10 is Only Triangular Number that is Sum of Consecutive Odd Squares
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Theorem
$10$ is the only triangular number which is the sum of two consecutive odd squares:
- $10 = 1^2 + 3^2$
Proof
The closed-form expression for the $n$th triangular number is:
- $\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$
for $n \in \Z_{\ge 0}$.
The expression for the $n$th odd square number is:
- $4 n^2 + 4 n + 1$
again, for $n \in \Z_{\ge 0}$.
Therefore the closed-form expression for the $n$th sum of two consecutive odd squares is:
- $4 n^2 + 4 n + 1 + 4 \paren {n + 1}^2 + 4 \paren {n + 1} + 1$
This simplifies to:
- $8 n^2 + 16 n + 10$
Equate the two with a variable replacing $n$:
- $8 x^2 + 16 x + 10 = \dfrac {y \paren {y + 1} } 2$
This simplifies to:
- $16 x^2 + 32 x + 20 = y^2 + y$
We then apply Solutions to Diophantine Equation $16 x^2 + 32 x + 20 = y^2 + y$:
The indeterminate Diophantine equation:
- $16x^2 + 32x + 20 = y^2 + y$
has exactly $4$ solutions:
- $\tuple {0, 4}, \tuple {-2, 4}, \tuple {0, -5}, \tuple {-2, -5}$
Due to the restrictions on the variables, solutions with negative inputs are invalid.
This leaves one solution:
- $\tuple {0, 4}$
as follows:
\(\ds 8 \paren 0^2 + 16 \paren 0 + 10\) | \(=\) | \(\ds \frac {\paren 4 \paren {\paren 4 + 1} } 2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 1^2 + 3^2\) | \(=\) | \(\ds 10\) |
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $10$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $10$