How to find in a more formal way $\lfloor{(2+\sqrt3)^4}\rfloor$

Question

Problem Statement:-

The largest integer which is less than or equal to $(2+\sqrt3)^4$ is

$\text{(A)}192\qquad\qquad \text{(B)}193\qquad\qquad \text{(C)}194\qquad\qquad\text{(C)}195\qquad\qquad$

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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.

Answer 1

If you know matrix multiplication, here's a fun trick. Arithmetic with $a+b\sqrt{N}$ can be performed as matrix arithmetic, under the representation

$$a+b\sqrt{N} \Leftrightarrow \begin{pmatrix} a & bN \\ b & a \end{pmatrix} $$

It's not necessarily any easier but I thought you might find it interesting. (It only works when all Ns are the same. If you have multiple square roots you need a larger matrix to represent all the cross terms.)

Regardless, expanding the power is a fine first step. I got $97+56\sqrt{3}$ same as you. We could bring the $56$ inside as 3136, to get $97 + \sqrt{9408}$. The largest natural number whose square is less than or equal to 9408 is 96. So we have $\sqrt{9408} = \sqrt{96^2 + 192}$. This means, whatever the square root of 9408 is, it's $96 + \delta$ for some $\delta < 1$.

So our answer should be $97+96 = 193$. The actual value is $193.99485\ldots$ so you can see how a poor estimation would give the wrong answer.

Answer 2

Hint: Find

$$(2+\sqrt3)^4 + (2-\sqrt3)^4 $$

(expand and cancel), and note that $0 < (2-\sqrt3) < 1$.

Answer 3

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*}

Answer 4

Consider the sequence: $T_n = (2+\sqrt{3})^n + (2-\sqrt{3})^n$.

Since $(\lambda - (2+\sqrt{3})(\lambda-(2-\sqrt{3})) = \lambda^2 - 4\lambda + 1$, this sequence satisfies a linear recurrence relation:

$$T_{n+2} = 4T_{n+1} - T_{n},\quad \forall n$$ Since $T_0 = 2$ and $T_1 = 4$ are integers, this recurrence relation implies all $T_n$ are integers.
Look at everything under modulus $4$, we obtain

$$T_4 \equiv -T_2 \equiv T_0 = 2 \pmod 4$$ Since $0 < 2-\sqrt{3} < 1$, we have $0 < (2-\sqrt{3} )^4 < 1$. This leads to $$\left\lfloor (2+\sqrt{3})^4 \right\rfloor = T_4 - 1 \equiv 1 \pmod 4$$ Among the $4$ given choices, only $193$ has the right remainder when you divide it by $4$.