Geometric Moments and Rational Lambert Values: Proofs and Further Results

Will Cook

Long paper

Geometric Moments and Rational Lambert Values: Proofs and Further Results

Erdős #1049 By Will Cook September 2026 Authorship, AI use and citation

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Summary. The record proves the fixed-base Hankel asymptotic for Zudilin's normalised moments and the rational-base irrationality criterion log b/log a<0.4056830213840605…. The determinant proof separates geometric node configurations from positive weights; the irrationality proof instead specialises Zudilin's 2004 polynomial forms after cancellation. Coefficient moments, evaluated lattices and congruence conditions distinguish integer divisibility from nonzero remainder decay. A broader ratio-limit argument and a rational-value model remain unformalised. None of these arguments settles $F (3 / 2)$ .

This paper covers the sharp fixed-base Hankel asymptotic and rational-base irrationality proofs, coefficient moments and arithmetic obstructions, with ordinary ratio-limit and rational-value models distinguished from checked results.

It does not establish the validity of claims tagged Lean, which belongs to the cited kernel-checked source, or a solution to Erdős #1049, which remains open.

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In this paper

Congruences after evaluation at $3 / 2$

We use congruences modulo powers of $2$ and $3$ to find combinations of evaluated polynomials with prescribed divisibility. The argument applies to arbitrary integral polynomials, including the forms of [8]. The denominator-clearing map must be the same for every row that will be added. We therefore fix a common truncation index $W \geq 0$ and, for $P (X) = \sum_{i} p_{i} X^{i} \in Z [X]$ , put

H_{W} (P) = \sum_{i = 0}^{W} p_{i} 3^{i} 2^{W - i} .

This is the denominator-cleared evaluation at $3 / 2$ . Each term $p_{i} X^{i}$ with $i \leq W$ contributes $p_{i} 3^{i} 2^{W - i}$ . Keep the index $W$ fixed when polynomials are added. When $W \geq \deg P$ , this is exactly the cleared numerator, since

\sum_{i = 0}^{W} p_{i} 3^{i} 2^{W - i} = 2^{W} \sum_{i = 0}^{W} p_{i} {(\frac{3}{2})}^{i} = 2^{W} P (\frac{3}{2}) .

Here a unit coefficient in $Z$ means $1$ or $- 1$ . Thus a monic polynomial of degree $W$ with constant term $\pm 1$ satisfies both unit conditions below. Those conditions are sufficient, not necessary: the congruences themselves only require that the relevant coefficient not be divisible by the relevant prime.

Theorem 5.1 (endpoint residues). Let $P = \sum_{i} p_{i} X^{i} \in Z [X]$ and let $W \geq 0$ . Then

H_{W} (P) \equiv p_{0} 2^{W} (\mod 3), H_{W} (P) \equiv p_{W} 3^{W} (\mod 2) .

Consequently a unit constant coefficient prevents divisibility by $3$ , and a unit coefficient at index $W$ prevents divisibility by $2$ .

Proof. Modulo $3$ , every summand with $i > 0$ vanishes; modulo $2$ , every summand with $i < W$ vanishes. The remaining powers are units in the corresponding residue fields. ◻

Since $2^{W}$ is invertible modulo $3$ and $3^{W}$ is invertible modulo $2$ , the two congruences say more than the stated consequence: divisibility of $H_{W} (P)$ by $3$ is decided by $p_{0}$ alone, and divisibility by $2$ by $p_{W}$ alone. The rest of the coefficient vector is invisible to both primes. The coefficient $p_{W}$ is the top coefficient of $P$ exactly when $\deg P = W$ , and is zero when $\deg P < W$ . The identity $H_{W} (P) = 2^{W} P (3 / 2)$ is guaranteed when $\deg P \leq W$ . Without a degree bound, increasing the truncation index gives

H_{W + 1} (P) = 2 H_{W} (P) + p_{W + 1} 3^{W + 1},

so the truncated sum doubles precisely when the extra coefficient is zero.

Example 5.2. Take $W = 2$ . The three polynomials below differ only at an endpoint.

