Divisibility of ceiling function of surd Show that:
The integer next greater than  $(\sqrt{7}+\sqrt{3})^{2n}$ is divisible by $4^n$
 A: Here is a hint. Often these kinds of equations with $\alpha^n$ nearly an integer of some kind boil down to finding some conjugate number $\beta\lt 1$ with $\alpha^n+\beta^n$ the solution to some simple recurrence relation. The recurrence can make the condition obvious.
For example, this works with the Fibonacci numbers.

Note for the discerning - there is a reason for choosing $n$ rather than $2n$ - the recurrence is simpler as revealed in @minar's excellent comment.
A: Hint: Show that
$$
(\sqrt7+\sqrt3)^{2n}+(\sqrt7-\sqrt3)^{2n}
$$
is an integer. Also $\sqrt7-\sqrt3<1$.
A: Since
$$
(\sqrt7+\sqrt3)^2+(\sqrt7-\sqrt3)^2=20\tag{1}
$$
and
$$
(\sqrt7+\sqrt3)^2(\sqrt7-\sqrt3)^2=16\tag{2}
$$
$(\sqrt7+\sqrt3)^2$ and $(\sqrt7-\sqrt3)^2$ satisfy the equation
$$
x^2-20x+16=0\tag{3}
$$
The recursion
$$
a_n=20a_{n-1}-16a_{n-2}\tag{4}
$$
where $a_0=2$ and $a_1=20$ is therefore satisfied by
$$
a_n=(\sqrt7+\sqrt3)^{2n}+(\sqrt7-\sqrt3)^{2n}\tag{5}
$$
Since $(\sqrt7-\sqrt3)^2\lt1$, we have
$$
a_n=\left\lceil(\sqrt7+\sqrt3)^{2n}\right\rceil\tag{6}
$$
Divide $(4)$ by $4^n$ to get
$$
\frac{a_n}{4^n}=5\frac{a_{n-1}}{4^{n-1}}-\frac{a_{n-2}}{4^{n-2}}\tag{7}
$$
Since $\dfrac{a_0}{4^0}=2$ and $\dfrac{a_1}{4^1}=5$, what does $(7)$ tell you about $a_n$? Combine with $(6)$.
A: A start: Show that the smallest integer greater than $(\sqrt{7}+\sqrt{3})^{2n}$ is
$$\left(\sqrt{7}+\sqrt{3}\right)^{2n}+\left(\sqrt{7}-\sqrt{3}\right)^{2n}.$$
To do this, expand by the Binomial Theorem (or at least imagine expanding).  You can see that we get an integer, since the the stuff that involves an odd power of $\sqrt{3}$, and therefore of $\sqrt{7}$, cancels. Your argument will also use the fact that $(\sqrt{7}-\sqrt{3})^{2n}$ is "small."
For the divisibility by $4^n$ part there are various approaches. Let's try induction. 
Let $\alpha=\sqrt{7}+\sqrt{3}$ and $\beta=\sqrt{7}-\sqrt{3}$. Note that
$$\alpha^{2n+2}+\beta^{2n+2}=(\alpha^{2n}+\beta^{2n})(\alpha^2+\beta^2)-\alpha^2\beta^2(\alpha^{2n-2}+\beta^{2n-2}).$$
