Prove that $a_n$ is a perfect square 
  
*
  
*Let $\,\,\,\left(a_{n}\right)_{\ n\ \in\ \mathbb{N}\,\,\,}$ be the sequence of integers defined recursively by
  $$
a_{1} = a_{2} = 1\,,\qquad\quad a_{n + 2} = 7a_{n + 1} -a_{n} - 2\quad
\mbox{for}\quad n \geq 1
$$
  
*Prove that $a_{n}$ is a perfect square for every $n$.
  

We have $a_{3} = 4, a_{4} = 25, a_{5} = 169,\ldots$

Is there a way we can simplify the recursion or get its closed form in order to get that it is a perfect square ?.

For first, we may translate our sequence in order to have some sequence fulfilling $b_{n+2}=7b_{n+1}-b_n$. For such a purpose, it is enough to set $a_n=b_{n}+\frac{2}{5}$, leading to $b_0=b_1=\frac{3}{5}$. Now the characteristic polynomial of the sequence $\{b_n\}_{n\geq 0}$ is
$$ p(x)=x^2-7x+1 $$
with roots given by $\frac{7\pm 3\sqrt{5}}{2}$, so the explicit form of $b_n$ is given by:
$$ b_n = A\left(\frac{7+3\sqrt{5}}{2}\right)^2+B\left(\frac{7-3\sqrt{5}}{2}\right)^n $$
where the values of the constants $A,B$ are fixed by $b_0$ and $b_1$:
$$ b_n = \frac{-\sqrt{5}+3}{10}\left(\frac{7+3\sqrt{5}}{2}\right)^2+\frac{\sqrt{5}+3}{10}\left(\frac{7-3\sqrt{5}}{2}\right)^n. $$
Now the identity:
$$\boxed{ a_n = b_n+\frac{2}{5} = F_{2n-1}^2 }$$
is a simple consequence of the closed formula for Fibonacci numbers.
