Sufficient Condition for Square of Product to be Triangular

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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

\(\ds \paren {m n}^2\) \(=\) \(\ds \paren {2 n^2 \pm 1} \times n^2\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {2 n^2 \pm 1} \paren {2 n^2} } 2\)


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.

$\blacksquare$


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