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

The arithmetic threshold for rational derivatives

We prove Theorem 2.2. The coefficient estimate and the divisibilities are both needed, including at the critical exponent.

Suppose the first $d$ derivatives at $1$ , starting with the value, are rational. Subtract their Taylor polynomial

P (z) = \sum_{r = 0}^{d - 1} \frac{f^{(r)} (1)}{r!} (z - 1)^{r}

and choose a positive integer $D$ clearing its coefficient denominators. Then $g = D (f - P) = \sum_{k \geq 0} g_{k} z^{k} / k!$ has integer coefficients, $| g_{k} | = O (k^{d})$ , and the same eventual divisibilities. Moreover $g^{(r)} (1) = 0$ for $0 \leq r < d$ . Thus $B (z) = g (z) / (z - 1)^{d}$ is entire. Expansion at zero gives

\begin{matrix} (Q) & B (z) = \sum_{n \geq 0} b_{n} \frac{z^{n}}{n!}, b_{n} = (- 1)^{d} \sum_{k = 0}^{n} (\binom{n - k + d - 1}{d - 1}) \frac{n!}{k!} g_{k} . \end{matrix}

In particular $b_{n}$ is an integer. Fix $q$ , and choose $K$ so that $q ∣ g_{k}$ for $k \geq K$ . Those terms in (Q) are divisible by $q$ . For each of the finitely many $k < K$ , the factor $n! / k!$ is divisible by $q$ once $n$ is sufficiently large. Hence $q ∣ b_{n}$ eventually as well.

The vanishing derivatives imply $\sum_{k \geq 0} R (k) g_{k} / k! = 0$ for every polynomial $R$ of degree less than $d$ , because the falling factorials of orders $0, \dots, d - 1$ form a basis. Apply this to $R (k) = (\binom{n - k + d - 1}{d - 1})$ , interpreted as a polynomial. Its values at $k = n + 1, \dots, n + d - 1$ vanish, and at $k = n + j$ , $j \geq d$ , they equal $(- 1)^{d - 1} (\binom{j - 1}{d - 1})$ . The complementary tail of (Q) therefore gives

\begin{matrix} (R) & b_{n} = n! \sum_{j \geq d} (\binom{j - 1}{d - 1}) \frac{g_{n + j}}{(n + j)!} . \end{matrix}

All series here converge absolutely. For large $n$ , the absolute value is bounded by $C$ times the sum of

v_{j} = (\binom{j - 1}{d - 1}) \frac{n! (n + j)^{d}}{(n + j)!}, j \geq d .

Their successive ratios satisfy

\frac{v_{j + 1}}{v_{j}} = \frac{j}{j - d + 1} {(\frac{n + j + 1}{n + j})}^{d} \frac{1}{n + j + 1} \leq \frac{d 2^{d}}{n + 1} .

For sufficiently large $n$ this is at most $1 / 2$ , whereas $v_{d} = n! (n + d)^{d} / (n + d)!$ is bounded. Thus $(b_{n})$ is bounded. Choose an integer $q$ larger than its eventual absolute bound. Eventual divisibility by this $q$ forces $b_{n} = 0$ eventually. It follows that $B$ , $g$ and $f$ are polynomials.

For the converse, when $c > d$ , Theorem 2.1 gives an open set of vectors realised by nonpolynomial members of $H_{c}$ . Every nonempty open subset of $R^{d}$ meets $Q^{d}$ . This completes the proof of the sharp threshold.

Remark (Why the congruences cannot be omitted). The function $(z - 1)^{d} e^{z}$ has its first $d$ derivatives zero at $1$ . Its integer factorial coefficients are

p_{d} (n) = \sum_{j = 0}^{d} (- 1)^{d - j} (\binom{d}{j}) (n)_{j} = n^{d} - \frac{d (d + 1)}{2} n^{d - 1} + O_{d} (n^{d - 2})

for $d \geq 2$ ; for $d = 1$ they are $n - 1$ . Thus $0 \leq p_{d} (n) \leq n^{d}$ eventually. Replacing the finitely many earlier coefficients, including the constant coefficient, by zero adds a rational polynomial. The resulting nonpolynomial function has nonnegative integer coefficients bounded by $n^{d}$ everywhere and all the specified derivatives rational. It fails the eventual divisibility assumption. Full dimension and even the growth bound alone therefore do not supply the arithmetic conclusion.

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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