End-symmetric continued fractions and quadratic congruences
  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 $\beta^2 \equiv \pm 1 \pmod{\alpha}$. 
Barry R. Smith
math.NT