$P$	$H_{2} (P)$	$3 ∣ H_{2} (P)$	$2 ∣ H_{2} (P)$
$X^{2} + 1$	$1 \cdot 4 + 0 \cdot 6 + 1 \cdot 9 = 13$	no	no
$X^{2} + 3$	$3 \cdot 4 + 0 \cdot 6 + 1 \cdot 9 = 21$	yes	no
$2 X^{2} + 1$	$1 \cdot 4 + 0 \cdot 6 + 2 \cdot 9 = 22$	no	yes

The first has both endpoints equal to $1$ and its evaluation, $13$ , is divisible by neither prime; as a check, $2^{2} ((3 / 2)^{2} + 1) = 13$ . The second and third show that each hypothesis is used: spoiling the constant endpoint admits $3$ , and spoiling the top endpoint admits $2$ .

Formal proofs cover the congruences modulo $3$ and modulo $2$ , together with the consequences for a unit constant coefficient and a unit coefficient of $X^{W}$ .

In the following proposition, a unit top endpoint means that the coefficient at index $W$ is $1$ or $- 1$ , with no degree bound imposed. The two conditions apply to different entries of the polynomial pair.

Proposition 5.3 (common divisor). Let $U, V \in Z [X]$ and let $W \geq 0$ . If $U$ has unit top endpoint, $V$ has unit constant endpoint, and an integer $c$ divides both $H_{W} (U)$ and $H_{W} (V)$ , then

2 ∤ c and 3 ∤ c .

Proof. If $2 ∣ c$ , then $2 ∣ H_{W} (U)$ , contrary to the unit coefficient at index $W$ and Theorem 5.1. Similarly, $3 ∣ c$ would imply $3 ∣ H_{W} (V)$ , contrary to the unit constant coefficient. ◻

Lean checks this common-divisor exclusion. The proposition does not say that the two evaluations are coprime; it says only that every common divisor is coprime to $6$ . It gives no restriction on the other prime factors of that divisor.

Example 5.4. Take $W = 2$ , $U = X^{2} + 3$ and $V = 5 X^{2} + 1$ . The top endpoint of $U$ and the constant endpoint of $V$ are both $1$ , and

H_{2} (U) = 3 \cdot 4 + 1 \cdot 9 = 21, H_{2} (V) = 1 \cdot 4 + 5 \cdot 9 = 49 .

Here $gcd (21, 49) = 7$ . The endpoint assumptions therefore allow a nontrivial common divisor, but that divisor is coprime to $6$ .

Corollary 5.5 (limits of rescaling and common-divisor cancellation at $3 / 2$ ). Under the endpoint hypotheses of Proposition 5.3, every common divisor of the unscaled evaluations $H_{W} (U)$ and $H_{W} (V)$ is coprime to $6$ . Multiplying two integer rows by nonzero integers $c_{n}, c_{m}$ multiplies their determinant by $c_{n} c_{m}$ and its absolute value by $| c_{n} c_{m} |$ . Cancelling this introduced scalar factor therefore leaves the original comparison between divisor and determinant size unchanged.

Proof. The common-divisor assertion is Proposition 5.3. By Theorem 4.1, rescaling the two rows by $c_{n}, c_{m}$ multiplies their errors by $c_{n}, c_{m}$ , respectively, and their $2 \times 2$ determinant by $c_{n} c_{m}$ . For nonzero scalars, the introduced divisor has the same absolute value as the introduced factor in that determinant. Dividing out that factor restores the original determinant, including its original divisor-to-size comparison. ◻

One further consequence of Theorem 5.1 is worth stating, because it bears on the most natural way one might hope to import an existing denominator reduction. Write $Φ_{m}$ for the $m$ th cyclotomic polynomial and, for coprime $a > b \geq 1$ , put $Φ_{m} (a, b) = b^{φ (m)} Φ_{m} (a / b)$ for its homogenisation, with exponent $φ (m) = \deg Φ_{m}$ .

Proposition 5.6 (coprimality of homogenised cyclotomic values). Let $a > b \geq 1$ with $gcd (a, b) = 1$ and let $m \geq 1$ . Then $gcd (Φ_{m} (a, b), a b) = 1$ . In particular $gcd (Φ_{m} (3, 2), 6) = 1$ for every $m$ .

