<s>[INST] 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
 [/INST] 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 $n\in\Bbb Z$.
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