Index of Square Triangular Number from Preceding

It has been suggested that this page be renamed.
In particular: Haven't a clue why I gave it the name I did. Suggest a rename.
To discuss this page in more detail, feel free to use 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

$\ds T_{4 n \paren {n + 1} }$

$=$

$\ds \frac {\paren {4 n \paren {n + 1} } \paren {4 n \paren {n + 1} + 1} } 2$

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

1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $15$
1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $15$