Proof. The polynomial $Φ_{m}$ is monic and $Φ_{m} (0) = \pm 1$ . The endpoint argument in Theorem 5.1 also works at $(a, b)$ : modulo a prime dividing $b$ , only $a^{φ (m)}$ survives, and modulo a prime dividing $a$ , only $Φ_{m} (0) b^{φ (m)}$ survives. Coprimality of $a$ and $b$ makes each surviving term nonzero. ◻

Rhin and Viola’s factorial-coset quotients [17] reduce denominators of rational forms in $ζ (2)$ . Zudilin’s $q$ -analogue has an order-twelve symmetry group for the normalised series; its denominator cancellation uses the order-six subgroup preserving the required sign condition [8]. Proposition 5.6 applies to these cyclotomic factors: each homogenised value at $(3, 2)$ , and hence any product of them, is divisible by neither $2$ nor $3$ . It makes no such assertion about an arbitrary integer or factorial factor, which can contain both primes. Cancelling a common cyclotomic factor can still reduce the real absolute values. That reduction is distinct from producing the powers of $2$ or $3$ sought in Section 10.

Scalar multiplication can introduce powers of $2$ and $3$ , but does not improve this comparison. Under the endpoint hypotheses, the unscaled pair has no common factor at either prime. Neither observation shows that a gain at these primes is necessary for a proof by linear forms at $3 / 2$ .

The operation studied next is different: take an integer combination of several rows and ask that the combination be divisible where the individual rows are not. The proposed divisor must come from cancellation between rows, not from multiplying a row by that divisor. The endpoint congruences are the first case of a divisibility condition that can be imposed to any depth, and it is that condition, read additively, which is counted below.

We first raise the two congruences to prime powers. Fix depths $R, S \geq 0$ . For $P \in Z [X]$ , let $J_{3, R} (P)$ and $J_{2, S} (P)$ be the residues of $H_{W} (P)$ modulo $3^{R}$ and $2^{S}$ , respectively. Theorem 5.1 computes the depth-one residues. At depth $R$ , terms with $i \geq R$ are invisible modulo $3^{R}$ ; at depth $S$ , terms with $W - i \geq S$ are invisible modulo $2^{S}$ . Thus deeper congruences inspect longer segments of the two ends of the coefficient list, always at the same fixed $W$ .

Lemma 5.7 (a residue and divisibility by $3^{R}$ ). Vanishing of the residue modulo $3^{R}$ is exactly divisibility by the corresponding power of three: $J_{3, R} (P) = 0$ if and only if $3^{R} ∣ H_{W} (P)$ .

Proof. By definition $J_{3, R} (P)$ is the image of the integer $H_{W} (P)$ in $Z / 3^{R} Z$ . The kernel of this quotient map is $3^{R} Z$ , which proves the equivalence, including $R = 0$ . ◻

The same argument gives $J_{2, S} (P) = 0 ⟺ 2^{S} ∣ H_{W} (P)$ for $S \geq 0$ .

The vector of four residues of a coefficient pair $(U, V)$ is then the quadruple

(J_{3, R} (U), J_{3, R} (V), J_{2, S} (U), J_{2, S} (V)) \in (Z / 3^{R} Z)^{2} \times (Z / 2^{S} Z)^{2},

two residues for each of the two primes, one from each entry of the pair. By the displayed criteria it vanishes exactly when $3^{R}$ divides both specialised entries and $2^{S}$ divides both.

Instead of requiring each input row to have a common divisor, we seek a small integer combination whose two entries are both divisible by $3^{R} 2^{S}$ . Since $H_{W}$ is additive, the four residues of the combination are the corresponding sums of the input residues. This allows a pigeonhole argument on subset sums.

Theorem 5.8 (equal residues for two subset sums). Fix a truncation index $W$ and depths $R, S$ , and let $(U_{j}, V_{j})_{j < M}$ be any $M$ pairs of integral polynomials. Represent each subset of ${0, \dots, M - 1}$ by its indicator vector in ${0, 1}^{M}$ . If the $2^{M}$ subsets outnumber the possible residue vectors in

(Z / 3^{R} Z)^{2} \times (Z / 2^{S} Z)^{2},

then two distinct subsets have the same residue vector. Subtracting their indicator vectors gives a nonzero coefficient vector in ${- 1, 0, 1}^{M}$ cancelling all four residues. The target has exact cardinality

(3^{R})^{2} (2^{S})^{2} .

In particular, if $R > 0$ and $4 R + 2 S \leq M$ , such a collision exists.

