---

Statement

Setting. $\mathcal U$ is the set of positive odd integers not of the form $p+2^k$ with $p$ prime and $k$ a positive integer (pp. 1--2). Conjecture A (p. 2, quoted): "The set $\mathcal U$ is the union of an infinite arithmetic progression of positive odd integers and a set of asymptotic density zero."

Theorem 3.1 (p. 13, quoted). "Conjecture A is false."

The paper notes that Theorem 1.1 already implies this; Section 3 gives a different proof, and Section 4 a third.

Lemma 3.3 (p. 14). $\{11184810s+992077:s=0,1,\ldots\}\subseteq\mathcal U$.

Lemma 3.4 (p. 15). $\{11184810s+3292241:s=0,1,\ldots\}\subseteq\mathcal U$.

Source. Yong-Gao Chen, A conjecture of Erdős on $p+2^k$, arXiv:2312.04120v3 (2024). Labels and pages are those of arXiv v3: Section 3 on pp. 13--17 (Theorem 3.1 and Lemma 3.2 on p. 13, Lemma 3.3 on p. 14, Lemma 3.4 on p. 15, the proof of Theorem 3.1 on p. 16, Remark 3.5 on p. 17), and the second proof on p. 25. 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 proofs were read but not checked step by step; the residue computations (3.4) and (3.7) were not rerun. Nothing here is independently reviewed.

Proof pointer

First proof, pp. 13--16. If $\mathcal U=\{m_0h+a_0\}\cup W$ with $W$ of density zero, any progression contained in $\mathcal U$ has modulus divisible by $m_0$ and residue congruent to $a_0$ (Lemma 3.2). Lemmas 3.3 and 3.4, each built from a covering of the exponents by six congruences (Erdős's method, with the primes $3,5,7,13,17,241$), give two progressions modulo $11184810$ whose residues differ by a number $d$ with $\gcd(11184810,d)=2$; so $m_0=2$, and $\mathcal U$ would contain all large odd integers, contradicting $2^n+3\notin\mathcal U$ for all $n\ge1$. Remark 3.5 notes the lemmas are needed only up to the density-zero set $\{2^k+p:k\in\mathbb N,p\in\{3,5,17,7,13,241\}\}$.

Second proof, p. 25. Under Conjecture A, $m_0\ge11184810$ by Theorem 1.3, so $\mathcal U$ would have density at most $11184810^{-1}$, while the 48 progressions of Theorem 1.5 give it more.

Dependencies

Lemmas 3.2--3.4; Theorem 1.3 and Theorem 1.5 for the second proof.

Bears on

• Problem 16: Conjecture A is the problem's question, with $k\ge1$ as the paper fixes it, and Theorem 3.1 answers it no.