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\centerline{\smalltt INTEGERS: 
 \smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 5(1) 
(2005), \#A25} 
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\begin{center}
\uppercase{\bf A QUESTION OF SIERPINSKI ON TRIANGULAR NUMBERS}
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{\bf Michael A. Bennett} \\
{\smallit Department of Mathematics, University of British Columbia, Vancouver BC, Canada} \\
{\tt bennett@math.ubc.edu} \\
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\centerline{\smallit Received: 8/30/05,
 Accepted: 11/6/05, Published: 11/22/05}
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\centerline{\bf Abstract}

\noindent We answer a question of Sierpinski by showing that there do not exist four distinct triangular number in geometric progression.

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\markright{\smalltt INTEGERS: \smallrm ELECTRONIC 
JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 5 (2005), \#A25\hfill}

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In D23 of \cite{Guy}, it is stated that Sierpinski asked the question of whether or not there exist four (distinct) triangular numbers in geometric progression and, further, that Szymiczek \cite{Szi} conjectured  the answer to this to be a negative one.  Recall that a triangular number is one of the form
$$
T_n = \frac{n(n+1)}{2}
$$
for $n \in \mathbb{Z}$. The problem of finding three such triangular numbers is readily reduced to finding solutions to a Pell equation (whereby, an old result of G\'erardin \cite{Ge} (see also \cite{Szi2}) implies that there are infinitely many such triples, the smallest of which is $(T_1,T_3,T_8)$). It is easy to show that the answer to Sierpinski's question is in the negative.
This is, in fact, an immediate consequence of the following.

\noindent {\bf Lemma}
 {\it If $a$ and $b$ are positive integers with $b > 1$ then at least one of $ab+1$ and $ab^3+1$ is not a perfect square.}

\noindent {\it Proof.}
Suppose that we have
$$
ab+1=x^2 \; \mbox{ and } \; ab^3+1 = y^2
$$
for positive integers $x$ and $y$.
From the theory of quadratic fields, since we may assume that $ab$ is not a perfect square, it follows that 
$$
y + b \sqrt{ab} = \left( x + \sqrt{ab} \right)^k
$$
for some positive integer $k$ (which, since $b > 1$, we may assume to be at least $2$). If $k \geq 3$, then we would have that
$$
y + b \sqrt{ab} \geq \left( x + \sqrt{ab} \right)^3
$$
and hence
$$
b \geq 3x^2+ab,
$$
a contradiction, since $a, x \geq 1$. It follows that $k=2$ and hence that $b=2x$. Since $ab+1=x^2$, we have $x=1$, contradicting the fact that $a$ and $b$ are positive integers. This completes the proof of our Lemma.

To connect this to Sierpinski's question, suppose that the four triangular numbers in geometric progression are
$$
T_x < T_y < T_u < T_v.
$$
Taking $a = 8 T_x$ and $b=T_y/T_x$, it follows that one of $8 T_y+1$ or $8 T_v+1$ cannot be a perfect square, contradicting the identity $8 T_n + 1 = (2n+1)^2$.


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\begin{thebibliography}{30}
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\bibitem{Ge}
A. G\'erardin.
\newblock {\em Sphinx-Oedipe} 9 (1914), 75, 145--146

\bibitem{Guy}
R. Guy.
\newblock Unsolved Problems in Number Theory, 3rd edition.
\newblock Springer Verlag, New York, 2004.
 
 \bibitem{Szi2}
K. Szimiczek.
\newblock L' \'equation $uv=w\sp{2}$ en nombres triangulaires. (French) 
\newblock {\em Publ. Inst. Math. (Beograd) (N.S.)} 3 (17) (1963), 139--141.

\bibitem{Szi}
K. Szimiczek.
\newblock The equation $(x^2-1)(y^2-1)=(z^2-1)^2$,
\newblock {\em Eureka} 35 (1972), 21--25.

\end{thebibliography}

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