Proof. Send each subset to the sum of the residue vectors of its members. The pigeonhole principle gives two distinct subsets with the same residue vector. The number of possible vectors is the product of the four moduli. For $R > 0$ ,

(3^{R})^{2} (2^{S})^{2} < (4^{R})^{2} (2^{S})^{2} = 2^{4 R + 2 S} \leq 2^{M},

which proves the stated sufficient threshold. ◻

The power bracket improves the generic coefficient $4 R$ when the depth $R$ is a positive integer multiple of $41$ . All depths and row counts in the following corollary, including $T$ , are integers.

Corollary 5.9 (the exact count at depth $41$ ). Let $T > 0$ . At modulus $3^{R}$ with $R = 41 T$ , any family of $M \geq 130 T + 2 S$ integral polynomial pairs has two distinct binary vectors with the same residue vector. For $T = 1$ the coefficient $130$ is exact for this counting argument:

2^{129 + 2 S} < | (Z / 3^{41} Z)^{2} \times (Z / 2^{S} Z)^{2} | .

No exact-optimality assertion is made here for $T > 1$ .

Proof. The upper power inequality gives

(3^{41 T})^{2} (2^{S})^{2} < (2^{65})^{2 T} (2^{S})^{2} = 2^{130 T + 2 S} \leq 2^{M},

so Theorem 5.8 applies. For $T = 1$ , direct integer evaluation gives $2^{129} < 3^{82}$ ; multiplying by $(2^{S})^{2}$ gives the displayed reverse count at $129 + 2 S$ . ◻

The formal statements check the sufficient count $130 T + 2 S$ and, when $T = 1$ , the failure of the count at $129 + 2 S$ .

For all depths at once the ambient-cardinality inequality $2^{M} > 3^{2 R} 2^{2 S}$ holds exactly when

M \geq ⌊ 2 R \log_{2} 3 + 2 S ⌋ + 1,

by the definition of the floor function; irrationality of $\log_{2} 3$ is not needed for this strict-inequality reformulation. At $R = 41$ this is $130 + 2 S$ , and at $R = 41 \cdot 31$ it is $4029 + 2 S$ , one below the uniform bound $4030 + 2 S$ of Corollary 5.9; the corollary trades exactness for a certificate that is a single integer comparison. Here $41$ is the exponent of $3$ in the modulus, not the number of input rows; at that depth the counting bound is $130 + 2 S$ rows.

The count produces two distinct subsets, not necessarily two distinct row sums or remainder values. Repeated rows already illustrate the first failure. To prevent the second, we bound how many subsets can have any specified remainder. The following finite-set statement isolates that requirement.

Theorem 5.10 (equal residues with different values). Let $A$ and $B$ be finite sets, let $f : A \to B$ , and let $g : A \to C$ be any map into a set $C$ . Suppose every fibre of $g$ has at most $k$ elements. If

| B | k < | A |,

then there exist distinct $x, y \in A$ such that

f (x) = f (y) and g (x) \neq g (y) .

Thus, when $f$ records the four residues and $g$ records the real remainder, a bound on the multiplicities of equal remainders guarantees a pair with equal residues and different remainders.

Proof. If every pair in a common $f$ -fibre also had the same $g$ -value, each $f$ -fibre would lie in one $g$ -fibre and hence have size at most $k$ . Summing over the at most $| B |$ fibres of $f$ would give $| A | \leq | B | k$ , contrary to the hypothesis. ◻

If $g$ is injective, the hypothesis holds with $k = 1$ . It need not hold with a useful small $k$ for subset sums: repeated rows produce many equal sums, even when every row is primitive. The theorem assumes a bound on every fibre of $g$ , not merely on the fibres of $(f, g)$ . It gives a nonzero difference but no upper bound on its real size. To bound the size as well, we partition the range of $g$ into small intervals. The count now uses residue class, interval and multiplicity. Only repetitions within one residue class matter, so the quantitative version assumes a bound for fibres of $(f, g)$ rather than for all fibres of $g$ .

Theorem 5.11 (equal residues with close, distinct values). Let $A, B, J$ be finite sets and let $f : A \to B$ , $g : A \to R$ and $ι : A \to J$ . Suppose that each simultaneous fibre of $(f, g)$ has at most $k$ elements and that, for some $δ > 0$ ,

ι (x) = ι (y) ⟹ | g (x) - g (y) | < δ .

