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I have a question which has fluxommed me and my pals for the past few days. Any help or solution is welcome

Show using Binomial theorem that the integer just after $(3^{1/2} + 1)^{2n}$ is divisble by $2^{n+1}$.

here n belongs to natural numbers (positive integers).

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  • $\begingroup$ Just for clarification: So you mean $\lceil (3^{1/2} + 1)^{2n} \rceil \mod 2^{n+1}=0$? $\endgroup$ – draks ... Apr 4 '12 at 6:54
  • $\begingroup$ By ‘the integer just after’ I assume that you mean the ceiling, $$\left\lceil(\sqrt3+1)^{2n}\right\rceil\;.$$ $\endgroup$ – Brian M. Scott Apr 4 '12 at 6:54
  • $\begingroup$ By "the integer just after" I mean the following : Suppose n=1 we get 7.46 and the integer just after is 8 which is divisible by 4 $\endgroup$ – Rohan Apr 4 '12 at 6:58
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    $\begingroup$ Actually, in order for it to work for $n=0$ we have to make it $$\left\lfloor(1+\sqrt3)^{2n}\right\rfloor+1\;.$$ $\endgroup$ – Brian M. Scott Apr 4 '12 at 7:02
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    $\begingroup$ @Rohan: There are many mathematicians who consider $0$ to be a natural number. So it is better to make it unambiguous by saying positive integer etc. $\endgroup$ – Aryabhata Apr 4 '12 at 7:20
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This is a classic, the standard proof is to consider

$$ (\sqrt{3} + 1)^{2n} = x_n + y_n \sqrt{3}$$

for integer $x_n$, $y_n$, (which you can see using Binomial theorem) and show that

$$ (\sqrt{3} - 1)^{2n} = x_n - y_n \sqrt{3}$$

again using Binomial theorem.

Now use the fact that $\sqrt{3} - 1 \lt 1$ and that

$$(\sqrt{3} + 1)^{2n} + (\sqrt{3} - 1)^{2n} = 2x_n$$

Thus the integer you are looking for is $2x_n$.

Get a recurrence for $x_n$ and $y_n$ and use induction.

To get the recurrence:

we have that $(\sqrt{3} + 1)^2 = 4 + 2\sqrt{3}$ and so

$$ x_{n+1} + y_{n+1} = (x_n + y_n \sqrt{3})(4 + 2 \sqrt{3}) = (4x_n + 6y_n) + (2x_n + 4y_n) \sqrt{3}$$

Thus

$$ x_{n+1} = 4x_n + 6y_n$$ $$ y_{n+1} = 2x_n + 4y_n$$

Write $$x_{n+2} = 4x_{n+1} + 6y_{n+1}$$ and eliminate $y_n$ and $y_{n+1}$

(I believe it comes to $x_{n+2} = 8x_{n+1} - 4 x_n$)

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  • $\begingroup$ well I accept your answer but can you please show a bit details like what you mean by "recurrence for x and y" I am really not as advanced as you are. Thnx for help $\endgroup$ – Rohan Apr 4 '12 at 7:06
  • $\begingroup$ A very big thanks man I had to rack my brains to understand your stuff but I finally got it !!! Could you also help me out with the induction $\endgroup$ – Rohan Apr 4 '12 at 8:03
  • $\begingroup$ @Rohan: I suggest you try it out first. $\endgroup$ – Aryabhata Apr 4 '12 at 15:25
  • $\begingroup$ @Aryabhata Emm! Why do you say this is a classic? Has it appeared before in some famous Math Magazines? If yes, do send me a source :) $\endgroup$ – crskhr Jun 29 '18 at 11:37
  • $\begingroup$ @crskhr: I have seen it in textbooks which were dated early 1900s. Sorry don't have a reference handy. $\endgroup$ – Aryabhata Jun 29 '18 at 21:00
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This similar to the Fibonacci sequence and is going to involve a recurrence.

Note that $1+\sqrt{3}$ is a solution of $x^2=2x+2$ and the other solution is $1-\sqrt{3}$.

So you can see that $$a_n=(1+\sqrt{3})^n + (1-\sqrt{3})^n$$ satisfies $$a_{n+2}=2a_{n+1}+2a_n.$$ If this is not instantly obvious then prove it by induction. The sequence $a_n$ starts $2,2,8,20,56,152,416,\ldots$. OEIS A080040

Since $(1-\sqrt{3})^{2n}$ is positive and is less than $1$ for positive $n$, you have $$a_{2n}=\left\lceil(\sqrt3+1)^{2n}\right\rceil\;$$ for positive $n$.

Since $a_0$ and $a_1$ are both divisible by $2$, it is an easy induction from the recurrence that $a_{2n}$ and $a_{2n+1}$ are both divisible by $2^{n+1}$.

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This is similar to Henry's answer, but things are simpler if we use $4+2\sqrt{3}=(1+\sqrt{3})^2$ and $4-2\sqrt{3}=(1-\sqrt{3})^2$ instead. These satisfy $x^2-8x+4=0$.

The sequence defined by $$ a_0=2\text{, }a_1=8\text{, and }a_n=8a_{n-1}-4a_{n-2}\tag{1} $$ has the solution $$ \begin{align} a_n &=(4+2\sqrt{3})^n+(4-2\sqrt{3})^n\\ &=(1+\sqrt{3})^{2n}+(1-\sqrt{3})^{2n}\\ &=\left\lceil(\sqrt{3}+1)^{2n}\right\rceil\text{ for }n>0\tag{2} \end{align} $$ Let $b_n=a_n2^{-n-1}$. Then, $(1)$ becomes $$ b_0=1\text{, }b_1=2\text{, and }b_n=4b_{n-1}-b_{n-2}\tag{3} $$ Recursion $(3)$ insures that $b_n\in\mathbb{Z}$ for all $n\ge0$. Thus, for $n>0$, $(2)$ yields $$ \begin{align} \left\lceil(\sqrt{3}+1)^{2n}\right\rceil &=a_n\\ &=2^{n+1}b_n\\ &\in2^{n+1}\mathbb{Z}\tag{4} \end{align} $$

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You can prove by induction method and easily get answer.

Obviously, the statement is true for n =1.

Suppose the statement is true for n=k;

i.e, (√3+1)^(2k) is divisible by 2^(k+1).

(√3+1)^(2k) = [2^(k+1)]*q, for some q.

Now we have to show that the statement is true for n=k+1;

(√3+1)^(2(k+1)) = (√3+1)^(2k + 2) 
                = [(√3+1)^(2k)]*[(√3+1)^(2)]
                = [2^(k+1)]*q*[3 + 1 + 2√3]
                = [2^(k+1)]*q*2*(2 +√3)
                = [2^(k+2)]*q*(2 +√3)  

There fore, (√3+1)^(2(k+1)) is divisible by 2^(k+2).

There fore the statement is true for n=k+1.

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  • $\begingroup$ Thnx Prasad G I appreciate ur efforts but I require A proof through Binomial Theorem $\endgroup$ – Rohan Apr 4 '12 at 10:05
  • $\begingroup$ @Rohan: Unfortunately, the edit I made to your question makes that unclear. So it is not Prasad's fault. Of course, this answer is incorrect for other reasons... $\endgroup$ – Aryabhata Apr 4 '12 at 15:25
  • $\begingroup$ What Prasad has done is incorrect as although his method his correct n=1 does't satisfy the equation and he was required to prove that the next integer is divisble which has not featured in his solution $\endgroup$ – Rohan Apr 6 '12 at 9:07

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