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Statement

Hypothesis (p. 452), the Hardy--Littlewood prime $k$-tuplets conjecture in the form the paper uses: if $a_1,\ldots,a_k,b_1,\ldots,b_k$ are integers with $(a_j,b_j)=1$ for each $j$, and for each prime $p\le k$ some integer $x$ makes none of $a_1x+b_1,\ldots,a_kx+b_k$ divisible by $p$, then there are arbitrarily large integers $x$ for which all of $a_1x+b_1,\ldots,a_kx+b_k$ are prime.

Theorem (p. 452). Let $c_1,c_2,\ldots,c_N$ be positive integers, let $g=\gcd(c_1,\ldots,c_N)$ and $d=\gcd(2g,\,c_1+c_2+\cdots+c_N)$, and assume the prime $k$-tuplets conjecture. Then one can construct infinite sets $A_1,A_2,\ldots,A_N$ of distinct odd primes such that every element of $\frac1d\{c_1A_1+\cdots+c_NA_N\}$ is prime.

Notation (the Remark, p. 452). $\{c_1A_1+\cdots+c_NA_N\}$ is the set of all sums of any $c_1$ elements of $A_1$, any $c_2$ elements of $A_2$, and so on up to any $c_N$ elements of $A_N$. The paper notes that every such element is divisible by $d$. The proof (p. 453) counts sums in which a newly added prime occurs $t$ times for $1\le t\le c_j$, so an element may be used more than once in a sum.

Consequences (unlabelled, p. 452), both under the same conjecture.

• With $N=2$, $c_1=1$, $c_2=2$, $A=A_1$ and $B=\{2a:a\in A_2\}$: there are infinite sets of integers $A$ and $B$ such that every element of $A+B$ is prime.
• With $N=1$ and $c_1=2$: there is an infinite set of integers $A$ such that $\frac12(a+a')$ is prime for any $a,a'\in A$.

These give infinite sets answering the question raised by the finite sets, chosen from $\{1,2,\ldots,N\}$, of Pomerance, Sárközy and Stewart (the paper's reference [2]): sets $A$ and $B$ with every element of $A+B$ prime, and a set of odd integers $A$ with $\frac12(a+a')$ prime for any $a\ne a'$ in $A$.

Proof pointer

Pp. 452--453. A lemma (p. 452) gives, for any $B>0$ and under the same hypothesis, distinct primes $a_1,\ldots,a_N$, all greater than $B$, with $\frac1d(c_1a_1+\cdots+c_Na_N)$ prime; it fixes residues by the Chinese Remainder Theorem, picks $a_2,\ldots,a_N$ by Dirichlet's theorem, and gets $a_1$ and the weighted sum prime together from the $k$-tuplets conjecture. The proof of the Theorem (p. 453) starts each set from the lemma with $B=e^{c_1+\cdots+c_N}$ and then adds one prime to each set in turn. A new prime for $A_j$ has the form $p=q+mx$, where $q$ is the last prime added to $A_j$ and $m$ is the product of the primes below $q/2$. Each new element, one divided by $d$ whose sum contains $p$, is then $r+tmx/d$ with $r>q/2$ prime and $1\le t\le c_j$, so it has no prime factor below $q/2$; the conjecture gives a large $x$ making $q+mx$ and all these new elements prime at once.

Read depth

Claims checked: the hypothesis, the Theorem, the Remark and the two consequences were read clause by clause on the page images of the print (pp. 452--453), and the proof was followed. Nothing here is independently reviewed.

Dependencies

None in the corpus. External inputs: the prime $k$-tuplets conjecture (assumed, unproved), Dirichlet's theorem on primes in arithmetic progressions and the Chinese Remainder Theorem.

Source. A. Granville, A note on sums of primes, Canad. Math. Bull. 33 (1990), no. 4, 452--454, doi:10.4153/CMB-1990-073-7; the edition read is named on the source card.

Bears on

• Problem 431: the first consequence gives, conditionally on the prime $k$-tuplets conjecture, infinite sets $A$ and $B$ with $A+B$ contained in the primes. The problem asks for $A+B$ to agree with the primes up to finitely many exceptions; the paper does not show that $A+B$ contains all but finitely many primes, and so does not address that question.