---

Claim. D. Hensley and I. Richards, Primes in intervals, Acta Arith. 25 (1973/74), 375--391, Theorem and Corollary (p. 380). Let $\varrho^*(x)$ be the largest size of an admissible tuple inside an interval of $x$ consecutive integers, a tuple being admissible when for every prime $p$ some residue class modulo $p$ contains none of its members. The Theorem proves, unconditionally, that $\varrho^*(x)-\pi(x)\to+\infty$, and more precisely that for every $\varepsilon>0$

$$\varrho^*(x)-\pi(x)\ \ge\ (\log2-\varepsilon)\frac{x}{(\log x)^2}$$

for all large $x$. The prime $k$-tuples conjecture (B) gives every admissible tuple infinitely many prime translates, so under (B) the value $\varrho^*(x)$ is attained by infinitely many intervals of primes. The Corollary reads: "The hypotheses (A) and (B) are incompatible. Moreover, if we assume (B), then we obtain: $(-A^*)$ For all sufficiently large $x$, there exist infinitely many $y$, such that $\pi(x+y)>\pi(x)+\pi(y)$", where (A) is the problem's inequality for all $x,y\ge2$. Through the Theorem, the excess $\pi(x+y)-\pi(x)-\pi(y)$ in $(-A^*)$ can be taken at least $(\log2-o(1))x/(\log x)^2$. So under (B) the answer to Problem 855 is no. The site's key [HeRi73], On the incompatibility of two conjectures concerning primes, Proc. Sympos. Pure Math. 24 (1973), 123--127, announces the same result. The paper is compiled at Hensley and Richards (1973/74), with the statement at its Theorem.

Hypothesis. The prime $k$-tuples conjecture (B): every admissible tuple $b_1<\cdots<b_k$ has infinitely many $n$ with all of $n+b_1,\ldots,n+b_k$ prime. It is unproved, so the claim gives no unconditional answer; the unconditional content is only that (A) and (B) cannot both hold.

Acceptance. The result is refereed in Acta Arithmetica 25. The symposium volume is not counted as refereed, and the site's commentary on a problem it labels open is not acceptance. The page is dated by the year of the site's key, the symposium paper of 1973.

Depends on. Nothing on this wiki; the argument is the paper's own.