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!

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.)
