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:

\(\displaystyle 8 \paren 0^2 + 16 \paren 0 + 10\) \(=\) \(\displaystyle \frac {\paren 4 \paren {\paren 4 + 1} } 2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 1^2 + 3^2\) \(=\) \(\displaystyle 10\)

$\blacksquare$


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