<s>[INST] 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!  
 [/INST] From the other answer,
$$5a_n=5F_n^2=a^n+b^n-2c^n$$ where $\displaystyle a=\frac{3+\sqrt5}2,b=\frac{3-\sqrt5}2,c=-1$
$$a^n+b^n-2c^n-5a_n=0\ \ \ \ (1)$$
So, $a,b$ are the roots of $t^2-3t+1=0$
$\implies a^2=3a-1,a^3=a(3a-1)=3a^2-a=3(3a-1)-a=8a-3$
Replacing $n$ with $n+1,n+2,n+3$ respectively we can form three more equations
$$a^{n+1}+b^{n+1}-2c^{n+1}-5a_{n+1}=0$$
 $$\iff a\cdot a^n+b \cdot b^n+2\cdot c^n-5a_{n+1}=0\ \ \ \ (2)$$
$$a^{n+2}+b^{n+2}-2c^{n+2}-5a_{n+1}=0$$
$$\iff(3a-1)a^n+(3b-1)b^n-2c^n-5a_{n+2}=0\ \ \ \ (3)$$
$$a^{n+3}+b^{n+3}-2c^{n+3}-5a_{n+3}=0$$
$$\iff(8a-3)a^n+(8b-3)b^n+2c^n-5a_{n+3}=0\ \ \ \ (4)$$
So, the eliminant of $a^n,b^n,c^n$ will be 
$\det\begin{pmatrix} 1 & 1 & -2 & -5a_n 
\\ a & b & -2 & -5a_{n+1} 
\\ 3a-1 & 3b-1 & 2 & -5a_{n+2} 
\\ 8a-3 & 8b-3 & 2 & -5a_{n+3}  
\end{pmatrix}=0$
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