Reciprocal Sums and the Sylvester Recurrence

A square forced by a zero numerator

At a zero of the numerator polynomial modulo a prime, the recurrence must still relate the two neighbouring nonzero numerators. Eliminating $v_n$ from two exact updates makes this constraint visible:

Whenever $u_k\equiv 0(\mathrm{mod}\ell )$, it follows that

Thus the negative product of the neighbouring numerators must be a square. Since our numerator is a polynomial, its roots modulo $\ell$ tell us exactly where to apply this observation.

Let $\ell \nmid 6m$ be prime and let $r\in {\mathbb{F}}_{\ell }$ satisfy $f(r)=0$. The constant $\eta =6c/m$ is nonzero modulo $\ell$, so $r\notin \{0,1,-1\}$. Choose a sufficiently late $k$ with $k+1\equiv r(\mathrm{mod}\ell )$. Then $u_k=\kappa f(k+1)\equiv 0$, whereas

Consequently, in ${\mathbb{F}}_{\ell }$,

Dividing by the nonzero square $9{\kappa }^2{r}^2$ gives

The condition holds for every root; primes with no root impose no condition. This quantifier matters in passing to a number field.

Let $\alpha$ be a root of $f$ and put $K=\mathbb{Q}(\alpha )$, a cubic field. Suppose ${\alpha }^2-1$ is a nonsquare in $K$, and choose $\beta$ with ${\beta }^2={\alpha }^2-1$. To contradict the modular condition, we seek a Frobenius that fixes $\alpha$ but exchanges $\beta$ and $-\beta$. The nontrivial automorphism of $K(\beta )/K$ does exactly this. Extend it to $\sigma \in \mathrm{Gal}(M/\mathbb{Q})$, where $M$ is a Galois closure of $K(\beta )/\mathbb{Q}$.

Chebotarev’s density theorem [9] supplies infinitely many unramified rational primes with Frobenius conjugate to $\sigma$; choosing a prime of $M$ above each makes it equal to $\sigma$. Omit residue characteristic two, primes dividing $6m$, and the finitely many primes where $\alpha$ or $\beta$ is nonintegral or $\beta$ reduces to zero. Frobenius acts by the $\ell$-th power on the residue field, so

Thus $r=\overline{\alpha }$ lies in ${\mathbb{F}}_{\ell }$ and satisfies $f(r)=0$, but neither root $\pm \overline{\beta }$ of $X^2-(r^2-1)$ lies in that field. The condition for every root in (7) applies to this $r$, giving the contradiction. Hence some $\beta \in K$ satisfies ${\beta }^2={\alpha }^2-1$.