Bulgarian Mathematical Olympiad 1987 The sequence $x_1,x_2,\dots$ is defined by the equalities $x_1=x_2=1$ and $$x_{n+2}=14x_{n+1}-x_n-4, n\geq 1$$
Prove that each number of the given sequence is a perfect square.

I used the standard way to solve recurrence relations and arrived at
$$x_n=\frac{1}{6}\bigg( (2+\sqrt{3})^{n-\frac{3}{2}} + (2-\sqrt{3})^{n-\frac{3}{2}} \bigg)^2$$
which I think should be quite close. Any idea how to proceed?

Long, but hopefully useful version ... Solving recurrence using characteristic polynomials (e.g. like here)
$$\color{red}{x_n}=\frac{1}{6}\left((26-15\sqrt{3})\cdot(7+4\sqrt{3})^n + (26+15\sqrt{3})\cdot(7-4\sqrt{3})^n + 2\right)=\\
\frac{1}{6}\left((26-15\sqrt{3})(2+\sqrt{3})^{2n} + (26+15\sqrt{3})(2-\sqrt{3})^{2n} + 2\right)=\\
\frac{1}{6}\left((2-\sqrt{3})(2+\sqrt{3})^{2(n-1)} + (2+\sqrt{3})(2-\sqrt{3})^{2(n-1)} + 2\right)=\\
\frac{1}{6}\left(\sqrt{2-\sqrt{3}}\cdot(2+\sqrt{3})^{n-1} + \sqrt{2+\sqrt{3}}\cdot(2-\sqrt{3})^{n-1}\right)^2=\\
\frac{1}{6}\left(\frac{\sqrt{3}-1}{\sqrt{2}}\cdot(2+\sqrt{3})^{n-1} + \frac{\sqrt{3}+1}{\sqrt{2}}\cdot(2-\sqrt{3})^{n-1}\right)^2=\\
\frac{1}{6}\left(\frac{3-\sqrt{3}}{\sqrt{6}}\cdot(2+\sqrt{3})^{n-1} + \frac{3+\sqrt{3}}{\sqrt{6}}\cdot(2-\sqrt{3})^{n-1}\right)^2=\\
\left(\frac{3-\sqrt{3}}{6}\cdot(2+\sqrt{3})^{n-1} + \frac{3+\sqrt{3}}{6}\cdot(2-\sqrt{3})^{n-1}\right)^2=\color{red}{(z_{n-1})^2}$$
where
$$z_{n+2}=4z_{n+1}-z_n, \space z_1=1, z_2=3 \tag{1}$$
Note that
$$\left(3-\sqrt{3}\right)\left(2+\sqrt{3}\right)=3+\sqrt{3}$$
and the recurrence in $(1)$ isn't really guessing, since both $2+\sqrt{3}$ and $2-\sqrt{3}$ are roots of $x^2-4x+1=0$, which is the  characteristic polynomial of $(1)$.
