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

Divisibility cuts at every algebraic base

A support need not be a chain for the prefix-clearing argument to work. What it needs is a place to cut: every earlier exponent divides one integer $L$ , while every later exponent is a multiple of a larger integer $M$ . The prefix then becomes a polynomial of degree at most $L$ , and the tail has a bounded number of possible initial coefficient patterns in powers of $t^{- M}$ . Keeping several of those powers as separate coordinates pays for the height of the prefix, including all its algebraic conjugates.

Theorem 6.6 (Lambert sums across divisibility cuts). Let $H$ be an infinite subset of the positive integers. Suppose that positive integers $L_{j}, M_{j}$ satisfy

L_{j} ⟶ \infty, M_{j} > L_{j}, M_{j} - L_{j} ⟶ \infty,

and, for every $j$ and $n \in H$ ,

n \leq L_{j} ⟹ n ∣ L_{j}, n > L_{j} ⟹ M_{j} ∣ n .

For every infinite $B \subseteq H$ , every bounded family of positive integer weights $(w_{n})_{n \in B}$ , and every real algebraic $t > 1$ , the sum

\sum_{n \in B} \frac{w_{n}}{t^{n} - 1}

is transcendental.

These hypotheses hold for every increasing divisibility chain, taking $L_{j} = n_{j}$ and $M_{j} = n_{j + 1}$ . They also allow arbitrarily large sets of pairwise incomparable exponents in one block, as the example below shows. They do not hold for the full positive support. The theorem therefore does not settle the rationality of the full Lambert series at $3 / 2$ .

The proof uses the number-field Subspace Theorem in the normalization of Evertse and Ferretti [44]. It extends the preceding argument based on the method of Corvaja and Zannier [23]. This is an ordinary proof; neither a Lean proof of the transcendence conclusion nor historical priority is asserted.

Proof. Write $ρ = t^{- 1} \in (0, 1)$ and $1 \leq w_{n} \leq W$ . The series converges by comparison with a geometric series. Suppose its value $x$ is algebraic, and put $K = Q (ρ, x)$ , $d = [K : Q]$ . Use absolute values $| \cdot |_{v}$ normalized by the product formula. At the specified real embedding $v_{0}$ , $| a |_{v_{0}} = | a |^{1 / d}$ . Let $S$ contain the archimedean places and every finite place where $ρ$ is not a unit. Then $ρ$ is an $S$ -unit. Write $h = \log H (ρ) > 0$ , where $H$ is the multiplicative projective height.

For one cut, abbreviate $L = L_{j}$ , $M = M_{j}$ , and set

s_{j} = \sum_{\begin{matrix} n \in B \\ n \leq L \end{matrix}} \frac{w_{n} ρ^{n}}{1 - ρ^{n}}, z_{j} = (1 - ρ^{L}) s_{j} = P_{j} (ρ) .

For $n ∣ L$ ,

(1 - X^{L}) \frac{X^{n}}{1 - X^{n}} = X^{n} + X^{2 n} + \dots + X^{L} .

Thus $P_{j} \in Z [X]$ , its degree is at most $L$ , and its coefficient sum is at most $B_{j} = W L^{2}$ . Consequently, at a finite place, $| z_{j} |_{v} \leq max (1, | ρ |_{v})^{L}$ ; at an archimedean place the additional factor is $B_{j}^{d_{v} / d}$ , where $d_{v}$ is its local degree. Since the archimedean exponents sum to one,

\begin{matrix} (2) & \prod_{v \in S ∖ {v_{0}}} | z_{j} |_{v} \leq B_{j} H (ρ)^{L} . \end{matrix}

All coordinates used below are $S$ -integers. Discard finitely many cuts so that the prefix is nonempty.

Every remaining exponent is a multiple of $M$ , so

x - s_{j} = \sum_{k \geq 1} c_{j} (k) ρ^{k M}, c_{j} (k) = \sum_{\begin{matrix} n \in B, n > L \\ n ∣ k M \end{matrix}} w_{n} .

Writing $n = M r$ shows $0 \leq c_{j} (k) \leq W τ (k) \leq W k$ . Choose a fixed integer $R \geq 2$ such that $γ = R (- \log ρ) / d - h > 0$ . On an infinite subsequence the vector $(c_{j} (1), \dots, c_{j} (R - 1))$ is constant; denote it by $(c_{1}, \dots, c_{R - 1})$ . The remaining tail satisfies

0 < E_{j} := x - s_{j} - \sum_{k = 1}^{R - 1} c_{k} ρ^{k M} \leq C_{R} ρ^{R M}, C_{R} = W (\frac{R}{1 - ρ} + \frac{ρ}{(1 - ρ)^{2}}) .

Strict positivity follows from the infinite positive support, even if some of the frozen coefficients are zero.

Consider the $2 R + 1$ coordinates

Y_{j} = (1, ρ^{L}, z_{j}, ρ^{M}, ρ^{L + M}, \dots, ρ^{(R - 1) M}, ρ^{L + (R - 1) M}) .

