$y_{n+1}>y_n$ $y_1=2$ $(2y_{n+1}-3y_n)^2=5y_n^2-4$ $y_{n+1}^2+y_n^2-3y_{n+1}y_n+1=0$ $y_{n+2}^2-3y_{n+2}y_{n+1}=y_n^2-3y_{n+1}y_n$ $(y_{n+2}-y_n)(y_{n+2}+y_n-3y_{n+1})=0 \to y_{n+2}=3y_{n+1}-y_n \to y_{n+2}$ is always integer