Does this sequence always give a square number? Question : Supposing that  a sequence $\{a_n\}$ is defined as 
$${a_{n+3}}^2=-{a_{n+2}}^2+2{a_{n+1}}^2+48a_{n+1}a_{n}+32{a_n}^2\ (n\ge 1)$$
$$a_1=a_2=a_3=1$$
then, is $a_n$ a square number for any $n$?
For example, we can see
$$\sqrt{a_n} : 1,1,1,3,1,5,7,3,17,11,23,45,1,91,89,93,271,85,457,627,287,1541,967,2115,\cdots$$
Motivation : I found the following question in a book without any proof.
Supposing that a sequence $\{b_n\}$ is defined as 
$$b_{n+3}=-b_{n+2}+2b_{n+1}+8b_n\ (n\ge 1)$$
$$b_1=b_2=b_3=1,$$
then, prove that $b_n$ is a square number for any $n$.
This is obvious by the following relational expression : 
$$(b_{n+3}-b_{n+2})^2=64b_{n+1}b_n,$$
which can be shown by induction on $n$. 
After solving this question, I've tried to find a similar sequence by using computer. Then, I reached the above expectation. The expectation seems true, but I can neither find any counterexample nor prove that the sequence always gives a square number. Can anyone help?
 A: Let $b_n = -b_{n-1} - 2b_{n-2}$, $b_1=1, b_2=-1$. 
I claim that $a_n = b_n^2$, which is obviously a perfect square, for all $n \in \Bbb{Z}^+$. (See footnote below)
Proof by induction: 
Base case: $a_1 = a_2 = a_3 = b_1^2 = b_2^2 = b_3^2 = 1$
Hypothesis: $a_n = b_n^2$ for $n \le k$
Induction step: Start by using the induction hypothesis on the expression for $a_{k+1}$:
$$\begin{align}
&a_{k+1}^2 = -a_k^2+2a_{k-1}^2+48a_{k-1}a_{k-2}+32a_{k-2}^2 = 
-b_k^4 + 2b_{k-1}^4+48b_{k-1}^2b_{k-2}^2+32b_{k-2}^4
\end{align}$$
We want to show that $a_{k+1}^2 = b_{k+1}^4$. Expanding $b_{k+1}^4$ gives:
$$\begin{align}
&b_{k+1}^4 = (-b_k - 2b_{k-1})^4 = b_k^4 + 8b_k^3b_{k-1}+24b_k^2b_{k-1}^2+32b_kb_{k-1}^3+16b_{k-1}^4 = \\\\
&-b_k^4+2b_{k-1}^4+48b_{k-1}^2b_{k-2}^2+32b_{k-2}^4 + \\
&{\color{red}{2b_k^4+8b_k^3b_{k-1}+24b_k^2b_{k-1}^2+32b_kb_{k-1}^3+14b_{k-1}^4-48b_{k-1}^2b_{k-2}^2-32b_{k-2}^4}}
\end{align}$$
If we now can show that the red part equals zero, we're done. This can be done by replacing $b_k$ by $(-b_{k-1}-2b_{k-2})$ and expanding. 

Motivation for my claim: A search on OEIS.

EDIT: I just realized that proof by induction is completely unnecessary. Showing that the expansion of $b_{n}^4$ is equal to the recursion formula for $a_n^2$ is enough, i.e. the details in the induction step above. I guess I was too involved in the algebraic manipulations to see the big picture at the time.
