<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] Using this on $\displaystyle a_{n+3}-2a_{n+2}-2a_{n+1}+a_n=0$
we have
$\displaystyle a_n=A(-1)^n+B\left(\frac{3+\sqrt5}2\right)^n+C\left(\frac{3-\sqrt5}2\right)^n\ \ \ \ (1)$ where $A,B,C$ are arbitrary constants 
Now using Binet's formula, the $n$th Fibonacci term $\displaystyle F_n=\frac{a^n-b^n}{a-b}$ where $a,b$ are the roots of $t^2-t-1=0$
$\displaystyle\implies a+b=1, ab=-1, (a-b)^2=(a+b)^2-4ab=5$
$\displaystyle\implies a^2=\frac{3+\sqrt5}2,b^2=\frac{3-\sqrt5}2;$
$$\displaystyle\implies F_n^2=\frac{a^{2n}+b^{2n}-2(ab)^n}{(a-b)^2}=\frac{(a^2)^n+(b^2)^n-2(-1)^n}5=\frac{\left(\frac{3+\sqrt5}2\right)^n+\left(\frac{3-\sqrt5}2\right)^n-2(-1)^n}5 $$ which is clearly a special form of $(1)$
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