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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The subseries across bases

Proof of Theorem 6.2. (a) Since $R_{n} \geq \sum_{k > n} t^{- k} = t^{- n} / (t - 1)$ and $w_{n} = t^{- n} / (1 - t^{- n})$ , the inequality $w_{n} \leq R_{n}$ holds as soon as $t^{- n} \leq 2 - t$ . Theorem 6.1(ii) with $D_{n} = 1$ gives, for each $F \subseteq {1, \dots, N_{0}}$ , the interval $[X_{F} (t), X_{F} (t) + R_{N_{0}}]$ , and every value $X_{S} (t)$ lies in the one with $F = S \cap [1, N_{0}]$ . For $t = a / b$ a finite subsum is $\sum_{n \in F} b^{n} / (a^{n} - b^{n})$ , whose denominator divides $\prod (a^{n} - b^{n})$ and is coprime to $a b$ . A rational whose denominator is not coprime to $a b$ is therefore never a finite subsum, and inside the interval it is a subsum. At $t = 3 / 2$ the inequality $t^{- n} \leq 1 / 2$ holds for $n \geq 2$ and $R_{1} > 1 / 2$ .

(c) For $t > 2$ , $2^{N} R_{N} \leq 2^{N} t^{- N} / ((t - 1) (1 - t^{- 1})) \to 0$ , and Theorem 6.1(i) applies. ◻

Proof of Theorem 6.3. For part (a), write $ρ = b / a$ , $D_{j} = a^{n_{j}} - b^{n_{j}}$ and $S_{j} = \sum_{i \leq j} b^{n_{i}} / (a^{n_{i}} - b^{n_{i}})$ . Because $n_{i} ∣ n_{j}$ for $i \leq j$ , each $a^{n_{i}} - b^{n_{i}}$ divides $D_{j}$ , so $D_{j} S_{j}$ is an integer. The tail satisfies

0 < X_{S} (t) - S_{j} \leq \sum_{n \geq n_{j + 1}} \frac{ρ^{n}}{1 - ρ^{n}} \leq \frac{ρ^{n_{j + 1}}}{(1 - ρ)^{2}} .

Each ratio $n_{j + 1} / n_{j}$ is an integer at least $2$ .

Suppose the ratio is at least $3$ for infinitely many $j$ . For those $j$ ,

0 < D_{j} (X_{S} (t) - S_{j}) \leq \frac{a^{n_{j}} ρ^{3 n_{j}}}{(1 - ρ)^{2}} = \frac{1}{(1 - ρ)^{2}} (\frac{b^{3}}{a^{2}})^{n_{j}} ⟶ 0

when $a^{2} > b^{3}$ . If $X_{S} (t) = p / q$ , then $q D_{j} (X_{S} (t) - S_{j})$ is a positive integer for every $j$ , a contradiction.

Otherwise the ratio is $2$ from some index on, so part (b) completes the proof of (a).

For part (b), discard a finite prefix and write the repeating ratio block as $r_{1}, \dots, r_{ℓ}$ , with every $r_{i} \geq 2$ . Put

Q = \prod_{i = 1}^{ℓ} r_{i}, e_{0} = 1, e_{i} = \prod_{h = 1}^{i} r_{h} (1 \leq i < ℓ) .

With $d$ the first exponent after that prefix, the remaining exponents are exactly $d e_{i} Q^{k}$ with $0 \leq i < ℓ$ and $k \geq 0$ . Indeed, one complete block multiplies an exponent by $Q$ , and its intermediate positions multiply it by the $e_{i}$ . These exponents are distinct since $1 = e_{0} < \dots < e_{ℓ - 1} < Q$ . Define

G (z) = \sum_{k \geq 0} \sum_{i = 0}^{ℓ - 1} \frac{z^{e_{i} Q^{k}}}{1 - z^{e_{i} Q^{k}}}, R (z) = \sum_{i = 0}^{ℓ - 1} \frac{z^{e_{i}}}{1 - z^{e_{i}}} .

