  We show that for a fixed integer $n \neq \pm2$, the congruence $x^2 + nx \pm
1 \equiv 0 \pmod{\alpha}$ has the solution $\beta$ with $0 < \beta < \alpha$ if
and only if $\alpha/\beta$ has a continued fraction expansion with sequence of
quotients having one of a finite number of possible asymmetry types. This
generalizes the old theorem that a rational number $\alpha/\beta > 1$ in lowest
terms has a symmetric continued fraction precisely when $\beta^2 \equiv \pm 1
\pmod{\alpha}$.
