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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.

Community
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user350331
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    The motivation is based on the fact that if we let $a_n=(2+\sqrt3)^n+(2-\sqrt3)^n$, then this sequence is recursively defined by $a_0=2$, $a_1=4$ and $a_{n+2}=4a_{n+1}-a_n$ and $a_n$ is an integer "close enough" to $(2+\sqrt3)^n$. This is a standard technique for handling problems like this. – Yuxiao Xie Apr 27 '17 at 15:11

4 Answers4

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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*}

quasi
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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 $4$.

achille hui
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  • The mod trick is nice (given competition time constraints). Alternatively, you can just find the actual value of $T_4$. – quasi Apr 27 '17 at 15:21
  • @quasi I didn't find $T_4$ on purpose. I'm presenting an approach which didn't need the value of $T_4$. The same logic works even when one replace $\left\lfloor (2+\sqrt{3})^4 \right\rfloor$ by monster like $\left\lfloor (2+\sqrt{3})^{4444} \right\rfloor$ ;-) – achille hui Apr 27 '17 at 15:29
  • I knew that you knew. I was just pointing out for the benefit of the OP, that -- given that it's only $T_4$, the same method you provided gets the exact value of $T_4$. – quasi Apr 27 '17 at 15:36
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Hint: Find

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

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

quasi
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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.

law-of-fives
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  • I do know matrix multiplication and this indeed seems very interesting but how did you get $$a+b\sqrt{n} \Leftrightarrow \begin{pmatrix} a & bN \ b & a \end{pmatrix}$$. – user350331 Apr 27 '17 at 15:06
  • @user350331 if you work out the arithmetic by hand you can see the correspondence for yourself. Also see https://math.stackexchange.com/questions/180849/why-is-the-complex-number-z-abi-equivalent-to-the-matrix-form-left-begins – law-of-fives Apr 27 '17 at 15:11
  • Okay this might seem dumb but I think I dont understand what is meant by$$a+b\sqrt{n} \Leftrightarrow \begin{pmatrix} a & bN \ b & a \end{pmatrix}$$ – user350331 Apr 27 '17 at 15:23
  • I mean if you multiply the matrix ((a bN)(b a)) by ((c dN)(d c)) you will find that the terms of the first column agree with $(a+b\sqrt{N})(c+d\sqrt{N})$, along with other arithmetic combinations (addition, subtraction, multiplicative inverses, associativity, distributivity...). With respect to arithmetic the matrix represents the number. – law-of-fives Apr 27 '17 at 23:19