The original problem : Given that $a_{1} = a_{2} = 1$ and $a_{n+2} - a_{n+1} + 4a_{n} = 0$ for $n \in \mathbb{N}$, find all $n$ such that $a_{n} = 1$.
I got the following expression for $a_{n}$ :
$$ a_{n} = \frac{2^{n+1}}{\sqrt{15}} \cdot \sin(n \arctan(\sqrt{15})) $$
for $n \in \mathbb{N}$. (Checked for $n = 1, 2$) Hence, if $a_{n} = 1$ then let $\theta = \arctan(\sqrt{15})$ and
$$ \frac{\sqrt{15}}{2^{n+1}} = \sin(n \arctan(\sqrt{15})) = \Im (\cos \theta + i \sin \theta)^{n} = \Im \left[ \sum_{r=0}^{n} \binom{n}{r} \left( \frac{1}{4} \right)^{n-r} \left( \frac{\sqrt{15}}{4} i \right)^{r} \right] $$
which is equal to :
$$ \frac{1}{4^{n}} \left( \binom{n}{1} \sqrt{15} - \binom{n}{3} \sqrt{15}^{3} + \binom{n}{5} \sqrt{15}^{5} - \cdots \right) $$
And now I'm stuck. The only thing I found online was this but we already know that it came from $\sin(n \theta)$ so it is not useful at all. I suspect $n = 1, 2$ as the only possible answer since $a_{n}$ seems to swerve violently with large values, not being able to be equal to $1$ after $n=2$. Can anyone solve it?
$$a_{n+2}-a_{n+1}+4a_n=0\implies a_{n+2}a_n-a_{n+1}^2=4(a_{n-1}a_{n+1}-a_n^2). $$
Hence $$a_{n+2}a_n-a_{n+1}^2=-4^n. $$
$$a_{n+1}=a_{n+2}+4a_n\implies a_{n+2}^2+7a_{n+2}a_n+16a_n^2=4^n. $$
If $a_n=1$, then $$(2a_{n+2}+7)^2=4^{n+1}-15. $$
$$15=(2^{n+1}+t)(2^{n+1}-t). \implies (t,n)=(7,2),(1,1).$$
Hence $a_4=0$ or $a_2=1$ and the first one is a conflict.