Square Product of Three Consecutive Triangular Numbers

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:

Thus we need to find $n$ such that $\dfrac {n \paren {n + 3} } 2$ is a square number.

We see that:

Thus $n = 3$ appears to satisfy the conditions.

It remains for us to check: