Find all $q \in \mathbb{N}$ such that $\frac{q(q+1)}{12}$ is a Perfect square.
Trivially we see that $q=3$ is the first candidate. Now let $$\frac{q(q+1)}{12}=r^2$$ $\implies$ $$q^2+q-12r^2=0$$ By Quadratic formula we get $$q=\frac{-1\pm\sqrt{48r^2+1}}{2}$$ Any help here?
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You correctly tagged this pell-type-equations.
Note that $\dfrac{q^2+q}{12}=r^2\iff (2q+1)^2-3(4r)^2=1$, which I got by completing the square.
You also have the first solution correct: $q=3$ and $r=1$.
As is known for Pell-type equations,
the other solutions are of the form $(2q+1)+\sqrt3 (4r)=(7+4\sqrt3)^n$,
for $n=2,3,4,5,...$.
The solutions for $q$ can be found in OEIS here.
It remains to find $r$ such that $48r^2+1=s^2$ or $s^2-48r^2=1$, which is Pell's equation with fundamental solution $s=7,r=1,q=3$ as you found. All other solutions are generated by the following recurrence: $$\begin{bmatrix}s_{k+1}\\r_{k+1}\end{bmatrix}=\begin{bmatrix}7&48\\1&7\end{bmatrix}\begin{bmatrix}s_k\\r_k\end{bmatrix}$$ This may be solved for $s=\sqrt{48r^2+1}$, giving $$s=\frac{(7+4\sqrt3)^k+(7-4\sqrt3)^k}2,k\in\mathbb N$$ Since we want only positive $q$, we take the positive sign in the formula for $q$, yielding $q=\frac{s-1}2$.