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.

Start here, by the editorial route

In this paper

Evidence.

Ordinary proof, with the $p$ -adic Subspace Theorem as an external premise, awaiting specialist review; it is not checked in Lean. Exact finite checks of the formulas used below are in research/experiments/chain_transcendence/ of the repository. They check the bookkeeping and form no part of the proof.

Proof. As in the proof of Theorem 6.3, write $ρ = b / a$ , $D_{j} = a^{n_{j}} - b^{n_{j}}$ and $S_{j} = \sum_{i \leq j} b^{n_{i}} / (a^{n_{i}} - b^{n_{i}})$ ; then $D_{j} S_{j}$ is a positive integer. Suppose that $x = X_{S} (a / b)$ is algebraic.

The tail. Fix $j$ and put $N = n_{j}$ and $M = n_{j + 1}$ . Each $n_{l}$ with $l > j$ is a multiple of $M$ , and $ρ^{n} / (1 - ρ^{n}) = \sum_{i \geq 1} ρ^{i n}$ , so

x - S_{j} = \sum_{k \geq 1} c_{j} (k) ρ^{k M}, c_{j} (k) = # {l > j : n_{l} ∣ k M} .

Since $n_{j + 1} = M$ and $n_{j + s} \geq 2^{s - 1} M$ , we have $1 \leq c_{j} (k) \leq 1 + \log_{2} k$ . For an integer $K \geq 1$ put $E_{j} = \sum_{k \geq K} c_{j} (k) ρ^{k M}$ . Then $0 < E_{j} \leq C_{K} ρ^{K M}$ , where $C_{K} = \sum_{i \geq 0} (1 + \log_{2} (K + i)) ρ^{i}$ does not depend on $j$ .

Choice of $K$ . The ratios $r_{j} = n_{j + 1} / n_{j}$ are integers at least $2$ . Choose an infinite set $J_{0}$ of indices on which either $r_{j}$ equals a constant $r$ , or $r_{j} \to \infty$ . In the first case fix $K \geq 1$ with $K r \log (a / b) > \log a$ ; at base $4 / 3$ with $r = 2$ the least such $K$ is $3$ . In the second case put $K = 1$ . The vector $(c_{j} (1), \dots, c_{j} (K - 1))$ takes only finitely many values, so it equals a fixed $(c_{1}, \dots, c_{K - 1})$ for all $j$ in an infinite set $J \subseteq J_{0}$ .

Integer points and one linear form. For $j \in J$ put $H = a^{N + (K - 1) M}$ and

Y_{j} = (H, H ρ^{N}, a^{(K - 1) M} D_{j} S_{j}, H ρ^{M}, H ρ^{N + M}, \dots, H ρ^{(K - 1) M}, H ρ^{N + (K - 1) M}) \in Z^{2 K + 1} .

Every coordinate except the third is an integer whose prime factors divide $a b$ ; for example $H ρ^{N + k M} = a^{(K - 1 - k) M} b^{N + k M}$ . For $y = (y_{1}, \dots, y_{2 K + 1})$ put

L (y) = x y_{1} - x y_{2} - y_{3} - \sum_{k = 1}^{K - 1} c_{k} (y_{2 k + 2} - y_{2 k + 3}) .

Since $a^{(K - 1) M} D_{j} S_{j} = H (1 - ρ^{N}) S_{j}$ , the expansion of the tail gives

L (Y_{j}) = H (1 - ρ^{N}) (x - S_{j} - \sum_{k = 1}^{K - 1} c_{k} ρ^{k M}) = H (1 - ρ^{N}) E_{j} .

