Prove all numbers in a sequence $a_{n+3}=15a_{n+2}-15a_{n+1}+a_n$ are perfect squares $a_{n+3}=15a_{n+2}-15a_{n+1}+a_n$, here $a_1=a_2=1,a_3=9$
Prove all numbers in a sequence are perfect squares.
My attempt is first to use the general formula of $a_n$.
It is $a_n=\frac{1}{6}\left((2+\sqrt3)^{2n-3}+(2-\sqrt3)^{2n-3}+2\right)$
But I still cannot prove every number in the sequence is a perfect square.

Consider the sequence $(b_n)$ defined by
$$ b_{n+2} = 4b_{n+1} - b_n, \qquad b_1 = b_2 = 1. $$
It is clear that every term of $(b_n)$ is an integer. To prove the claim, it is thus sufficient to establish:

Claim. $a_n = b_n^2$ for all $n$.

Indeed, it is easy to check that the claim is true for $n = 1, 2, 3$. Now suppose the claim is true for $n$, $n+1$, $n+2$. Then
\begin{align*}
a_{n+3}
&= 15a_{n+2} - 15a_{n+1} + a_n \\
&= 15b_{n+2}^2 - 15b_{n+1}^2 + b_n^2 \\
&= 15b_{n+2}^2 - 15b_{n+1}^2 + (-b_{n+2} + 4b_{n+1})^2 \\
&= 16b_{n+2}^2 + b_{n+1}^2 - 8b_{n+1}b_{n+2} \\
&= (4b_{n+2} - b_{n+1})^2 \\
&= b_{n+3}^2,
\end{align*}
and so, the claim is true for $n+3$ as well. Therefore the claim follows by the principle of mathematical induciton.

Motivation. The choice of the sequence $(b_n)$ is motivated by the fact that OP's formula for $(a_n)$ simplifies to
$$ a_{n} = \left[ \frac{\sqrt{3}-1}{2\sqrt{3}}(2+\sqrt{3})^{n-1} + \frac{\sqrt{3}+1}{2\sqrt{3}}(2 - \sqrt{3})^{n-1} \right]^2. $$
Using this, I reverse-engineered the recurrence relation with the characteristic equation having $2\pm\sqrt{3}$ as roots, namely $x^2 - 4x + 1 = 0$.
