How find this positive integer numbers $m$,such $a_{n}=2a_{n-1}+a_{n-2}$,if $2^{2011}|a_{m}$ let sequence 
$$a_{0}=0,a_{1}=1,a_{n}=2a_{n-1}+a_{n-2},(n\ge 2)$$
Find all positive integer numbers $m$ ,such $2^{2011}|a_{m}$
My try:since $a_{0}=0,a_{1}=1$,then
$$a_{n}=\dfrac{(1+\sqrt{2})^n-(1-\sqrt{2})^n}{2\sqrt{2}}$$
then I can't. maybe this idea is not usefull.It is said this can use :Binary number
But I can't.
Thank you
 A: Let $\lambda_+ = 1 + \sqrt{2}$ and $\lambda_- = 1 - \sqrt{2}$, so that $$a_n = \frac{\lambda_+^n - \lambda_-^n}{2\sqrt{2}}.$$ One thing you can observe immediately is that $$a_{2n} = \frac{\lambda_+^{2n} - \lambda_-^{2n}}{2\sqrt{2}} = a_n \cdot(\lambda_+^n + \lambda_-^n).$$ By simply expanding, $$\lambda_+^n + \lambda_-^n = \sum_{k=0,\,\,even}^n\binom{n}{k}2(\sqrt{2})^k,$$ and from this we see that $\lambda_+^n + \lambda_-^n$ is an integer whose $2$-adic valuation is exactly $1$. Thus we conclude that $$\nu_2(a_{2n}) = 1 + \nu_{2}(a_n).$$ By induction, we similarly have $\nu_2(a_{2^mn}) = m + \nu_2(a_n)$. 
To prove that $\nu_2(a_n) = \nu_2(n)$, it therefore suffices to show that $\nu_2(a_n) = 0$ when $n$ is odd. We can do this by induction on $n\geq 1$, $n$ odd. Indeed, $\nu_2(a_1) = \nu_2(1) = 0$. Now, if $n>1$ is odd and we know that $a_{m}$ is odd for all odd $m<n$, then by the recursion relation and induction, we have that $a_n = 2a_{n-1} + a_{n-2}$ is odd, so $\nu_2(a_n) = 0$. This completes the proof that $\nu_2(a_n) = \nu_2(n)$. 
In particular, $2^{2011}\mid a_n$ if and only if $2^{2011}\mid n$.
A: Any 2-term linear recursion can be represented as a power of a matrix (in this case, $[a_{n+1}, a_n; a_n, a_{n-1}] = [2,1;1,0]^n$). This means that the identities $a_{2n} = a_{n}*(a_{n-1} + a_{n+1})$ and $a_{2n+1} = a_{n+1}^2 + a_{n}^2$ hold. Given the base {0, 1, 2}, the 2-adic evaluation follows inductively.
