Reading Eight Erdős Problems Together

Will Cook

Cross-problem paper

Reading Eight Erdős Problems Together

Eight Erdős problems read together By Will Cook 22 September 2026 Authorship, AI use and citation

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Précis. One account of the mathematics developed across the programmes: an exact tail-capacity criterion under eventual congruences, its sharp factorial support-gap threshold, rational and irrational Lambert subsums across bases, and the limits of finite greedy tests and other irrationality methods. Full arguments, exact counterexamples, unsuccessful routes and attribution are retained together. The principal proofs are ordinary mathematics; cited Lean ingredients have their own stated scope. Historical novelty and independent expert review are not established.

This paper owns the synthesis exposition and ordinary proofs across the covered problems: capacity and congruence constructions, Lambert subsums, rational-point counts, method limits and their research record.

It is not authority for a solution to any original Erdős target, historical novelty, independent expert review, or a full Lean proof of the analytic capacity criterion or Lambert-chain theorem.

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In this paper

Choices against contraction

Proof of Theorem 6.1. (i) A choice of $ε_{1}, \dots, ε_{N}$ fixes $\sum_{n \leq N} ε_{n} u_{n}$ , and the rest of the sum lies in $[0, C_{N}]$ . So $V$ is covered by at most $\prod_{n \leq N} (D_{n} + 1)$ intervals of length $C_{N}$ .

(ii) Subtracting $μ$ , it suffices to reach every $y \in [0, C_{N_{0}}]$ with the levels $n > N_{0}$ . Put $y_{N_{0}} = y$ and, for $n > N_{0}$ , let $ε_{n} = min (D_{n}, ⌊ y_{n - 1} / u_{n} ⌋)$ and $y_{n} = y_{n - 1} - ε_{n} u_{n}$ . If $0 \leq y_{n - 1} \leq C_{n - 1} = D_{n} u_{n} + C_{n}$ then $0 \leq y_{n} \leq C_{n}$ : when $ε_{n} = D_{n}$ this is a subtraction, and when $ε_{n} < D_{n}$ it holds because $y_{n} < u_{n} \leq C_{n}$ . Since $C_{n} \to 0$ , the sum of the $ε_{n} u_{n}$ is $y$ .

(iii) Here $C_{n} = D β^{- n} / (β - 1)$ . The product in (i) is a constant multiple of $((D + 1) / β)^{N}$ , and $u_{n} \leq C_{n}$ says $β - 1 \leq D$ . ◻

The reading that matters for irrationality is immediate. Suppose a family of series has values $x_{0} + \sum ε_{n} u_{n}$ with the $ε_{n}$ free as in (ii). The values then fill an interval, which contains rationals and irrationals, so a property that all members of the family share implies neither. A proof of irrationality for one member has to use something that distinguishes it inside the family. Two constructions in the companion notes are families of this kind.

Problem #251. Proposition 1 of [32] (ordinary proof), applied in its Corollary 2 to the prime gaps, changes the gaps on a set of upper Banach density zero, by nonnegative integers below any prescribed function tending to infinity, keeping every congruence modulo every fixed $q$ from some point on, and reaches every real in an interval. The resulting positions $P_{n}$ satisfy $P_{n} \sim n \log n$ and are not asserted to be prime. In the notation above the $j$ -th block has $u_{j} = M_{j} 2^{- n_{j} - 2}$ and $D_{j} = 2^{s_{j}} - 1$ , and the inequality verified there is $u_{j} \leq C_{j}$ . The lemma is credited there to [11], [12] and [13]. The note also proves that a bounded allowance is impossible under the stated congruences. Thus every allowance tending to infinity suffices, while no bounded allowance does. The specified growth, eventual fixed-modulus congruences and empirical distributions of unnormalised blocks therefore do not suffice to prove $\sum p_{n} 2^{- n}$ irrational; neither primality nor every quantitative correlation is preserved.

Problem #249. Section 5 of [31] gives an integer sequence $c$ with $c (n) = φ (n)$ for odd $n$ , $| c (n) - φ (n) | \leq 2$ for even $n$ , $0 \leq c (n) \leq n$ and $\sum c (n) 2^{- n} = 5 / 4$ (Lean-checked there). Changing even indices by at most $2$ is the case $u_{n} = 2^{- n}$ for even $n$ , $D_{n} = 4$ , where $C_{n} \geq \frac{4}{3} 2^{- n} > u_{n}$ .

The same comparison appears in Kovač and Tao’s theorem that for integers $2 \leq t_{1} < \dots < t_{m}$ with $\sum 1 / (t_{k} - 1) > 1$ there are sets $S_{k}$ , one of them infinite, with $\sum_{k} X_{S_{k}} (t_{k})$ rational [13]: merging the weights of the $m$ bases, the sum of all weights below a given weight $u$ is at least $(1 - O (u)) u \sum 1 / (t_{k} - 1)$ , which exceeds $u$ once $u$ is small.

About this paper

Authorship and AI use. Will Cook built and directed the research infrastructure and maintains the public release. He reviewed claims when he could. AI agents did most of the research and drafting. Cook did not independently verify every claim.

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