10 is Only Triangular Number that is Sum of Consecutive Odd Squares

Theorem
$10$ is the only triangular number which is the sum of two consecutive odd squares:


 * $10 = 1^2 + 3^2$

Proof
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 $16 x^2 + 32 x + 20 = y^2 + y$:

Due to the restrictions on the variables, solutions with negative inputs are invalid.

This leaves one solution:


 * $\tuple {0, 4}$

as follows: