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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Statement

Setting (pp. 1--4). U\mathcal U is the set of positive odd integers not of the form p+2kp+2^k with pp prime and kk a positive integer. In Problem 1.8 (p. 4), AiA_i (i∈Ii\in I) is the collection of all infinite arithmetic progressions of positive odd integers none of which is a prime plus a power of two, that is, all infinite progressions contained in U\mathcal U.

Theorem 1.9 (p. 4). Let a∈Ua\in\mathcal U. Then a∈⋃i∈IAia\in\bigcup_{i\in I}A_i if and only if there is an integer m>1m>1 with (a−2k,m)>1(a-2^k,m)>1 for every positive integer kk.

Corollary 1.10 (p. 4). a∈U∖⋃i∈IAia\in\mathcal U\setminus\bigcup_{i\in I}A_i if and only if a∈Ua\in\mathcal U and p(a−2k)p(a-2^k) is unbounded (in kk), where p(n)p(n) is the least prime divisor of nn and p(±1)=+∞p(\pm1)=+\infty by convention. The paper says this is easily seen to be equivalent to Theorem 1.9.

Before the theorem (p. 4) the paper shows that 1,3,1271,3,127, each of the form 2k−12^k-1, lie in U\mathcal U but in no AiA_i. After it (p. 5) it records that 509203∈U509203\in\mathcal U and that {11184810h+509203:h≥0}⊆U\{11184810h+509203:h\ge0\}\subseteq\mathcal U, so that the least element e1e_1 of ⋃iAi\bigcup_iA_i satisfies 149≤e1≤509203149\le e_1\le509203; Problems 1.12 and 1.13 ask for the exact least values.

Source. Yong-Gao Chen, A conjecture of Erdős on p+2kp+2^k, arXiv:2312.04120v3 (2024). Labels and pages are those of arXiv v3: the statements on p. 4, the proof in Section 5 on p. 26. 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

Page 26. If aa lies in a progression {mh+b}⊆U\{mh+b\}\subseteq\mathcal U, Sun's positive-proportion result (Lemma 4.4, p. 23) gives (b−2k,m)>1(b-2^k,m)>1, hence (a−2k,m)>1(a-2^k,m)>1, for all k≥1k\ge1. Conversely, given such mm, choose k0k_0 with 2k0−1>max⁡{a,m}2^{k_0-1}>\max\{a,m\}; any n=p+2kn=p+2^k in {2k0mh+a}\{2^{k_0}mh+a\} would force p∣mp\mid m, and comparing residues modulo 2k02^{k_0} then gives a=p+2ka=p+2^k or a=pa=p, both impossible, so the progression lies in U\mathcal U.

Dependencies

Lemma 4.4 (X.-G. Sun's positive-proportion theorem, p. 23).

Bears on

  • Problem 16: context only. The theorem describes which elements of the problem's set are covered by infinite progressions contained in it; the paper's answer to the problem does not use it.