Put $D = L + (R - 1) M$ . The coordinates $1, ρ^{D}$ give the lower height bound; the polynomial estimate for $z_{j}$ gives the upper bound:

\begin{matrix} (3) & H (ρ)^{D} \leq H (Y_{j}) \leq B_{j} H (ρ)^{D} . \end{matrix}

At every place of $S$ use the coordinate linear forms, except that at $v_{0}$ replace the $z$ -coordinate form by

F (Y) = x Y_{0} - x Y_{1} - Y_{z} - \sum_{k = 1}^{R - 1} c_{k} (Y_{k, 0} - Y_{k, 1}) .

Its coefficient of $Y_{z}$ is $- 1$ , so the forms remain independent, and $F (Y_{j}) = (1 - ρ^{L}) E_{j}$ . Every other coordinate is an $S$ -unit and its product of absolute values over $S$ is one. Hence

\prod_{v \in S} \prod_{i} | F_{v, i} (Y_{j}) |_{v} \leq C_{R}^{1 / d} B_{j} \exp (- γ M) .

Since $L < M$ , $\log B_{j} = o (M)$ and $\log H (Y_{j}) \leq (R h + 1) M$ eventually. The last product is at most $H (Y_{j})^{- ϵ}$ for some fixed $ϵ > 0$ . Outside $S$ the local vector norm is exactly one, because the coordinates are integral and the first is one. Dividing the displayed product by $H (Y_{j})^{2 R + 1}$ therefore gives precisely the normalized hypothesis of the Subspace Theorem. Infinitely many of the $Y_{j}$ lie in one proper $K$ -linear subspace, and hence in one fixed nonzero hyperplane.

Substitute

z_{j} = (1 - ρ^{L}) (x - \sum_{k = 1}^{R - 1} c_{k} ρ^{k M} - E_{j})

in that hyperplane equation. It becomes a fixed linear combination of the monomials with exponents

0, L, M, L + M, \dots, (R - 1) M, L + (R - 1) M

equal to a fixed multiple of $(1 - ρ^{L}) E_{j}$ . Successive exponent gaps are $L$ or $M - L$ , both tending to infinity; the remainder starts at $R M$ , also a gap $M - L$ beyond the last exponent. If any monomial coefficient were nonzero, divide by the first such monomial and let $j$ tend to infinity. All other terms tend to zero, a contradiction. Thus all these coefficients vanish. The strict inequality $E_{j} > 0$ then forces the hyperplane’s $z$ coefficient to vanish, and substitution forces every other coefficient to vanish. This contradicts the chosen nonzero hyperplane and proves the theorem. ◻

Corollary 6.7 (A host of unbounded divisibility width). Define

\begin{matrix} N_{0} = 1, m_{j} = 2^{N_{j}}, L_{j} = N_{j} lcm (1, \dots, m_{j}), N_{j + 1} = 2 L_{j}, \\ H_{*} = ⋃_{j \geq 0} {N_{j}, 2 N_{j}, \dots, m_{j} N_{j}} . \end{matrix}

Then $\sum_{n \in H_{*}} 1 / n = \infty$ , and $H_{*}$ is not contained in any finite union of divisibility chains. Nevertheless every infinite subset of $H_{*}$ , with bounded positive integer weights, has a transcendental Lambert sum at every real algebraic base greater than one.

Proof. Every exponent in the first $j + 1$ blocks divides $L_{j}$ ; all later exponents are multiples of $N_{j + 1} = 2 L_{j}$ . The blocks are disjoint and ordered. These are the required cuts, with $M_{j} = 2 L_{j}$ . Dyadic grouping gives $\sum_{k \leq 2^{N_{j}}} 1 / k \geq N_{j} / 2$ , so each block contributes at least $1 / 2$ to the reciprocal sum. In the upper half of the $j$ th block, two different coefficients have ratio less than two and neither divides the other. The resulting antichains have unbounded cardinality. A union of finitely many chains cannot contain them. Apply the theorem. ◻

This example also lies in the one-prime weighted class used for #257. For $h (n) = 2^{v_{2} (n)}$ and every real $t > 1$ , its weighted mass obeys

\sum_{n \in H_{*}} \frac{h (n)}{n (t^{h (n)} - 1)} \leq \frac{2 t^{2}}{(t - 1)^{3}} .

Indeed, $h_{j} = h (N_{j})$ at least doubles. For $n = N_{j} k$ , the inequality $t^{h_{j} h (k)} - 1 \geq h (k) (t^{h_{j}} - 1)$ bounds one block by $[h_{j} / (t^{h_{j}} - 1)] \sum_{k \leq 2^{N_{j}}} 1 / (N_{j} k)$ , which is at most $2 h_{j} / (t^{h_{j}} - 1)$ . Summing over distinct positive integers $h_{j}$ and using $t^{r} - 1 \geq (t - 1) t^{r - 1}$ proves the displayed bound. Thus the new conclusion here is algebraic-base transcendence on every infinite subset, not merely another instance of the existing integer-base irrationality criterion.

For a smaller example take $H = ⋃_{j \geq 0} {12^{j}, 2 \cdot 12^{j}, 3 \cdot 12^{j}}$ , with cuts $L_{j} = 6 \cdot 12^{j}$ , $M_{j} = 12^{j + 1}$ . Every infinite thinning satisfies the theorem, including the terms $2 \cdot 12^{p}, 3 \cdot 12^{p}$ for prime $p$ at base $3 / 2$ . The theorem is hereditary under thinning because the same cuts continue to work. Arbitrary supports do not have this property; the earlier rational subsums below base two remain a necessary warning.

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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