Tricky question involving binomial expansion

Question

For given $m$, what is the highest power of $2$ that divides $[(\sqrt3 +1)^m]+1$? where $[x]$ denotes the greatest integer less than or equal to $x$.

I have no clue how to proceed.

Answer 1

A collection of hints.

For any $m$ sufficiently large ($m\geq 3$), the closest integer to $(\sqrt{3}+1)^m$ is given by $$a_m=(1+\sqrt{3})^m+(1-\sqrt{3})^m\tag{1}$$ and the sequence $\{a_m\}_{m\geq 0}$ fulfills the recurrence relation: $$ a_{m+2} = 2a_{m+1} +2a_m \tag{2}$$ since the characteristic polynomial is $(x-1-\sqrt{3})(x-1+\sqrt{3})=x^2-2x-2$.
Due to the coefficients $2$ appearing in the RHS of $(2)$, it is not difficult to study $$ \nu_2(a_m) = \max\left\{n\in\mathbb{N}: 2^n\mid a_m\right\}.\tag{3} $$ For sure, $(2)+\text{induction}$ imply that $\nu_2(a_m)\geq\frac{m}{2}$.