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