If $| B | | J | k < | A |$ , then some distinct $x, y \in A$ satisfy

f (x) = f (y), 0 < | g (x) - g (y) | < δ .

Proof. Suppose that no such pair exists. Two points in a fibre of $(f, ι)$ have equal $f$ -values and $g$ -values less than $δ$ apart, so their $g$ -values must be equal. Thus each fibre of $(f, ι)$ lies in a fibre of $(f, g)$ and contains at most $k$ points. Summing over its at most $| B | | J |$ fibres gives $| A | \leq | B | | J | k$ , a contradiction. ◻

For an application, let $M$ integral coefficient rows $(A_{j}, B_{j})$ have real remainders $e_{j}$ at the same target. Let $D$ be a positive integer, and suppose that the coefficient-pair subset sums $\sum_{j \in I} (A_{j}, B_{j})$ take $Q$ values in $(Z / D Z)^{2}$ . Put $T = \sum_{j} | e_{j} |$ . The least and greatest subset remainders are respectively

\sum_{e_{j} < 0} e_{j} and \sum_{e_{j} > 0} e_{j},

so the interval between them has length $T$ . Divide it into half-open intervals of width $D / n$ , using a final interval for the right endpoint if necessary. At most $⌊ n T / D ⌋ + 1$ intervals are needed. If each simultaneous residue-and-remainder fibre has size at most $k$ , the inequality

2^{M} > Q k (⌊ n T / D ⌋ + 1)

therefore gives equal residues with remainder difference strictly between $0$ and $D / n$ in absolute value. Subtracting the two subset sums and dividing both coefficients by $D$ gives an integral form of nonzero absolute value less than $1 / n$ . For a prescribed family, the number $Q$ , the width $T$ and the multiplicity bound $k$ must all be justified. We have no such bounds giving the required forms for the Zudilin family at $3 / 2$ .

The next improvement needs a much stronger hypothesis than primitivity: adjacent determinants must vanish modulo the chosen modulus. Unit multiples of one unimodular row satisfy it. Arbitrary primitive rows do not; for example, $(1, 0)$ and $(0, 1)$ have determinant $1$ . Under this hypothesis the rows lie in the free rank-one submodule generated by a unimodular row. A change of coordinates makes their second coordinates zero, so only one residue coordinate must be counted.

Theorem 5.12 (vanishing minors and a residue count). Let $R_{0}$ be a commutative ring and let $w_{n} = (A_{n}, B_{n}) \in R_{0}^{2}$ . Suppose that each row is unimodular, meaning that $u_{n} A_{n} + v_{n} B_{n} = 1$ for some $u_{n}, v_{n} \in R_{0}$ , and that every adjacent minor vanishes:

A_{n} B_{n + 1} - B_{n} A_{n + 1} = 0 (n \geq 0) .

Then every pairwise minor $A_{i} B_{j} - B_{i} A_{j}$ vanishes. In particular, take $R_{0} = Z / (2^{S} 3^{R}) Z$ with $R > 0$ . If $S + 2 R \leq k$ , there are two distinct binary vectors $s, t \in {0, 1}^{k}$ such that

\sum_{i < k} s_{i} w_{i} = \sum_{i < k} t_{i} w_{i} .

Thus $S + 2 R$ rows suffice. The ambient two-coordinate argument gives the sufficient bound $2 S + 4 R$ .

Proof. If $(a, b)$ is unimodular, say $u a + v b = 1$ , and $a y - b x = 0$ , then

(x, y) = (u x + v y) (a, b) .

