The solution states:
Let $x_n = (2 + \sqrt3)^{n} + (2 - \sqrt3)^{n}$
Then $x_{n+2} = 4x_{n+1} - x_n$
How has the second line of the solution been found? Can you derive the equality through algebraic manipulation? Or has it been derived inductively? I can not see an easy rearrangement.
More generally, it is a consequence of the identity $$a^{n+1} + b^{n+1} = (a+b)(a^n + b^n) - ab(a^{n-1} + b^{n-1}),$$ with the choice $a = 2 + \sqrt{3}$, $b = 2 - \sqrt{3}$, thus $a+b = 4$, and $ab = 2^2 - 3 = 1$. But we can also use this identity even when $a$ and $b$ do not have the special property that their sum and their product are integers.
Notice that $2\pm\sqrt{3}$ are roots of $x^2-4x+1$, indeed: $$(2+\sqrt{3})+(2-\sqrt{3})=4,(2+\sqrt{3})(2-\sqrt{3})=1.$$ Therefore, $(x_n)_{n}$ is solution of $x_{n+1}=4x_{n+1}-x_n$.
Indeed, $(x_n)_n$ is solution of $x_{n+2}=ax_{n+1}+bx_{n}$ if and only $X_{n+1}=AX_n$, where: $$X_n:=\begin{pmatrix}x_n\\x_{n+1}\end{pmatrix},A:=\begin{pmatrix}0 & 1\\b & a\end{pmatrix}.$$ Therefore, using induction $X_n=A^nX$. Then, the powers of $A$ are related to the roots of $X^2-aX-b$. Actually, if it admits two distinct roots, say $\varphi_1$, $\varphi_2$, then $A$ is diagonalizable and one gets: $$A^n=P\begin{pmatrix}{\varphi_1}^n&0\\0&{\varphi_2}^n\end{pmatrix}P^{-1}.$$ Finally, $(x_n)_n$ is a solution if and only if $x_n$ is a linear combination of the $n$-th power of $\varphi_1$ and $\varphi_2$.
$\alpha = 2+\sqrt{3}$ and $\bar \alpha = 2-\sqrt{3}$ are roots of the equation $x^2-4x+1$. This equation gives rise to the recurrence $x_{n+2} = 4x_{n+1} - x_n$ The term $-4$ comes from $\alpha + \bar \alpha$ and $1$ follows from $\alpha \cdot \bar \alpha$
Let $t = 2 + \sqrt{3}$, and notice $\frac{1}{t} = 2 - \sqrt{3}$, so $x_n = t^n + t^{-n}$. We can check that $t$ and $\frac{1}{t}$ are the roots of the quadratic $u^2 - 4u + 1=0$, so$$\begin{align}x_{n+2} &= t^{n+2} + t^{-n-2} \\ &= t^n (4t-1)+t^{-n}(4t^{-1}-1) \\ &=4(t^{n+1}+t^{-(n+1)}) - (t^n + t^{-n}) \\ &=4x_{n+1} - x_n\end{align}$$
