Reciprocal Sums and the Sylvester Recurrence

Traces in the cubic field

We now need the square relation to constrain the coefficient $\eta$. Factoring it produces reciprocal elements, avoiding a separate expansion of the square root in a basis of $K$:

Put $z=\alpha +\beta$. Then $z^{-1}=\alpha -\beta$ and $2\alpha =z+z^{-1}$. Since $z\in K$ and $\alpha \in \mathbb{Q}(z)$, we have $\mathbb{Q}(z)=K$. Write the minimal polynomial of $z$ as

The minimal polynomial of $z^{-1}$ is the monic reciprocal polynomial

We compute the same three traces from $f$ and from the reciprocal pair $z,z^{-1}$. All traces are from $K$ to $\mathbb{Q}$. The cubic for $\alpha$ gives:

Newton’s identities for $p$ and its reciprocal give

For completeness, the first equality follows from $0=2\mathrm{Tr}(\alpha )=-b-d/w$. After substituting $d=-bw$, Newton’s identities give

The first line is $4\mathrm{Tr}({\alpha }^2)-6=2$. The second is $8\mathrm{Tr}({\alpha }^3)=-24\eta$, because the mixed terms in $(z+z^{-1}{)}^3$ have trace $3\mathrm{Tr}(z+z^{-1})=0$. These give the other two equalities in (8).

Multiplying the middle equation by $w$ and bringing all terms to one side gives $(bw-1)(b+w)=0$. Write $w=r/s$ in lowest terms, with $r\ne 0$, $s>0$. Substituting $b=-w$ or $b=1/w$ into the last equation and using $\eta =6c/m$ gives respectively

Here the integer scale $m$ is decisive: coprimality $gcd(r^2+s^2,rs)=1$ prevents cancellation with the variable factors in either numerator, so $(r^2+s^2{)}^2$ divides $48$. Hence $r^2+s^2$ is $1$, $2$ or $4$. Both $r$ and $s$ are nonzero, which leaves only $2$: a sum of two nonzero integer squares cannot equal $1$ or $4$. Thus $|r|=s=1$, and positivity of $m$ gives $m=12$. The only remaining numerators are