Square Sum 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 $T_n + T_{n + 1} + T_{n + 2}$ be a square number.


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

$n = 5$


Proof

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

We have:

\(\ds T_n + T_{n + 1} + T_{n + 2}\) \(=\) \(\ds \dfrac {n \paren {n + 1} } 2 + \dfrac {\paren {n + 1} \paren {n + 2} } 2 + \dfrac {\paren {n + 2} \paren {n + 3} } 2\)
\(\ds \) \(=\) \(\ds \dfrac {n \paren {n + 1} + \paren {n + 1} \paren {n + 2} + \paren {n + 2} \paren {n + 3} } 8\)
\(\ds \) \(=\) \(\ds \dfrac {n^2 + n + n^2 + 3 n + 2 + n^2 + 5 n + 6} 2\) multiplying out
\(\ds \) \(=\) \(\ds \dfrac {3 n^2 + 9 n + 8} 2\) multiplying out

Thus we need to find $n$ such that $\dfrac {3 n^2 + 9 n + 8} 2$ is a square number.


We see that:

\(\ds \dfrac {3 \times 5^2 + 9 \times 5 + 8} 2\) \(=\) \(\ds \dfrac {75 + 45 + 8} 2\)
\(\ds \) \(=\) \(\ds \dfrac {128} 2\)
\(\ds \) \(=\) \(\ds 64\)


Thus $n$ appears to satisfy the conditions.

It remains for us to check:

\(\ds T_5 + T_6 + T_7\) \(=\) \(\ds 15 + 21 + 28\)
\(\ds \) \(=\) \(\ds 64\)
\(\ds \) \(=\) \(\ds 8^2\)

$\blacksquare$


Sources

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