EN
We show that for a fixed integer n ≠ ±2, the congruence x² + nx ± 1 ≡ 0 (mod α) has the solution β with 0 < β < α if and only if α/β 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 α/β > 1 in lowest terms has a symmetric continued fraction precisely when β² ≡ ±1(mod α ).