Let $n\geq2$. Are there sets $A, B \subseteq \mathbb{N}$ such that $|A|=|B|=n$ and all numbers in $A+B$ are primes? A well-known conjecture is that there are infinite set $A$ and finite set $|B|=n$ such that all numbers in $A+B$ are primes.
Are there infinite sets $A, B \subseteq \mathbb{N}$ such that all numbers in $A+B$ are primes? I conjecture that the sets do not exist. Someone told me this question seems to be open.
Yes, conditional on the Hardy-Littlewood prime tuples conjecture.
Let $A$ and $B$ be two finite sets such that $A +B$ consists of primes, and such that for all primes $p$, there are residue classes $x_p$ and $y_p$ mod $p$ with $x_p+y_p \neq 0 \mod p$ such that $A$ does not contain any numbers congruent to $x_p$ mod $p$ and $B$ does not contain any numbers congruent to $y_p$ mod $p$.
Conditionally on the Hardy-Littlewood conjecture, it is possible to add one new element to $A$ or $B$, whichever you prefer, while preserving all these properties.
Given this, the existence of infinite $A$, $B$ follows by starting with the empty set, adding an element to $A$, then an element to $B$, then an element to $A$, then an element to $B$, etc. So it suffices to prove this step.
Without loss of generality, we may assume that we are trying to increase $B$.
First, enlarge $A$ to a set $\overline{A}$ such that $\overline{A}$ still does not contain any numbers congruent to $x_p$ mod $p$, and, in addition, for all $p< |B|+3$, $\overline{A}$ contains all residue classes modulo $p$ except $x_p$. This is easy to do with the Chinese remainder theorem (and may require adjusting $x_p$ for $p$ large).
Then $\overline{A}$ is an admissible tuple so, by the Hardy-Littlewood prime tuples conjecture, there is some large $z$ such that $a +z$ is prime for all $a\in \overline{A}$.
Setting $B' = B \cup \{z\}$, we can see that $A = B'$ consists of primes. For each prime $p$, if $p \geq |B|+3$ then there are at least $2$ choices of residue class $y_p$ such that $B'$ does not contain any numbers congruent to $y_p$ mod $p$, so at least $1$ such class satisfying $x_p +y_p \neq 0$ mod $p$. If $p < |B|+3$ then, taking $z$ to be sufficiently large, for every $a\in \overline{A}$, $a + z$ is not $p$ and thus, because it is prime, is not a multiple of $p$. Thus, by assumption on $\overline{A}$, $z$ is congruent to $-x_p$ mod $p$. So for $p <|B|+3$, the same $y_p$ works for $B'$ as worked for $B$.
So indeed $A,B'$ satisfy all the conditions, completing the induction step.
This implication was earlier established by Andrew Granville in A Note on Sums of Primes.
