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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Square specialisation

For the proposed cubic $f (T) = T^{3} - T + 6 c / m$ , the recurrence will force $r^{2} - 1$ to be a square whenever $r$ is a root modulo a good prime. We need the same conclusion for an algebraic root $α$ in $Q (α)$ . The issue is that a putative square root could lie in a quadratic extension of this field. Its nontrivial automorphism would fix $α$ and change the sign of the square root. Chebotarev supplies a prime with exactly that behaviour, contradicting the finite-field condition. The following lemma isolates this passage; in the cubic application we take $H (T) = T^{2} - 1$ .

Lemma 2.7 (square specialisation). Let $f \in Q [T]$ be irreducible with root $α$ , and let $H \in Q [T]$ satisfy $H (α) \neq 0$ . If for all but finitely many primes $ℓ$ every root $r \in F_{ℓ}$ of the reduction of $f$ has $H (r)$ a square in $F_{ℓ}^{\times}$ , then $H (α)$ is a square in $Q (α)^{\times}$ .

Proof. Suppose that $H (α)$ is a nonsquare in $K = Q (α)$ , and choose $β$ with $β^{2} = H (α)$ . The nontrivial automorphism of the quadratic extension $K (β) / K$ extends to an element $σ \in Gal (L / K)$ for a finite Galois extension $L / Q$ containing $K (β)$ . In particular $σ α = α$ and $σ β = - β$ .

By Chebotarev’s density theorem in the form stated by Stevenhagen and Lenstra [10], infinitely many unramified rational primes have Frobenius conjugate to $σ$ . Only infinitude is needed. We discard the exceptional primes in the hypothesis and the finitely many primes at which coefficients, $α$ or $β$ have denominators, together with $2$ and the primes at which $β$ reduces to zero. Above each remaining prime $ℓ$ , choose a prime of $L$ whose Frobenius is $σ$ . Its residue field then satisfies

{\overset{―}{α}}^{ℓ} = \overset{―}{α}, {\overset{―}{β}}^{ℓ} = - \overset{―}{β} \neq \overset{―}{β} .

Thus $\overset{―}{α} \in F_{ℓ}$ is a root of the reduction of $f$ . Neither of the two roots $\pm \overset{―}{β}$ of $X^{2} - H (\overset{―}{α})$ is fixed by Frobenius, so neither belongs to $F_{ℓ}$ . Hence $H (\overset{―}{α})$ is a nonsquare there, contradicting the hypothesis. ◻

An alternative proof of Lemma 2.7 uses zeta functions. A nonsquare algebraic integer $b$ in a number field $K$ is nonsquare modulo infinitely many degree-one primes of $K$ . The underlying zeta-function argument first gives zero or a nonsquare. The zero case is excluded beyond the prime divisors of the nonzero constant coefficient of the minimal polynomial of $b$ . If $H (α)$ is not a square in $K = Q (α)$ , this applies to $b = m^{2} H (α)$ , where the nonzero integer $m$ makes $b$ an algebraic integer, and the residue map at such a prime, of norm $ℓ$ , sends $α$ to a root $r \in F_{ℓ}$ of the reduction of $f$ at which $H (r)$ is a nonsquare. The prime-reduction argument (nonsquares modulo primes) is proved from the simple pole of the Dedekind zeta function at $s = 1$ , which is in Mathlib. If $b$ were a nonzero square modulo all but finitely many primes of degree one, these primes would split in $K (\sqrt{b})$ , and for real $s > 1$ near $1$ the zeta function of $K (\sqrt{b})$ would be at least a positive constant times the square of the zeta function of $K$ , which the two simple poles forbid.

Both arguments use the condition at every root for all but finitely many primes. Testing finitely many primes supplies no such condition. The zeta-function argument is the one used in the linked formal proof.

Proposition 2.8 (the square forced by the numerator recurrence). Write $κ = m / 6$ and $η = 6 c / m$ , so that $Q_{m, c} (n) = κ f (n + 1)$ for $f (T) = T^{3} - T + η$ . Then $α^{2} - 1$ is a square in $Q (α)^{\times}$ for a root $α$ of $f$ .

Proof. Eliminating $v_{n}$ from (2) gives

\begin{matrix} (4) & u_{n + 2} = (a_{n} + a_{n + 1}) u_{n + 1} - a_{n}^{2} u_{n} . \end{matrix}

Let $ℓ$ be a prime and $k$ an index with $ℓ ∣ u_{k}$ . Reading (4) at $n = k - 1$ modulo $ℓ$ gives $u_{k + 1} \equiv - a_{k - 1}^{2} u_{k - 1}$ , hence

\begin{matrix} (5) & - u_{k - 1} u_{k + 1} \equiv (a_{k - 1} u_{k - 1})^{2} (\mod ℓ), \end{matrix}

so $- u_{k - 1} u_{k + 1}$ is a square modulo $ℓ$ .

Let $ℓ ∤ 6 m$ and let $r \in F_{ℓ}$ be a root of $f$ . Then $r \neq 0$ and $r \neq \pm 1$ , since $f (0) = η \neq 0$ and $f (\pm 1) = η$ . Using $η = r - r^{3}$ ,

f (r - 1) = - 3 r (r - 1), f (r + 1) = 3 r (r + 1),

so at an index $k \equiv r - 1 (\mod ℓ)$ we may reduce the polynomial values in $F_{ℓ}$ . All equalities in the following calculation are in that field: $Q (k) = 0$ , $Q (k - 1) = - 3 κ r (r - 1)$ , $Q (k + 1) = 3 κ r (r + 1)$ , and

\begin{matrix} (6) & - Q (k - 1) Q (k + 1) = 9 κ^{2} r^{2} (r^{2} - 1) . \end{matrix}

If $r^{2} - 1$ were a nonsquare modulo $ℓ$ , then so would be the right side of (6), because $9 κ^{2} r^{2}$ is a nonzero square; agreement at all three of $k - 1, k, k + 1$ would then contradict (5). That prohibition recurs at every index of the class $r - 1$ modulo $ℓ$ , and Lemma 2.4 with $L = 3$ , $h = ℓ$ would give $\underset{―}{d} (S) \geq 1 / (3 ℓ)$ . So $r^{2} - 1$ is a square in $F_{ℓ}^{\times}$ at every root of every good reduction, and Lemma 2.7 applies with $H (T) = T^{2} - 1$ . ◻

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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