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

Which dyadic shifts detect irrationality?

Consider a real sequence satisfying

T_{N + 1} = 2 T_{N} - g_{N + 1}, g_{N + 1} \in Z .

Write $‖ x ‖ = dist (x, Z)$ . Iterating the recurrence shows that $T_{N} = 2^{N} T_{0} - a_{N}$ for some integers $a_{N}$ , and hence

\begin{matrix} (4) & ‖ T_{N + h} - T_{N} ‖ = ‖ 2^{N} (2^{h} - 1) T_{0} ‖ . \end{matrix}

For $H \subseteq Z_{> 0}$ and $c \in R$ , say that $H$ detects $T$ at threshold $c$ when

\begin{matrix} (5) & \forall h \in H \forall N_{0} \exists N \geq N_{0} : ‖ T_{N + h} - T_{N} ‖ \geq c . \end{matrix}

The witnessing index may depend on the shift.

The Lean-checked transfer from the #251 tail classifier and the #269 bounded-radix escape theorem gives the implication from irrationality in (5) for every positive shift when $c \leq 1 / 3$ ; the research record keeps the original $1 / 31$ experiment and the later sharp-constant argument distinct. Dubickas’s theorem, in the form stated by Akiyama and Kaneko [4], gives

\underset{N \to \infty}{lim sup} ‖ 2^{N} ξ ‖ \geq τ (ξ \notin Q), τ = \sum_{n \geq 0} \frac{t_{n}}{2^{n + 1}} = 0.412454 \dots,

where $t_{n}$ is the parity of the binary digit sum of $n$ . This cited input extends the implication to every $c < τ$ . The endpoint argument below is ordinary mathematics; the sharp bound itself is not formalised here.

For fixed $c$ and $H$ , call (5) the selected-shift test. The next theorem asks when that test detects irrationality for every integer-digit dyadic recurrence, rather than for one chosen orbit.

Theorem 7.1 (Restricted dyadic shifts). Fix $c \in R$ and $H \subseteq Z_{> 0}$ . The following conditions are equivalent:

For every integer-digit dyadic recurrence $T$ , the initial value $T_{0}$ is irrational if and only if $T$ passes the selected-shift test.
The parameters satisfy $0 < c < τ$ , and every positive integer $d$ divides some shift $h \in H$ .

Proof. Suppose first that these two conditions hold. For irrational $T_{0}$ and fixed $h > 0$ , $(2^{h} - 1) T_{0}$ is irrational. Equation (4) and Dubickas’s bound give indices as late as desired with distance at least $c$ . Conversely, let $T_{0} = p / q$ with $q = 2^{s} r$ and $r$ odd. Some $d > 0$ satisfies $2^{d} \equiv 1 (\mod r)$ ; take $d = 1$ if $r = 1$ . Choose $h \in H$ divisible by $d$ . For every $N \geq s$ , $2^{N} (2^{h} - 1) p / q$ is an integer, so (5) fails.

For necessity of the divisibility condition, suppose no member of $H$ is divisible by some $d \geq 2$ . Put $L = 3 d$ , $q = 2^{L} - 1$ , and $T_{N} = {2^{N} / q}$ . This bounded rational orbit has digits in ${0, 1}$ . For a tested shift $h$ , let $r \in {1, \dots, L - 1}$ be its residue modulo $L$ . Since $2^{L} \equiv 1 (\mod q)$ , the distance sequence for $h$ is periodic and agrees with that for $r$ . The sequences for $r$ and $L - r$ agree up to a cyclic shift, because $2^{r} (2^{L - r} - 1) \equiv - (2^{r} - 1) (\mod q)$ . We may therefore use $m = max (r, L - r) \geq L / 2 \geq 3$ . At the index $N = L - m - 1$ , one distance is

v = \frac{2^{L - 1} - 2^{L - m - 1}}{2^{L} - 1} = \frac{1 - 2^{- m}}{2 (1 - 2^{- L})}, \frac{7}{16} \leq v < \frac{1}{2} .

It recurs every $L$ indices. The first six Thue–Morse digits are $011010$ , so $τ < 27 / 64 < 7 / 16$ . Thus this rational orbit passes every tested shift at every $0 < c < τ$ .

If $c \leq 0$ , the zero orbit passes. If $H$ is empty, every orbit passes. Finally, for $c \geq τ$ and any $h_{0} \in H$ , take $T_{0} = τ / (2^{h_{0}} - 1)$ and $g_{N} = 0$ . The cited sharp-bound construction identifies $τ$ as irrational [4]. The strict endpoint inequality $‖ 2^{N} τ ‖ < τ$ for every $N \geq 1$ follows from the Thue–Morse shift argument just below. Equation (4) makes the test fail at $h_{0}$ . ◻

Here is the strict endpoint step used in the proof. Write the Thue–Morse word as $t = 011010 \dots$ , its bitwise complement as $\bar{t}$ , and let $μ$ be the order-preserving substitution $0 \mapsto 01$ , $1 \mapsto 10$ . The word $t$ is fixed by $μ$ . An odd-indexed suffix of $t$ starts with $001$ or $010$ when its first bit is $0$ , both strictly below the prefix $011$ of $t$ . When its first bit is $1$ , it starts with $101$ or $110$ , both strictly above the prefix $100$ of $\bar{t}$ . An even-indexed suffix is the image under $μ$ of a shorter suffix, so induction and order preservation give the same strict comparisons at every positive index. The binary value of each suffix is therefore below $τ$ or above $1 - τ$ , according to its first bit. This proves $‖ 2^{N} τ ‖ < τ$ for $N \geq 1$ . The research record retains the longer historical calculation. Related extremal-word constructions appear in Allouche, Clarke and Sidorov [5], whose published bibliography points to earlier work of Allouche and Cosnard. The linked research record gives the longer nearest-integer calculation; no historical priority for the specific formulation above is asserted.

