Finding integer part of $(3+\sqrt{3})^4$ I need to find the integer part of $(3+\sqrt{3})^4$, ie the floor of it and I am unable to proceed.
I can find integer part of $(2+\sqrt{3})^4$ using the binomial theorem as $2-\sqrt{3}<1$.
Please give a hint. Thanks!
 A: HINT
$$
\left(3+\sqrt3\right)^2 = 3^2 + 3 + 6\sqrt3 = 6\left(2+ \sqrt3\right)
$$
and
$$
\left(3+\sqrt3\right)^4 = \left(\left(3+\sqrt3\right)^2\right)^2
$$
A: I used my recollection that $\sqrt3=1.732\dots$
Note that $\left(3-\sqrt3\right)^4\doteq1.268^4\in(2,3)$ since $1.2^4=2.0736$ and $1.3^4=2.8561$.
$$
a_n=\left(3+\sqrt3\right)^n+\left(3-\sqrt3\right)^n
$$
satisfies $a_n=6a_{n-1}-6a_{n-2}$ and starts out $2,6,24,108,\color{#C00}{504},\dots$
Thus, $\left\lfloor\left(3+\sqrt3\right)^4\right\rfloor=501$.
A: One possibility is to look at $f(n) = (3+\sqrt{3})^n + (3-\sqrt{3})^n$.  It's easy to see that $f(4) = 504$.  You have
$$(3 + \sqrt{3})^4 = 3^4 + 4 \cdot 3^3 \cdot \sqrt{3} + 6 \cdot 3^2 \cdot 3 + 4 \cdot 3 \cdot 3 (\sqrt{3}) + 9 = 252 + 144 \sqrt{3}$$
and when you expand out $(3-\sqrt{3})^4$ similarly the terms with $\sqrt{3}$ come out negated, so you get $(3 - \sqrt{3})^4 = 252 - 144 \sqrt{3}$.
Now what else can you say about $(3-\sqrt{3})^4$?  In particular, can you find its integer part? (This might require a bit of hand calculation.)
A: Unless you have restrictions on the methods to use the simplest one is to use your calculator (!).
$$\lfloor (3+\sqrt3)^4\rfloor \approx\lfloor 501.415\rfloor = 501.$$
