Square of Odd Number as Difference between Triangular Numbers

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Theorem

Let $n \in \Z_{\ge 0}$ be a positive integer.

Then:

$\exists a, b \in \Z_{\ge 0}: \paren {2 n + 1}^2 = T_a - T_b$

where:

$T_a$ and $T_b$ are triangular numbers
$T_a$ and $T_b$ are coprime.


That is, the square of every odd number is the difference between two coprime triangular numbers.


Proof

\(\ds T_a - T_b\) \(=\) \(\ds \dfrac {a^2 + a} 2 - \dfrac {b^2 + b} 2\) Closed Form for Triangular Numbers

Let $a = 3b + 1$

\(\ds T_{3 b + 1} - T_b\) \(=\) \(\ds \dfrac {\paren {3 b + 1}^2 + 3 b + 1} 2 - \dfrac {b^2 + b} 2\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {3 b + 1}^2 + 3 b + 1 - b^2 + b} 2\)
\(\ds \) \(=\) \(\ds \dfrac {9 b^2 + 6 b + 1 + 3 b + 1 - b^2 - b} 2\)
\(\ds \) \(=\) \(\ds \dfrac {8 b^2 + 8 b + 2} 2\)
\(\ds \) \(=\) \(\ds 4 b^2 + 4 b + 1\)
\(\ds \) \(=\) \(\ds \paren {2 b + 1}^2\)

The square of every odd number can be made through this method.


\(\ds T_{3 b + 1}\) \(=\) \(\ds \dfrac {\paren {3 b + 1} ^ 2 + 3 b + 1} 2\)
\(\ds \) \(=\) \(\ds \dfrac {9 b^2 + 9 b + 2} 2\)
\(\ds \) \(=\) \(\ds 9 \paren {\dfrac {b^2 + b} 2} + 1\)
\(\ds \leadsto \ \ \) \(\ds T_{3 b + 1} - 9 T_b\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds \gcd \set {T_{3 b + 1}, T_b}\) \(=\) \(\ds 1\) Bézout's Identity

Both triangular numbers are coprime.

$\blacksquare$


Sources

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