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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.

Authawich Narissayaporn
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    Where did you encounter this beauty? – Daniel W. Farlow Mar 09 '15 at 06:31
  • ^ I want to know that too – Hasan Saad Mar 09 '15 at 06:34
  • I'm not sure. It's the one of grade 9th math competition problem. – Authawich Narissayaporn Mar 09 '15 at 06:54
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    How did it get 10 votes in 37 minutes? – Asaf Karagila Mar 09 '15 at 06:55
  • @AsafKaragila Because it's a fascinating/fun little problem? :) – Daniel W. Farlow Mar 09 '15 at 06:56
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    @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. – Asaf Karagila Mar 09 '15 at 06:57
  • @AuthawichNarissayaporn Alright mate, thanks. I solved it and my answer is down there. But I must say, it's a cool one. – Hasan Saad Mar 09 '15 at 06:58
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    @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. – Daniel W. Farlow Mar 09 '15 at 07:00
  • @HasanSaad Thank you. I still don't know the way this result comes.T_T – Authawich Narissayaporn Mar 09 '15 at 07:01
  • Well, the appropriate way is to comment on my answer so I can help you there. Tell me what part you're struggling with and I'll either edit my answer to clarify or explain to you in the comment – Hasan Saad Mar 09 '15 at 07:03
  • @AsafKaragila Thank you for your opinion. Now, I just edited the name of title to make more informative. :D – Authawich Narissayaporn Mar 09 '15 at 07:24
  • here is more hard problem http://math.stackexchange.com/questions/464346/why-does-this-ratio-of-sums-of-square-roots-equal-1-sqrt2-sqrt42-sqrt2-co?lq=1 – math110 Mar 09 '15 at 08:14

2 Answers2

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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.

Hasan Saad
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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 $$a_{3}=2\sqrt{2}a_{2}-a_{1}=16-2=14$$ $$a_{4}=2\sqrt{2}a_{3}-a_{2}=28\sqrt{2}-4\sqrt{2}=24\sqrt{2}$$ $$a_{5}=2\sqrt{2}a_{4}-a_{3}=96-14=82$$ $$\cdots$$

math110
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