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Let $A=\sqrt{13+\sqrt{1}}+\sqrt{13+\sqrt{2}}+\sqrt{13+\sqrt{3}}+\cdots+\sqrt{13+\sqrt{168}}$ and $B=\sqrt{13-\sqrt{1}}+\sqrt{13-\sqrt{2}}+\sqrt{13-\sqrt{3}}+\cdots+\sqrt{13-\sqrt{168}}$.

Evaluate $(\frac{A}{B})^{13}-(\frac{B}{A})^{13}$.

By Calculator, I have $\frac{A}{B}=\sqrt{2}+1$ and $\frac{B}{A}=\sqrt{2}-1$.

But, I don't know how. Has someone any idea about this.

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    $\begingroup$ Where did you encounter this beauty? $\endgroup$ – Daniel W. Farlow Mar 9 '15 at 6:31
  • $\begingroup$ ^ I want to know that too $\endgroup$ – Hasan Saad Mar 9 '15 at 6:34
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    $\begingroup$ How did it get 10 votes in 37 minutes? $\endgroup$ – Asaf Karagila Mar 9 '15 at 6:55
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    $\begingroup$ @crash: Uninformative title with subjective difficulties qualifications? Question without showing any efforts? Those usually garner between two downvotes to two upvotes in a span of days. Certainly not 10 upvotes in less than an hour. $\endgroup$ – Asaf Karagila Mar 9 '15 at 6:57
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    $\begingroup$ @AsafKaragila Usually is the problem in what you just wrote--the most upvoted questions on MSE are the ones that often get closed in the matter of minutes and garner downvotes themselves. I cannot explain the voting patterns of users here. I imagine what happened to most people, like myself, was they opened a question expecting to find a really easy precalc problem, but then it turned out to be something rather intriguing, hence the upvotes. $\endgroup$ – Daniel W. Farlow Mar 9 '15 at 7:00
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This took me some time to solve. Here you go:

First, we find this:

$$\begin{aligned} (\sqrt{13+\sqrt{a}}-\sqrt{13-\sqrt{a}})^2 &=13+\sqrt{a}+13-\sqrt{a}-2\sqrt{13+\sqrt{a}}\sqrt{13-\sqrt{a}}\\ &=2(13-\sqrt{169-a}) \end{aligned}$$

So,

$$\sqrt{13+\sqrt{a}}-\sqrt{13-\sqrt{a}}=\sqrt{2}\sqrt{13-\sqrt{169-a}}$$

By what we have, we write,

$A-B=\sqrt{2}B$

What I used here is the fact that I've summed over all $a$ from $1$ to $168$, and that summing with $\sqrt{169-a}$ is the same as summing with $\sqrt{a}$ in this question.

Now, we have $A=(1+\sqrt{2})B$

Thus, $\frac{A}{B}=1+\sqrt{2}$ and $\frac{B}{A}=\sqrt{2}-1$

We just calculate $(\frac{A}{B})^{13}$ and $(\frac{B}{A})^{13}$ which I believe is okay to be done using calculator. Else, comment so I can edit my answer.

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  • $\begingroup$ Nice work. But "which I believe is okay to be done using calculator"--can you show how to reach the integer solution $94642$ manually through algebraic manipulation? $\endgroup$ – Daniel W. Farlow Mar 9 '15 at 7:04
  • $\begingroup$ I'll give it a try. Please wait a bit. $\endgroup$ – Hasan Saad Mar 9 '15 at 7:05
  • $\begingroup$ Cool. Will do. :) Because I think, @AuthawichNarissayaporn correct me if I'm wrong, that computing that is actually part of the problem. $\endgroup$ – Daniel W. Farlow Mar 9 '15 at 7:06
  • $\begingroup$ I'm not really sure I can do it. The best way after my work is to use @math110 solution. He has it pretty nice there to be honest. $\endgroup$ – Hasan Saad Mar 9 '15 at 7:10
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    $\begingroup$ @crash Sorry, I hit "enter" at the wrong time. Take another look at my message, which shows you how to get to 94642 by algebraic manipulation without using a calculator, as you asked for a couple of days ago. $\endgroup$ – BudgieJane Mar 11 '15 at 17:26
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Let $$A=\sum_{n=1}^{168}\sqrt{13+\sqrt{n}},B=\sum_{n=1}^{168}\sqrt{13-\sqrt{n}}$$ since $$\sqrt{2}A=\sum_{n=1}^{168}\sqrt{26+2\sqrt{n}}=\sum_{n=1}^{168}\left(\sqrt{13+\sqrt{169-n}}+\sqrt{13-\sqrt{169-n}}\right)=A+B$$ so we have $x=\dfrac{A}{B}=\sqrt{2}$,then we have $$x=\sqrt{2}+1,\dfrac{1}{x}=\sqrt{2}-1\Longrightarrow x+\dfrac{1}{x}=2\sqrt{2}$$ let $$a_{n}=x^n-x^{-n}$$ use this well know indentity $$a_{n+2}=(x+\dfrac{1}{x})a_{n+1}-a_{n}\Longrightarrow a_{n+2}=2\sqrt{2}a_{n+1}-a_{n}$$ $$a_{1}=2,a_{2}=4\sqrt{2}$$ so $$a_{3}=2\sqrt{2}a_{2}-a_{1}=16-2=14$$ $$a_{4}=2\sqrt{2}a_{3}-a_{2}=28\sqrt{2}-4\sqrt{2}=24\sqrt{2}$$ $$a_{5}=2\sqrt{2}a_{4}-a_{3}=96-14=82$$ $$\cdots$$

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    $\begingroup$ But, without calculator, we don't know $x=\sqrt{2}+1$. $\endgroup$ – Authawich Narissayaporn Mar 9 '15 at 6:56
  • $\begingroup$ can see @Hasan Saad solution $\endgroup$ – math110 Mar 9 '15 at 6:58
  • $\begingroup$ Thank you for making a solution perfect :) $\endgroup$ – Authawich Narissayaporn Mar 9 '15 at 7:14

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