Provided a recurrence relationship, I converted the equation assuming $T(n) = r(n)$, and found the roots to the quadratic equation.
root 1: $r_1 = 2+\sqrt{3}$
root 2: $r_2 = 2-\sqrt{3}$
Resulting in the equation with the form $T(n) = c_1(r_1)^n + c_2(r_2)^n$
$$T(1): 1 = c_1(2+\sqrt{3})^1 + c_2(2-\sqrt{3})^1$$
$$T(3): 3 = c_1(2+\sqrt{3})^3 + c_2(2-\sqrt{3})^3$$
How do I go about solving for $c_1$ and $c_2$ in this particular instance. I'm used to an initial condition being 0, eliminating one of the $c$'s, then plugging the remaining into the second equation. But I am stuck here.
My end goal is to result with a closed form solution to a recurrence relationship from $T(n)$.
For brevity, take
\begin{align} \alpha &= 2+\sqrt{3} \\ \beta &= 2-\sqrt{3} \\ A &= c_1 \\ B &= c_2 \end{align}
Now,
\begin{align} \alpha+\beta &= 4 \\ \alpha-\beta &= 2\sqrt{3} \\ \alpha \beta &= 1 \\ A\alpha+B\beta &= 1 \tag{1} \\ A\alpha^3+B\beta^3 &= 3 \tag{2} \end{align}
$(2)-\beta^2 \times (1)$,
\begin{align} A\alpha (\alpha^2-\beta^2) &= 3-\beta^2 \\ A\alpha (\alpha+\beta)(\alpha-\beta) &= 3-(4-4\sqrt{3}+3) \\ 8A\alpha\sqrt{3} &= 4(\sqrt{3}-1) \\ A &= \frac{\sqrt{3}-1}{2\sqrt{3}(2+\sqrt{3})} \\ &= \frac{(\sqrt{3}-1)(2-\sqrt{3})}{2\sqrt{3}} \\ &= \frac{9-5\sqrt{3}}{6} \end{align}
$\alpha^2 \times (1)-(2)$,
\begin{align} B\beta (\alpha^2-\beta^2) &= \alpha^2-1 \\ B\beta (\alpha+\beta)(\alpha-\beta) &= (4+4\sqrt{3}+3)-3 \\ 8B\beta\sqrt{3} &= 4(\sqrt{3}+1) \\ B &= \frac{9+5\sqrt{3}}{6} \end{align}