The product over places. Let $Σ$ consist of the archimedean place and the primes dividing $a b$ , with $| p |_{p} = p^{- 1}$ . At the archimedean place take the $2 K + 1$ forms $L$ and $y_{i}$ for $i \neq 3$ ; at each prime $p ∣ a b$ take the $2 K + 1$ coordinate forms. The coordinate forms are linearly independent, and so are the forms at the archimedean place, because the coefficient of $y_{3}$ in $L$ is $- 1$ . By the product formula, $\prod_{v \in Σ} | y |_{v} = 1$ for every nonzero integer $y$ whose prime factors divide $a b$ . Hence, in the product of $| F (Y_{j}) |_{v}$ over all $v \in Σ$ and all forms $F$ chosen at $v$ , only $L$ and the third coordinate at the primes remain:

Π_{j} = | L (Y_{j}) | \prod_{p ∣ a b} | a^{(K - 1) M} D_{j} S_{j} |_{p} \leq H E_{j} a^{- (K - 1) M} \leq C_{K} a^{N} ρ^{K M},

by $| D_{j} S_{j} |_{p} \leq 1$ and $\prod_{p ∣ a b} | a |_{p} = a^{- 1}$ . Also $‖ Y_{j} ‖ = max_{i} | Y_{j, i} | \leq max (1, x) a^{N + (K - 1) M}$ . In the first case $M = r N$ and $a^{N} ρ^{K M} = a^{- δ N}$ with $δ = K r \log (a / b) / \log a - 1 > 0$ , so $Π_{j} \leq ‖ Y_{j} ‖^{- ε}$ for all large $j \in J$ , where $ε = δ / (2 + 2 (K - 1) r)$ . In the second case $K = 1$ , $‖ Y_{j} ‖ \leq max (1, x) a^{N}$ and $a^{N} ρ^{M} \leq a^{- N}$ once $r_{j} \log (a / b) \geq 2 \log a$ , so $Π_{j} \leq ‖ Y_{j} ‖^{- 1 / 2}$ for all large $j \in J$ .

The Subspace Theorem. We use Schmidt’s Subspace Theorem in the $p$ -adic form due to Schlickewei [20][21], as stated in [22]. Let $Σ$ be a finite set of places of $Q$ containing the archimedean one and, for each $v \in Σ$ , let $F_{v, 1}, \dots, F_{v, m}$ be linearly independent linear forms in $m$ variables with algebraic coefficients. Then for every $ε > 0$ the vectors $y \in Z^{m}$ with $\prod_{v \in Σ} \prod_{i = 1}^{m} | F_{v, i} (y) |_{v} \leq ‖ y ‖^{- ε}$ lie in finitely many proper linear subspaces of $Q^{m}$ . Here every form has rational coefficients except possibly $L$ , at the archimedean place, whose coefficients lie in $Q (x)$ with $x$ real algebraic. The vectors $Y_{j}$ are pairwise distinct, so there are $λ \in Q^{2 K + 1} ∖ {0}$ and an infinite $J^{'} \subseteq J$ with $λ \cdot Y_{j} = 0$ for every $j \in J^{'}$ .

No fixed relation. Divide $λ \cdot Y_{j} = 0$ by $H$ and put $z = ρ^{n_{j}}$ , which tends to $0$ along $J^{'}$ . In the first case $ρ^{k M} = z^{k r}$ , and substituting $S_{j} = x - \sum_{k < K} c_{k} z^{k r} - E_{j}$ gives $Q (z) = λ_{3} (1 - z) E_{j}$ for the polynomial

Q (z) = λ_{1} + λ_{3} x + (λ_{2} - λ_{3} x) z + \sum_{k = 1}^{K - 1} ((λ_{2 k + 2} - λ_{3} c_{k}) z^{k r} + (λ_{2 k + 3} + λ_{3} c_{k}) z^{1 + k r}) .