The factorial family $H = {j! : j \geq 1}$ satisfies the divisibility condition: $d ∣ d!$ . The power-of-two family does not, since none of its members is divisible by $3$ . An especially small counterexample for the latter is the rational orbit $T_{N} = {2^{N} / 7}$ : for each tested shift its distances cycle through $1 / 7$ , $2 / 7$ , and $3 / 7$ , so it passes at every $c < τ$ . No finite shift family suffices. Conversely, excluding all multiples of a large $d$ leaves a family of density $1 - 1 / d$ that fails the test; factorial shifts have density zero and succeed. This criterion does not establish irrationality for the actual prime-gap tail in #251.

Divisor coverage also implies that $H \cap d Z_{> 0}$ is unbounded for every $d > 0$ : apply coverage to the multiples $k d$ as $k$ grows. Consequently deleting finitely many shifts from a working family preserves the criterion.

Polynomially selected shifts

For a polynomial $P \in Z [x]$ with positive leading coefficient, set

H_{P} = {P (n) : n \geq 0, P (n) > 0}, H_{P}^{prime} = {P (p) : p prime, P (p) > 0} .

The argument of $P$ in the second family is prime; this is different from checking whether $P$ has a root modulo every prime.

Corollary 7.2 (Polynomial shift families). Fix $0 < c < τ$ . The $H_{P}$ -selected test detects irrationality for every integer-digit dyadic recurrence if and only if $P$ has a root modulo every positive integer. The $H_{P}^{prime}$ -selected test has this property if and only if, for every positive integer $d$ , there is a root $r$ of $P$ modulo $d$ with $gcd (r, d) = 1$ .

Proof. By Theorem 7.1, each assertion reduces to whether every $d > 0$ divides a member of the selected family. If $P (r) \equiv 0 (\mod d)$ , all sufficiently large integers $n \equiv r (\mod d)$ give positive multiples $P (n)$ of $d$ . This proves the first assertion in both directions.

For the prime-argument family, a root $r$ coprime to $d$ gives arbitrarily large primes $p \equiv r (\mod d)$ by Dirichlet’s theorem, and hence positive multiples $P (p)$ of $d$ . Conversely, suppose there is no unit root modulo some $d$ . Every prime $p$ for which $d ∣ P (p)$ then satisfies $gcd (p, d) > 1$ , so $p$ is one of the finitely many prime divisors of $d$ . Thus $H_{P}^{prime} \cap d Z$ is bounded. Divisor coverage would make this intersection unbounded, since for every $k > 0$ it supplies a member divisible by $k d$ . This is a contradiction. ◻

The first condition is the usual intersectivity condition [1]. The unit-root condition for prime arguments is $P$ -intersectivity, also called intersectivity of the second kind [2]. For example, $P (n) = n^{2}$ works with integer arguments, while $n^{2} + 1$ fails modulo $3$ . At prime arguments, $P (p) = p$ fails already modulo $6$ , while $p - 1$ and $p^{2} - 1$ work: the residue $1$ is a unit root modulo every $d$ .

Intersectivity need not come from an integer root. Mishra lists $F (x) = (x^{2} - 13) (x^{2} - 17) (x^{2} - 221)$ as a polynomial with a root modulo every positive integer but no rational root [3]. In fact, it also has a unit root modulo every positive integer. For odd primes other than $13$ and $17$ , at least one of $13, 17, 221$ is a nonzero quadratic residue, since $221 = 13 \cdot 17$ ; its root lifts to every prime power. Modulo powers of $13$ , use $x^{2} - 17$ with $x \equiv 2 (\mod 13)$ ; modulo powers of $17$ , use $x^{2} - 13$ with $x \equiv 8 (\mod 17)$ . At powers of $2$ , the unit $17 \equiv 1 (\mod 8)$ has a square root. The Chinese remainder theorem supplies unit roots modulo arbitrary $d$ . Thus both $H_{F}$ and $H_{F}^{prime}$ pass the criterion, without relying on a single global root.

Prime moduli alone do not suffice for the first condition. Let $Q (x) = (x^{2} - 2) (x^{2} - 3) (x^{2} - 6)$ . It has a root modulo every prime: for odd primes not dividing $6$ , if neither $2$ nor $3$ is a square, their product $6$ is; the primes $2$ and $3$ are immediate. But $Q (n) \equiv 4$ when $n$ is even and $Q (n) \equiv 6$ when $n$ is odd, modulo $8$ . Therefore neither $H_{Q}$ nor its prime-argument subfamily contains a multiple of $8$ . Lê’s cited arXiv v1 introduction lists this $Q$ as intersective [1]; the modulo- $8$ calculation corrects that example, without affecting the local-root criterion stated there. The failure is visible without the general counterexample construction: take the rational orbit $T_{N} = {2^{N} / 255}$ . Since $2^{8} \equiv 1 (\mod 255)$ , for every positive shift $h = Q (n)$ , indices $N \equiv 3 (\mod 8)$ when $h \equiv 4 (\mod 8)$ give $‖ T_{N + h} - T_{N} ‖ = 120 / 255$ , and indices $N \equiv 1 (\mod 8)$ when $h \equiv 6 (\mod 8)$ give $126 / 255$ . Both distances exceed $7 / 16 > τ$ , so this rational orbit passes every $Q$ -selected test at $0 < c < τ$ . The modular root and orbit calculations are ordinary proofs; this polynomial extension is not claimed as Lean checked.

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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