# 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 $2-\sqrt{3}<1$.

• Did you try expanding $(3+\sqrt{3})^4$? Commented Jun 19, 2017 at 20:43
• Observe that $\sqrt{a^2 + r}$, where $a^2 < a^2 + r < (a+1)^2$, satisfies $a < \sqrt{a^2 + r} < a +1$ - so it's clear that the integer part of $\sqrt{a^2 + r}$ will be $a$. Commented Jun 19, 2017 at 20:48
• @fractal1729 Sorry and thanks! never bother me i am an idiot. Commented Jun 19, 2017 at 20:51

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

• I think its not possible without some approximation, I too was getting a result as $504-t$ where $t \in (0,10)$ Commented Jun 20, 2017 at 5:35

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

• How do we use $2-\sqrt{3}<1$ ? Commented Jun 19, 2017 at 20:50
• Simple and sweet. I understand it! Commented Jun 19, 2017 at 20:56
• Wait I didn't quite get the answer! We have this equal to $36(2+\sqrt{3})^2$. Letting $(2+\sqrt{3})^2 = x$, we have now $36(x) = 36(\lfloor x\rfloor + (x-\lfloor x\rfloor))$. Now what do we do? Commented Jun 19, 2017 at 21:03
• I would then say $36(2+\sqrt 3)^2=36(7+4\sqrt 3)=36\cdot 7+ 144\sqrt 3$ and just multiply $144 \cdot 1.732$ by hand. If it came out too close to an integer I would be in trouble, but it doesn't here. Commented Jun 19, 2017 at 21:18

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.)

• robjohn's solution fills in some of the details of mine. Commented Jun 20, 2017 at 13:43

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

• Sorry I forgot to mention but calculators were not allowed! Commented Jun 19, 2017 at 20:57