The series for $G$ converges normally on compact subsets of $| z | < 1$ : on $| z | \leq r < 1$ its absolute sum is at most $ℓ (1 - r)^{- 1} \sum_{k \geq 0} r^{Q^{k}} < \infty$ . Its Taylor coefficients are integers, and removing the $k = 0$ block gives $G (z) = R (z) + G (z^{Q})$ . For the one-term block $(2)$ this is the function $g (z) = \sum_{k \geq 0} z^{2^{k}} / (1 - z^{2^{k}})$ used below.

The function $G$ is transcendental over $C (z)$ . To see this, fix a primitive root of unity $ζ$ of order $Q^{j}$ , with $j \geq 1$ , and approach it along $r ζ$ as $r ↑ 1$ . For every $k \geq j$ , all terms in the $k$ th block are positive real numbers. Already the term $k = j, i = 0$ tends to $+ \infty$ . Among the finitely many earlier terms, those with $ζ^{e_{i} Q^{k}} = 1$ are also positive, and all the others remain bounded. Thus $Re G (r ζ) \to + \infty$ . The infinitely many distinct orders $Q^{j}$ give infinitely many singularities, whereas an algebraic function over $C (z)$ has only finitely many.

We use Nishioka’s value theorem for Mahler systems [19], in the precise form quoted by Adamczewski and Faverjon [18]. At an algebraic regular point it equates the transcendence degree of the function values with that of the functions over $\overset{―}{Q} (z)$ . Here the system is

(\begin{matrix} G (z) \\ 1 \end{matrix}) = (\begin{matrix} 1 & R (z) \\ 0 & 1 \end{matrix}) (\begin{matrix} G (z^{Q}) \\ 1 \end{matrix}) .

For algebraic real $t > 1$ , the point $α = t^{- d}$ is algebraic and lies in $(0, 1)$ . The matrix and its inverse have poles only at roots of unity, so none of $α^{Q^{k}}$ is a pole and $α$ is regular. Nishioka’s theorem gives ${trdeg}_{\overset{―}{Q}} (G (α), 1) = 1$ . The discarded finite sum is algebraic, so adding it to $G (α)$ proves part (b). ◻

The accompanying exact coefficient probe checks the proposed functional equation through degree $100,000$ for five ratio blocks, including $(2, 3)$ . For that alternating chain, the doubling equation $G (z) - G (z^{2}) = z / (1 - z)$ already fails at degree four. The probe also rejects the $(2, 3)$ block’s equation on an explicit nonperiodic ratio word. The probe is in research/experiments/interestingness/periodic_chain_probe.py. Those finite checks test the formulas; the block decomposition above proves the equation at every degree.

Theorem 6.5 (divisibility chains at every rational base). Let $a > b \geq 1$ be coprime integers and let $S = {n_{1} < n_{2} < \dots}$ be infinite with $n_{j} ∣ n_{j + 1}$ for every $j$ . Then $X_{S} (a / b)$ is transcendental.

Theorem 6.5 strengthens the rational-base conclusion of Theorem 6.3. It removes the hypothesis $a^{2} > b^{3}$ from part (a) and strengthens the conclusion there from irrationality to transcendence. In particular every divisibility chain at bases $4 / 3$ and $5 / 4$ has a transcendental sum, including the chains at base $4 / 3$ whose ratios are $2$ except for infinitely many $3$ s, the first case left open by Theorem 6.3(a). At rational bases it also gives part (b) without Mahler’s method; part (b) also covers irrational algebraic bases. The proof of Theorem 6.3(a) compares one partial sum with the whole tail, and at base $4 / 3$ with ratio $n_{j + 1} / n_{j} = 2$ that comparison fails: $D_{j} (X_{S} (4 / 3) - S_{j})$ grows like $(9 / 4)^{n_{j}}$ . The proof below keeps the first few terms of the tail as separate coordinates of an integer vector. Those coordinates are products of powers of $a$ and $b$ , and by the product formula they contribute nothing to the product of absolute values over the archimedean place and the primes dividing $a b$ . Only one linear form and the coordinate carrying the partial sum remain, and the Subspace Theorem applies.

Evidence.

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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