Sequence $(a_n)_{n\geq 0}$ is defined as $a_{0}=0, a_1=1, a_2=2, a_3=6$,
and $ a_{n+4}=2a_{n+3}+a_{n+2}-2a_{n+1}-a_n, n\geq 0$.
Prove that $n^2$ divides $a_n$ for infinite $n$.
(Romania)
Characteristic polynomial:
$p(\lambda)=\lambda^4-2\lambda^3-\lambda^2+2\lambda+1=(\lambda^2-\lambda-1)^2$
So $a_n=(An+B)\left(\frac{1+\sqrt{5}}{2}\right)^n+(Cn+D)\left(\frac{1-\sqrt{5}}{2}\right)^n$
Using initial conditions gives $A=\frac{1}{\sqrt{5}}$, $C=-\frac{1}{\sqrt{5}}$, $B=D=0$.
$a_n=\frac{n}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right]$.
Using Binet's formula for Fibonacci numbers gives $a_n=nF_n$.
So the problem reduces to showing $n|F_n$ for infinitely many $n$.
Exercise: Show that $n|F_n$ for all $n=5^k$. See Self-Fibonacci numbers.