Is it possible for a natural number $k$ to exist such that for all primes $p$:
$$(p+1)k-1$$ is composite?


(i.e., can all $3k-1$, $4k-1$, $6k-1$, $8k-1$, $12k-1$, $14k-1$, $18k-1$, $20k-1$, $24k-1$, $30k-1$, $32k-1$, $38k-1$, $42k-1$, $44k-1$, $48k-1$, ... be composite?)



A question related to Dirichlet's theorem
Seva proved that Dickson's conjecture implies there is no such $k$. 
One only needs a weak form of this conjecture. Such a weak form could potentially 
be much easier than for example the problem on 
infinitely Sophie Germain primes, as one does not need to know whether 
for two fixed forms, such as 
$f\_1(n)=n$ and $f\_2(n)=kn+(k-1)$ are
prime simulanteously, infinitely often, but just needs to know they produce one prime pair.


One can prove the following:
For every integer $k\geq 2$, there exists some power $k^r$ (where $r$ may depend on $k$)
such that there exists some prime $p$ with $k^r(p+1)-1$ also prime.
(This statement does not rule out the possibilty that some $k$ could exist.
for which all $k(+1)-1$ are composite, but it illustrates the view that some unconditional version
of a weak form of Dickson's-type conjectures could potentially solve the problem.


Proof: Results of Maynard-Tao 
(see e.g. James Maynard: Dense clusters of primes in subsets, <https://arxiv.org/abs/1405.2593>)
imply that two of the polynomials $f\_i(t)=k^i t-1$ take prime values, simultaneously, infinitely often.
(Actually, we do not even need "infinitely often", we just need one such prime pair.)
With $p\_1=k^i t-1, p\_2=(p\_1+1)k^{j-i}-1=k^j t-1$.
there exists a pair of primes $p, k^r (p+1)-1$. (We do not have good 
control over $r=j-i$ but that does not matter. Some upper bound on $r$ is possible in view of the 
quantitatve nature of Maynard's results).


Hence for $k^r$ the expression $k^r(p+1)-1$ is not always composite.


In view of quantitatve versions of Linnik's theorem on primes in progression,
or Dirichlet's theorem:
for every $p+1$ there are many values $k$ making $(p+1)k-1$ a prime. 
In this way one could sieve out the possible values for $k\leq N$ in an interval.  

Probabilistically,
if one uses more than about $(\log N)^2$ linear forms, then  

there should be very few $k \leq N$ left.
This is, as $(1- \frac{1}{\log N})^{(\log N)^2}\approx \frac{1}{N}$.




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As Seve remarked that one can study $k=\frac{p+1}{q+1}$.
This topic has been extensively studied by Peter Elliott.
(Mathscinet: "Elliott" combined with "rationals" gives many papers).


In particular:
 The multiplicative group of rationals generated by the shifted primes, I.
by P.D.T.A. Elliott, Journal für die reine und angewandte Mathematik (1995), 463, page 169-216.
<https://eudml.org/doc/153727>


Here Elliott explicitly conjectures:
(Conjecture 1): Every positive rational $r$ has a representation $r=\frac{p+1}{q+1}$, $p$ and $q$ primes.


He comments that the strengthening has infinitely many representations is an analogue to Schinzel's hypothesis, and would follow from Dickson's conjectures. He studies the group of rationals generated by these shifted primes and achieves many deep results.


