Square Product of Three Consecutive Triangular Numbers

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Theorem

Let $T_n$ denote the $n$th triangular number for $n \in \Z_{>0}$ a (strictly) positive integer.

Let $T_n \times T_{n + 1} \times T_{n + 2}$ be a square number.


Then at least one value of $n$ fulfils this condition:

$n = 3$


Proof

Let $T_n \times T_{n + 1} \times T_{n + 2} = m^2$ for some $m \in \Z_{>0}$.

We have:

\(\ds T_n \times T_{n + 1} \times T_{n + 2}\) \(=\) \(\ds \dfrac {n \paren {n + 1} } 2 \dfrac {\paren {n + 1} \paren {n + 2} } 2 \dfrac {\paren {n + 2} \paren {n + 3} } 2\)
\(\ds \) \(=\) \(\ds \dfrac {n \paren {n + 1}^2 \paren {n + 2}^2 \paren {n + 3} } 8\)
\(\ds \) \(=\) \(\ds \paren {\dfrac {\paren {n + 1} \paren {n + 2} } 2}^2 \dfrac {n \paren {n + 3} } 2\)

Thus we need to find $n$ such that $\dfrac {n \paren {n + 3} } 2$ is a square number.


We see that:

\(\ds 3 \times \paren {3 + 3}\) \(=\) \(\ds 3 \times 6\)
\(\ds \) \(=\) \(\ds 2 \times 3^2\)


Thus $n = 3$ appears to satisfy the conditions.

It remains for us to check:

\(\ds T_3 \times T_4 \times T_5\) \(=\) \(\ds 6 \times 10 \times 15\)
\(\ds \) \(=\) \(\ds \paren {2 \times 3} \times {2 \times 5} \times {3 \times 5}\)
\(\ds \) \(=\) \(\ds \paren {2 \times 3 \times 5}^2\)
\(\ds \) \(=\) \(\ds 30^2\)
\(\ds \) \(=\) \(\ds 900\)

$\blacksquare$


Sources