<s>[INST] Find all values of $ r $ such that $ a_n = r^n $ for $ n \in \mathbb{Z}_{\ge2} $ in a recurrence relation I am trying to solve the following problem:
Given the following recurrence relation:
\begin{equation}
a_n = 6a_{n-1} - 8a_{n-2} \ \text{ for } \ n \in \mathbb{Z}_{\ge2}
\end{equation}
Find all values of $ r $ so that when $ a_0 = 1 $ and $ a_1 = r $ then $ a_n = r^n \ \ \forall n \in \mathbb{N} $
My attempt:
\begin{align}
\text{By distinct roots theorem,} \\ \\
\text{Characteristic equation: }
& t^2 - 6t + 8 = 0 \\
& (t-3)^2 + 8 - 9 = 0 \\
& (t-3)^2 = 1 \\
& t = 4 \text{ or } t = 2
\end{align}
\begin{align}
\text{Let } a_0 = 1 = C(4^0) + D(2^0) \text{ where C, D } \in \mathbb{R}
\end{align}
\begin{align}
1 = C + D \\
C = 1 - D
\end{align}
\begin{align}
\text{Let } a_1 = r &= C(4) + D(2) \\
r &= 4C + 2D \\
r &= 4(1-D) + 2D \quad \text{(By substitution)} \\
r &= 4 - 4D + 2D \\
r &= 4 - 2D
\end{align}
It seems that I must have gone wrong somewhere as $ r < 0 $ for $ D > 2$. Hence the answer for $ a_n $ would be very strange (i.e. $ a_n < 0 $ when $ r = -4 $). Could anyone please assist me?
 [/INST] Both roots give solutions: $r=4$ for $a_n = 4^n$ ($C=0, D=1$) and $r=2$ for $a_n = 2^n$ ($C=1, D=0$).
</s>