Product of Two Triangular Numbers to make Square

Theorem

Then there is an infinite number of $m \in \Z_{>0}$ such that $T_n \times T_m$ is a square number.

Since $n^2 < n \paren {n + 1} < \paren {n + 1}^2$, $n \paren {n + 1}$ cannot be a square number.

$x^2 - n \paren {n + 1} y^2 = 1$

and for each solution:

$\ds T_n T_{x^2 - 1}$

$=$

$\ds \frac {n \paren {n + 1} } 2 \times \frac {x^2 \paren {x^2 - 1} } 2$

$\ds $

$=$

$\ds \frac {n \paren {n + 1} } 2 \times \frac {x^2 n \paren {n + 1} y^2 } 2$

$\ds $

$=$

$\ds \paren {\frac {x y n \paren {n + 1} } 2}^2$

Hence the result.

$\blacksquare$

$T_2 \times T_{24} = 30^2$

$T_3 \times T_{48} = 84^2$