Let $k \geq 2$ be an integer. Put $[k] = \{1, \cdots, k\}$. Let $\mathcal{P} = \{p_1, \cdots, p_k\}$ be a set of $k$ primes. For every subset $S \subseteq [k]$ put $d_S = \prod_{j \in S} p_j$. The empty product is equal to $1$, by convention.
For every $k \geq 2$, does there always exist a set of primes $\mathcal{P}$ of cardinality equal to $k$ such that for every partition $A \sqcup B = [k]$ we have $d_A + d_B = 2^k q$, where $q$ is an odd prime?
This would follow from the Dickson's conjecture or Schinzel's hypothesis H.
Let $M:=2^k \prod_\limits{\text{prime }p\leq 2^{k-1}+1} p$ and let us fix any primes $p_1<p_2<\dots<p_{k-1}$, each of which $\equiv 1\pmod{M}$. We will be looking for $p_k$ of the form $f(x):=2^kx-1$ for some integer $x$. In addition, consider $2^{k-1}$ polynomials indexed by subsets $A\subseteq [k-1]$: $$g_A(x) := \frac1{2^k}\bigg(f(x) \prod_{i\in A} p_i + \prod_{j\in [k-1]\setminus A} p_j\bigg).$$ It can be seen that each such polynomial $g_A(x)$ has integer and coprime coefficients, and its free term is divisible by each prime below $2^{k-1}+1$. This makes the set of polynomials $F:=\{f(x)\}\cup\{g_A(x)\ :\ A\subseteq [k-1]\}$ to miss any "fixed divisor" as defined in Schinzel's hypothesis H, which then would imply that there are infinitely many values of $x$ such that all polynomials from $F$ simultaneously take prime values. Hence, we can take a value of $x=x_0$ such that $p_k:=f(x_0) > p_{k-1}$ to have a required collection of primes.
