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Statement

Setting (pp. 1--5). $\mathcal U$ is the set of positive odd integers not of the form $p+2^k$ with $p$ prime and $k$ a positive integer, and $A_i$ ($i\in I$) are all infinite arithmetic progressions contained in $\mathcal U$ (Problem 1.8, p. 4). $J$ is the set of $i$ for which the common difference of $A_i$ is a power of 2 times a squarefree odd integer, and $K$ the set of $i$ for which it is squarefree (p. 5).

Theorem 1.15 (p. 6).

• (i) $\bigcup_{i\in I}A_i=\bigcup_{i\in J}A_i$.
• (ii) If there are infinitely many Mersenne primes, then $\bigcup_{i\in I}A_i=\bigcup_{i\in K}A_i$.

Corollary 1.16 (p. 6). Either Problem 1.11 has an affirmative answer, or $\bigcup_{i\in I}A_i=\bigcup_{i\in K}A_i$, or both. Problem 1.11 (p. 5) asks whether some positive odd integer $a$ has $2^\ell-a$ composite for all large $\ell$ while no integer $m>1$ has $(a-2^k,m)>1$ for every positive integer $k$; the paper notes that $a=1$ would do if there are only finitely many Mersenne primes.

Source. Yong-Gao Chen, A conjecture of Erdős on $p+2^k$, arXiv:2312.04120v3 (2024). Labels and pages are those of arXiv v3: the definitions of $J$ and $K$ on p. 5, the statements on p. 6, the proof in Section 5 on pp. 26--28. The edition read is identified on the source card.

Read depth. Claims checked: the statements were read clause by clause on the printed pages. The proof was read but not checked step by step. Nothing here is independently reviewed.

Proof pointer

Pages 26--28. For $a$ in some $A_i$, Theorem 1.9 gives $m>1$, which may be taken squarefree and odd, with $(a-2^k,m)>1$ for all $k\ge1$; the argument of Theorem 1.9 puts $a$ in $\{2^{k_0}mh+a\}\subseteq\mathcal U$, proving (i). For (ii), a Mersenne prime $q=2^r-1$ with $2^{r-2}>\max\{a,m\}$ replaces $2^{k_0}$: reducing $2^k$ modulo $q$ shows $\{mqh+a\}\subseteq\mathcal U$, and $mq$ is squarefree.

Dependencies

Theorem 1.9 and its proof.

Bears on

• Problem 16: context only. The theorem concerns the infinite progressions contained in the problem's set; the paper's answer to the problem does not use it.