<s>[INST] Prove this form $2^{n+2}-7a^2_{n}$ always is square numbers. Let sequence $\{a_{n}\}$ such $a_{1}=1,a_{2}=-1$,and 
$$a_{n}+a_{n-1}+2a_{n-2}=0,n\ge 3$$
show that 
$2^{n+2}-7a^2_{n}$ is square numbers.
such as
$$2^3-7=1,n=1$$
$$2^4-7=9=3^2,n=2$$
$a_{3}=-1$
$$2^{5}-7=25=5^2,n=3$$
$a_{4}=3$
$$2^6-7\cdot 9=1^2,n=4$$
 [/INST] One finds 
$$a_n=\frac{1}{\sqrt 7i}\left(A^n-B^n\right)\ \ (n\ge 1)$$ where $A=-\frac 12-\frac{\sqrt 7}{2i},B=-\frac 12+\frac{\sqrt 7}{2i}$.
Noting that $AB=2$, one finds
$$\begin{align}2^{n+2}-7{a_n}^2&=2^{n+2}-7\cdot\frac{1}{-7}\left(A^{2n}+B^{2n}-2\cdot 2^n\right)\\&=2^{n+2}+A^{2n}+B^{2n}-2^{n+1}\\&=2^{n+1}(2-1)+A^{2n}+B^{2n}\\&=\left(A^n+B^n\right)^2.\end{align}$$
Now it should be easy to prove that $A^n+B^n$ is an integer for every $n\ge 1$.
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