Question: For $k>3$ does there exist an odd prime $q\_k$ such that $p\_k=2^kq\_k+1$ is prime and $p\_k$ divides $a\_k=\dfrac{3^{2^{k-1}}+1}{2}$?\


If $k=3$ the answer is Yes because for $q\_3=5$ we get $p\_3=a\_3=41$. \


$a\_4=3281=17\cdot 193$ but neither $17=2^4\cdot 1+1$ nor $193=2^4\cdot 12+1$ qualifies to be $p\_4$ because $1$ and $12$ are not (odd) prime numbers.\


$a\_5,a\_6$ and $a\_7$ turn out to be prime numbers, so, the answer to the question is No (see the recursive definition below).
$a\_8$ has $61$ digits and none of its factors qualifies to be $p\_8$. Unfortunately $a\_9$ has approximately $121$ digits and $a\_{10}$ has approximately 243 digits. I would like to see a proof for the No answer for $k>3$ or see the condition on $k$ for which the answer is Yes.\


There is also a recursive definition for $a\_k$ : $a\_2=5,a\_k=2^k(a\_2\cdots a\_{k-1})+1,k>2,$ which makes it very clear why $k=3$ is a Yes answer. ($a\_1$ doesn't really matter but a meaningful definition for it is $a\_1=2.$)



For $k>3$ does there exist an odd prime $q_k$ such that $p_k=2^kq_k+1$ is prime and $p_k$ divides $a_k=\dfrac{3^{2^{k-1}}+1}{2}$?
This isn't a complete answer, but a heuristic argument which seems to indicate that (as Christian Elsholtz suggests), there is no serious obstacle to there being infinitely many such triples. For ease of notation, I'll just write $p$ and $q$ for odd primes $p$ and $q$ such that $p = 2^{k}q+1$ and $p$ divides $\frac{3^{2^{k-1}}+1}{2}$, where $k >3.$ Notice that $3$ has multiplicative order $2^{k}$ in the multiplicative group $\left( \mathbb{Z}/p\mathbb{Z} \right)^{\times}.$ In particular, $3$ is a quadratic non-residue (mod $p$). Since $p \equiv 1$ (mod $4$), we must have $p \equiv 2$ (mod 3). If $k$ is even,we conclude that we have $p \equiv 2^{k}+1$ (mod $3.2^{k}$) while if $k$ is odd, we have $p \equiv 2^{k+1}+1$
(mod $3.2^{k}$). Hence when $k$ is even, we have $q \equiv 1$ (mod $3$) and when $k$ is odd, we have $q \equiv 2$ (mod $3$). Note also that $3^{q}$ is a quadratic non-residue (mod $p$) since $q$ is odd. 


Hence one procedure for producing triples $(k,q,p)$ (which does produce all solutions) is as follows:
Choose an integer $k > 3.$ Choose a prime $q \equiv 2^{k}$ (mod $3$). If $p = 2^{k}q + 1$ is also prime, check whether $3$ lies in the cyclic subgroup generated by $3^{q}$ in $\left( \mathbb{Z}/p\mathbb{Z} \right)^{\times}.$ If it does, then $(k,q,p)$ is a solution. If not, it isn't.
The justification for the last step is that given that the previous criteria are satisfied, $3^{q}$ has multiplicative order exactly $2^{k}$ in $\left(\mathbb{Z}/p\mathbb{Z} \right)^{\times}$ so is a generator of the (cyclic) Sylow $2$-subgroup of the multiplicative group $\left(\mathbb{Z}/p\mathbb{Z} \right)^{\times}.$ On the other hand, if $(k,q,p)$ is a solution, then $3$ must have multiplicative order exactly $2^{k}$ in $\left(\mathbb{Z}/p\mathbb{Z} \right)^{\times}.$ Hence it must be a power (in the unique Sylow $2$-subgroup of that group) of the generator $3^{q}.$ In fact, if $3$ is a power of $3^{q}$ at all, it is necessarily a generator of that Sylow $2$-subgroup, since $3^{q}$ is already a generator.