Because $r \geq 2$ , the exponents $0$ , $1$ , $k r$ and $1 + k r$ with $1 \leq k < K$ are distinct and smaller than $K r$ . The polynomial $Q$ does not depend on $j$ and $| Q (z) | \leq | λ_{3} | C_{K} z^{K r}$ along $J^{'}$ , so every coefficient of $Q$ vanishes; otherwise its lowest nonzero term would dominate as $z \to 0$ . Then $λ_{3} (1 - z) E_{j} = 0$ , and $E_{j} > 0$ gives $λ_{3} = 0$ . The coefficients of $Q$ are now $λ_{1}$ , $λ_{2}$ , $λ_{2 k + 2}$ and $λ_{2 k + 3}$ , so $λ = 0$ , a contradiction. In the second case the same substitution gives $λ_{1} + λ_{3} x + (λ_{2} - λ_{3} x) z = λ_{3} (1 - z) E_{j}$ with $0 < E_{j} \leq C_{1} z^{r_{j}} \leq C_{1} z^{2}$ , and the same comparison gives $λ = 0$ . Therefore $x$ is transcendental. ◻

The last step uses that the ratios of the chain are at least $2$ . The identity

\sum_{j \geq 0} \frac{z^{2^{j}}}{1 - z^{2^{j + 1}}} = \frac{z}{1 - z} (| z | < 1)

shows what happens without this. Let $μ_{j}$ be its $j$ th partial sum and $w = z^{2^{j + 1}}$ . Then $(1 - w) μ_{j}$ is a polynomial in $z$ with integer coefficients of degree less than $2^{j + 1}$ , so at $z = b / a$ it becomes an integer after multiplication by $a^{2^{j + 1}}$ , as $D_{j} S_{j}$ does above. The tail, however, begins at the exponent of $w$ itself: $(1 - w) (z / (1 - z) - μ_{j}) = w$ exactly. At $z = b / a$ this is a fixed linear relation between $a^{2^{j + 1}}$ , $b^{2^{j + 1}}$ and the cleared partial sum, the situation that the last step excludes when the ratios are at least $2$ , and the value is rational.

The same proof applies when the terms of $X_{S} (a / b)$ carry bounded positive integer weights, since the tail coefficients $c_{j} (k)$ then remain positive integers bounded in terms of $k$ .

The argument is an instance of the method of Corvaja and Zannier, who apply the Subspace Theorem when a fixed linear combination of numbers composed of finitely many fixed primes approximates an integer [23]. Here those numbers are products of powers of $a$ and $b$ , and divisibility along the chain supplies the integer $a^{(K - 1) M} D_{j} S_{j}$ . Two older methods bear on parts of Theorem 6.5. When the ratios $r_{j}$ are unbounded, Roth’s theorem [25] suffices: once $r_{j} \log (a / b) \geq 3 \log a$ , the rational $S_{j}$ , of height at most $max (1, X_{S} (a / b)) a^{n_{j}}$ , satisfies $0 < X_{S} (a / b) - S_{j} \leq C a^{- 3 n_{j}}$ with $C$ independent of $j$ , and irrationality follows as in the proof of Theorem 6.3. When the ratios are bounded, the extension of Mahler’s method to chains of functional equations by Loxton and van der Poorten [24] applies to $\sum_{h} φ (α^{n_{h}})$ with $φ (z) = z / (1 - z)$ at algebraic $α$ with $0 < | α | < 1$ , provided the functions $f_{k} (z) = \sum_{h \geq k} φ (z^{n_{h} / n_{k}})$ satisfy their hypothesis of strong transcendence, a transcendence condition uniform in $k$ . That hypothesis fails for unbounded ratios, since $(1 - z) f_{k} (z) - z$ vanishes at $0$ to order $n_{k + 1} / n_{k}$ . We have not checked it for bounded ratios; if it holds, their theorem gives such chains at every real algebraic base greater than $1$ . We did not find Theorem 6.5 stated in the literature. The search, with locators, is recorded in research/experiments/chain_transcendence/README.md and ended on 26 September 2026.

The preceding integer-vector argument uses a rational base. The next number-field argument controls the conjugates explicitly and extends the conclusion to every real algebraic base greater than one. It also permits non-chain blocks with separated divisibility cuts. Arbitrary supports, including the full support at $3 / 2$ , remain outside its hypotheses.

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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