recurrence relation for squares of fibonacci numbers I have a problem finding a proof that the squares of the Fibonacci numbers satisfy the recurrence relation $a_{n+3} - 2*a_{n+2} - 2*a_{n+1} + a_n = 0$ and solving this recurrence relation. 
Some help would be great!  

Let $(u_n)$ denote the sequence of Fibonacci numbers. By definition,
$$
u_{n+2} = u_{n+1} + u_n\quad\text{and}\quad u_{n+3} = u_{n+2} + u_{n+1} = 2u_{n+1} + u_n,
$$
so that using $(x+y)^2 = x^2 + y^2 + 2xy$,
$$
a_{n+2} = a_{n+1} + a_n + 2u_n u_{n+1},\quad\text{and}\quad
a_{n+3} = 4a_{n+1} + a_n + 4u_nu_{n+1}.
$$
Finally, for all $n$ one has:
$$
a_{n+3} - 2a_{n+2} - 2a_{n+1} + a_n = (4-2-2)a_{n+1} + (1-2+1)a_n + (4-4)u_nu_{n+1}.
$$
