Take $a_n = b_n + \frac {1} {3}$; then $b_{n+1} = 14b_n - b_{n-1}$ is solved by classical methods considering the characteristic polynomial $\lambda^2 - 14\lambda +1 = 0$, of roots $7\pm 4\sqrt{3}$. This yields \[a_n = \dfrac {2-\sqrt{3}} {6} \left ( 7 + 4\sqrt{3}\right )^n + \dfrac {2+\sqrt{3}} {6} \left ( 7 - 4\sqrt{3}\right )^n + \dfrac {2} {6}.\] Consider now the integer sequence $(s_n)_{n\geq 0}$ given by $s_0=s_1 = 1$ and $s_{n+1} = 4s_n - s_{n-1}$, also solved by classical methods considering the characteristic polynomial $\lambda^2 - 4\lambda +1 = 0$, of roots $2\pm \sqrt{3}$. This yields \[s_n = \dfrac {3-\sqrt{3}} {6} \left ( 2 + \sqrt{3}\right )^n + \dfrac {3+\sqrt{3}} {6} \left ( 2 - \sqrt{3}\right )^n.\] But $(2\pm \sqrt{3})^2 = 7\pm 4\sqrt{3}$, and so it is seen that $a_n = s_n^2$.

let $ b_{n+1}=4b_n-b_{n-1} $(*) and $b_0=b_1=1$. We show that $a_n=b_n^2.$ It's easy from mathematical induction. At $n=0,1$ - triaval. Since (*), we see not difficult $ b_n^2-4b_nb_{n-1}+b_{n-1} ^2= -2 $(**) for all $n\in N$. So, since(**) we see not difficult $ b_{n+1} ^2=14b_n ^2-b_{n-1} ^2 -4 $. Hence, \[ a_n=b_n ^2.\]

All "not difficult" parts you mention are all the nontrivial parts of your proof. Can you elaborate?

@mathuz: Could you explain how you get to (**)?

hey @FlorianK, we have that $b_{n+1}=4b_n-b_{n-1}$ $ \Rightarrow $ $(b_{n+1}-b_{n-1})(b_{n+1}+b_{n-1}-4b_n)=0$ and \[ b_{n+1}^2-4b_{n+1}b_n+b_n^2=b_n^2-4b_nb_{n-1}+b_{n-1}^2 . \]

mathuz wrote: let $ b_{n+1}=4b_n-b_{n-1} $(*) and $b_0=b_1=1$. We show that $a_n=b_n^2.$ It's easy from mathematical induction. At $n=0,1$ - triaval. Since (*), we see not difficult $ b_n^2-4b_nb_{n-1}+b_{n-1} ^2= -2 $(**) for all $n\in N$. So, since(**) we see not difficult $ b_{n+1} ^2=14b_n ^2-b_{n-1} ^2 -4 $. Hence, \[ a_n=b_n ^2.\] Hello, your solution is just perfect for me, but i have one question: what thing did you prompt to take a sequence $ b_{n+1}=4b_n-b_{n-1} $ and then construct $a_{n}$ from it(i think at first you began counting the first terms of $a_{n}$ and then observed its properties, didn't you?)?

Because the roots of $x^2 - 14x + 1$ are the squares of the roots of $x^2 - 4x + 1$.

fractals wrote: Because the roots of $x^2 - 14x + 1$ are the squares of the roots of $x^2 - 4x + 1$. How did you know that at once...?

Also by induction we can prove $a_{n-1}\cdot a_{n+1}=(a_n +2)^2$

You can see https://www.artofproblemsolving.com/community/u241685h150791p6595068

mathuz have that because casio example cal x(n) and x(n+2)=ax(n+1)+bx(n) ==> a=..b=... by x1,x2,x3,x4,...