Let $\alpha$ be a positive integer. Let $b_n$, $n=1, 2, 3 \ldots$ be the sequence given by the recurrence relation $b_{n+2}=2\alpha^2b_{n+1}-\alpha^4 b_{n}$, $n=1, 2, 3 \ldots$ with initial conditions $b_1=5\alpha^2$, $b_2=9\alpha^4$. Prove that for all positive integers $\gamma$ and $j$, we have that $4\gamma+3$ is not equal to $b_j$.
What I have so far:
the characteristic equation is $r^2 - 2\alpha^2r + \alpha^4 = (r-\alpha^2)^2 = 0$ So $\alpha^2$ is a root with multiplicity 2. Therefore, the general solution to this relation is as follows
$b_n = (x_1 + x_2n)(\alpha^2)^n = (x_1 + x_2n)(\alpha^{2n})$
using the intitial conditions we get
$5α^2 = (x_1 + x_2(1))(\alpha^{2(1)}) = (x_1 + x_2)(\alpha^2)$
$5 = x1 + x2$
$9\alpha^4 = (x_1 + x_2(2))(\alpha^{2(2)}) = (x_1 + 2x_2)(\alpha^4)$
$9 = (x_1 + 2x_2) = (5 - x_2 + 2x_2) = 5 + x_2$
Therefore $x_2 = 4$ and $x_1 = 1$
The solution is $(1 + 4n)\alpha^{2n}$
From here I am not sure what to do. I was thinking I could show that $(1 + 4n)\alpha^{2n} \not \equiv 3 \text{ (mod 4)}$ and therefore $a_i \not = 4\gamma + 3$, but I have not made much progress. Any help on this one is appreciated.