Reciprocal Sums and the Sylvester Recurrence: Further Results and Proofs

Will Cook

Long paper

Reciprocal Sums and the Sylvester Recurrence: Further Results and Proofs

Erdős #243 By Will Cook September 2026 Authorship, AI use and citation

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Summary. Integer finite differences exclude nonintegral regular rates and reduce the cubic rate to a polynomial. A square condition in the cubic field and congruences modulo seven then prove irrationality. For rational reciprocal sums with $a_{n + 1} / a_{n}^{2}$ tending to one, the record derives sufficient Sylvester-recurrence criteria from persistent divisors, bounded numerator increases and weighted crossings of new maxima. The arithmetic estimates needed to apply these criteria to every rational reciprocal sum remain open.

This paper covers cubic and nonintegral regular-rate irrationality, integer-tail arithmetic and sufficient Sylvester-recurrence criteria, with their separate hypotheses and remaining estimates.

It does not establish the validity of claims tagged Lean, which belongs to the cited kernel-checked source, or a solution to Erdős #243, which remains open.

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In this paper

The square condition forces $m = 12$

Lemma 2.9 (classification of the scales). Let $m$ be a positive integer, let $c \in {- 1, 1}$ and $η = 6 c / m$ , and suppose $T^{3} - T + η$ is irreducible over $Q$ with a root $α$ such that $α^{2} - 1$ is a square in $Q (α)^{\times}$ . Then $m = 12$ .

Proof. Choose $β \in Q (α)$ with $β^{2} = α^{2} - 1$ . The identity $(α + β) (α - β) = 1$ gives a reciprocal pair in the cubic field. Taking $z = α + β$ , we obtain

\begin{matrix} (7) & z^{- 1} = α - β, α = \frac{1}{2} (z + z^{- 1}), \end{matrix}

and $z$ generates $Q (α)$ . Write its minimal polynomial as $p (Z) = Z^{3} + b Z^{2} + d Z + w$ with $b, d, w \in Q$ and $w \neq 0$ . The minimal polynomial of $z^{- 1}$ is

\frac{Z^{3} p (Z^{- 1})}{w} = Z^{3} + \frac{d}{w} Z^{2} + \frac{b}{w} Z + \frac{1}{w} .

Thus Newton’s identities apply directly to both $z$ and its inverse. We use the first two traces of $α$ to restrict these coefficients, then the third trace to impose $η = 6 c / m$ . All traces below are from $Q (α)$ to $Q$ . The polynomial $T^{3} - T + η$ gives

\begin{matrix} (8) & Tr α = 0, Tr α^{2} = 2, Tr α^{3} = - 3 η . \end{matrix}

Newton’s identities give $Tr z = - b$ and $Tr z^{- 1} = - d / w$ , so the first equation in (8) and (7) give

\begin{matrix} (9) & d = - b w . \end{matrix}

For the second trace, Newton’s identities and $d = - b w$ give

Tr z^{2} = b^{2} + 2 b w, Tr z^{- 2} = b^{2} - 2 b / w .

Since the trace of $1$ is $3$ , squaring (7) yields $4 Tr α^{2} = Tr z^{2} + 6 + Tr z^{- 2}$ . Substituting $Tr α^{2} = 2$ gives $b^{2} + b (w - w^{- 1}) = 1$ . Multiplication by $w$ factors this equation:

\begin{matrix} (10) & (b w - 1) (b + w) = 0 . \end{matrix}

For the third trace, the same identities give

\begin{aligned} Tr z^{3} & = - b^{3} - 3 b^{2} w - 3 w, \\ Tr z^{- 3} & = b^{3} - 3 b^{2} / w - 3 / w . \end{aligned}

The cross terms in $(z + z^{- 1})^{3}$ have trace $3 (Tr z + Tr z^{- 1}) = 0$ by (9). Thus $8 Tr α^{3} = Tr z^{3} + Tr z^{- 3}$ . Using $Tr α^{3} = - 3 η$ now gives

\begin{matrix} (11) & η = \frac{(b^{2} + 1) (w + w^{- 1})}{8} . \end{matrix}

By (10), either $b = - w$ or $b = w^{- 1}$ , and (11) becomes

η = \frac{(w^{2} + 1)^{2}}{8 w} or η = \frac{(w^{2} + 1)^{2}}{8 w^{3}} .

Write $w = r / s$ with $r \in Z \ {0}$ , $s \in Z_{> 0}$ and $gcd (r, s) = 1$ . Since $η = 6 c / m$ ,

\begin{matrix} (12) & m = \frac{48 c r s^{3}}{(r^{2} + s^{2})^{2}} or m = \frac{48 c r^{3} s}{(r^{2} + s^{2})^{2}} . \end{matrix}

Now $gcd (r^{2} + s^{2}, r s) = 1$ , so integrality of $m$ and $| c | = 1$ force $(r^{2} + s^{2})^{2} ∣ 48$ , hence $r^{2} + s^{2} \in {1, 2, 4}$ . With $r \neq 0$ and $s \geq 1$ the value $1$ is too small, and $4$ is not a sum of two nonzero squares, so $r^{2} + s^{2} = 2$ and $| r | = s = 1$ . Then (12) gives $m = 12 c r$ , and $m > 0$ gives $m = 12$ . ◻

The square condition leaves the two possible cubics

\begin{matrix} (13) & Q_{12, \pm 1} (n) = 2 n (n + 1) (n + 2) \pm 1 . \end{matrix}

The square condition uses three consecutive numerators. The last step also uses the preceding denominator update, so it tests four consecutive numerators instead.

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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The square condition forces 𝑚 =12m=12

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The square condition forces $m = 12$