Indeed, the first coordinate follows by replacing $a y$ with $b x$ , and the second by the reverse substitution. Apply this identity to consecutive rows: each next row is a scalar multiple of the current one. Induction places the entire tail on the line through $w_{0}$ and proves the pairwise-minor assertion. Right multiplication by

(\begin{matrix} u_{0} & - B_{0} \\ v_{0} & A_{0} \end{matrix})

has determinant $u_{0} A_{0} + v_{0} B_{0} = 1$ and sends $w_{0}$ to $(1, 0)$ . All transformed subset sums therefore have second coordinate zero and occupy at most $2^{S} 3^{R}$ values. Finally

2^{S} 3^{R} < 2^{S} 4^{R} = 2^{S + 2 R} \leq 2^{k},

so the pigeonhole principle gives two distinct subsets with the same sum. ◻

No particular coordinate needs to be invertible: modulo six, $(2, 3)$ is unimodular because $- 2 + 3 = 1$ , although neither entry is a unit. Some nondegeneracy is essential. The rows $(1, 0), (0, 0), (0, 1)$ have zero adjacent minors but outer minor one. For the counting conclusion, vanishing is required only in the quotient ring. The integer rows $(1, 0), (1, 6)$ used in Section 10 are independent over $Q$ , with determinant $6$ , but coincide modulo $6$ . Thus dependence modulo the modulus neither follows from primitivity nor implies dependence of the integer rows.

This conditional theorem is stronger than the ambient count of possible residue vectors only after its minor-vanishing hypothesis has been established. No such all-tail hypothesis is proved here for an actual $q$ -Apéry or Zudilin family, and the theorem says nothing about whether the difference of the subset sums has nonzero analytic remainder.

The count in Corollary 5.9 is sharp at $T = 1$ for comparison with the full residue space. To obtain different remainders, Theorem 5.10 additionally needs a bound on how often the same real value occurs. Such a bound is not proved here for the $q$ -Apéry or Zudilin remainder family.

Lean checks the cardinality of the residue space, the pigeonhole argument for the residue map and the sufficient row count for equal residues. The pigeonhole argument gives divisibility without an independence hypothesis. Further information about the family is needed to ensure that the nonzero difference of the indicator vectors gives a nonzero polynomial pair and a nonzero analytic remainder. None of these statements supplies such a family or proves either nonvanishing conclusion.

Example 5.13. At depths $R = S = 1$ the space of four residue coordinates is $(Z / 3 Z)^{2} \times (Z / 2 Z)^{2}$ , of cardinality $9 \cdot 4 = 36$ , and the threshold reads $M \geq 4 \cdot 1 + 2 \cdot 1 = 6$ . With six pairs there are $2^{6} = 64$ subsets and only $36$ possible residue vectors. Two subsets therefore have equal residues. The difference of their indicator vectors is a nonzero vector in ${- 1, 0, 1}^{6}$ whose sum vanishes in all four coordinates.

The following elementary comparison concerns only the two scalar exponents of Zudilin’s construction; it makes no assertion about the coefficient polynomials or their real remainders.

Theorem 5.14 (a restriction on the two scalar exponents). Let $C_{1} > 0$ . If $C_{0} \leq 0$ or $2 C_{0} \leq C_{1}$ , then

C_{0} \log 3 - C_{1} \log 2 < 0.

In the positive branch $C_{0} > 0$ and $2 C_{0} \leq C_{1}$ , the stronger estimate is

C_{0} \log 3 - C_{1} \log 2 < - \frac{17}{41} C_{0} \log 2.

Proof. If $C_{0} \leq 0$ , then $C_{0} \log 3 - C_{1} \log 2 \leq - C_{1} \log 2 < 0$ . If $C_{0} > 0$ and $2 C_{0} \leq C_{1}$ , use $3^{41} < 2^{65}$ to obtain

C_{0} \log 3 - C_{1} \log 2 \leq C_{0} (\log 3 - 2 \log 2) < - \frac{17}{41} C_{0} \log 2 < 0.

◻

Written multiplicatively, the conclusion is $3^{C_{0}} < 2^{C_{1}}$ . Lean checks the inequality and the $17 / 41$ margin of the positive branch. The scope of this elementary inequality matters. The primary Zudilin theorem [8] supplies an integer-base irrationality-exponent estimate on its parameter cone, and the elementary inequality $μ \geq 2$ then forces $2 C_{0} \leq C_{1}$ whenever $C_{0} > 0$ ; Lean checks that implication separately. The primary 2004 theorem is stated for an integer inverse base. The rational specialisation proved earlier in this record has separate proofs for the constructed forms. The scalar statement in this paragraph is only the displayed inequality; it neither constructs a new family at $3 / 2$ nor formalises a universal exclusion of linear-form methods.

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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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Congruences after evaluation at 3/23/2

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Congruences after evaluation at $3 / 2$