Question

Let $x_1=1,x_3=4$ and $x_{n+3}=2x_{n+2}+2x_{n+2}-x_n$ for all $n\ge 1.$ Prove that $x_n$ is a square for all $n\ge 1.$

Answer 1

We first notice that $x_1=x_2=1^2,x_3=2^2,x_4=3^2,x_5=5^2,x_6=8^2,$ and so forth. This leads to the assumption that $x_n=f_n^2,$ where $f_n,$ is the nth term of the Fibonacci sequece defined by $F_1=F_2=1$ and $F_{n+1}=F_n+F_{n-1},$ for all $n\ge 2.$ We prove this assertion inductively. Suppose it is true for all $k\ge n+2.$ Then $x_{n+3}=2F_{n+2}^2+2F_{n+1}^2-F_n^2=2F_{n+2}^2+2F_{n+1}^2-{(F_{n+2}-F_{n+1})}^2$ $=F_{n+2}^2+F_{n+1}^2+2F_{n+1}{F}_{n+2}={(F_{n+2}+F_{n+1})}^2=F_{n+3}^2,$ and the assertion is proved.