Product of $3$ consecutive triangular numbers is a perfect square When is the product of $3$ consecutive triangular numbers a perfect square
My try:
$k^2 = n(n+1)^2(n+2)^2\frac{n+3}{  8}$ where $n$ and $k$ are integers, then
$\frac{n(n+3)}{2}$ must be a square number
$n(n+3) =2x^2$ , $n^2+3n-2x^2=0$
then the discriminant must also be a perfect square.
$9+8x^2 = y^2$ but I don't know how to solve this in integers
 A: You have a Pell-type equation
$$y^2-8x^2=9.$$
This implies that $y$ and $x$ are multiples of $3$, so
$$(y/3)^2-8(x/3)^2=1$$
which is a genuine Pell equation. Its solution is
$$(y+2x\sqrt2)/3=\pm(3+2\sqrt2)^n$$
for $n\in\Bbb Z$.
A: You can start by using a familiar result that is the product of two consecutive triangular numbers. I write it personally as the sum of two triangular numbers whose indices are the former two minus $1$ each.
$$T_nT_{n-1}=T_{T_n-1}+T_{T_{n-1}-1} \iff 2T_nT_{n-1}=T_{n^2-1}$$
For $n=3$, two times the product of those two triangular numbers multiplied by each other is already a square, more specifically $2\times3\times6=36$, but $2$ is not a triangular number unless we multiply the whole expression by $25$, the reason being that it's another square, which you can split and rearrange into a product of three consecutive triangular numbers like this:
$$6\left (2\times5\right)\left(3\times5\right)=900 \iff 6\times10\times15=30^2 \implies n=5$$
In the following expression:
$$T_nT_{n-1}T_{n-2}=\frac{T_{n^2-1}T_{n-2}}{2}$$
I hope that I helped.
