Square Product of Three Consecutive Triangular Numbers
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Theorem
Let $T_n$ denote the $n$th triangular number for $n \in \Z_{>0}$ a (strictly) positive integer.
Let $T_n \times T_{n + 1} \times T_{n + 2}$ be a square number.
Then at least one value of $n$ fulfils this condition:
- $n = 3$
Proof
Let $T_n \times T_{n + 1} \times T_{n + 2} = m^2$ for some $m \in \Z_{>0}$.
We have:
\(\ds T_n \times T_{n + 1} \times T_{n + 2}\) | \(=\) | \(\ds \dfrac {n \paren {n + 1} } 2 \dfrac {\paren {n + 1} \paren {n + 2} } 2 \dfrac {\paren {n + 2} \paren {n + 3} } 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {n \paren {n + 1}^2 \paren {n + 2}^2 \paren {n + 3} } 8\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\dfrac {\paren {n + 1} \paren {n + 2} } 2}^2 \dfrac {n \paren {n + 3} } 2\) |
Thus we need to find $n$ such that $\dfrac {n \paren {n + 3} } 2$ is a square number.
We see that:
\(\ds 3 \times \paren {3 + 3}\) | \(=\) | \(\ds 3 \times 6\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \times 3^2\) |
Thus $n = 3$ appears to satisfy the conditions.
It remains for us to check:
\(\ds T_3 \times T_4 \times T_5\) | \(=\) | \(\ds 6 \times 10 \times 15\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2 \times 3} \times {2 \times 5} \times {3 \times 5}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2 \times 3 \times 5}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 30^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 900\) |
$\blacksquare$
Sources
- 1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $1$: Some Preliminary Considerations: $1.3$ Early Number Theory: Problems $1.3$: $5 \ \text {(b)}$