<s>[INST] How to prove each element of the following sequence is a perfect square? Sequence $\{a_n\}$ satisfies the following formula:
$a_{n+2}=14a_{n+1}-a_n+12$,
and $a_1=1, a_2=1$.
It is easy to check that $a_3=25$ and $a_4=361$.
The question is how to prove each element of the sequence $\{a_n\}$ is a perfect square?
 [/INST] As Umberto P. commented, it is sufficient to prove by induction that
$$a_{n+2}=b_n^2\tag1$$
where $b_{n+2}=4b_{n+1}-b_{n}$ with $b_0=1, b_1=5$.
The base case : $a_2=1=b_0^2$ and $a_3=25=b_1^2$
Suppose that $a_{n+2}=b_n^2$ and $a_{n+1}=b_{n-1}^2$. 
Then, using that 
$$b_n^2-4b_nb_{n-1}+b_{n-1}^2=6\tag2$$
we get
$$\begin{align}a_{n+3}&=14a_{n+2}-a_{n+1}+12\\&=14b_{n}^2-b_{n-1}^2+12\\&=16b_n^2-8b_nb_{n-1}+b_{n-1}^2-2b_n^2+8b_nb_{n-1}-2b_{n-1}^2+12\\&=(4b_n-b_{n-1})^2-2(b_n^2-4b_nb_{n-1}+b_{n-1}^2-6)\\&=b_{n+1}^2\qquad\blacksquare\end{align}$$

Let us prove $(2)$ by induction.
The base case : $b_1^2-4b_1b_0+b_0^2=25-20+1=6$
Suppose that $b_n^2-4b_nb_{n-1}+b_{n-1}^2=6$. Then,
$$\begin{align}b_{n+1}^2-4b_{n+1}b_{n}+b_{n}^2&=(4b_{n}-b_{n-1})^2-4(4b_{n}-b_{n-1})b_n+b_n^2\\&=16b_{n}^2-8b_{n}b_{n-1}+b_{n-1}^2-16b_n^2+4b_nb_{n-1}+b_n^2\\&=b_n^2-4b_nb_{n-1}+b_{n-1}^2\\&=6\qquad\blacksquare\end{align}$$
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