TITLE: Divisors of $s(2s+1)$ and the primes $s^2+1$
QUESTION [1 upvotes]: I need help proving that $s^2+1$ is prime in the following claim. 

claim:  If $s$ is any positive integer I write $f_s =s(2s+1)$. I will need the divisor counting function $\tau$. Suppose that  $3s+1$
  is prime and $\tau(f_s)=8$ then: $2^{-1}s$, $2s+1$ and $s^2+1$ are prime.

Let $q=2^{-1}s$ then I can write out the eight divisors as:
\begin{align}
 1,2,q,s,2s+1,q^{-1}s(2s+1),2^{-1}s(2s+1),s(2s+1) \end{align}
I can write $f_s =2q(2s+1)$. Surely $(2,2s+1)=1$ and$(2,q)=1$ then $\tau(2q(2s+1))=\tau(2)\tau(q(2s+1))=2\tau(q(2s+1))=8$. Consequently $\tau(q(2s+1))=4$. If $(q,(2s+1))>1$ then $f_s>8$ a contradiction. In particular $\tau(q(2s+1))=\tau(q)\tau(2s+1)=4$. Both $q$ and $2s+1$ are greater than $1$ so I have that $\tau(q)$ and $\tau(2s+1)$ are both equal to $2$ and so $q$ and $2s+1$ are prime. Note that $4q+1=2s+1$ and $6q+1=3s+1$.  Currently I cannot get to the primality of $s^2+1$. Check out the table below:
\begin{array}{| l | l | l | l |l|}
\hline
s & f_{s} & q & 2s+1& 3s+1 & s^2+1\\ \hline
6 & 78 &  3 & 13 & 19 & 37\\ 
14 & 406 &  7 & 29 & 43 & 197\\ 
26 & 1378 &  13 & 53 & 79 & 677\\ 
74 & 11026 & 37 & 149 & 223 & 5477\\ 
146 & 42778 & 73 & 293 & 439 & 21317\\ 
    \hline
    \end{array} 
We can compare $q$ to A186721 and $2s+1$ to A090866. 
Update: The above claim is false by computational counter example see the comment below and the Sage Program. We do have the following result though. 

claim:  If $s$ is any positive integer I write $f_s =s(2s+1)$. I will  need the divisor counting function $\tau$. Suppose that  $3s+1$
  is prime and $\tau(f_s)=8$ then: $2^{-1}s$ and $2s+1$ are prime.

REPLY [1 votes]: In your setting, we have 
1. $q$ is prime
2. $4q+1$ is prime
3. $6q+1$ is prime
Your questions is whether $4q^2 +1$ is prime. 
Running the following program in SAGE, we find that $4q^2+1$ is not prime for $q=277$. 
for p in primes(100000):
    if is_prime(4*p+1)==1:
        if is_prime(6*p+1)==1:
            if is_prime(4*p^2+1)==0:
                print p;

This gives the following result:
277
373
577
727
853
1033
1423
1453
1777
1987
2293
2677
2767
3037
3163
3457
...

It looks like there would be infinitely many such primes, but this will be of difficulty as at least Sophie Germain Primes problem.
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