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.
$\blacksquare$
Sources
- 1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $1$: Some Preliminary Considerations: $1.3$ Early Number Theory: Problems $1.3$: $6 \ \text {(a)}$