Sufficient Condition for Square of Product to be Triangular

Theorem
Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Let $2 n^2 \pm 1 = m^2$ be a square number.

Then $\paren {m n}^2$ is a triangular number.

Proof
That is, either:


 * $\paren {m n}^2 = \dfrac {\paren {2 n^2 - 1} \paren {2 n^2} } 2$

and so:
 * $\paren {m n}^2 = T_{2 n^2 - 1}$

or:


 * $\paren {m n}^2 = \dfrac {\paren {2 n^2} \paren {2 n^2 + 1} } 2$

and so:
 * $\paren {m n}^2 = T_{2 n^2}$

Hence the result.