Finding integer part of $(3+\sqrt{3})^4$

Question

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!

Answer 1

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, $\left\lfloor\left(3+\sqrt3\right)^4\right\rfloor=501$.

Answer 2

HINT $$ \left(3+\sqrt3\right)^2 = 3^2 + 3 + 6\sqrt3 = 6\left(2+ \sqrt3\right) $$ and $$ \left(3+\sqrt3\right)^4 = \left(\left(3+\sqrt3\right)^2\right)^2 $$

Answer 3

Unless you have restrictions on the methods to use the simplest one is to use your calculator (!). $$\lfloor (3+\sqrt3)^4\rfloor \approx\lfloor 501.415\rfloor = 501.$$

Answer 4

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.)