Question
The integer next above $(\sqrt3+1)^{2m}$ contains $2^x$ as a factor. Find $x$.
Attempt 1
$(\sqrt3+1)^{2m}=(4+2\sqrt3)^m=2^m(2+\sqrt3)^m=I+f$, where $I$ is the integral value and $f$ is the fractional value of the expression.
Since $0\lt\sqrt3-1\lt1\implies0\lt(\sqrt3-1)^m\lt1$
Let $(\sqrt3-1)^m=f'$
$(\sqrt3+1)^{2m}+(\sqrt3-1)^{2m}=2^m\left((2+\sqrt3)^m+(2-\sqrt3)^m\right)=2^{m+1}(^mC_0\cdot2^m+^mC_2\cdot2^{m-2}\cdot3+...)$
LHS is $I+f+f'$. RHS is even integer. So, $I+f+f'$ is integer. And $f+f'$ can only be $1$.
So, I got $x$ as at least $m+1$.
Can we say something about the odd/even nature of $(^mC_0\cdot2^m+^mC_2\cdot2^{m-2}\cdot3+...)?$
Attempt 2
Let $(\sqrt3+1)^{2m}=I+f, I$ being the integral value and $f$, the fractional value.
Let $(\sqrt3-1)^{2m}=f'$, where $0\lt f'\lt1$
$(\sqrt3+1)^{2m}+(\sqrt3-1)^{2m}=I+f+f'$
LHS$=2(^{2m}C_0\cdot3^m+^{2m}C_2\cdot3^{m-1}+...+^{2m}C_{2m})$
This time, I am just getting $2^1 $ as the factor. What's wrong here?
Let $a_m=(\sqrt3+1)^{2m}+(-\sqrt3+1)^{2m}=\alpha^m+\beta^m$, where $\alpha=(\sqrt3+1)^2$ and $\beta=(-\sqrt3+1)^2$.
Since $0 < \beta < 1$, we have $(\sqrt3+1)^{2m} < a_m < (\sqrt3+1)^{2m}+1$ and so $a_m = \lceil (\sqrt3+1)^{2m} \rceil$.
Now, $\alpha$ and $\beta$ are roots of $x^2 = 8 x - 4$ and so $a_{m+2}=8a_{m+1} -4a_m$.
Write $a_m=2^{e(m)}b_m$ with $b_m$ odd. Then, $$a_{m+2}=8a_{m+1} -4a_m=2^{e(m)+2}(2^{e(m+1)-e(m)+1}b_{m+1}-b_m) \quad (*)$$ and so $e(m+2)=e(m)+2$. Therefore, $e(m)$ for $m=0,1,2,\dots$ is $ 1,3,3,5,5,7,7,\dots $
$(*)$ provided $e(m)\le e(m+1)$, which I expect can be proved by induction.