# 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:

- $\displaystyle T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$

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): $10$