Compute the floor of $(\sqrt{3}+ 1)^5$, or of $(x+ 1)^n$ with $x$ irrational How to find $\lfloor(x+ 1)^n\rfloor$ where $x \in\Bbb R - \Bbb Q$ and $n \in \Bbb Z_+$ without using a calculating or calculating the series by hand.
For example, what is $\lfloor(\sqrt{3}+ 1)^5\rfloor$ ?

I thought I would use Bernoulli's inequality, $ 1+ 5\sqrt{3} \le(\sqrt{3}+ 1)^5$, but it is too weak to give a correct answer.
I also tried, $2 &lt; \sqrt{3} + 1 &lt; 3 \implies 32 &lt; (\sqrt{3} + 1)^5 &lt; 243$ but this is also too weak.
I can use $\lfloor x \rfloor = x - \{ x \}$ where $\{x\}$ is fractional part function. Here I need to find $\{x\}$. I don't have any idea to do that either.
You have chosen a very special example.  $1+\sqrt 3$ is a Perron number in that its absolute value exceeds $1$, clearly, but the absolute value of its conjugate, $1-\sqrt 3$, is less than $1$.

To work with this example, start with the minimal polynomial $x^2-2x-2$.  That leads us to consider the Fibonacci-like recursion $$a_{n+2}=2a_{n+1}+2a_n$$  Of course such a recursion will always have solutions of the form $a_n=A(1+\sqrt 3)^n+B(1-\sqrt 3)^n$ for suitable constants $A,B$.  For our purposes we want $A=B=1$ so we get the initial conditions $a_0=2=a_1$.

We use the recursion to compute $a_5=152$.  This means that $$(1+\sqrt 3)^5=152-(1-\sqrt 3)^5$$  thus $$\boxed {\lfloor (1+\sqrt 3)^5\rfloor =a_5=152}\quad \&amp; \quad  \boxed {\{(1+\sqrt 3)^5\}=-(1-\sqrt 3)^5}$$

Of course, this holds for all odd exponents.  For even exponents you get $$\lfloor (1+\sqrt 3)^{2n}\rfloor = a_{2n}-1\quad \&amp; \quad \{(1+\sqrt 3)^{2n}\}=1-(1-\sqrt 3)^{2n}$$ 

Note:  this trick will only work with Perron Numbers.
First consider $(\sqrt {3}+1)^5=a + b \sqrt{3}$ . Now consider $(\sqrt {3}+1)^5-(\sqrt {3}-1)^5=2a , (\sqrt {3}+1)^5+(\sqrt {3}-1)^5=2b\sqrt {3} $ . I think you know why this is true. Now $\lfloor(\sqrt {3}+1)^5\rfloor=\lfloor a+b\sqrt {3}\rfloor $