Problem

Source: Romanian IMO Team Selection Test TST 1988, problem 13

Tags: quadratics, algebra proposed, algebra



Let a be a positive integer. The sequence {xn}n1 is defined by x1=1, x2=a and xn+2=axn+1+xn for all n1. Prove that (y,x) is a solution of the equation |y2axyx2|=1 if and only if there exists a rank k such that (y,x)=(xk+1,xk). Serban Buzeteanu