Problem

Source: Romanian TST, Day 1, P9

Tags: quadratics, algebra, Sequences, Periodic sequence



A sequence $ \left( x_n\right)_{n\ge 0} $ of real numbers satisfies $ x_0>1=x_{n+1}\left( x_n-\left\lfloor x_n\right\rfloor\right) , $ for each $ n\ge 1. $ Prove that if $ \left( x_n\right)_{n\ge 0} $ is periodic, then $ x_0 $ is a root of a quadratic equation. Study the converse.