}$. Then
$$8T_{n_2}=aq,\quad 8T_{n_3}=aq^2,\quad 8T_{n_4}=aq^3.$$
Let $m_i=2n_i+1$ $(i=1,\, 2,\, 3,\, 4)$. Then
$$a+1=m_1^2, \quad aq+1=m_2^2, \quad aq^2+1=m_3^2, \quad aq^3+1=m_4^2.\eqno{(1)}$$
Let $$q=\frac{b_1}{a_1},\quad  a_1, b_1\in \mathbb{Z}, \,
(a_1,b_1)=1,\, a_1\geq 1.$$  Because $aq^3$ is positive integer,
we have $a_1^3\mid ab_1^3$. Noting that $(a_1,b_1)=1$, we have
$a_1^3\mid a$. Let $a=a_1^3a_0$, $a_0\in \mathbb{N}$. By (1) we
have
$$m_1^2-a_1^3a_0=1,\quad  m_3^2-b_1^2a_1a_0=1.\eqno{(2)}$$
Because $a=m_1^2-1$ and $a=a_1^3a_0\in \mathbb{N}$,  we have
$a_1a_0$ is not a perfect square.

Let $x_0+y_0 \sqrt{a_0 a_1}$ be the basic solution of Pell
equation $x^2-a_0a_1y^2=1$. Then by (2) and the theory of Pell
equations, we have $$m_1+a_1 \sqrt{a_0 a_1}=(x_0+y_0 \sqrt{a_0
a_1})^k,$$
      $$ m_3+b_1 \sqrt{a_0 a_1}=(x_0+y_0 \sqrt{a_0 a_1})^l.$$
where $k,l$ are all positive integers. By $0<q<1$ and (1) we have
$m_1>m_3$ and $a_1>b_1$. So $k>l\geq1$.

If $k=2$, then $m_1+a_1 \sqrt{a_0 a_1}=(x_0+y_0 \sqrt{a_0
a_1})^2.$ Thus we have $a_1=2x_0 y_0$. So $x_0\mid a_1$. Since
$x_0^2-a_0a_1y_0^2=1$, we have $x_0=1$, a contradiction with
$x_0+y_0 \sqrt{a_0 a_1}$ being the basic solution of Pell equation
$x^2-a_0a_1y^2=1$. If $k\geq 3$, then $m_1+a_1 \sqrt{a_0
a_1}=(x_0+y_0 \sqrt{a_0 a_1})^3.$  Thus $a_1> \binom{k}{3}
x_0^{k-3}a_1 a_0 y_0^3$, which is obviously impossible.
\hfill $\Box$




\section*{References}
\footnotesize
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[1] M. A. Bennett, A Question of Sierpinski on Triangular
Numbers, Integers: Electronic Journal of Combinatorial Number
Theory {\bf 5(1)} (2005).

[2] L. E. Dickson, History of the Theory of Numbers, Vol.II, p.
36, Carnegie Inst., Washington, D. C. 1920.

[3] A. Gerardin, Sphinx-Oedipe {\bf 9} (1914),75,145-146.

[4] R. K. Guy, Unsolved Problems in Number Theory, 3rd ed.
Springer Verlag, New York, 2004.

[5] K. Szymiczek, L'equation $uv=w^2$ en Nombres Triangulaires
(French), Publ. Inst. Math. (Beograd) (N.S.) {\bf 3(17)} (1963), 139-141.

[6] K. Szymiczek, The Equation $(x^2-1)(y^2-1)=(z^2-1)^2,$ Eureka
{\bf 35} (1972), 21-25.


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