How to solve the following Diophantine equation $(s+1)(3s+2)(2s+1)^2=r^2?$

Question

How to solve the following equation $(s+1)(3s+2)(2s+1)^2=r^2,$ where $s,\space r \in \mathbb{Z},\space s>0,\space r>0$? If it is not possible to solve it, probably something can be said about $s,\space r$ in terms of divisibility?

Answer 1

If the left hand side is a perfect square, then $(s+1)(3s+2)$ is a perfect square. However, $s+1$ and $3s+2$ are relatively prime, and so they have no factors in common, so the only way $(s+1)(3s+2)$ is a perfect square is if both $(s+1)$ and $(3s+2)$ are perfect squares.

Therefore, we wish to solve $s+1=x^2, 3s+2=y^2=3x^2-1$.

Now, consider the equation modulo $3$.