The primes $s$ of the form $6m+1$ and the greatest common divisor of $2s(s-1)$

Question

I had trouble coming up with a good title.

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 if $s$ is prime and $(\psi(s),\psi(s+2))=12$ we have that $s=6m+1$ for some positive integer $m$. Conversely if $s=6m+1$ is prime then $(\psi(s),\psi(s+2))=12$.

I would like to solve this problem or find a counterexample? I do know that $(\psi(s),\psi(s+2))$ is even.

Dirichlet tells us that there are infinitely many primes $6m+1$ so there should be infinitely many $s$ that satisfy the above statement?

Answer 1

Well, $\psi(s+2) = 2 (s+2) (s+1), \psi(s) = 2 s (s-1).$ The $2$ factors, so we have $((s+2)(s+1), s (s-1)) = 6.$ If $s$ is prime, then if $s \neq 2,$ then $((s+2)(s+1), s(s-1)) = ((s+2)(s+1), s-1) = 6.$ Since $s$ is odd, $s+2$ is odd also, while the highest power of $2$ dividing $((s-1), (s+1))$ is always just $2.$ Now, if $3 \mid s-1,$ then the highest power of $3$ dividing $((s+2)(s+1), (s-1)$ is $3.$ Any other prime dividing $s-1$ does not divide $(s+2)$ or $s+1,$ so the gcd is $6.$ If $3 \nmid s-1,$ then $3$ does not divide the gcd. So, the prime $s$ has to be of the form $3m+1.$ Now, $((s+2)(s+1), s-1) = 3 ((m+1)(3m+2), m).$ If $m$ is odd, then the gcd of $((m+1)(3m+2), m)$ is odd, so $m$ is even. The other direction is easy.

Answer 2

$\psi(s+2)=2(s+1)(s+2)$, so $(\psi(s),\psi(s+2))=2((s-1)s,(s+1)(s+2))$. Let $d=((s-1)s,(s+1)(s+2))$. If $s=2$ then $d=2$. So if $d=6$ then $s$ is odd. Hence, $s$ is coprime with $s+1$ and $s+2$. Since these two numbers are also coprime, $d=(s-1,s+1)\cdot(s-1,s+2)$. The first factor is clearly $2$, so the second is $3$. That is, $s-1$ is an even multiple of $3$. Thus, $s=6m+1$ for some $m$.

Conversely, if $s=6m+1$ and $s$ is prime, $s$ has no common factor with $s+1$ or $s+2$, and $s-1$ is a multiple of $6$, $(s-1,s+1)=2$ and $(s-1,s+2)=3$. The solution follows immediately.

Answer 3

The gcd of $\psi(s)$ and $\psi(s+2)$ is $12$ when $s=6m+1$ regardless of whether $s$ is prime. It is straightforward to show that $4$ and $3$ (and therefore $12$) divide both, and on the other hand

$$12 = (2s+5)\psi(s) - (2s-3)\psi(s+2)$$ shows that the gcd divides $12$.