<s>[INST] How to find in a more formal way $\lfloor{(2+\sqrt3)^4}\rfloor$ 
Problem Statement:-
The largest integer which is less than or equal to $(2+\sqrt3)^4$ is
$\text{(A)}192\qquad\qquad \text{(B)}193\qquad\qquad \text{(C)}194\qquad\qquad\text{(C)}195\qquad\qquad$


My Solution:-
I started with the binomial expansion but as I was not able to think of any manipulation which would make my life easier so I just did the following.
$$(2+\sqrt3)^2=\binom{4}{0}2^4+\binom{4}{1}(2^3)(\sqrt3)+\binom{4}{2}(2)^2(\sqrt3)^2+\binom{4}{3}(2)(\sqrt3)^3+\binom{4}{4}(\sqrt3)^4$$
On arranging the integers and the irrational terms separately I ended up with $$(2+\sqrt3)^4=97+56\sqrt3=97+56+56(\sqrt3-1)=153+56(\sqrt3-1)$$
At this point still no manipulation was obvious to me that I can do that would give me answer to $\lfloor{(2+\sqrt3)^4}\rfloor$, so I just used a rough estimate of $(\sqrt3-1)\approx0.732$ to get $56(\sqrt3-1)=40.992$.
The rough estimate of $56(\sqrt3-1)\approx40.992$ is dangerously close to $41$ so at this point I was pretty confused as to how to ascertain that $56(\sqrt3-1)\lt 41$ or $56(\sqrt3-1)\gt 41$, which had left me with the answer as one of the options out of (C) and (D).
Your help is needed for resolving this doubt of mine.

Also, what would have been a more subjective approach.

 [/INST] Another way . . .

Let $a = 2 + \sqrt{3},\;\;b = 2 - \sqrt{3}$.
Then 
$$a + b = 4$$
$$ab = 1$$
so
$$a^2 + b^2 = (a+b)^2 - 2ab = 4^2-2(1) = 14$$
hence
$$a^4 + b^4 = (a^2 + b^2)^2 - 2(ab)^2 = 14^2 - 2(1)^2 = 194\\[0pt]$$
Then
\begin{align*}
&a^4 + b^4 = 194\\[6pt]
\implies\; &193 < a^4 < 194\qquad\text{[since $0 < b < 1$]}\\[6pt]
\implies\; &\lfloor{a^4}\rfloor=193
\end{align*}
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