This problem is related to the question asked here: The primes $s$ of the form $6m+1$ and the greatest common divisor of $2s(s-1)$
I would like to prove the following. If $s$ is prime Let $\psi(s) = 2s(s-1)$. I write $(\psi(s),\psi(s+2))$ to be the greatest common divisor of $\psi(s)$ and $\psi(s+2)$. Then
$ (\psi(s),\psi(s+2))= \begin{cases} 4 &\mbox{if } s \equiv \{0,2\} \pmod{3} \\ 12 & \mbox{if } s = 6m+1 \end{cases}.$
The case equal to 12 has already been shown. Suppose that $s$ is congruent to $0$ or $2$ modulo $3$. If $s=3$ then $(\psi(3), \psi(5)) = (12,40) = 4$. With the exception of $3$ I am thinking we can write every other $s$ as $3n-1$ for some positive integer $n$. Then $\psi(s+2) =\psi(3n-1+2) = \psi(3n+1)$. So we are computing $((\psi(3n-1),\psi(3n+1))$? Is this the right approach?