Question on binomial expansion - divisibility

Question

I am not able to solve this problem

Show that the integer just greater than $( \sqrt{3} + 1 )^{2m}$ contains $2^{m+1}$ as a factor.

Any help would be thoroughly appreciated

Answer 1

You can write the expression as $\lceil (\sqrt 3+1)^2 \rceil ^m = (\sqrt 3^2 +2\sqrt 3+1^2 )^m=(4+2 \sqrt 3)^m$ and then you can extrac a factor $2$ out of the $m$ power: $(4+2\sqrt 3)^m=2^m(2+\sqrt 3)^m$. Now you have a $2^m$ factor, and you only have to see that the first integer that is bigger than $(2+\sqrt 3)^m$ is always even because the first digit of $(2+\sqrt 3)^m$ is always odd. So, the integer bigger than $(2+\sqrt 3)^m$, $T$, is of the form $T=2q$ and $2^m T=2^m 2q = 2^{m+1}q$.

Answer 2

This is a Pisot numbers issue.

Let $u_n=(4+2\sqrt{3})^n+(4-2\sqrt{3})^n$. Then we have the recurrence relation $u_{n+2}=8u_{n+1}-4u_n$, and $u_0=2,u_1=8$.

It is easy then to see by induction on $n$ that $u_n$ is an integer multiple of $2^{n+1}$ for every $n$.

Further, $0 \leq (4-2\sqrt{3})^n \leq \frac{1}{2}$, so that $\lceil (4+2\sqrt{3})^n \rceil$ is in fact $u_n$, and we are done.