Index of Square Triangular Number from Preceding

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Theorem

Let $T_n$ be the $n$th triangular number.

Let $T_n$ be square.


Then $T_{4 n \paren {n + 1} }$ is also square.


Proof

\(\ds T_{4 n \paren {n + 1} }\) \(=\) \(\ds \frac {\paren {4 n \paren {n + 1} } \paren {4 n \paren {n + 1} + 1} } 2\) Closed Form for Triangular Numbers
\(\ds \leadsto \ \ \) \(\ds T_{4 n \paren {n + 1} }\) \(=\) \(\ds \frac {n \paren {n + 1} } 2 \times 4 \paren {4 n \paren {n + 1} + 1}\) Extract $T_n$
\(\ds \leadsto \ \ \) \(\ds T_{4 n \paren {n + 1} }\) \(=\) \(\ds T_n \times \paren {16 n^2 + 16 n + 4}\) Simplify
\(\ds \leadsto \ \ \) \(\ds T_{4 n \paren {n + 1} }\) \(=\) \(\ds T_n \times \paren {4 n + 2}^2\) Factorise


The product of two squares is also square.

Let $T_n$ be square.

Therefore $T_{4 n \paren {n + 1} }$ is also square.

$\blacksquare$


Sources

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