# Index of Square Triangular Number from Preceding

 It has been suggested that this article or section be renamed: Haven't a clue why I gave it the name I did. Suggest a rename. One may discuss this suggestion on the talk page.

## 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

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

The product of two squares is also square.

Let $T_n$ be square.

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

$\blacksquare$